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N, a.“ ‘ , m “raw - -- .xwva- Y 1‘ - -\ _ j 1 ,fgvw-"' ' I , r’ ~a \ _ “I, I 4 r~m-zs ' {$79 "fin-fer 53: d3" 1: ¢ - 1.1-— , ; . at? ‘Ek-"i'f ‘ ‘Z‘ifig‘i' - w ' .- ‘ ,, ' in» . > ' Yu 2' “I ) M . I ;g M VP 3' I 1?“. ,9: 4 #égf " . V a ' - w H.{SQ:~:~'$§:T%H»~*~=~; fiw 1-5-=" ~ ~ =* ,7 3% :- "d ' xvii ~- Fug“ '1“! ~ : - L‘3w’.“ J . &_?;*-’#?'~" " * "PM §%F’"+€:if . v FKgg-r " ‘*" II of “Ii {In-‘4 .gz. v , ’-_A_1_“_ e235? - __ ~ pkI I; r ' ~ ‘1“ ’w , ‘3'“, ; ‘ :5“ m -: 1‘ k “i t.“ 1 o 0-. ‘I‘I' t'\-"{>I~.)Id 6‘2',I‘,~I>_ , . t; u ;.}'§RU\-Z:JL-‘J\- ' 5-ffgffii'é f i; w ; a ' ~ * '- ‘ 2;34§¥‘§3l1"r7i3l}: *klw'é 1' "- W' it h sum} illus! .J . ;_ ‘1» _..,. . ‘ ' 3 * .:‘.‘.".'i.~.- "s $f .31? "a - - f ; :- ; s... _ . ‘ l‘ . " “ 5“3"<‘1‘ "0 » “~ ' J: ".5 “.1;. If— Q’QO-z-vfitOa I -' ‘ _ ‘ v _- - ‘— ' - _ F 3‘ A ' - , - , 59'5"}? if; Arntis "Elm 55‘ . _t ' "311$: mmr “W‘s OFTHE . nsnrorltuonm-.-. b :1 ’ ,/ . ' - _-' ‘1; ; / I _ 'I I); i “ fiQk ' I 1 ~ ¢:' ‘ *1)»- M v ".- . -t l 4 t I 1 ‘ i" ‘- 1 n ‘ ":11"; 'i/i Ii ' I, "rim 1: ,fi» ' " " '\' 5 3;; Ill _ a I I" -__I ‘ I 7- 3: ‘0?’ a 7' _. i if ~.' i '— ‘ ‘ I1 1 ' ‘ £H'|‘;;!,.H.U’fii h~_;*")‘_,n'* I, I ) , I"; l \X; ' g V "r _--- i i l ' Camus-1'"? ~ ~ ‘ i i“ - l i NW '16P“: """R‘ 3"“; - a» :.0_ the ' I' .3- Anv ulna! insr un ermtu! will he foullil l important M1 tit-r the hoolr in The firatfof ti mantargl' tihtmt turo mu theaifi -_ r? a_ _ ,_.r -_~,,,o*o, , itnmrtunt prothn-tions of U u' lirocl? mind are briefly rljanuaaotl and ohnrarterised, :50 M- 1.0 form n compon- ilioua history of litoratura and the arts among tho Greek“: The around lrlition cotwlslf of 187 wood I i: :- :0 z . ' _.\ , .~ _.-w.;.'..t-. .' . ~o ~ . & "1.x. a '11 .,4~ 4 “34- g ‘I' 'a Q t ‘.\' I P’J _ “1’84? '. . " ~“.“ W 3 7 ,.~ 1“ t. a,“ ‘- ’1 h. H \,“'| .1“...— 0_4' h" 5. w b ‘ - l engraving», either 00 sfrom the uni |ue(portrult-a. “The new edition 0! Dr. Schmitz's Uirloru of Creme containqaix supplementary char-tors devoted excluaively to the historv of literature and arts among the Greeks. e edition is copiouslyillustrated with beautifully executed t-ngruviuza by Mr. George Scharl‘. from Sir William Gcll‘a original sketches. in the llriliah Muaeum,anclfrom m urns in llu." same repository. '7‘ If "I 3- 11-. . ‘ol;.-.J" ' ‘t' '— v w 1 _ *— . r ‘ ' , 'lvnx‘ ~” ' wimp-n“ ’ l i qr“ I |N9r urirr In a, “ For a long period t1 [a want of an ably tli-Iyzsictl history of Greece, for the use of y oliug 436110111 rs, was lcli' (-0 ho liObh a 51mm; In our litoraiiuw and n disparage- merit to our learning. h'litford’a'iiolumea are too long for tho purpose; and much knowledge lms been acquired ainco they were writtvn. \Vithiu the last fifty years moro has hem done both by English and tory than at any period since the rorival of learning; and the result. ofthvac labours is the acquisition of two English works - upon the subject. with as no other nation can boast of. In (irrmauy we have div» acrtat-ionst treatises, cisnya; in England consecutive history. it, remained onlI that Dr. 'l‘llit'lwall’s work should bo con- densed, and that the dot-ails so copiously elnboratei? by him ahOuld he aualyaatl and roduccrl to a compendioua manual, foreign aolmlura to elucidate Grecian his? '00: (be Mommm Pos'r. l l l Q1 llihllllllllllIllllllllllllll mm l m ' ! ro rmcntatlonn I: men. The nginulnlaotchaa, m. ruul all hum , (l. Bchnrt', jun. been favourably ho "Jilli- "11' 4. ,fi‘, it _, _ ,t o ,, fl ,, let. a makathu volume more attractive to the eye, will ncreauelt-s unel’ulnoaa hi hi) in r the baaialqf a sound knowledge of the history an orta of ancient Greece, and in aiding the young scholar in trace their influence on modern oivilisa ‘ on. The author, referring to the wonderful profici- ency 01 the Greeks in iheirknowledge and ap- preciation of art, attributes that perfection in their great innate genius, their religion, and their social and political institutions, which were all peculiarly favourable to the develop- ment of intellectual accomplishments." ‘From Me J OHN BULL adapted to avholastic aoquimmont, with- out iletrantion from literary graoa. Three editions of Dr. L. summit, history PM.) its acceptability; but, not satisfied with ntteniiou to the rrquiremonts previous ly alluded l». tho author has added to the fourth edition i-ru ,iidomcnlary chapters on the lilt’t‘aluru and tho arts of the ancient, Greeks ; including admirable exposition“ of the language and dialects, of poetry and the drama, as well as dismrmtimig upon the progress of historical and mien. tific study. In a small compass the scholar has here before Hill] a full View of tho wholr subject, his courav through which is rnlii'nnu‘ and infomicd by beau- tiful illuatr'ttions drawn from actual lmul-zrupua and anciunl; n-maina,--illus- trutnms which inv in volume remarkable for historical ooourucv w ll'llqll tho 0 utmo- tiona of artist'waiki '. l ." . London: LONGMAN, BROWN, GREEN, JONGlllANS, and RUIi'lli'l‘S. YIIW'Zii ‘0 .‘Sa‘ ‘ iii-4 “'52.. THE ELEMENTS OF ALGEBRA. ‘m?‘\ \o \Qs1 @ambtih g2 : PRINTED BY 0. J. CLAY, ILA. AT THE UNIVERSITY PRESS. THE ELEMENTS OF ALGEBRA DESIGNED FOR THE USE OF STUDENTS IN THE UNIVERSITY. BY /7@0—|%3q ( THE LATE)JAMES XVOOD, D.D. DEAN OF ELY, AND MASTER OF ST JOHN’S COLIEGE, UALIBRIDGE. O i C o o 0 00° .01 FIFTEENTH EDITION, \ CAREFULLY REVISED AND MUCH ENLARGED BY THOMAS LUNDWBll LATE FELLOW AND SADLERIAN LEOTURER OF ST JOHN'S COLLEGE, CAMBRIDGE. LONDON: LONGMAN, BROWN, GREEN, LONGMANS, AND ROBERTS, PATERNOSTERQROW. 1857. ADVERTISEMENT TO THE FIFTEENTH EDITION. THIS Edition has been revised throughout, and some new matter carefully introduced by the Rev. J. R. Lunn, Fellow of St John’s College, whose assistance, through continued ill health, I was compelled to seek. By this circumstance the public at least, I am persuaded, will be gainers, inasmuch as a youthful freshness has been imparted to the book which I could scarcely hope to have given it myself. Besides various other improvements the Articles, to which duplicate numbers with an asterisk are prefixed, and the NOTES at the end, are insertions entirely due to Mr Lunn. The Rev. Gr. F. Reyner also, one of the most able and experienced Sadlerian Lecturers in the University, has lent his valuable aid in per- fecting the work; and has supplied me with a new series of original Problems of a very superior class. And, lastly, through the kindness of the Rev. J. (l. Ebden, formerly Fellow and Tutor of Trinity Hall, a collection of Notes has been placed in my hands, the work of a deceased friend of his, the late Rev. R. B. Wildig, of Gaius College, which amount to a complete recension of the last Edition, and exhibit an uncommon degree of acuteness and depth of thought. These Notes have been found most useful. With help of such an extent and character I may fairly assert, that the present Edition will issue from the Press under great advantages and with many solid improvements; although I have resisted extensive altera- tions, suggested by others, out of a due regard to the stability which, in the main, I conceive ought to belong to such a book. As before, that for which Dr. Wood is answerable is printed in the larger type, so as to be easily distinguishable at a glance. T. L. M orton Rectwy, near Alfreton, Oct. 6, 1857. “ AT the latter end of the last century Dr Wood, the present learned and venerable Master of St John’s College, Cambridge, in conjunction with the late Professor Vince, undertook the publication of a series of elementary works on analysis, and on the application of mathematics to different branches of natural philosophy, principally with a view to the benefit of students at the Univer~ sities. The works of the latter of these two writers have already fallen into very general neglect, in consequence partly of their want of elegance, and partly in consequence of their total unfitness to teach the more modern and improved forms of those different branches of science. But the works of his colleague in this undertaking have continued to increase in circulation, and are likely to exercise for many years a considerable influence upon our national system of education; for they possess in a very eminent degree the great requisites of simplicity and elegance, both in their composition and in their design. The propositions are clearly stated and demonstrated, and are not incumbered with unnecessary explanations and illustrations. There is no attempt to bring promis nently forward the peculiar views and researches of the author, and the different parts of the subjects discussed are made to bear a proper subordination to each other. It is the union of all these qualities which has given to his works, and par- ticularly to his Algebra, so great a degree of popularity, and which has secured, and is likely to continue to secure, their adoption as text-books for lectures and instruction, notwithstanding the absence of very profound and philosophical views of the first principles, and their want of adaptation, in many important particulars, to the methods which have been followed by the great continental writers. “In later times a great number of elementary works on Algebra, possessing various degrees of merit, have been published. Those, however, which have been written for purposes of instruction only, without any reference to the advancement of new views, either of the principles of the science, or to the extension of its applications, have generally failed in those great and essential requisites of sim- plicity, and of adequate, but not excessive, illustration, for which the work of Dr Wood is so remarkably distinguished; whilst other works, which have pos- sessed a more ambitious character, have been generally devoted too exclusively to the development of some peculiar views of their authors, and have consequently not been entitled to be generally adopted as text-books in a system of acade- mical or national education.”—-Professor Peacock’s Report to the British Ass0~ citation “ On Certain Branches of Analysis.” TR a“. \e. hem-ram seaw- CONTENTS. \ c 7,. e PAGE VULGAR FRACTIONS . . . . . . . . . . 1 Decimal Emotions . . . . . . . . . . 17 Definitions and Signs used in Algebra . . . . . . 29 Addition of Algebraical Quantities . . . . . . 34 Subtraction . . . . . . . . . . . 36 Addition and Subtraction by Brackets . . . . . 37 Multiplication . . . . . . . . . . . 39 Division . . . . . . . . . . . . 45 Algebraical Fractions . . . . . . . . . 49 Greatest Common Measure . . . . . . . . 50 Least Common Multiple . . _. . . . . . . 57 Addition and Subtraction of Fraction . . . . . 61 Multiplication and Division . . . . . . . . 63 Involution and Evolution . . . . . . . . 67 Square Root . . . . . . . . . . . 73 Cube Root . . . . . . . . . . . 79 Theory of Indices . . . . . . . . . . 84 Surds . . . . . . . . . . . . 87 Simple Equations . . . . . . . 95 Problems which produce Simple Equations . . . . . 111 Quadratic Equations . . . . . . . . . . 116 Problems producing Quadratic Equations . . . . . 131 Inequalities . . . . . . . . . . . 136 Ratios . . , . ' . . . . . . . . . 141 Proportion . . . . . . . . . . . . 145 Vaériation . . . . . . . . . . . 155 Arithmetical Progression . . . . . . . . . 161 Geometrical Progression . . . . . . . . 165 Harmonical Progression . . . . . . . . . 170 Permutations and Combinations . . . . . . . 174 Binomial Theorem . . . . . . . . . . 180 Exponential Theorem . . . . . . . . . 193 Multinomial Theorem . . . . . . . . . 194 Evolution of Surds . . . . . . . . . . 199 Indeterminate Coefficients . . . . . . . . . 206 Continued Fractions . . . . . . . . . 220 CONTENTS. PAGE Indeterminate Equations and Unlimited Problems . 230 Scales of Notation . 242 Properties of Numbers 250 Vanishing Fractions 256 Infinite Series 258 Recurring Series 262 Logarithms 266 Simple Interest 270 Compound Interest 270 Discount . . 275 Equation of Payments 276 Annuities 277 Renewal of Leases 281 Chances or Probabilities 283 Life Annuities . . . . . . 295 Discussion and Interpretation of Anomalous Results 302 Maxima and Minima . . 307 Application of Algebra to Geometry 308 APPENDIX. Solution of Equations by the Application of Peculiar Artifices 314—347 Problems . . . . . 347—361 Single and Double Position 361 Piles of Shot or Cannon Balls 362 EXAMPLES . 365—486 COLLEGE EXAMINATION PArEEs 487 Miscellaneous Examples, Third Series 501 EASY EXERCISES 511 ANSWERS TO THE EXEncIsEs 548 NOTES 569 ERRATA. ____.'_._ Page 83, line 19, after 2n+1 insert the word “digits.” . l 1 260, lme 3 from bottom, for ——-read —, n—l 92+] 324, Ex. 26. The denominator of the first fraction should he N/a*+\/h'§:a-5 instead of \/._'+ VIZ-IQ. . . . , I 4 1‘” 3552, Ex. 39. The answer .r=\/5(\/2-— 1) should be x=\/-5 (\/2 — l). 350, Prob. 5. For all after the second paragraph substitute the following: These hours are those indicated by the clock: and since the clock gains 01‘ in 24 hours, 24x60xt30’ + 0'1s of clock time =24><60+ 0'1 1 hour f e If e: O r a 1‘“ 24x60x60 hours of clock time; - s4><60><60’ 0‘s (2,, _ . . . l 0'0002 the diurnal gain of the watch 1s 24.72, or m+4'8+T , i. 8. 40000055... seconds. 359, line 3 from bottom, for A 864 17. To reduce a continued fraction to a simple one. Apply the rule (Art. 14) for reducing a mixed number to an improper fraction, commencing at the lowest extremity of the continued fraction, and proceeding gradually upwards until the whole is reduced to a simple fraction. But as this operation re- quires the use of a rule not yet proved, the example is deferred to rt. 38. 18. To reduce a fraction to lower terms. RULE. Whenever the numerator and denominator of a fraction have a common measure, that is, a number ‘which di- vides each of them without remainder, greater than unity, the fraction may be reduced to lower terms by dividing both the numerator and denominator by this common measure. _ 105 , 21 . . . Ex. — is reduced to ~22 by d1V1dmg both the numerator . 21 , . 7 . and denominator by 5; and 51. IS again reduced to g by d1- viding its numerator and denominator by 3. That the value of the fraction is not altered appears from Art. 12, Con. 168 84 28 4 In the same manner —-- = -___=. __ =.- -. 210 105 35 5 19. The “ Greatest Common Measure” of two numbers is found by dividing the greater by the less, and the preceding" 6‘ VULGAR FRACTIONS. \ divisor by the remainder, continually, till nothing is left: the last divisor is the greatest common measure required. DEF. The Greatest Common Measure of two or more numbers is the greatest number which will divide each of them without remainder. Ex. To find the greatest common measure of 189 and 224. 189)224.(1 189 3; )189(5 175 ~12 )sste as -_7_)14(2 14 _—. O -__| By proceeding according to the rule, it appears that 7 is the last divisor, or the Greatest Common Measure sought. The proof of this rule will be given hereafter. See Art. 103. 20. A fraction is reduced to its lowest terms by dividing its numerator and denominator by their greatest Common Measure. Ex. 385 , To reduce 59—6- to its lowest terms. By the Rule given in the last Art. the greatest Common Measure of the numerator and denominator is found to be 11; and therefore 5% is the fraction m its lowest terms. Con. If unity be the greatest Common Measure of the numerator and denominator, the fraction is already in its lowest terms. . 21. To reduce any number of- fractions to a common denominator. VULGAR FRACTIONS. 7 RULE. Having reduced, if there be any, compound frac- tions to simple ones, and mixed numbers to improper fractions, multiply each numerator by all the denominators except its own for the new numerator, and all the denominators together for a common denominator. l 2 3 , Ex. 1. Reduce g , g , and Z to a common denominator. 1x3x4 2x2x4 2x3x3 12 16 18 ,——~—, an -—-; or—, -—, an -— 2x3x4 Qx3x4 2x3x4 24 24 24 are the fractions required. These fractions are respectively equal to the former, the numerator and denominator in each case having been multiplied by the same numbers, namely, the product of the denominators of the rest (Art. 12), 1x3x4s 1 2x2x4s 2 2x3x3 3 I —~ , -—' ._ an -- .’ 2x3x4 2’ 2x3><4 a 2x3x4 4 2 3 Ex. 2. Reduce g of Z and 4%.- to a common denominator. 13 3 13 130 -—— , or —- and -- ; therefore 2- and -—-—- 3 10 3 3O 30 are the fractions required. These are —6- and 20 22. If the denominator of one of two fractions contain the denominator of the other a certain number of times ewactly, multiply the numerator and denominator of the latter by that number, and it will be reduced to the same denominator with the former. 5 2 . Ex. Reduce T2- and g to a common denominator. Since 12 contains 3 four times exactly, multiply both the . 2 . 8 numerator and denominator of g by 4, and It becomes 15 , a . . . . 5 fractlon havmg the same denominator With T2 . In the reduction of fractions to a 'common denominator the following rule is frequently required, in order that the reduced fractions may be in their lowest terms :— 23. To find the “Least Common Multiple” of any numbers. - DEF. The “Least Common Multiple” of any numbers is the least number which is divisible by each of them without remainder. 8 VULGAR rnxcrross; RULE. To find it, write down in one line the numbers of which the least common multiple is required, separating them by some mark, as a comma. Divide all those which have a common measure by that common measure*, and bring down the other numbers placed in a line with the quotients, separated as before; and repeat this pro- cess as long as any common measure exists between two or more of them. The Least Common Multiple required will be the continued product of the divisors and of the final quotients. Ex. Required the Least Common Multiple of 8, 12, and 18. 2 8, 12, 18 2 A 4, 6, 9 a 2, s, 9 A 2, 1, 3 Least Com. Mult. is 2x2><3><2><1x3 or 72. The proof of this rule will be given hereafter. See Art. 115. N.B. In finding the Least Common Multiple of any numbers care must be taken to follow the Rule strictly, viz. to “divide all those which have a common measure by that common measure.” Thus, as above, the Least Common Multiple of 8, 12, and 18 is Correctly found to be 72; but the operation might be carelessly attempted as folloWs: 6 8’ 12’ 18 first taking for a divisOr 6, the common' 2 8, Q, 3 measure of 12 and 18, instead of 2, the ..___________. common measure of all; 4, I, 3 from which we should conclude that the Least Common Multiple required is 6x2x4xlx3, or 144; which is twice as great as the true Least Common Multiple. Also, it is obvious that in any proposed case those numbers may be entirely omitted in the operation which are contained in any of the others. Thus, to find the Least Common Multiple (of 6, 7, 12, and 14, we observe that 6 is contained in 12, and 7 in 14; there- fore it remains only to find the Least Common Multiple of 12 and 14, which is 84. 24. To reduce fractions to a common denominator, in their lowest terms, find the “Least Common M ultiple" of all the denominators, and make that the common denominator by multiplying both the numerator and denominator of each fraction by the quotient of Least Common Multiple divided by the denominator. 2 3 12 3 ' . 1 Ex. Reduce to a common denom n t r — - —— nd — . " 18° s’7’1a’21a 4 * That is, the divisor must be prime to those numbers which it does not measure. \Ve are sure we comply with the direction, if we divide by prime mun. bers only. I VULGAR FRACTIONS; 9‘ The Least Common Multiple of the denominators by Art. 23 is found to be 84; and therefore the required fractions are QSXI IQXQ 6X3 4X12 21x3 581—53— ’ W’ m’ M’ 5551’ 28 24 18 48 63 52 ’ 8—4 ’ 374* ’ 8—4, ’ YE ' 25. Con. By reducing fractions to a common denomi- nator their values may be compared. 01‘ . 4* 7 . 'lhus ; and i5, when reduced to a common denomlnator, 48 49 . .. . are E; and 8—4;; that 1s, the fractlons have the same relatlve values that 48 and 49 have. 26. By reducing fractions to a common numerator also their values may be compared. - - 3 4 Thus -—- and — when reduced to a common numerator, are 13 17 ’ 12 12 . . . . 5—2 and H; and smce the former of these fractlons slgmfies that the unit is divided into 52 equal parts of which 12 are taken, and the latter signifies that the unit is divided into 51 equal parts of which 12 are taken, it is obvious that the latter fraction is the greater of the two, or that which has the smaller denominator. 27. Tofind the value of a fraction of a proposed deno- mination in terms of a lower denomination. RULE. Multiply the fraction by the number of integers of the lower denomination contained in one integer of the higher, and the product is the value required. The value of any fractional part of the lower denomination may be obtained in the same manner, till we come to the lowest. Ex. I. What is the value of g of £1 P . 5 . 5 . r ‘ Flrst, 77- of £1 1s; of 20 shlllmgs, 5 20 i “100 i . . or — of — shillings = —- = 14-?- shillmgs; 7 1 7 2 , 2 12 Next, 5 of a shilling = E of -i- pence, 24 = 77- pence = 3-73 pence; 10 VULGAR FRACTIONS. 3 3 . 3 4 Lastly, E of a penny = I; of It farthmgs = -7_ of I , 12 . . or -;- farthlngs = 1% farthmgs: 8. 5 . d. q. hence, - of a pound 1s 14. 3 . 1%. 7 The operation is usually performed in the following manner: £5 20 7)100 14"“‘280 12 7)24 3--3d. 4 7)12 1 --5g. 8. 6. 9. Ass. 14 . a . 1%. EX. 2. What is the value of g of a crown? 5C 5 9)25 2-"780 12 9)84 9--ad. 4 9)12 1--3q. ’ a. d. 9. Ans. 2 . 9 . 1%. VULGAR FRACTIONS. 11 28. To reduce a quantity to a fraction of any denomi- nation. RULE. Make the given quantity the numerator, and the number of integers of its denomination in one of the proposed denomination the denominator, and the fraction required is determined. 8. d. . Ex. What fraction of a pound is 12 . 7 . g P a. d. q. q. Q- 12 . 7 . 3 = 607; and one pound = 960; 60 therefore -—7 960 is the fraction sought: because the integer being a. d. q. divided into 960 equal parts, 12 . 7 . 3 contains 607 such parts. 29. In the last example we were obliged to reduce the whole to farthings; and in general, if the higher denomina-= tion do not contain the lower an exact number of times, reduce them to a common denomination, and proceed as before. Ex. Here sixpence is the greatest common denomination, of which a guinea contains 42, and half a crown 5, therefore What fraction of a guinea is half a croWn P 5 , . . E IS the fraction required. Any common denomination would ansWer the purpose; but, if the greatest be taken, the resulting fraction is in the lowest terms. 30. RULE. Find what fraction of the proposed denomination an nteger of the denomination of the giVen fraction is, and the fraction required will be found by Art. 16. To reduce a fraction to any denomination. Ex. 1. What fraction of a pound is g of a shilling? . . . 1 2 . . . Q 1 shilling 1s '22?) of a pound, therefore 5 of 1 shlllmg 1s 5 1 2 1 of —— of a ound = —=— = —— of a ound. 20 P ’ 60 so p 7 ,MM 12 VULGAR FRACTIONS: ' 5 Ex. 2. What fraction of a yard is E of an inch? . 1 5 , , 5 1 1 inch 1s — of a yard, therefore ~ of an inch is — of —— 36 7 7 36 5 of a yard, = 2—5—2- of a yard. , 4 Ex. 3. What fraction of a guinea 1s -9- of a pound? O 4 1 pound is 2—1- of a guinea (Art. 29); hence 5 of a pound _ 20 _ 80 . 1s — of — of a guinea = -——- of a gulnea. 9 21 189 ADDITION OF FRACTIONS. 31. Tojind the sum of two or more fractions. RULE I. If fractions have a common denominator, their sum is found by taking the sum of the numerators, and sub- joining the common denominator. 1 2 3 , . . . . ‘ Thus -5- + g = g . For, if an integer be d1V1ded 1nto five equal parts, one of those parts, together with two parts of the same kind, must make three such parts. 32. RULE II. If the fractions have not a common de-i nominator, reduce them to a common denominator, and pro- ceed as before. . 2 3 4 Ex. Requlred the sum of 5 , Z, and g . , 4O 45 These reduced to a common denominator are 66 , E.- , and 48 . 133 13 6-6 , whose sum 1s 6—6— , or 250. 33. When mixed numbers are to be added, to the sum of the fractional parts, found as before, add the sum'of the integers. yULeAR FRACTIONS. 13 2 Ex. Add together 5%, 631 and g of 71; . e 1 2 315 14.0 24. 479 159 4+3+35_420+420+420'4120- 420’ therefore the whole sum required is 5 + 6 + 155-0, or 125230. SUBTRAGTION. 34. To find the difl’erence of two fi'actions'. RULE I. The diference of two fractions which have a common denominator is found by taking the diference of their numerators, and subjoining the common denominator. 4 3 1 . . ' Thus -5--—5-=E. For, 1f the umt be supposed to be divided into five equal parts, and three of those parts be taken . 1 from four, the remalnder must be one of the parts, or g . 35. "RULE II. If the fractions have not a common de~ nominator, let them be reduced to a common denominator, and then take the diference as before. 4 Ex. 1. From 2- take —- . 11 5 11 3 1 Ex. 2. From __ of _ take -_ of Z. 12 5 3 8 11 a as 1 7 7 --o -_=—-, and -of-=——; 12 5 60 s s 24 as ‘ 66 35 31- __-l-_-_-,_-__ (Art. 24.) = . 60 24 120 120 120 When mixed numbers are to be subtracted, the integers may be subtracted separately, and then the fractional parts. Thus, 3é—2711—=3—2+-§-—%=1+i-=171i. And if the fractional part of the mixed number to be subtracted is greater than that of the other, deduct a unit from the greater number and add it in a fractional form to the smaller fractional part; then proceed as before. Thus, 65-35;-5§-3g=5-3+%-g=2+g=2;. 14 VULGAR FRACTIONS. MULTIPLIOATION. 36. DEF. To multiply one fraction by another is to take such part or parts of the former as the latter empresses. RULE. This is done by multiplying the numerators of the two fractions together for a new numerator, and the de- nominators for a new denominator. , 3 5 15 3 , Thus 2 x;=2—é; for ; multiplied by g is, accordlng to 5 3 15 the definition of multiplication, :7- of Z, or Ice—é , (Art. 16). (See ‘ Companion,’ p. If there be more than two fractions to be multiplied together, a similar rule applies :--multiply all the numerators together for a new numerator, and all the denominators for a new denominator. 1 2 3 6 1 1 2 1 1 3 1 —- — —=-—-=--' — f—=-- -- —-=—, Thus, 2x3><4$ 24! 4!, forgo 3 3, andgofli 4 Compound fractions must be reduced to simple ones, and mixed numbers to improper fractions, and they may then be multiplied as before. 2 Ex. 1. Multiply g of 195 by 7%, 2 9 18 57 —of-—=-——; and 7—1-=-—' 5 1s 65 8 8’ therefore their product is 18 57 1026 506 ,5, axr=sn=lr= 2 Ex. 2. Multiply 51—7- by 7. 2 2x7 2 .__x = ____. 217 217 31 fl Hence it appears, that a fraction may be multiplied by a whole number by dividing the denominator by that number, when this division can take place. VULGAR FRACTIONS. 15 Much trouble is frequently saved by observing what multipliers are common to the new numerator and denominator, and striking them out (Art. 12, COR.) before the multiplication is efected. 1 . Thus the product of 5, g, 2, and %, 1s, by the rule, ‘ 1222324 2232425’ which we see at once to be :2, by striking out the quantity 2x3x4 common to the numerator and denominator. Again, if in the proposed fractions, to be multiplied together, there be any numerator and any denominator which have a common measure, divide them both by that common measure, and use the resulting quotients for the purpose of forming the required product. 18 57 9 57 513 Thus -6—5><—8—=-63>< quot1ent 3 =;; let these equal quantities be multiplied by the same ‘16 VULGAR raaorrons. . . 5 quant1ty %, and the products must be equal; that 1s, ;x;>< _ 3 35 . 21 35 quot1ent = —><~7-, or -—-x quot1ent = —; but —= 1; therefore the 4 5 e5 , 2o 55 21 ' quotient=;5, as was found by the rule. And the same method of proof is applicable to all cases. Compound fractions must be reduced to simple ones, and mixed numbers to improper fractions, before the rule can be applied. 5 Ex. 2. Divide 5 of; by 331-. 5 4 20 10 — of—=-—~, and 3%=_; 9 7 63 3 2O 3 2 therefore the uotient re uired is -—x—— = —. q 63 IO 21 38. It will often happen in practice that fractions present themselves which require the application, not of one single rule only, as of .Addition, or Subtraction, or Multiplication, &c., but of several rules in one operation. Thus, Ex. 1. Required to find the single, fraction which is equi~ valent to 5 2 7%} 7x{100 8 0f 100+271£ . . 2 1 100 First, 100—5 of 100=~§ of 100-? , 22 7i 3 22 9 88 an 271,- g a 4 27’ 4 therefore the whole quantity is equivalent to §x{.1_02+§§ 7 3 27}, _5xgoo+ss ""7 27 ’ 3,932.40 '7 27 '7 189 ’ DECIMAL FRACTIONS. 17 1 . . Ex. 2; Reduce ---i to a simple fractlon. 2+ ——-- 3 +3- 4 I 13 l 1 4 , Here 34-74:;- , and Q_fi-E. I‘herefore 4 4 2+iT=2+g = g, and the fraction required is 516 or g , 4' 15 DECIMAL Fnac'rrolvs. 39. In order to lessen the trouble which in many cases attends the use of vulgar fractions, decimal fractions have been introduced, which differ from the former in this respect, that their denominators are always 10 or some power of 10, as 100, 1000, 10000, &c. and instead of writing the denomi- fiator finder the numerator, it is e.vpressed by pointing off from the right of the numerator as many figures as there are cyphers in the denominator; . - 2 thus '2 si nifies -- g 10’ 23 ‘23 00.00 _, 100 127 127 00.0! 13 00000 '__'_é’ 10000 437 ' 4.3.7 0.... or“‘ 10 40. Con. 1. The value of each figure in a decimal decreases from the left to the right in a tenfold proportion, that is, each figure is ten times as great as if it were removed one place ,to the right, as in whole numbers; thus 2 2 2 ~ '2=-—; but -02 =---, and .002 = -—, &c.; 10 100 1000 and the decimal '127 is one tenth, two hundredths, and seven thousandths, of an unit. 2 18 DECIMAL rnac'rmns. 41. Con. 2. ‘Adding cyphers to the right of a decimal does not alter its value; thus _ the numerator and denominator having been multiplied by the same number. (Art. 12.) 42. Con. 3. Decimals may be reduced to a common denominator by adding cyphers to the right, where it is neces- sary, till the number of decimal places is the same in all. Ex. '5, '01, and '311, reduced to a common denominator, are '500, '010 and '311; 500 10 d 311 that is —~, , an . 1000 1000 1000 43. Con. 4. Hence in all complicated numerical reductions decimal fractions possess great advantages over vulgar fractions. For, 1st, the denominators of the former being always 10 or some power of 10, their reduction to a common denominator is easily effected, and consequently all operations requiring that previous reduction are facilitated: 2nd, the numerators and denominators of decimal fractions being usually written in one line, and the value of each figure decreasing in a tenfold proportion, from left to right, as in whole numbers, the common rules of Arithmetic are im- mediately applicable to such fractions, care only being taken, by means of rules for that purpose, to mark of correctly the decimal result. - As decimals are only fractions of a particular description, their operations must depend upon the principles already laid down. ADDITION OF DEOIMALS. 44. RULE. Tojind the sum of any number of decimals place the figures in such a manner that those of the same de- nomination may stand under each other ; add them together as in whole numbers, and place the decimal point 'in the. sum under the other points. . DECIMAL FRACTIONS. 19 Ex. Add together 79, 5143 and '0118. These, when reduced to a common denominator, are 79000 514300 and '0118; and proceeding according to the rule, 7’9000 51'4300 '0118 593418 = the sum required. (Art. 31.) In the operation the cyphers may be omitted, if the seve- ral decimal points stand exactly under each other thus, 7'9 51'43 “0118 593418 SUBTRAO'I‘ION. 45. RULE. To find the difierenoe of two decimals place the figures of the same denomination under each other ; then subtract as in whole numbers, and place the decimal point under the other points. Ex. From 6113 take 42'012. These, reduced to a common denominator, are 61300 and 42012; therefore their difference is 19288 (Art. 34). In the operation the cyphers may be omitted thus, 61'3 42'012 19'288 MULTIPLIOATION. 46. RULE. To multiply one decimal by another multiply the figures as in whole numbers, and point of as many decimal places in the product as there are in the multiplier and mul- tiplicand together. Ex. 51-3246 = 23598. 513 46 235 8 . . . or -—-~><-- = --9_- = (accordlng to the decimal notation) 10 10 100 235'98. And a similar proof may be given in all other cases. 2—2 20 DECIMAL FRACTIONS; 47. When there are fewer figures in the product than there are decimals in the multiplier and multiplicand together, cyphers must be annexed to the left of the product, that the. decimal places may be properly represented. 25 3 75 Ex. '25x'3 = '075; for -— -—- x— = = (according to the 100 10 1000- decimal notation) '075. DIVISION. 48. RULE. Division in decimals is performed as in whole numbers, observing to point q?" as many decimals in the quotient as the number of decimal places in the dividend ewceeds the number in the divisor. EX. Divide 77922 by 3'7. 77'922 = 21-06: here there are three decimals in the dividend and one in the divisor; therefore, there are two in the quotient. The truth of this rule is apparent from the nature of mul- tiplication; for the product of the divisor and quotient is the dividend; there are, therefore, as many places of decimals in the dividend, as there are in the divisor and quotient together (Art. 46); consequently there are as many in the quotient as the number in the dividend exceeds the number in the divisor. 49.. If figures be wanting in the quotient to make up‘ the proper number of decimal places, cyphers must be added to the left. Ex. Divide '336 by 42. 336 ~— _I 42 and as the quotient of 3% divided by 42 must containithree decimal places, that quotient is '008. ' 0 Q 8 _____ , or __ 42000 1000 cording to the decimal notation) '008. For 160—0 d1V1ded by 42 1s , that IS, (ac-, 50. When the dividend does not contain as many deci- mals as the divisor, cyphers must be added to the right of the decimals in the dividend, till that is the case. (Art. 41.) DECIMAL FRACTIONS. 21 Ex. Divide 36 by '012. 36 = 36000, and 36000 divided by '012 is 3000, according to the rule. REDUCTION. 51. To reduce a vulgar fraction to a decimal. RULE. Add cyphers at pleasure, as decimals, in the numerator, and divide by the denominator according to the rule for the division of decimals. The truth of this rule is evident from Art. 11. -. s ' s-oo Ex. 1. —= -_-= '75. 4 _ 7 -000 Ex. 2. - = Z— = -s75. A 8 8 .. - 4 40000 ‘ EXQ 30 M = = .006é0 625 625 - 1 1000 8:0. ' EX. 4‘. ‘3- : w = 8Z0. 4 40000 &c. ' EX. 5. '_ = m = &co as as 52. In some cases, as in the last two examples, the vulgar fraction cannot exactly be made up of tenths, hundredths, &0. but the decimal will go on without ever coming to an end*, the '* It is evident that no vulgar fraction can be exactly expressed by a decimal, unless it either has, or can be reduced to another which has, 10 or some power of 10, for its denominator, (Art. 39). Thus, reverting to the Exs. of the last Art. 3 3x52 75 _ Z_~22x52—I0§=075’ 7 7553 875 ,_ 4 4x2‘ 64 523347217101” “"64, each of which Vulgar fractions is expressed decimally with perfect exactness. But 3 , since they cannot be expressed by equivalent fractions with a denominator of 10 or some power of 10, are not capable of being expressed by terminating decimals. Also since 10, and its powers, are divisible by 2 and 5 only, and their powers, it follows that no vulgar fraction can be expressed by a terminating decimal unless, when it is in its lowest terms, its denominator is divisible by one or both of the num- bers, 2 and 5, or their powers, and by no other number. ED. the following Exs. viz. 1, and 22 DECIMAL raac'rrons. same figure or figures recurring in the same c'rcleriIE ; but though we cannot represent the exact value of the vulgar fraction, yet, by increasing the number of decimal places, we may approach . 1 to it as near as we please. Thus 5 = '1111 &c. Now '1, or 1 ' 1 a a t 1 b 1 11 11 i t 0 -— IS 883 an e rue V3 ne '_ ' ' or —' S O 10’ y 90’ ’ 100’ 1 little by 506; and so on. Again é4g§=°123123 &c., the figures 123 being repeated without 4: end - —-—='148148 &c.; , 27 ='138888 &c.; and so on. i 36 Decimals of this kind are called recurring or circulating decimals. Hence, although some vulgar fractions cannot be accurately represented by decimals, this afi'ords no objection ‘to the use of deci- mals, because for such fractions equivalent decimals can be found approximating to the true value as nearly as we please 'l‘. ' This is easily shewn by a particular instance; and it may thence be seen to be 4 5i 0 The required decimal is found by dividing 4'00000 Sac. by 27 ; and if the quotient. does not terminate, after each division there will be a remainder less than 27. Therefore, under the most unfavourable circumstances, at least after the quotient has reached to 26 figures, one of the remainders l, 2, 3, &c....26 must recur; and conse- quently after that the figures in the quotient will recur. In the case proposed the remainder for one figure in the quotient is 13; for two figures, 22 ; for three figures,4; which is a recurrence of the original figure : consequently the decimal is 0148148, &c. the figures 148 being repeated in infinitum. Similarly also in other cases. ED. + “ The addition, subtraction, multiplication, and division, of decimal fractions, are much easier than those of common fractions; and though we cannot reduce all common fractions to decimals, yet we can find decimal fractions so near to each of them, that the error arising from using the decimal instead of the common fraction will not be perceptible. For example, if we suppose an inch to be divided into ten million of equal parts, one of those parts by itself will not be visible to the eye. Therefore, in finding a length, an error of a ten-millionth part of an inch is of no consequence, even where the finest measurement is necessary.” “ In applying Arithmetic to practice, nothing can be measured so accurately as to be represented in numbers without any error whatever, whether it be length, weight, or any other species of magnitude. It is therefore unnecessary to use any other than decimal fractions; since, by means of them, any quantity may be re- presented with as much correctness as by any other method." De Morgan’s Arithmetic, pp. 63-9. ' true in all cases. Thus, suppose it is required to find the decimal equivalent to nncnmr. FRACTIONS“. 23 53. The method of reducing a terminating decimal t'o a‘vulgar fraction is pointed out in Art. 39. The following method will serve for converting recurring decimals into their equivalent vulgar frac- tions.-- It appears, by actual division, that l —=O'll ll .......... .. q I 5 I =0'010101 .........m [0-0 £0 EDI =0'001001 .......... .. £0 99 @— 55%:0‘000100Q1 ....... .. and so on; where the recurring part of the decimal is always 1, preceded by as many cyphers as make the number of recurring digits equal to the number of 9’s recurring in the denominator of the fraction. If then, for instance, the vulgar fraction equivalent to 01212 8w. be required, we have 1 12 4: 031212 &C.=0‘0101 SEC. 12=—> 5, add 1 to the last remaining figure. Thus, if 2'7182818 be the decimal under consideration, 2'72 is nearer to the true value than 2'7], for 2'7182818—2'71, is 00082818; and 2'72—2'7182818 is 00017182, which is considerably less than the former difference. Also 2'7183 is nearer to the true value than 27182, as may be shown in‘a similar manner. It may also be observed here, that in the multiplication of deci- mals some caution is requisite in taking the product as correct to a certain number of places of decimals, when either the multiplicand or multiplier is only approximately correct. Thus, if 312 express a certain length in inchesfand is known to be correct within the thousandth part of an inch, the true length may be any thing between 3'12-1-1—6166 and 312—10100, that is, between 3121 and 3'119; and if the proposed number is to be multiplied by 10, for example, the product is 31'2 ; whereas it may be any thing between 31°21 and 3149, and therefore may not be correct even to one decimal place. ' 55. To find the value of a decimal of one denomination in terms of a lower denomination. ) This may be done by the rule laid down in Art. 27. Ex. Required the value of '615625 £. ~615625 £ 20 ~ * 125312500 shillings 3- 12 53-7500 pence A , 3'00 farthings. , a. d. (1. The value required is 12 . 3 . 3. First, '61o625 45’ Next, . '3125 s. = 3'75....pence. Lastly, . '75 d. r = 3'.'.....farthings. 12'3125" shillings. DECIMAL FRACTIONS. 25 56. To reduce a quantity to a decimal of a superior denomination. RULE. Divide the quantity by the number of integers of its denomination contained in one of the superior denomination, and the quotient is the decimal required. Ex. 1. What decimal of a shilling is threepence? 12)a-00 __ '25 Ans. For in the denomination shillings its numerical value must 1 . . . . be Z of 1ts value 1n the denomlnatlon pence. s. d. q. Ex. 2. “What decimal of~a pound is 13 . 4 . 3? 4) am ‘ 12) 4-75 20)1a-3958e33 &c. '66979166 &c. First, we find what decimal of a penny g is; this, by the d. . rule, is '75; then, what decimal of a shilling 4 . g or 4'75d; is; this is found in the same manner to be '3958333 820.; lastly, we find, by the same rule, what decimal of a pound 13'3958333 &0. sh. is, which appears _t0_be 366979166 &c. The conclusic'm 'will be the same if we reduce the quantity to a vulgar fraction (Art. 28), and this fraction to a decimal (Art. 51). ' 57. It will often happen in practice that a whole series of' vulgar fractions, instead of a single one, is to be reduced to a de- cimal, and in such cases considerable trouble may frequently be saved by making each fraction, when reduced, subservient to the reduction of some one or more of the others. Thus, Ex. 1. Required to reduce to a single deciinal having 5 decimal places the following series offractz'oizs :-- 2+ I + 1 + I +&c 1 1x2 1‘x2xa 1x2><8><4 ' 26 DECIMAL FRACTIONS. Here i 21% =2'00o000 ‘ 1% = 1.200 = '500000 TEE? = 3292 = '166667 (Art. 54). A m ==l§gg§l= 041667 m =lgé駧lz= '008333 m =Eg?§i== '001389 1x2><3><7 =loigigg= ’000198 ” 1x2x3x4:5x6x7x8 =.002198= '000025 1 .000025== '000003 1xQ><3x1£x5><6x7x8x9= 9! Sum=2'718282 Hence the decimal required is 2'71828. Each single fraction is calculated to six places of decimals, that the figure which occupies the 5th place in their sum may be correct; and no more terms of the series need be added, because the first fiVe places of decimals are not afl'ected by them. Ex. 2. Required to convert into a single decimal having five decimal places the following series :--- } 7 37“ 57“ 77’ ' Here -;—= 01428571 is = W = 0'0204081 7 7 I . 5, = 0 00291 54. 1 U 7, = 0 0094164. " Some trouble in writing may be saved in {his and similai examples by making the symbol L‘istand for 1x2x3, Lifer lx2x3x4, Li for Ix2x3x4x5; and so on, DECIMAL FRACTIONS. 27 1 . -,-7—5 = 00000594 1 o g =0 0000084. 1 77-; = 00000012. Since 5 decimal places only are required, it is not necessary to add any more terms of the series. Therefore we have —1- =0'1428571 x = 00009718 \dal'd <0 =0'000011 8 X 5 lr-t QIH x =O'OOOOOOI 7 'QIH Qwlv-I boll—l q Sum=01438408 Mult. by i 2 _ 0-2876816 Therefore the required decimal is 028768. The proofs of the rules for the management of vulgar and decimal fractions here given are necessarily confined to parti- cular instances, though the same reasoning may be applied in every case; and by using general signs the proofs mav be made general. But this requires a knowledge of Algebra. The Student is recommended to accustom himself to reason out the place of the decimal point. He will find the subject admirably treated in De Morgan’s Arithmetic, Section VI, EXAMPLES. . 173 4d 2001 Reduce to mixed numbers F , 4—6, W . Reduce to improper fractions 9%, 14%, and 10,35. What is g of}; of 2;? Ans. 1,213,. 5184 7631 Q36432 R d 1 t . e uce to owest erms 6912, 26415, 2347432 3 " 13 14 (1) Ans. Z. (2) Ans. 4—6. Ans. 28 , EXAMrLEs %, and 2520 1680 ‘1260 1008 840 720 ' 5040’ 5040’ 5040’ 5040’ 5040’ 5040' 6. Reduce to a common denominator 21, 5% , g— and 1;. 336 27 10 24 Ans-‘13, ,16’ To" To" 7. Find the Least Common Mult. of 1, 2, 3, 4, 5, 6, 7, 8, and 9. , Ans. 2520. 8. Find the Least Com. Mult. of 21, 22, 23, and 24. Ans. 42504. 9. Find the Least Com. Mult. of 24, 7, 4, 21, and 14. Ans. 168. . 1 3 4 5 7 10. Reduce to a common denominator 7, g, 5, 9-21., 72. 72 1_8_9 224 105 49 ' 504’ 504’ 504’ 504’ 504' 5. Reduce to a common denominator %, g, i, 21;, Ans Ans 11. What is the value of 1% of half a guinea? Ans. 2s. 3d. 12. What is Z— of'a day—g- of an hour 3’ ‘ Ans. 9b., 105m. 13. What fraction of half a crown is :5;— of 6s. 8d? Ans. 1%. 1 2 3 4 5 14. Add together a , g , 1, g , and -6— . Ans. 3,35, 15. Add together 387;, 285}, 394%, and g of 3704. Ans. 25483,." 16. ,From 201% take 978. Ans. 103% . . 1 1 1 1 1 1 1 1 1 1 l . _ __ _ __ __ _. _. __ ..._ 17. D1V1de 1,, by 1;, and 2+4+6+8+10 by 3+5+9+15 . (1) Ans. 1%. (2) Ans. 15g. 18. Convert 0'08, 000125, and 00078125 into their equivalent - 2 1 1 25’ 866’ 128' 19. Divide 3'1 by 00025; and 365 by 018349 to 6"pla_ces of decimals. _ (1) Ans. 1240. (2) Ans. 1989209221. 20. Reduce 2.451 112d. to the decimal of £1. Ans. 56014895833810. 21. Find the value of 007 of £2. 10s. and expressethe result as the decimal of £1. Ans. 3s. 6d. or £0175. 221 Prove that 0'304565821 is more nearly represented by 030457 than by 0'30456. ' ' vulgar fractions. _ ' Ans. THE ELEMENTS 0F ALGEBRA. DEFINITIONS AND EXPLANATION OF SIGNS. 58. THE method of representing the relation of abstract’ ,, ; quantities by letters and characters, which are made the signs of " such quantities and their relations, is called ALGEBRA. ~ Algebra is the science of generalisation as regards number and mag- nitude. Thus, for example, whilst Arithmetic teaches that the sum of, the two numbers 6 and 4 multiplied by their difference is equal to the difference of their squares, Algebra teaches that the same is true for any two numbers whatever, whole or fractional. ’ 59. Known or determined quantities are usually represented by the first letters of the alphabet, a, b, c, d, &c. and unknown or undetermined quantities, by the last, .0, y, a', w, &c. It must be observed, that this is simply a matter of agreement amongst Algebraical writers, for the sake of convenience, and not essen- tial to the subject. Also, sometimes the letters of the Greek alphabet are used; sometimes A, B, C, D, &c. X, Y, Z; and sometimes others, according to circumstances or the will of the writer. The following signs or symbols are made use of to express the relations which the quantities bear to each other :— 60. + (which is read Plus) signifies that the quantity to' which it is prefixed must be added. Thus a + b signifies that the \’ quantity represented by b is to be added to the quantity repre- sented by a; if a represent 5, and b represent 7, then a+ [2 represents 12. Also a+b+c signifies that the sum of the quantities represented by a, b, and c, is to be taken. ’ If no sign be placed before a quantity, the sign + is under- stood: thus “a signifies +a. Such quantities are called positive, quantities. 61. - (which is read Minus) signifies that the quantity to "which it‘ is prefixed must be subtracted. Thus a-b signifies 3O DEFINITIONS. 4 that I) must be taken from a; if a be 7, and b be 5, a- b ex- presses 7 diminished by 5, or 2.. Quantities to which the sign — is prefixed are called negative quantities. ~ *, or ¥, (the former of which is read plus or minus, the latter minus or plus) signifies that the quantity to which it is prefixed may be either added or subtracted. Thus 6=1=4 is either 10 or 2. 62. x (which is read Into) signifies that the quautities'between which it' stands are to be multiplied together. Thus axb signi- fies that the quantity represented by a is to be multiplied by the quantity represented by bl“. This sign is frequently omitted; thus abc signifies axbxc. Or a full point is used instead of it; thus 1><2><3, and 1.2.3, signify the same thing. . ' But the sign must never be omitted, for obvious reasons, when two or more numerals are to be multiplied together. ' Any quantity which, as a multiplier, serves to make up a product, is called a factor of that product. Thus, of the product 3abc, each of the quantities, 3, a, b, c is a factor; as also each of the quantities 3a, 3b, 3c, 3ab, 3ac, 3bc, ab, ac, be, abc; the former being called simple factors, and the latter compound factors. 63. If in multiplication the same quantity be repeated as a factor any number of times, the product is usually expressed by placing, above the quantity, the number which represents how often it is repeated; thus a, ma, axaxa, axaxaxa, 8:0. have respec-- tively the same signification as a1, a2, a3, a", 8:0. These quantities are called powers,- thus a1, or a, is called the first power of a; a2 thesecond power, or square, of a; a3 the third power, or cube, of a, 8:0. The numbers 1, 2, 3, 8:0. (thus affixed to a) are called the indices of a, or emponents of the powers of a. Likewise a‘, a”, a”, 8:0. are said to be of one, two, three, 8:0. dimensions respectively; and, in general, any product is said to be of 12 dimensions, if the sum of the indices of its several literal factors is equal to n. Thus ab, that is a‘b‘, is of two dimensions; 3:126" is of five dimensions; and $0 on. ’ 64. + (which is read Divided by) signifies that the former of the quantities between which it is placed is to be divided by the latter. Thus a-5-b signifies , that the quantity a is to be divided by b. - * By quantities we understand such magnitudes as can be represented by numbers; we may therefore without impropriety speak of the multiplication, division, &0. of quantities by each other: ' ' ' ‘ ' DEFINITIONS; 3.1 The division of one quantity by another is frequently repre-‘ sented by placing the dividend over the divisor with a line between them, in which case the expression is called a Fraction. Thus g signifies a divided by b*; a b + c _ , + signifies that a, b, e+f+g , and 0 added together, are to be divided by e, f, and g added together. 65. A power in the denominator of a fraction is also expressed byJplacing it in the numerator, and prefixing the negative sign, “ 1 1 1 Es ’ iii ’ 5 respectively; these are called the negative powers of a. 66. The sign ~ between two quantities signifies that their diference is to be taken. Thus a~a', is a - a', or a: — a, accord- ing as a or or is the greater; and aim signifies that the sum or diiference of a and a: is to be taken. 67. When several quantities are to be taken collectively, they are enclosed by brackets, as ( ), { }, Thus (a—b+c)‘ ><(d— e) signifies that the quantity represented by a —- 6 +0 is to. be multiplied by the quantity represented by d—e. Leta stand for 6; b, 5; c, 4; d, 3; and e, 1; then a—-b+c is 6—5+4, or 5; and d—e is 3-1, or 2; therefore (a — b + c)><(d - e) is 5x2, or 10. the denominator, of the fraction; also 1 O O . -1 _2 _3 — O D to Its Index, (thus a , a , a , a 1', slgmfy 51-, Also (ab — cd)><(ab - cd), or (ab — 0d,)2, signifies that the quan-q tity represented by ab - cd is to be multiplied by itself. Sometimes a line, called a oinculum, isdra'wn over quantities, when taken collectively. as (a — b + c)><(d - e). 68. = (which is read Equals or is Equal to) signifies that the quantities betWeen which it is placed are equal to each other; " 511100 5 has already received a dtstmct slgmfication by Definition m Art. 2, It seems scarcely allowable to define it again, as the Author has done here, without showing that the. a '5 proved. It was shewn to be true in a particular case in Art. 11. The general proof will be given hereafter. ED. ‘ two Definitions are coincident. It is true that is equal to a-I-b, but it requires to be and a is the numerator, and 6‘ Thus a - b + Cxd - 8 means the same 32 DEFINITIONS. f": :‘-" P "4 I l p q 1 l I thus am — by = cd + ad signifies that the quantity as -i by is canal to the quantity cd + ad. I 69. The sign > between two quantities signifies that the former is greater than the latter, and the sign < that the former is less than the latter, ‘ The sign signifies tlzertg‘bre, and since or because. ' .70. The square root of any proposed quantity is that quan- tity- whose square, or second power, gives the proposed quantity; The cube root is that quantity whose cube gives the proposed quantity; and so on. ~ “ \ - ' The nth root is that quantity whose n'Lh power gives the proposed quantity The signs \/ or \2/, \3/, \7, 8:0. \7 are used to express the square, cube, biquadrate, &c. nth roots respectively of the quantities before which they are placed. 2 o 3 —— a —‘- fl=a, m=a, \7a‘=a, &c. \/a"=a. , 1 1 1 These roots are also represented by the fractions —, -, ~ , 8:0. ‘ 2 3 4 placed a little above the quantities, to the right. Thus a5“, ai, ai, I - y . » .- a", represent the square, cube, fourth, and nth, ‘root of a, re-. - , 5 7 g spectlvely; as, afi, (15, represent the square root of the fifth power, the cube root of the seventh power, the fifth root of the cube,- of a, respectively. 71. If any of these roots cannot be exactly determined, the: ‘ quantities are said to be irrational, or are called surds. The quantities are called rational, when the roots expressed can be exactly determined. " 72. Certain points are made use of to denote proportion; thus a : b ::- c : d signifies that a bears the same proportion to b that c bears to 61. Sometimes this is written thus, a : b :0 : d. 73. The number prefixed to any quantity, as a factor, and which shews how often the quantity is to be taken, is called its coefi‘icient. Thus, in 2a, which signifies twice a, the coefficient of a is 2; and in the quantities 7am, Gby, 3dz, the numerals 7, 6 and S are called the coefiicients of are, by, and dz, respectively. When no number is prefixed, the quantity is to be taken once, or the coefficient 1 is understood. ~ A fraction is not excluded from being ascqflicz'ent. Thus in gs: the x I . I coeficzent of a 1s 5. DEFINITIONS. 33 These numbers are sometimes represented by letters, which are also called coeflicients. Thus in the quantities pa“, qr”, rm, we call p, q, and r the coefii- cients of was”, and a, respectively; since they may be read p times a“, q times x2, and r times a, respectively. In fact coeflicient simply means co-factor; so that in the product ab, for example, a is the coefficient of b, and b is the coefficient of a. 74. Similar or like algebraical quantities are such as differ only in their coefficients; 4a, 6ab, 9a2, 3a2bc, &c. are respectively similar to 15a, Bab, 12a2, 15a'2bc, 81cc. ' Unlike quantities are different combinations of letters; thus ab, agb, abc, &c. are unlike. . But a distinction must be made in those cases where letters are taken to represent coqfl‘icients; for as“ and pa:3 are like quantities when a and p' are coefliczents of a“. ' - --\. 75. A quantity is said to be a multiple of another, when it contains it a certain number of times exactly; thus 16a is a multiple of 4a, as it contains it exactly four times. ' 76. A quantity is called a measure of another, when the former is contained in the latter a certain number of times exactly; thus 4a is a measure of 16a. When two numbers have no common measure but unity, they are said to be prime to each other. ' Thus 3 is prime to 7; 13 to 31'; and so on. ‘A prime number is one which is prime to every other number. 78. A simple algebraical quantity is one which consists'of a single term; as 4a, or- aibc, or 6mg, &c. _A.compound algebraical quantity consists of more terms than one—- the number of terms meaning the number of quantities connected together by + or —; as a+b, or 2a—3x+4_y, &c. A binomial is a quantity consisting of two terms, as a+b, or 2a - 3ba'. A trinom-ial is a quantity consisting of three terms, as 2a + bd - 30. A polynomial or multinomial is a quantity consisting of many terms, as a+bx+ca’+d._r3+&c. The following examples will serve to illustrate the method of representing quantities algebraically. Let “a=8_, b'=7, c==6, d=5, angle=1; then "-J...“_..punf ,/ 4‘4 84 , 'ADDI’l‘ION. Ex.1. sa-2b‘+4c-e=24-Ia+24-1=sa. Ex.2. ’ab+ce—bd=56+6“-35=27._ a+b sb-ec 8+7 21-12 15 9 ' c-e+a-d_6-1+s-_-5_5+e Ex. 4. d2><(a - c) -- 3ce2+ d3= 25x2 - 18 +125, =5o-Is+125=157. Ex. 5. ,/a2—3d><,3/63—03—2€=~/I§X,3/125=7X5=35. [The student is recommended at this stage to test the accuracy of his ' knowledge of the preceding Definitions by working out the EaercisesA, placed at the end of the book] _ AXIOMS. 79. If equal quantities be added to equal quantities, the sums will be equal. ' 80. If equal quantities be taken from equal quantities, the remainders will be equal. 81. If equal quantities be multiplied by the same, or equal quantities, the products will be equal. ‘ 82. If equal quantities be divided by the same, or equal quan, tities, the quotients will be equal. 83. If the same quantity be added to and subtracted from another, the value of the latter will not be altered. 84. If a quantity be both multiplied and divided by another, its value will not be altered. ADDITION OF ALGEBRAIGAL QUANTITIES. 85. RULE. The addition of algebraical quantities is per- formed by connecting those that are unlike with their proper signs, and collecting those that are similar into one sum. Add together the following unlike quantities; Ex. 1. as." +e2 Ex. 2. a +‘2b-c -ed d-5e+f Sum=aw—by+e2—ed Sum=a+ab_-¢+d_5e+ffl ADDITION. 35 It is immaterial in what order the quantities are set down, if we take care to prefix to each its proper sign. " Generally speaking, however, it is convenient to arrange algebraical quantities in the order in which the letters occur in the alphabet. When any terms are similar, they may be incorporated, and the general expression for the sum shortened. 15‘. When similar quantities have the same sign, their sum is found by taking the sum of the coefficients with that sign, and annexing the common letters. Ex. 3. 5a— 3b Ex. 4. aaic-lObde I 4a- 7b Gaic- Qbde llagc- 3bde Sum = 9a - 10b “w Sum = Qla‘ic - QQbde The reason is evident; 5a to be added (Ex. 3), together with 4a to be added, makes 9a to be added; and 3b to be subtracted, together with 7b to be subtracted, is 10b to be subtracted. 2“. If similar quantities have diferent signs, their sum is found by taking the difference of the coefficients with the sign of the greater, and annexing the common letters as before. Ex._5. 7a+3b EX.6_ in??? -5a-9b 5 2 -———-— _.lx+l Sum =2a—6b 5 4y '3 1 Sum_5.r-—Z_y In the first part of the operation 5) we have 7 times a to add, and 5 times a to take away; therefore upon the whole we have 2a to add. In the latter part, we have 3 times b to add, and 9 times b to take away; therefore we have upon the whole 6 times b to take away; and thus the sum of all the quantities is flat - 6b. Ex. 7. a+ b - Ex. 8. l—x-m’ 1 +.2x-2.r’ a-b Sum-~=2+.r--._r2 Sum = 2a 36 SUBTRACTION. It must be borne in mind that when any quantity, as a, or a, or .r’, has no‘coefficient expressed, the coefficient 1 is understood. ' 36‘. If several similar quantities are to be added together, some with positive and some with negative signs, take the difference between the sum of the positive and the sum of the negative coeffi- cients, prefix the sign of the greater sum, and annex the common letters. Ex. 9. 3a2+4bc- e2+10 —-5a2+6bc+ 262—15 -4a2-9b0-1062+21 Sum =—6a2+ bc— 9e2+16 The method of reasoning in this case is the same as in' Ex. 5. Ex. 10. _, 4ac-15bd+ ea' ‘ 11ac+ 7b" -19e:c -411a2+ 6bd— 7de Sum =15ac —41a2— 9bd + 7b2 - 18cm - 7de Ex. 11. p.r3-g.r2-ra' as:3 .- bar:2 - a: Sum =(p+a)ar°'- (q+b)a:2-(r+1)a' In this example, letters are taken to represent coefficients, and the coeflicients of like powers of a are enclosed within brackets ; for it is evi- dent that p times .223 together with a times as is the same as (p+a) times as; also q times a2 to be subtracted together with b times a” to be sub- tracted is the same as (q+b) times .222 to be subtracted; and r times a: to be subtracted together with x, or 1m, to be subtracted, is (r +1) times a: to be subtracted. ‘ [Exercises B.] SUBTRACTION. 86. RULE.“ Subtraction, or the' taking away of one quantity from another, is performed by changing the sign of the quantityto be subtracted,. and then adding it to the other. by the rules laid down in Art. 85. Ex. 1. A “Fr0m_2b.r take cy, and the'difference is properly represented by Qba-cy; because the - prefixed to cy shews that ADDITION AND SUBTRACTION BY BRACKETS. 37 it is to be subtracted from the other; and Qbar—cy is the sum of QM? and —-—cy, (Art. 85). ' Ex. 2. Again, from Qba' take -cy, and the difference is Qbtv'i- cy; because 2bw=2b.c+cy—cy, (Art. 83); take away -cy from these equal quantities, and the differences will be equal (Art. 80); that is, the difference between Qba' and -cy is 2b.r+cy, the quantity which arises from adding +cy to Qba’. ‘ Ex. 3. From a + b ' Ex. 4. From 6a - 12b take a - b take - 5a — 10b Difference = + 2b Diff. = 1 1a - 2b Ex. 5. From 5a2 + 4ab .- EX. 6. From 70—26 +410—2 take 1‘] a2 _+ Gab - My take fia—fib +4c- 1 v i = a+4b-1 Diff. = -6a2-2ab—2.ry. Ex. '2'. From 4a-3b+60—11 Ems, From x+1y+1 take‘ IOa'+a—l5—Qy, 2 1 1 -. k *- - Diff. =-10a*+3a-3b+4+60+2y ta 6 2x+y+2 . 1 1 1 Dlfi‘. -§$—-éy+'2- EX. 9- @8173 "" [£172 + (I? take p.103 - q.c2*+ ra' Diff. = (a —p)ar3 — (b — q)at'2 + (1 - HI? In this example the coefficients of like powers of a: are bracketed, for reasons similar to those given in Ex. 11, Art. 85. ' [Exercises (3.] ADDITION AND SUBTRACTION BY BRACKETS. In actual practice it seldom happens that either Addition or Subtraction of Algebraical quantities is presented to us as in the Examples, Art. 85, and 86. All the quantities concerned are more commonly in one line, and are so retained through the whole operation, for the sake of convenience. . This arrangement renders necessary the frequent use of Brackets. Thus Ex. 3. Art. 85 would stand (5a—3b)+(4a-7b)=9a-106. Ex. 3. Art. 86 ............. .. (a+b)-(a-b)=2b. ' 38 ADDITION -AND SUBTRACTION BY BRACKETS. -~ In the management of Brackets much care is needed,'and the follow- ing rules are to be observed:— 87. RULE I. If any number of quantities, enclosed within Brackets, be preceded by the sign + , the brackets may be struck out, as of no value or signification. RULE II. If any number of quantities, enclosed within Brackets, be preceded by the sign-, the brackets may be struck out, the signs of all the quantities within the brackets be changed, namely + into -, and -- into +. Rule I. is obviously ‘true; for in this case all that is meant is, that a number of quantities are to be added; and it can clearly m'ake noldifference whether they be added collectively or separately. - Thus a+(b+c) is equiva- lent to a+b+c; for the former signifies that the sum of b and c is to be added to a, which is evidently the sum of a, b, and c. Also a+(b-c) is equivalent to a+b—c; . for the former signifies that a quantity is to be added to a less than b by the quantity 0 ; and the latter, that when b has been added to a, 0 must be subtracted, which is evidently the same thing. Rule II. is proved thus :— Let a, b, 0 represent any Algebraical quantities, simple or compound, of ‘which b+c is to be subtracted from a ; this will be expressed by a—(b+c). Now if from a the portion b be taken, the result is a—b ; but there is not enough subtracted from a by the quantity 0, since b+c was to be sub- tracted. Therefore 0 must also be subtracted, which leaves the result a-b—c; that is, ' a—(b+c) =a—b—c. Again, if b-c is to be taken from a, this will be expressed by a—(b—c). . _ , Now if from a the quantity b be taken, the result is a-b, but there has been too much taken away by the quantity 0, since b—c only was to be subtracted; therefore 0 must be added, and the result becomes a—b+c ; that is, a—(b—-c)=a-b+c. The preceding rules apply also to quantities held together by a vincue lum, since a vinculum serves the same purpose as brackets. Art. 67. N.B. It is immaterial whether a vinculum or brackets be used in any case, that being dependent solely upon the will of the writer: but in some cases it is necessary, for distinction’s sake, to use both at the same time, or else two kinds of brackets. Thus, to express a times the difference between b and c— d, we must write either a.(b—c—d), or a.{b-(c—d)}. Conse- quently it requires to be especially noted, that in all cases when the Student meets with ( or { or [, he must look, whatever may intervene, for the counterpart ) or } or] respectively; and all that is included within the complete bracket must be treated, irrespective of other brackets or vincula, as the sign which precedes it directs. SO that in striking out brackets by Rules I and II, each pair of symbols, as (), { }, [ 1, must be Struck out separately, ‘and not all confusedly‘ and‘at once. MULTIPLICATION. 39 A few examples will make this clearer. Ex. 1. Perform the addition expressed by (a+ b) +(a—b). (a+b)+(a-b)=a+b+a—~b, by Rule I, :26!- Ex. 2. Perform the subtraction expressed by (a+b)-(a—b). (a+b)-(a—b)=a+b-~(a—-b), by Rule I, =a+b—-a+b, by Rule II, =2b. Ex. 3. Simplify a—(x—a)-{.r-(a—.r)}. a-(x—a)—{.r-(a-.r)}==a—.r+a—x+(a-.r), =a—m+a-x+a—x, =3a-3x. Ex. 4. Simplify 1—{1—(1—Tl—d)}. __ 1-{1-(1-i:5.~)}=1-1 +(1-T-17v), =l—1+1——I:‘d’, = 1—1 +1 -1 +11. =.r. Ex. 5. Simplify -[—{-(—-a)}]. _[_.{_(_a)}]=+{—(—a)}, by Rule II, =-(-a), by Rule I, =a, by Rule II. MULTIPLIOATION. 88. The multiplication of simple algebraical quantities must be represented according to the notation pointed out in Art. 62. Thus axb, or ab, represents the product of a multiplied by b; abc the product of the three quantities a, b, and c; and so on. It is also indifferent in what order they are placed, axb and bxa being equal. , For 1><1; and c><+'b=+ab; because in this case a is to‘be taken. positively b times; therefore the product ab must be positive. 2d. —ax+bl= —ab; because -a is to be taken b times; that is, we must take -ab. 3d. +ax—b=—ab; for a quantity is said .to be multiplied by a negative number —b, if it be subtracted b times; and a sub- tracted b times is -ab. . This also appears from Art. 92. Ex. 2. 4th. -ax-b=+ab. Here -a is to be subtracted b times, that is, — ab is to be subtracted; but subtracting —ab is the same as adding +ab (Art. 86); therefore we have to add +ab. The 2d. and 4th cases may be thus proved; a - a=0, multiply these equals by b, then ab together with -—_a><5b=15ab; because 3xax5xb=3x5xwxb=l5ab (Art. 83). Again, arex-11y= -~M.vy; —9bx-d5c=+45bc; ' and - 6dx4m = - 24md. ~ 91f- The powers of the same quantity are multiplied together ;by adding the indices; thus a2xa3=a5, for aaxaaa=aaaaa. In :the same manner a6xa1°= a1“; and --3a2rr”><5a.vy2 = -15a3a4y'"’.' . “‘ It is a common mistake of beginners to say that an algebraical expression which .appears under the form axO is equal to a, by supposing it to, signify a not multiplied at all; whereas, since axb {signifies a taken b times, in the same manner ax0 signifies, a‘ taken 0 times, and is therefore equal to 0.--ED. ‘ ‘ MULTIPLI-cATION. 41. T o prove generally, that amxa“=am+”, m and n being any positive integers. . By Def. Art. 63. am=axaxaxax &c. continued to m factors; i also a"=axaxaxax &c. ............. .. n ...... ..; a amx a”=axaxa...to m factorsxaxaxa...to n factors, =axaxa...to M factors, _=a"‘+“, by Def. Art. 63. _ It will be proved hereafter that the same rule holds when and n are either fractional or negative. (Arts. 132, 162.) Also a'“b"xaPb’ =a’". a”. b". b9=am+Pb“+9. 92. If the multiplier or multiplicand consist of several terms, each term of the latter must be multiplied by every term of the former, and the sum of all the products taken for the whole product of the two quantities. _ = ' 2 Ex. 1'. Mult. a'+b‘ ~~ - by c+d l Prodt. = ac + bc + ad + bd _ Here a +b is to be taken 0 + d times, that is, 0 times and d times, or (a + b)c + (a + b)d. Ex. 2. Mult. a+ b by c- d : Prodt. = ac + be - ad _- bd Here a + b is to be taken 0 - d times, that is, 0 times wanting d times; or 0 times positively and d times negatively; that is, (a + b)c - (a+ b)d, or ac + be - (ad+ bd), or ac + be - ad - bd, Art. 87f. Ex. 3. Mult; a +b Ex. 4. Multi a +‘b‘ by a + b _ by a -b Prod‘. by a = a2 + ab a2+ ab ... by +b-----f+ab+b2 _ —ab-b" Whole prodt. = a2 + 2ab + b2 Prodt. a2 — b2 42 MULTIPLIOATION. Ex. 5. ' Malt. ’ ea2 - 5bd '8 EX. 6. Mult- w+a by — 5a2 + 4bd by - :——4—_—2 marl-ax _ 115a + 25a bd 2 v + bx} ab 4‘ 12a2bd "‘ 20b d2 Prodt. =a'2-j-. (11+ b)a'+ab Prodt. = —15a‘ + 37a2bd ~20b2d“ Ex. 7. Mult. a2+2ab +b8 Ex. 8. Mult. a’"+.r ' by ag—Qab +b2 by h 4"“ ———-——-—--., Prod‘.=.z:"‘+”+ a?“ a4 + 2a3b + ab2 ~——-—-—— - 2a3b -'4.a‘~’b2- flab“ + a2b2+ 2ab"+ b“ Prodt. = a4 — 2a2b2+ b‘ EX. 9. 1 ‘— w + 092 --ll/3 EX. 10. ha'ult. I s. by + w - by 2x—3y 1-.r+.v2;-a23 ‘ ,__2_x +tv-sc2+a'3-.r‘ x 3 y 3 2 Prod“. = 1 — .r4 " 5*”! +9 Prod". = .r’— .ry +y’ Ex. 11. Mult. a2-pa+q My by w + a .503 -pa;2 + gar + an2 — apm + aq 'Prodt. == mi — (p - a).v2 + (q - ap).v + aq Here the coefficients of .222 and a' are bracketed, since -'(p-- a).v'*' '= - pa? + aa"; and (q — ap).v = gar — aprc. Ex. 12. , Mult. and” + na” by no" + ma" mnag’" + nga’w‘ + m”a”‘+"+ mna’" ~ Prod“. = mna’m + (m2 + ng)a'"+" + mnag" . [Exercises D.:| MULTIPLICATION. 43‘ i It may be useful to exhibit the Rules for Multiplication as follows :— RULES. p(a+b)=pa +pb ................... ..(l), p(a —b)=pa-pb...l ................ ..(2), ~(a+b)(c+d)=ac+bc+ad+bd ......... (a+b)(c-d)=ac+ bc—ad—bd ........ ..(4~), (a—b)(c- d)=ac-bc—ad+bd ........ Assuming ( l) and (2), which are too obvious to need a proof, to prove (3), let a+b=m, then, (a+ b)(c+d) =m(c+d), =mc+md, by (1), =(a+b)c+(a+b)d,~ =ac+bc+ad+bd, by (1). Similarly for To prove (5) let a-b=m, then e-vc-o=ms-o. =mc—md, by (2), =(a—b)c-(a-b)d, =ac—bc—(ad—bd), =ac—bc—ad+bd, by Art. 87. N.B. The Rules for the management of Brackets, given in Art. 87, apply only to the Addition and Subtraction of quantities so enclosed. If a collection of quantities within brackets is to be multiplied or divided by any quantity or collection of quantities, the brackets must not be struck out until the multiplication or division is actually performed. Thus (a+b)>< (c+d) signifies that a+b is to be taken c+d times, and is obviously not the same as either a+b(c+d), or (a+b)c+d. Again, (a+b)—:-(c+d) is not equbivalent to either a+b+(c+d), or (a+b)+c+d; but it may be written a+ m, a vinculum to both. . The learner would do well to practise multiplication of quantities by means of brackets as early as possible. Thus, Ex. 1. (a-b)(c-—d)=(a—b)c-(a-b)d, =ac—bc—(ad— bd), =ac—bc—ad-I-bd. Ex. 2. (a: + a)(.r+b)=(x+a).r+(x+a) b, ' =a2+aw+bx+ab, =x’+(a+b).v+ab. Ex. 3. (x+ 1)(Ir+2)(a:+3) =(eJIIa-t2Xa-t 2). (Ex. 2.) = (13+ 3x+2)x+(a2+31r+2)3 5 == a"+' 3x2+ 21‘ +3x2+ 9x + 6, “ =aa+ 6a:2+1l.r+6. the line which separates the numerator and denominator serving as 44 MULTIPLIOATION. It is also useful to commit to memory at an early stage the three fol- lowing results; as will appear from the subjoined Examples :— (A +B). (A +8) or (A +B)’=A’+B2+ 2A B ....... . . (i) (A -B).(A -B) or (A —B)2=A’+B’—2AB ....... . . (ii) (A +B).(A —B)= A2- B2 ............. . . (iii) whatever quantities A and B may represent, simple or compound. Ex. 1. (ar+by)(a.r+by)=(a.r)2+(by)2+2.a.r.by, by (i), ‘ =a2m2+ b2y2+ Qabxy. Ex. 2. (ax—by)(ar—by)=(a.r)’+(by)2——2.a.r.by, by (ii), =a’w’+b2y2—-2ab.ry. Ex. 3. (ax+by)(a.r-by)=(aa)”-(by)2, by (iii), =a2x2—b2y2. Ex. 4. (ax+b+cy+d)2=(dm+W)2, = (as: + b)’+ (cy + d)2+ 2(ax+ b)(cy + d), = a2x2+ b2+ Qabx + c”y’+ d9+ Qcdy +2acny+2adr+2bcy+2bd Ex. 5. (a+b+c)(a+b—c)=(d1b+c)(efib—c), =(a+b)’-c*, by (iii), = a2+ b2+2ab —cf. Ex. 6. (b+c—a)(c+a—b)=(c-h:b)(c+d:b), =09—(a—b)“, by (iii), “Lana-an), =02— a2— b2+ 2a b. SCHOLIUM. The method of determining the sign of a product from the consideration of abstract quantities has been found fault with by some algebraical writers, who contend that --a, without reference to other quantities, is imaginary, and consequently. not the object of reason or demonstration. In answer to this objection we may observe, that whenever we make use of the notation —a, and say it signifies a quantity to be subtracted, we make a tacit‘ reference to other quantities. Thus, in numbers, -a represents a number to be subtracted from those with which it is connected; and when we suppose —a to be taken b times, we must understand that a is to be taken b times from some other numbers. In estimating lines, or distances, -a represents a line, or distance, in a particular direction. The ative sign does not render quantities imaginary or impossible, but points out the relation of real quantities to others with which they are concerned. DIVISION. 45 DIVISION. 93. To divide one quantity by another is to determine how often the latter is contained in the former, or what quantity ‘mnl; tiplied by the latter will produce the former. ‘ _ 6 Thus to divide ab by a is to determine how often a must be taken to make up ab, that is, what quantity multiplied by a will give ab; which we know is b. I From this consideration are derived all the rules for the division of algebraical quantities. The words Dividend, Divisor, and Quotient, have the same meaning here as in common Arithmetic. 94. I f the divisor and dividend be affected with like signs’, the sign of the quotient is ..+ ; but if their signs be unlike, thesign of the quotientis —. Thus, ’ If -ab be divided by - a, the quotient is + b; because -a x+b gives -ab; and a similar proof may be given in the other cases. 95. To divide one SIMPLE quantity by another. RULE. In the division of' simple quantities, if the'divisor be found as-a factor in the dividend, the other part of the dividend, with the sign determined by the last rule, is the quotient. “ This is obvious, since QuotientxDivisor='Dividend in all cases. Thus —7b+b=- 7; -ax+—a==x; 14ab+7b=2a.7b+7b=2a. Also abc—Z-ab=c; because ab multiplied by 0 gives abc. If we first divide by a, and then by b, the result will be the safne; for a50+a=b¢3 and be-;-b=c, as~before. " a ' (30R. If any POWER of a quantity be divided by any other POWER of the same quantity, the quotient is the same quantity with an indem which is found by taking the indem of the divisor from the indew of the dividend. Thus a5-.'.-a3=a2>n, a'"+a"=a’”“"xa"-:-a"-=a”“". Similarly 6a456+3a’b’=261253><361263+3a263=2a263; and (a’"b")+(a*’b’)= a’”‘Pb"‘Q. apbq -:- aPZfl = rim-Pb”. DIVISION. 96. LEMMA. To shew that a-z-b is equal to the fraction %. According to the definition of a ‘fraction’ the unit is divided into 6 equal parts, and a of them are taken, to make the quantity represented by %. Now, each of these parts is clearly the b'h part of the~unit, and is equal to a times the b‘h part of 1, =2, suppose. Multiplying these equals by I), (Art. 81), bxa times the b‘h part of 1, or axb times the 6‘“ part of 1=b.v ; but 6 times the b“ part of 1 is clearly 1, a=bx. Now let a+b=y, then bypD‘efinition (Art. 98), a=by, bx=by, or w=y, (Art. 82) ; that is -Z-=a—:-b. 97. In dividing one simple quantity by another, if only apart of the product which forms the divisor be contained in the dividend, the division must be represented by a fraction according to the direction in the last Art., and the factors which are common to the divisor and dividend expunged. 15a3bgc - 5abc W _ a: For, 1st, divide by -Saib, and the quotient is -5abc; this quantity is still to be divided by as (Art. 95), and as a' is not con- tained in it, the division can only be represented in the usual way ‘; Thus 15a3b20 -I— — 3a2bw = . c . . that Is, Is the quot1ent. l 98. To divide a quantity of two or more terms by a simple quantity. RULE. If the dividend consist of several terms, and the divisor be a simple quantity, every term of the dividend must be separately divided by it. Thus to divide a3a2 - 5aba3+ 6aa'4 'by ad”, , asa'2 5aba'3 6am4 Quotlent = '2 2 = a2 - 5ba' + 6002. am" am am a b c I 1 1 azfiatar'atzzta- 99. To divide one quantity by another when the DIVISOR consists of two or more terms. Also (a+b+c)+abc= RULE. When the divisor consists of .several terms, arrange both the divisor and dividend according to the powers of some one letter516 contained in them, (that is, beginning with the highest g r The operation will be shortest when that letter is chosen whose highest power in the ‘dividend comes nearest to the highest power of the same letter in the divisor; and the same arrangement according to the powers of that letter must be kept up throughout the whole operation.--ED. ' DIVISION. power and going regularly down to the lowest, or vice versii,) then find how often the first term of the divisor is contained in the first term of the dividend, and write down this quantity for the first . term in the quotient; multiply the whole divisor by it,.subtract the product from the dividend and bring down to the remainder as many other terms of the dividend, as the case may require, and repeat the operation till all the terms are brought down. Ex. 1. If a2 - Qab + b2 be divided by a — b, the operation will be as follows: i i a - b)a"’-.2ab+ biba- b ag- ab m -_- ab + b2 The reason of this and the foregoing rule is, that as the whole dividend is made up of all its parts, the divisor is contained in the whole as often as it is contained in all the parts. In the preceding operation we inquire first, how often a is contained in a2, which gives a for the first term of the quotient; then multiplying the whole divisor by it, we have a2— ab to be subtracted from the dividend, and the remainder is - ab + b”, with which we are to proceed as before. The whole quantity a“z — flab + b2 is in reality divided into two parts by the process, each of which is divided by a - b; therefore the true quotient is obtained. Ex. 2. a+b)ac+ad+bc+bdke+d ac + be ad + bd ad + bd Ex.3. 1-.v)1 k1+a'+.v2-i-.v3+&c.+il—Eil-IiEIZv—dgE 1 -- a' +66 +w—fi +w2 +w2—a'3 +ab“ +£3'-a/4 . -" '~ .+.v4&c. 48 DIVISION. Ex. 4. y—I)y3—1(y2+y+1 213—21“ +f—1 +¢—y ‘ +y-1 +y—1 From this example it appears that ya—l is divisible by y—Al without remainder, the quotient being y’+y+1. It may be shewn in the same manner that sci-a“ is divisible by x—a, the quotient being .r’+a.r+a2; and that ems is divisible by x+a, the quotient being w2—aa:+a’. These results are worth remembering. > _Ex. 5. To divide eab3+51a’b3+10a‘—48a3b—1§b‘ by 4ab—5a’+3b’. First arrange the terms of both dividend and divisor according to, the powers of a, beginning with the highest. -5a”+ 4ab +3b’) 10a‘-48ai’b + 51a”b’+ tabs—1 5b‘ (-2a’+ 8ab- 5b2 10a‘—- Saab -— 6a262 - 40:23!) + 57 a262+ Ital)8 —40aab+32a’b’+ 24:16“ s. 256262— 20053—1 55‘ 250252—20053—1 512‘ EX, 6, -ymkwm-1 +Iwm—2y+wm—3y2...+y’m-1 w’" - arm-13,1 + wm-ly _ym + mm-ly _al,m-2y2 + {cm-23]? — mm’3y3 This division will obviously terminate without remainder for any proposed integral and positive value of m, when the quotient has reached to m terms, the last term being 31”“. Hence we have, (f—y‘)+(w—y) =w’+ we +y’. (w‘—y‘)+(=v—y)=w’+w’ywere“: ' (x‘-y‘)-:—(x-y)=x‘+.r3y+x’y’+ay°+y‘; and so on. FRACTIONS. 49 Ex. 7. .v—a)a13—pa'2+q.v-r &v2+(a—p).r+a2-pa+q :03 — aa'2 (a new" + as (a "10);"2 - (a2 - W)” + (a2 --pa +q).v —r (a2 - pa + q)a" — (a3 -pa2 + qa) Remainder a3 —pa2 + qa — r [Exercises See NOTE 1. TRANSFORMATION OF FRACTIONS TO OTHERS OF EQUAL VALUE. 100. If the signs of all the terms both in the numerator and denominator of a fraction be changed, its value will not be altered. _ b +ab For a =-ab-:--a(Art.96)=+b=+ab—}-+a= ; -a +a I) _ -—a +a Ex a—a' -a+.r x—a b—w=—b+x=x—b' 101. If the numerator and denominator of a fraction be both multiplied, or both divided, by the same quantity, its value is not altered. For in g- the unit is divided into 6 equal parts, and a of them are taken, in g- . . . . . . . . . . . . . . . . . . . . .bc . . . . . . . . . 0 times as many as before, 0 and a of them taken; therefore in the former case each part is 0 times as great as in the latter. But if 0 times as many of the smaller as of the greater parts be taken it is obvious that the result in either case will be the same—that is, that ac parts of the latter are equal to a parts of the ac a former, or -__-=__, bcb Hence a fraction is reduced to its lowest terms by dividing both the numerator and denominator by the greatest quantity that measures them both. This quantity is called the Greatest Common Measure, or Highest Common Divisor, of the numerator and denominator. 102. A fraction which has either its numerator or denominator a simple algebraical quantity is easily reduced to lowest terms; for the great- est common divisor of the numerator and denominator is at once found 4 5O GREATEST COMMON MEASURE. by inspection, that is, by observing at sight what factors are common. a” c . . Thus to reduce W to Its lowest terms, we see that a2b Is the greatest common measure of the numerator and denominator; therefore the fraction re uired is 30 q 5abd' 2a 4- 5am 2 + 5.1: a2 Ao‘ain -—‘-—- is at once reduced t ' nd -—-—— to -—- . b ’ 7a 0 7 ’a a2+ab a+b But in the case of fractions having for both numerator and denomi- nator a compound algebraical quantity the following rule is often, though not always, needed. 103. The Greatest Common ZlIeasure of two compound alge- braical quantities is found by arranging them according to the powers of some letter (as in division), and then dividing the greater by the less, and the preceding divisor always by the last remainder, till the remainder is nothing; the last divisor is the greatest common measure required. Let a and b represent the two quantities, and b)abp let b be contained p times in a, with a remainder pb 0; again, let 0 be contained q times in b, with a Ub(q remainder d; and so on, till nothing remains; let qc d be the last divisor, and it will be the greatest ECU. common measure of a and b. ml 104. The truth of this rule depends upon 1)— these two principles; 1“. If one quantity measure another, it will also measure any multiple of that quantity. Let a' measure y by the units in n, then it will measure my by the units in mn; for since y=n.v, my=mnw(Art. 81) =mn.w, that is, a; is contained mn times in my, or measures my by the units in mn. 2‘1. If a quantity measure two others, it will measure their sum or difference. Let a be contained m times in a', and n times in y; then ma=a~, and na=y; therefore aiy=maina= (m&n)a; that is, a is contained m =l=n times in w=!=y, or it measures .v-J-=y by the units in min. 105. Now it appears from the operation (Art. 103) that a—pb=c, and b—qc= d; every quantity therefore, which mea- sures a and b, measures pb, and a—pb, or 0; hence also it measures qc, and b—qc, or d; that is, every common measure of a and b measures d. GREATEST COMMON MEASURE. 51 It appears also from the division that a=pb+c, b=qc+d, c=rd; therefore d measures 0, and qc, and qc+d, or b; hence it measures pb, and pb+c, or a. Every common measure then of a and b measures d, and d measures a and b; therefore d is their greatest common measurei“. Ex. 1. To find the Greatest Common Measure of a2+2a+1 and I a'i-t-Qa +1 . a3+2a2+2a+1; and to reduce T—-,———-—— to Its lowest terms. a +2a +2a +1 ag+2a+lj a3+2a2+2a+1 (a a3+2a2+a a+1 a+1) a2+2a+1 (a+1 a2+a a+1 a+1 a+1 is therefore the Greatest Common Measure of the two quantities ; and if they be respectively divided by it, the fraction is .reduced to 721211 , and is in its lowest terms. Ex. 2. To find the Greatest Common Measure of a4—w‘ and 3 2 2 3 a -a.v-a.v +a _ a3-a2.v—a.v2+.v3; and to reduce 4 4 to us lowest a -a terms. a3-a2a—aa'2+.v3) a4 ---.v4 ka + a' a4 - a3a - agar:2 + are3 a3a' + agar2 - are3 - .v‘ asa’ — agar2 - are3 + .v‘ 2a2.v2 - 2:04 = 2.222(a2 -- we), leaving out 2:02, the next divisor is a2 — .vg. ag-ay as—aim— amz-i-ws (pi-w a3 - as?2 - a2x + a'3 - aew + .v3 I " This conclusion is more obvious when stated thus :-every common divisor of a and b is a divisor of d, but no quantity can be a divisor of d which is greater than d, therefore every common divisor of a and b is not greater'than d; and since dis one of them, therefore d is the greatest. ED. ' 4-2 52 GREATEST COMMON MEASURE. Therefore ai—a:2 is the e.c. M. required; and the fraction Is 2 2 a reduced to , and is in its lowest terms. The quantity 2a”, found as a factor in every term of one of the diViSOI'S, Qagwf —2w4, but not in every term of the dividend, aa-aa’w—aw2+a'3, must be left out; otherwise the quotient will be fractional, which is contrary to the supposition made in the proof of the rule: and by omitting this part, 2.122, no common measure of the divisor and dividend is left out, because, by the supposition, no part of 2m2 is found in all the terms of the dividend. [Exercises 106. The proof of the rule for finding the G.C.M. of two algebraical quantities (Art. 103—5) excludes every case in which any one of the quotients p, q, r, is fractional: for mb is not considered a “multiple” of 6, unless m be either a whole number, or free from terms in a fractional form. But the fractional quotients may always be avoided by rejecting certain factors which can be detected by inspection. (1) Let such factors be common to the proposed quantities; so that a= ha'; b=kb', where It contains all such common factors. Then, as all factors that are common to a and b, excepting k, are also common to a' and b’, and all that are common to a' and b' are also common to a and b, the G.C.M. ofa and b will = hxG.c.M. of a' and b’. Whence we see that such a common factor as k may be neglected, provided we multiply it into the G.o.M. that will afterwards be obtained. (2) Let such factors be not common; and let a' = ma”, b’ = nb”, where m and n contain all the factors that can be detected by inspection; a" and b” consequently have no such factors; therefore m and n have no factors in common with b” and a”, and by hypothesis they have none in common themselves; therefore all factors common to a' and b' are also common to a" and b", and conversely: therefore the G.C.M. of a’ and b’=G.c.M. of a" and b". The factors m and n therefore may be entirely rejected. From this also it is evident that such factors as the above can be introduced into one of the proposed quantities, provided that they do not contain any factor that already appears in the other. If then we dispose at once of the factors that we can detect by inspec- tion, according to the foregoing remarks, we can proceed to find the G.C.M. as follows : b) a tr pb c = mc' suppose 0') 6 Le 22' d = nd' suppose d') c’br rd: 0 GREATEST COMMON MEASURE. 53 where m and n consist entirely of such factors as have been considered, or are not either of them a “measure” of the succeeding dividend. Now every common measure of a and b measures a-pb or c, and therefore 1s a common measure of b and 0; also every common measure of b and 0 measures c+pb or a, and therefore is a common measure of a and 6. Therefore the G.C.M. of a and b is also the G.C.M. of b and 0. But as m contains no factor that appears in b, the G.C.M. of b and 0 will be also the G.C.M. of b and c’, by what has been said above. And this, by the same reasoning as before, is also the G.C.M. of c' and d, and therefore also of c' and d’; and so on: d’ being the last divisor, is evidently the G.C.M. of c’ and d’; and therefore it is the G.C.M. of a and b. All this reasoning will equally hold good, if we introduce such factors as m and n in the course of the operation, provided only that by such introduction they do not become common to the divisor and dividend; for otherwise the G.C.M. of the divisor and dividend, which has been proved to be the same as that of the proposed quantities, would be increased by such common factors, and the result would consequently be erroneous. Care must be taken therefore, lest in rejecting a factor, we reject a part of the G.C.M.; as also in introducing a factor, lest we introduce one which will increase the common factors of the proposed quantities. From what has been said, it will be seen that every common measure of a and b will measure c', and by the same reasoning, WIll measure d’, and so on: i. e. every common measure of a and b will measure their G.C.M. Ex. 1. Required the G. C. M. of 9x3+ 53x2—9x-18 and x2+11x+30. xii-11.2; +30) 9003+ 53xi—9a2—18 (gr—46 Qx3+ 99x2+ 270.2: — 46.135 279x—1 8 -—4.~6.t‘2—- 506.17—1380 227x+1362=927(w+6); 227 is evidently not a common divisor of the two proposed quantities, and may therefore be rejected. .r+6) x2+11x+30 (aw-5 x2+ 6.1: 5x+ 30 5.2: +30 '. .r+6 is the o.C.M. required. Ex. 2. Required the G.C. M. of 2x3+ .rz— 8x+5 and 7a’—-12.r+5. To avoid a fractional quotient in the first step the former quantity must be multiplied by 7, which is not a factor of the other quantity, and will therefore not affect their G.c.M. GREATEST COMMON MEASURE. 2x3+ w’ —8.r+5 7 7a:’-12ar+5) 14w3+ 7x2—56x+35 k2ax+4 1 4x3— 24w2+1 Ox 31x2—66x+35 28x2—43x+20 3x2—1 8x +15=3(x2—-6x+ 5); 3 is evidently not a common divisor of the proposed quantities, and may therefore be rejected. x’—6.r+5) 7x’—12w+ 5 k7 7x2—42x+35 30x—30=30(x—1); reject the factor 30; -—1) x2—6x+5 Kai—5 wi—a: —5x+5 ~5x+5 x—l is the G.C.M. required. EX. 3. Required the mom of 36a"—18a"‘—27a“+9aa and 270552—1 8a4b 2— 9613192. Here SGaG—l 8a5—- 27a4+ 9a3= 9a3(4r a3— Qaz— 8a +1), and 27a562—1 8a‘b2-- 9a362= 9a3b2(3 a2— 2a —1) ; 9a3 is a factor of the G.C-M., and the factor 62 in the latter quantity may be rejected. Proceeding with the other factors, 4a3—2a2- 3a +1 3 3a2~2a -—l ) 12a3— Gag—9a + 3k4a 12a3—8a2—-4a QaL 5a+3=2a(a-1)-3(a—-1), =<2a-3>; Qa—S may be easily seen not to he a common divisor of the proposed quantities, and may therefore be rejected: a-1) 3a2~2a~1 k3a+1 3a2~3a a-1 a~=1 _— 5°, a===1 is the G.c. M. of 4a3= 2a2=3a+1 and 3a’—=2a *1. Hence 9a3(a-—1) is the G.c.M. required. GREATEST COMMON MEASURE. 55 107. To find the Greatest Common Measure of three quan- tities, a, b, 0, find (1 the Greatest Common Measure of a and b; and the Greatest Common Measure of d and c is the Greatest Common Measure required. Because every common measure of a, 6, and 0, measures d and c; and every measure of d and 0 measures a, b, and 0 (Art. 105); therefore, the Greatest Common Measure of d and 0 must be the Greatest Common Measure of a, b, and c. 108. In the same manner the Greatest Common Measure of four or more quantities may be found. The Greatest Common Measure of four quantities, a, b, c, d, may be found by finding a: the Greatest Common Measure of a and b, and y the Greatest Common Measure of c and d; then the Greatest Common Measure of a? and y will be the common measure required. 109. It should be borne in mind that the Greatest Common Measure of two or more algebraical quantities found as above is not necessarily the Aril/zmetz'cal Greatest Common Measure of the same quantities when numerical values are given to the letters contained in them,- and the reason is this :--there may be factors in the several quantities which, in their algebraical state, are prime to one another, but become divisible by some common number, when certain values are given to the letters contained in them. For example, the factors .r—7, and .r—4, have no common mea- sure greater than 1, as algebraical expressions; but if 10 be put for w, they become 3, and 6, which have a common measure, 3. Hence it is plain, that if we exclude certain factors from the Greatest Common Measure as having no common divisor, and aflermards change their form so as to make them to have a common divisor, the Greatest Common Measure obtained on the former supposition cannot possibly agree with the Greatest Common Measure obtained on the latter supposition. In fact, the phrases ‘ Greatest Common Measure,’ ‘Lomest Terms,’ ‘Least Common Multiple,’ applied to algebraical quantities, do not regard comparative magnitude. 110. In practice the Greatest Common Measure of two or more algebraical quantities is frequently found by a more expeditious method than the preceding, as follows. Taking Ex. 2, Art. 105. Ex. Required the G.C. M. of a4—ac4 and a3-aga:—-ax2+.r3. First, a‘—— 1:“: (0.2+ m2)(a’—a:’). Also (13- agar— ax2+ .223 = a2(a —x) — x2(a —-x), = (“2~x2)(“—w) ; therefore a2~x2 is a common factor or divisor of the proposed quantities ; and since the other factors a”+a:2 and a—x have no common measure greater than 1, therefore a'Z—a:2 is the Greatest Common Measure required. 56 GREATEST comron MEASURE“. 111. In the same manner fractions are usually. reduced to their lowest terms without the application of the Rule for finding the G. c. M. of the numerator and denominator. x2+1 1x + 30 9w3+ 53.1;2- 9.1,- -1 s ‘ x2+11x+30 _ w(x+6)+ 5(a:+6) 919+ 53x2—9x—1 8 _ 9x2(a: + 6) -— (w2+ 9x+18) ’ _ (1H- 5)(~'" +6) _ 9x”(a:+6)—(x+3)(w+6) ’ _ (a: + 5)(.z: + 6) _ (9x2—w—3)(x+ 6) ’ a: + 5 9x2—x—3 ' Ex. 1. Reduce to lowest terms (dividg. mum“. and denomr. by x+6)= a12+(a + c)x+ac EX. 2. REdllCe W to its lowest terms. x’+(a +c)x+ac=x2+ ax+cx+ac, =:v(x+a)+c(.r+a), ={x+ c)(x+a); also w2+ (b +c).r+bc=(.r+ c) (x+b); '. the fraction becomes , (a: + cXa: + b) x + a x+b' and in lowest terms is Ex. 3. Reduce 3‘13- fagb+ £162;— bs to its low est terms. 4a — 5ab +6 3a3— 3a’b + a62- 63 __ 3a2(a -b)+ 52(a—b) 4a2— 5ab+b2 _ 4a(a-b) —b(a—b) ’ __ (8a2+ 62) (a— 6) (4m -b)(a —5) ’ _ 3a2+ b2 _ 4a —b ' 2a are + b)’—@"’} 46209— (a 2— b2- 02)2 (a+b){(a+b)z—cg} __ (a+b)(a-t-b+c)(a+b—c) 4b202— (612— 0’— c2)2 _ (260 + a”— 62— 02) (260— a2+ 62+ 02) ’ a: (a+ b)(a+b+c)(a+_b-c) {‘12- (b - °)"}{(b + C)"- “2} ’ at (a+b)(a+b+c)(a+b-c) (a+ b—c)(a+c~b)(a +b+c)(b+c—a) ’ EX. 4. Reduce to its lowest terms. LEAST COMMON MULTIPLE. 57 a+b = (a+c—b)(b+c-a) ’ __ a+b Tc"- (a—b)2' [Exercises G.] 112. Fractions may be changed to others of equal value, with a common denominator, by multiplying each numerator by every denominator except its own, for the new numerator, and all the denominators together for the common denominator. d b bd Let g, 2, E, be the proposed fractions; then 62—65, 22;, Zdf, fractions of the same value respectively with the former, having the d b bd common denominator bdf. For Z—di;= %; Z—Zl§= 3; an ig7=§ (Art. 101); the numerator and denominator of each fraction having been multiplied by the same quantity, viz. the product of the deno~ minators of all the other fractions. are 113. When the denominators of the proposed fractions are not prime to each other, find their Greatest Common Measure; multiply both the numerator and denominator of each fraction by the deno- minators of all the rest, divided respectively by their Greatest Common Measure; and the fractions will be reduced to a common denominator in lower terms than they would have been by proceed~ ing according to the former rule. a b c . Thus —--, —, —, reduced to a common denomlnator, are ma: my me ayz baz cmy mwyz ’ mwyz ’ mmyz ' 114:. To obtain them in the lowest terms each must be reduced to another of equal value, with the denominator which is the Least Common Multiple, or Lowest Common Multiple, of all the denomi— nators. It becomes necessary therefore to investigate a rule for finding the Least Common Multiple of two or more quantities. And first, 115. T 0 find the Least Common Multiple of any number of simple quantities. ' To do this, observe the combinations of letters which form the several quantities, and resolve each quantity, by inspection, into its simple factors. The object then will be, to construct such a quantity as shall contain all the different factors found in the proposed quantities, but no factor repeated 58 LEAST COMMON MULTIPLE. which is not also similarly repeated in some one of them; for thus we shall obviously form a quantity, and the least quantity, which is divisible by each of the proposed quantities without remainder, that is, the Least Common Multiple of them all. To this end, detach from each quantity all the factors, which are common to two or more of them, until the quantities are left prime to each other. The continued product of these common factors and prime results will be the Least Common Multiple required. Thus, let the L.C.M. of 2a, Gab, and Sub be required. We see that the three quantities have a common factor 20:, which being detached leaves the quantities 1, 3b, and 4b: of these again, the two latter have a common factor b, which being detached leaves the quantities 1, 3, and 4; and these are prime to each other. Therefore the L.C.M. required is 2axbx1x3x4, or Qéab. OBS. Since the detaching of the Common Factors is the same thing as dividing the quantities by their Greatest Common Measures, it is clear that this method coincides with the arithmetical rule given in Art. 23. It may also be observed that the preceding method is applicable to compound quantities, as well as simple, provided that each of the quantities can be readily resolved into its component factors. Thus, if the L.C.M. of ab+ad, and ab—ad be required, we see that the quantities have a com- mon factor a, and when stripped of this become b +d, and b— d, which are prime to each other. Therefore the L.c.M. required is a(b + d)(b - d) or ab2 — adg. The following method is generally applicable to all quantities Simple or Compound. 116. To find the Least Common Multiple of two quantities, or the least quantity which is divisible by each of them without remainder. Let a and b be the two quantities, m their greatest common measure, m their least common multiple, and let m contain a, p times, and b, q times, that is, let m=pa=qb; then dividing the two latter equal quantities by pb (Art. 82), %=2; and since m P is the least possible, p and q are the least possible; therefore 2 P \ Q a Q I * a is the fraction 5 1n 1ts lowest terms , and consequently q=— ; hence a' m= qb=c—Zxb. a; I . ' . a . . . * For, if not, let some other fractlon % be the fraction 3 in its lowest terms 5 then smce q; = 5;, multiplying these equal quantities by p’b (Art. 81), q'b=p'a, or there are common P multiples of a and b less than pa and qb, which is impossible, since pa and qb are the least. ED. LEAST COMMON MULTIPLE. 59 The rule here proved may be thus enunciated :— Find the G.C.M. of the two proposed quantities; divide one of them by this G.C.M.; and multiply the quotient thus obtained by the other quantity. The product is the Least Common Multiple required. Ex. Required the Least Common Multiple of a4—x‘ and aa-a’x —a..132+x8. The G.C.M. of these two quantities (See Art. 105, Ex. 2), is ag—x’; and (a*—x‘)-:-(a’—.x2)=a’+.r’. Therefore the Least Common Multiple required = (a’+ a22).(a3— azx— axial-.223), =a5—a‘x—ax‘+a:5. 117. Every other common multiple of a and b is a multiple . of m. Let n be any other common multiple of the two quantities ; and, if possible, let m be contained r times in n, with a remainder s, which is less than m; then n-rm=s; and since a and b measure n and rm, they measure n—rm, or 8 (Art. 104); that is, they have a common multiple less than m, which is contrary to the supposition. 118. To find the Least Common Multiple of three quantities a, b, c, find m the Least Common Multiple of a and b, and n the Least Common Multiple of m and c ; then n is the Least Common Multiple sought. For every common multiple of a and b is a multiple of m (Art. 117); therefore every common multiple of a, b, and c is a multiple of m and 0:, also every multiple of m and c is a multiple of a, b, and c; consequently the Least Common luultiple of m and c is the Least Common Multiple of a, b, and 0. And similarly if there be four or more quantities of which the Least Common Multiple is required. Ex. Required the Least Com. Mult. of w“— a’x—ax2+a3, w4—a‘, and aw3+ ass: — azxi— a4. Here axa-l- aax— aixg— a4: a(a:3+ aix - axg—as) ; to find the G.C.M. of this quantity and the first, reject the factor a; x3- agar — aw2+ a3) a3+ agx— aaf— a8 (1 wa—aQx—ax9+a3 Qa’a: - 2a”: 2a2(.r —a), a: —- a) .1.”- a’a: — ax2+ a3 (x2— 0.13 x3— ax’ — a2x + a3 - a’a' + a3 60 LEAST COMMON MULTIPLE. .r-a is the G.C.M. of the first and last of the proposed quantities; and their least com. mult. is (axa+ aax— agwg— a‘)(a:”- a2) . . . . . . The other quantity is (x2+ a2)(.:v’—-a’) . . . . . . . . . The G.C.M. of (1) and is (r2—a’)x the G.C.M. of ax3+ a’x—a’x’-a‘ and x’+ a”. Rejecting the factor a in the former quantity, m'+ a”) .:r3+a"’a.*-~a.:r:"‘—a3 (a-a w3+ aim -- axg— a3 —- ax2— a3 the G.C.M. of (1) and (2) is (x’+ a2)(a:”—a’) ; least com. mult. required is (a .r"+ aax -a"’x’- a‘) (.r’- a“), or axs— a2x4— a‘x+a6. 119. A more expeditious method of applying the preceding rule to find the Least Com. Mult., when it can readily be done, is that of resolving each quantity into its component factors, as follows :-taking the last Example, (1 ) x“-- a%— ax2+ a3: x2(a:— a) —a2(.r—a), =(.r”-a’)(a:-a). (2) w‘- a4= (.r”+ a2)(.r2- a”) = (mg-t a’)(.r— a) (a: + a). (3) aa“+a3.v—a2.r’-a‘=a.r’(x-a)+a3(.z:-a), =a(.r2+a2)(.r—a). Now the G.C.M. of (2) and is (a2+a2)(x—a); least com. mult. of (2) and (3) is a(.r‘-a‘) . . . . . Again, the G.C.M. of (1) and (4) is w2—a’; least com. mult. required is a(.x‘-a‘)(.x—a), or ax5—a2w‘—a5.r+a6. 119*. The G.C.M. of two or more quantities is the L.C.M. of all the common measures. For the L.C.M. of all the common measures contains all the factors that appear in them, and therefore contains all the factors common to the proposed quantities: but their G.C.M. contains all these common factors: therefore the L.C.M. in question is either equal to, or a multiple of the G.C.M. But since every common measure of the proposed quantities measures their GC.M., i.e. their G.C.M. is a common multiple of all their common measures, therefore (Art. 117) this G.C.M. is either equal to, or a multiple of the L.C.M. of all the common measures. But these two condi- tions cannot be both satisfied unless the above G.O.M. and L.C.M. are equal, therefore they are equal. [Exercises FRACTIONS. 61 ADDITION AND SUBTRACTION OF FRACTIONS. 120. If the fractions to be added together have a common denominator, their sum is found by adding the numerators together for a new numerator and retaining the common denominator. a c a+c h —- - == ——-. T us b + b b This follows from the principle laid down in Art. 99, Ex. 1. Or thus; in each of the fractions the unit is divided into the same a '5': a+c , must be a+c such parts, or -—b— . number, 6, of equal parts ,- and it is plain that a of these parts, or toge- ther with c of the same parts, or Z— 121. If the fractions have not a common denominator they must be transformed to others of the same value, which have a common denominator (Art. 112...114), and then the addition may take place as before. E I a 0 ad bc ad+bc "" Hafiz-a+1;- bd 1 1 -b b Ex.2. a a+ a+b+a—b==a*‘--b2 az-bz’ a-b+a+b 2a a2—b2 _a“-b"i e of e af+e EX.3. a+-=-—+-= . f f f f Here a is considered as a fraction whose denominator is 1. a+b a—b 2a2--2b2 a2+2ab+b2 Ex. 4. 2+ = -. a-b+a+b a2--b~ a"’--b2 az- eab + b2 2a2- 252+ a2+ eab + b2+ a2— 2ab + 62 a2__b2 = a2__ be ’ 4a2 = a2__ b2 ' a b c a z baa car Eli-5. "—+-—-+~—= + + y: me my me mxyz mxyz mxyz ayz + bxz + my _ mwyz 62 FRACTIONS. 2 3 m and By Art. 116 or Art. 119 the Least Com. Mult. of the denominators is found to be w4-1 ,' therefore the sum required is 2(02-1) 3(a+1) .r‘—1 x4—1 ’ 2x—2+3r+3 = .r“-1 ’ 5x+1 2 x*—-1 ' Ex. 6. Required the sum of 122. If two fractions have a common denominator, their difi‘erence is found by taking the diference of the numerators for a new numerator and retaining the common denominator. (See Art. 99, Ex. 1.) a c a--c Thusb—5= b Or, the same reasoning will apply as that in Art. 120. 123. If they have not a common denominator, they must be transformed to others of the same value, which have a common denominator, and then the subtraction may take place as before. EX_ 1_ g_g=q_i_i_§g =M. b d bd bd bd ed ab cd ab - cd 'F=T_i= b ' Ex 3 a __c_i-_t_l __ ac—ad _bc+bd _ ac—ad—bc—bd " ' b c—d— bc—bd bc-bd_ bC—bd The sign of bd in the numerator is negative, because every part of the latter fraction is to be taken from the former. (See Art. 87.) a+b a—b a2+2ab+b2 a2--2ab+b2 Ex. 2. Ex. 4. ————~ - ————— = - a—b a+b a2--b2 ag—b'i ’ a2+ Qab + bg- a2+ Qab - b2 4ab = 0.2-1.2 3+Qx 2 _3+2x 2(1+:c) Ex. 5. —-——- --—- , l-w" l—a l—w‘ l—w __ 8+2x—2—2a:__ 1 _ l—x’ — l-a: [Exercises 1.] 2 ’ FRACTIONS. 63 MULTIPLICATION AND DIVISION OF FRACTIONS. 124. To multiply a fraction by any quantity multiply the numerator by that quantity and retain the denominator. a ac Tl — =——, 1118 bxc b _ . (2 ac . . . . . For In both the fractions 5, -b—, the unit 1s dIVIded Into the same number of equal parts (since the denominators are the same), and 0 times as many of these equal parts are taken in the latter as in the former, therefore the latter fraction is 0 times as great as the former. a db . . . . 125. COR. 1. Zxb=€ =a; that 1s, 11" a fraction be multi- plied by its denominator, the product is the numerator. 126. COR. 2. The result is the same, whether the numerator be multiplied by a given quantity, or the denominator divided by it. d Let the fraction be 25— , and let its numerator be multiplied by c '_ adc ad . . . c, the result 1s T , or —b- (Art. 101), the quantity which arlses c from the division of its denominator by c. 127. To divide a fraction by any quantity multiply the deno- minator by that quantity, and retain the numerator. a ac —-= —, and a 0th be The fraction i; divided by c is g; . Because b part of this is g: the quantity to be divided being a 0th part of what it was before, and the divisor the same. (Art. 96.) 128. Con. The result is the same, whether the denominator is multiplied by the quantity, or the numerator divided by it. Let the fraction be 2%; if the denominator be multiplied by . U O O I I O c, It becomes — or the quantity WhlCll arises from the dIVIs1on a bdc bd ’ of the numerator by c. 64 FRACTIONS. 129. To prove the Rule for the .Multiplication of Fractionsl“. RULE. The product of two fractions is found by multiplying the numerators together for a new numerator, and the denominators for a new denominator. a c . a 0 ac Let — and - be the two fractions ' then ~x— = ~ . b d ’ b d but One. In the strict sense of the word Multiplication, which supposes a quantity to be added to itself a certain number of times, to multiply by a fraction would be impossible : the operation must therefore be understood in an extended sense. Since to multiply a by b is the same thing as to take b of it, we shall easily perceive that the extension of the sense of multiplication will lead us to conclude that to multiply 5;; by 9‘2 will be the same thing as to take 2 d But the dth part of ths of it; i. e. to take the dbh part of it, and then to take 0 of such parts. a . a . . a . ~6- 1s 5;, (Art. 127), and c of such quantities as-b—d belng taken will produce 2i;— (Art. 124) : this therefore is the result required. %X%=%=Z—:Zci, which proves the Rule. 130. To prove the Rule for the Division of Fractions. RULE. To divide one fraction by another, invert the numerator and denominator of the divisor, and use it as a multiplier accord- ing to the rule for multiplication. . D d d Let (I; and f-lbe the two fractions; then g—I—g = gx; = g; . OBS. Since ordinary division is the inverse of ordinary Multiplica- tion, the signification of division here must be extended to mean the inverse of multiplication in all cases. Then the Quotient will always be such a quantity as when multiplied by the Di-visor will produce the Dividend; therefore if the Dividend can be put into two factors, one of which is the Divisor, the other must be the Quotient. * The proofs of the Rules for Multiplication and Division of Fractions given in former Editions of this work are very objectionable. The fallacy consists in assuming bc'.dy=bd.a:y and {gig = '5 , which have only been proved true for integral values of the symbols. ED. FRACTIONS. 65 a . . Now — the Dlvulend, b, a>n, and Dividend 1) _ a 1) 1 '—‘-_-_~________ _ Divisor _ 10'" ' 10" d '10m'"’ R 1 . . . . . . . = (2+? . W , 1f'd1s contamed Q tlmes mD With RemamderR, Q R 1 ::—-————-—+ ."—_" low a 10mm Q =1_0,-n_-n, if 12:0, or n + a fraction which < R < d always, _ 1 _ 10”“ 10M ’ Quotient is either actually equal to 1 or approaches to 16%;; as near 0722-1: 5 can be made as small as we please by increasing ' l as we please, SIDCBW 5 66 FRACTIONS. m—n, that is, by adding ciphers to the proposed dividend. Hence the Rule, viz. Aflia: ciphers to the right of the dividend, wanted, so that it may have a few more decimal places than the divisor, proceed as in whole numbers, taking no notice of the decimal points, and then mark of as many decimals in the quotient as the number of decimal places in the dividend (including the ciphers used) exceeds the number in the divisor. N.B. The fraction 10%,, expressed decimally will always be the . . . 1 correct quot1ent, as far as it goes, because no quantity less than F can 1 add 1 to the last figure of a decimal with (m—n) decimal places*. Con. Since a vulgar fraction represents the quotient of the numera- tor + denominator, the preceding rule may be applied to find its equivalent decimal fraction by actually performing the division; in this case n=0 or the divisor is an integer, viz. the denominator of the given fraction, and the dividend, i.e. the numerator of the fraction, can be made to have as many decimal places as may be necessary, by putting a decimal point at the end of it, and adding ciphers: and the number of decimal places in the quotient will here be m, i. e. there will be as many decimal places as we have added ciphers. a 131*. If 6 be a fraction in its lowest terms, by the operation of m a. b-10”" m being the number of ciphers added: and if the equivalent decimal is a terminating one a'lO'" is divisible by b exactly. But by hypothesis a is not so; therefore 10m is exactly divisible by b. But the prime factors of 10 are 2 and 5, and no others. Therefore b must contain no factors but 2 and 5, i. e. it must be of the form 2P.59. And when this is the case, if m be taken equal to the greater of the two p and q, 10”‘—:-(2P.5q) is an integer, and is not so for any less value of m. Therefore the fractions that will produce terminating decimals are those only whose denominators (when they are reduced to their lowest terms) contain as factors 2 and 5 and no others, and the number of decimal places will be equal to the greatest number of times either one of those factors is repeated. All other fractions will therefore produce non-terminating decimals. If the division in these cases be performed as far as the factors 2 and 5 are concerned, we have a terminating decimal to be divided by the product of the other factors of the denominator, c suppose. As the decimal does not terminate there must always be a remainder, which of course is less than c, i. e. is one of the numbers 1, 2, 0—1; after 0—1 places of figures then have been obtained, to take the most unfavourable case, when the remainders are at first all different, the next remainder must be the same as some one of the preceding, and then the whole operation is repeated, and the figures in the quotient recur. There can therefore be no more adding ciphers as above after a decimal point, we transform it into * The proof of the Rule for the Division of Decimals, found in most Treatises on Algebra, is unsatisfactory and imperfect, because it includes only the particular case when, neglecting the decimal point, the Dividend is an exact multiple of the Divisor.-ED. FRACTIONS. 67 recurring figures than c-l; and all fractions that do not produce termi- nating decimals will produce recurring ones. 132. The rule for multiplying the powers of the same quantity (Art. 91) will hold when one or both of the indices are negative. 1 am Thus amxa‘"=a’""'; for amxa-“=a’”><—; (Art. 65) = E; =am’”, a 3 .v 1 (Al‘t- 95, Con.). In the same manner mam-5:7 = 7,: w‘z. a: a: Again, a""><-— (Art. 65) = =a . am an am-rn am a =1; therefore a°=1, according to the notation adopted in Arts. 63, 65. 134:. The rule for dividing any power of a quantity by any other power of the same quantity (Art. 95, Con.) holds, whether those powers are positive or negative. 1 Thus a’”—;- a-"= am”? 7; (Art. 65) = amxa" = a"‘“‘ = elm-("W a . 1 1 a" Again, a”" -1- 06"" =—— + —— = —~ (Art. 130) = a""’"= a'm_('”l. a”) an am 135. Con. Hence it appears, that a factor may be transferred from the numerator of a fraction to the denominator, and vice versa, by changing the sign of its index. am‘ an am am am.a - n ThUS = ' and = b? bPa-n ’ a”b1’ b? 4ab“’c.v'3 _ 4a.a’cy _ 4a30y sit-{alas _ straw “ stare [Exercises J =a’".a"’.b‘P. Also INVOLUTION AND EVOLU '1‘ION. 136. If a quantity be continually multiplied by itself, it is said to be involved, or raised; and the power to which it is raised_ is expressed by the number of .times the quantity has been em- ployed in the multiplication. 5--2 68 INVOLUTION AND EVOLUTION. Thus axa, or a2, is called the second power of a; axaxa, or as, the third power; axa...t0 n factors, or a”, the n.tb power of a. 137. If the quantity to be involved be negative, the signs of the even powers will be positive, and the signs of the odd powers negative. For -a><--a=a2, -a><-a><-a=—a3, &c. 138. A simple quantity is raised to any power by multiplying the index of every factor in the quantity by the exponent of that power, and prefixing the proper sign determined by the last Article. Thus a’" raised to the nth power is am"; because amxamxam...to n factors, by the rule of multiplication, is am+m+&°'t°”terms, or am". Also (ab)"=ab><(:/a+x—f/a-x), by (2), =2w—3.,3/a+w.,3/a—x.{:/a+a—f/a—x}. 143. EVOLUTION, or the extraction of roots, is the method of determining a quantity which raised to a proposed power will produce a given quantity. 1414.4. Since the nth power of a'” is a’"" (Art. 138), the nth root of am” must be a’”; that is, to extract any root of a simple quantity, we must divide the index of that quantity by the index of the root required. Thus j/Zé-ar, ,4/a‘2=a3,- 8zc. ,”/a"”‘=a’". 145. When the index of the quantity is not exactly divisible by the number which expresses the root to be extracted, that root must be represented according to the notation pointed out in Art. 70. Thus the square, cube, fourth, nth, root of a2+ar2,‘ are respectively represented by 1 (a2 + avg)? (a2 + .r‘ifi, (a2 + .v2)%, (0&2 + $2)"; 72 EVOLUTION. 1 the same roots of m, or (a2 +102)“, are represented by -_1. (a2+m2)—%, (a2+w2)_%’ (a9+w2)—%, (a2+w2) 71.. 146. If the root to be extracted be expressed by an odd number, the sign of the root will be the same with the sign of the proposed quantity, as appears by Art. 137. Thus J3 is 2; f/rg is —2; &c. 1117. If the root to be extracted be expressed by an even number, and the quantity proposed be positive, the root may be either positive or negative. Because either a positive or negative quantity raised to such a power is positive (Art. 137). Thus Jr? is =‘=a; ,:/(a+.r)8 is =-*=(a+.rc)’,' &c. 148. If the root proposed to be extracted be expressed by an even number, and the sign of the proposed quantity be negative, the root cannot be extracted ; because no quantity raised to an even power can produce a negative result. Such roots are called impos- sible. 149. Any root of a product may be found by taking that root of each factor, and multiplying the roots, so taken, together. 1 1 1 Thus (ab)"=a">; and therefore a+ b=~/dx\/1+2(Art. 149) ; (Z we have by the last Example, putting-g in the place of x, / b b 62 1+z=1+2—a—é—a2+&(3. __ z Ja+b=ai+ b,--i§ +&c. 2a2 8a . . we 152. It appears from Ex. 2, that a trinomlal, a2—ax+——, 4 in which four times the product of the first and last terms is equal to the square of the middle term, is a complete square. The same relation is found to subsist between the parts of all com- plete squares of three terms arranged according to the powers of some one letter. Thus 4.‘.:s"+440x+c2 is a complete square, viz. (2x+c)2, because 4x4nc’xc” =1602x2=(40x)2. . . p2 P 2 Slmilarly x’-px+-Z is a complete square, viz. (iv-é) , because P2 ixx’x Z=pgxg= (~px)’. 153. The method of extracting the cube root is discovered in the same manner as that for the square root. The cube root of a3+.‘3’agb-1-3ab2+b3 is a+ b (Arts. 140, 143); and to obtain a+b from this compound quantity, arrange the terms 76 EVOLUTION. as before, and the cube root a a 2 3 of the first term, as", is a, the a3+ 3a b +3ab +b Kall- 6 first term in the root; sub- a tract its cube from the whole 301,2) 3a?!) +3a52+193 quantity, and divide the first 3a2b+3ab2+b3 term of the remainder by Sag, the result is b, the second term in the root; then subtract 3a2b+3ab2+ b3 from the remainder, and the whole cube of a+b has been subtracted. If any quantity be left, proceed with a+ b as a new a, and divide the last remainder by 3(a+b)2 for a third term in the root; and thus any number of terms may be Obtained*. Ex. To extract the cube root of 8x3+ 6xy9—12x2y—y3. 8x3—1 2yx9+ 6y’x --y8 (2x - y as: 8.1:3 3112:1241:2 ) —1 2yx2+ 6y2x—y3 -—12yx2 = 3a’b + 6y2x = 3ab2 _ya_= be 2.1: -y is the cube root required. [Exercises SCHOLIUM. 154. The rules above laid down for the extraction of the roots of compound quantities are but little used in algebraical operations; but it was necessary to give them at full length, for the purpose of investigating rules for the extraction of the square and cube roots in numbers. The square root of 100 is 10, of 10000 is 100, of 1000000 is 1000, &c. from which consideration it follows, that the square root of a number less than 100 must consist of only one figure, of a number between 100 and 10000 of two places of figures, of any * That the rule may be thus extended will be obvious from comparing the form of the cubes of a+b+c, a+b+c+d, &c., with that of a+b from which the Rule was deduced ; For (a +b+c)3=(a+b)3+3(a+b)gc+3(a+b) 02-1-03, =a3+(3a2+3ab+b2) b+{3(a+b)g+3(a+b)c+cz}e. Similarly (a +b+c+d)3= a3+(3a2+ 3ab + b2)b+{3(a + b)‘~’ +3(a+b) c+c‘~’} e +{3(a+b+c)2+3(a+b+c)d+d2}d; and so on.--En. SQUARE ROOT. 77 number from 10000 to 1000000, of three places of figures, &c. If then a point be made over every second figure in any number, beginning with the units, the number of points will shew the number of figures, or places in the square root. Thus the square root of 4357 consists of two figures, the square root of 56478, of three figures, 8Z6. NOTE 2. Ex. 1. Let the square root of 4356 be required. Having pointed it according to the direction, it appears that 4356k00+6 01' 66 the root consists of two places 3600 of figures; let a+b be the root, 120+6) 756 where a is the value of the figure or 126] 756 in the tens’ place, and b of that in the units”; then is a the nearest square root of 4.300, which does not exceed the true root*; this appears to be 60; subtract the square of 60, (a2), from the given number, and the remainder is 756; divide this remainder by 120, (2a), and the quotient is 6, (the value of b), and the subtrahend, or quantity to be taken from the last remainder 756, is 126x6, (2a+ b)b, or 756. Hence 66 is the root required. It is said that a must be the greatest number whose square does not exceed 4300]”: it evidently cannot be a greater numberi than this; and if possible let it be some quantity, :0, less than this; then since as is in the tens” place and b in the units’, x+b is less than a; therefore the square of x+b, whatever he the value of b, must be less than a2, and consequently x+b less than the true root. If the root consist of three places of figures, let a represent the hundreds, and b the tens; then having obtained a and b as before, let the new value of a be the hundreds and tens toge- ther, and find a new value of b for the units: and thus the process may be continued when there are more places of figures in the root. * It will be clearer to read “ a the greatest multiple of 10 whose square does not exceed 4300.” 1' Or, the greatest multiple of 10 whose square does not exceed 4300. 1 Multiple. 78 SQUARE ROOT. 155. The ciphers being omitted for the sake of expedition, the following rule is obtained from the foregoing process. Point every second figure beginning ' ' . . . 4356 66 with the units’ place, dimdmg by thls pro- k cess the whole number into several periods; 36 find the greatest number whose square is 126) 756 contained in the first period, this is the 756 first figure in the root; subtract its square from the first period, and to the remainder bring down the next period; divide this quantity, omitting the last figure, by twice the part of the root already obtained, and annex the result to the root and also to the divisor; then multiply the divisor, as it now stands, by the part of the root last obtained, for the subtrahend. If there be more periods to be brought down, the operation must be re- peated. Ex. 2. Let the square root of 611525 be required. 61'15'25K782 49 _-—-__- 148)1215 1184 1562)3125 3124 '1 remainder. The remainder in this example shews that we have not obtained the number which is the exact square root of the proposed quantity; but 782 is a near approximation to the square root, being in fact the square root of 611524; and 783 is too great, being the square root of 613089. 156. In extracting the square root of a decimal, the pointing must be made the contrary way, beginning with the second place of decimals, and the integral part must be pointed as before, be- ginning with the units’ place: or, if the rule be applied as in whole numbers, care must be taken to have an even number of decimal places, by annexing ciphers to the right (Art. 41); because, if the root have 1, 2, 3, 4, 8:0. decimal places, the square must have 2, 4, 6, 8, 8:0. places (Art. 46). CUBE ROOT. 79 EX. 3. To extract the square root of 64853. 64-3530t8-053 8:0. 64 1605 ) 8530 8025 16103) 50500 48309 2191 The remainder in this example appears to be great; but if the decimal point were retained throughout the operation, it would easily be seen that its real value is very small, and that it becomes smaller for every figure that is added to the root. For every pair of ciphers which we suppose annexed to the decimal another figure is obtained in the root. And in this and similar cases, when ciphers are added, the root can never terminate, because no figure multiplied by itself can produce a cipher in the units’ place. 157. The cube root of 1000 is 10, of 1000000 is 100, &c. there- fore the cube root of a number less than 1000 consists of one figure, of any number between 1000 and 1000000, of two places of figures, &c. If then a point be made over every third figure contained in any number, beginning with the units, the number of points will shew the number of places in its cube root. NOTE 2. Ex. 1. Let the cube root of 405224 be required. 405224t70+4 a3 = 343000 302 = 14700 ) 62224 the first remainder. 58800 = 3a2b 3360 = sub2 64 = b3 62224 subtrahend. By pointing the number according to the direction, it appears that the root consists of two places; let a be the value of the figure 80 0088 ROOT. in the tens’ place, and b of that in the units’. Then a is the great- est number* whose cube is contained in 405000, that is, 70; subtract its cube from the whole quantity, and the remainder is 62224; divide this remainder by 30.2, or 14700, and the quotient 4, or b, is the second term in the root: then subtract the cube of 74 from the original number, and as the remainder is nothing, 74 is the cube root required. Observe that the ciphers may be omitted in the opera- tion; and that as a3 was at first subtracted, if from the first remainder 8aib+3abg+ b3 be taken, the whole cube of a+b will be taken from the original quantity. 158. In extracting the cube root of a decimal care must be taken that the decimal places be three, or some multiple of three, before the operation is begun, by annexing ciphers to the right (Art. 41); because there are three times as many decimal places in the cube as there are in the root (Art. 46). Ex. 2. Required the cube root of 31189791. 81i89i'910(67-8 216.. . = a3 3a2= 108.. ) 95897 first remainder. 756. . = Bail) 882. = sab‘3 343 = 03 84763 subtrahend. 80'? = 13467.. ) 11184910 second remainder. The new value of' a is 670, or, omitting the cipher, 67; and 3a2, the new divisor, is 13467..hence 8 is the next figure in the root; and .107736..= 301.2?) 12864.=8ab2 512=b3 10902752 subtrahend 282158 the third remainder. It appears from the pointing, that there is one decimal place in the root; .therefore 67'8 is the root required nearly. If three more " It will be clearer to read “ a is the greatest multiple of 10 8:0.” CUBE ROOT. 81 ciphers be annexed to the decimal, another decimal place is obtained in the root; and thus approximation may be made to the true root of the proposed number to any required degree of accuracy. 159. Since the first remainder is 3agb+3rzb2+ b3, the exact value of b is not obtained by dividing by 302; and if upon trial the subtrahend be found to be greater than the first remainder, the value assumed for b is too great, and a less number must be tried. The greater or is with respect to b, the more nearly is the true value obtained by division. 3 2 For the first remainder divided by 302 gives 0+ %+ :71, for the quo~ . . . 62 03 . . tlent; and if this be adopted for b, the error: 2 +322; which for a glven value of b is evidently less as a is greater. 160. In extracting the square or cube root of a vulgar fraction the rule stated in Art. 150 may be followed; but it is generally preferable to convert the vulgar fraction into a decimal, and then extract the root. ‘ ll . Thus let the cube root of 5%, or 7)— be required. Now, if the rule of Art. 150 be applied to this case, the cube root of 11, and the cube root of 2, must be found to a certain number of places of' decimals, and then the long division of the one root by the other must be effected: whereas, if 5% be, first of all, converted into a decimal, viz. 5'5, one single extraction of the cube root completes the whole process. Another method is, to multiply the numerator and denominator by such a quantity as will make the latter a perfect cube, and then apply the rule of Art. 150. 11 44 f/4. 4 1 —— Thus the cube root of 5%, or T2- , or —8-, = ~78 =éi/44. 161. In extracting either the square or cube root of any number, When a certain number of figures in the root have been obtained by the common rule, that number may be nearly doubled by division only. I. The square root of any number may be found by using the common Rule for extracting the square root until one more than half the number of digits in the root is obtained; then the rest of the digits in the root may be determined by Division. For, let N represent the number whose square root, consisting of 212 +1 digits, is required; a . . . . . . . . . the first n+1 digits of the root found by the common Rule, with n ciphers annexed; 6 82 SQUARE ROOT. x . . . . . . . .. the remaining part: so that ~/TV=a+.r. Then N =a2+2ax +82 (Art. 81); N—a.2 2a that is, N --a’, (which is the remainder after n+1 digits of the root are found) divided by 2a will give the rest of the root required, 8‘, increased 2 x2 = - .A.t.80 82 ' x+2a( rs 7 )2 by 5;. Now, since .1: contains 22 digits, .22 has 212 at most (Art. 154). But, by the supposition, a is a number of 2n +1 digits, and 2a has 212 +1 digits at least; therefore 11:” . x2< 2a, or —2—a is a proper fractlon, or<1 ; that is, if the quotient of (N—a2)+2a be taken for x, the error is less than 1. Hence it appears, that if n+1 digits of a square root are obtained bylthe common Rule, 12 digits more may be correctly obtained by Division on v Ex. Required the square root of 2 to 6 places of decimals. 2-0000...(1=414 1 F- 24)100 96 --_~—_- 281)400 281 2824)11900 11296 2828)604000(21e 5656 3840 2828 10120 8484 1636 the root required is 1414213 .... .. When only one figure in the root has been obtained, a, which repre- sents the part already obtained, may be as small as 10, and x, the next 2 dlgit, may be as great as 9; the error 1n the quotient g, may therefore CUBE ROOT. 83 be easily greater than 1, unless a be as great as 50, 2'. e. the first figure in the root as great as 5, and we should then obtain too large a quotient; this is not unfrequently observed to happen at the first division, but from the foregoing proposition it appears that this error cannot take place in any subsequent division. II. In the extraction of a cube root, when 171-1 digits have been found by the ordinary rule, 12 more can be correctly obtained by dividing by the trial divisor. Let a+b be the cube root, when a consists of 17:1 digits, and n ciphers, b . . . . 12 digits. aa+8azb+3abi+ 63 the quantity whose root is required. Then after a has been found, we have remainder = 3a2b + 3002+ {)3 ; trial divisor = 8a2 ; I)“ 03 quotient=b+—+;—,—. a all If this be adopted as the value of b, ' b” 53 the error = 7‘- + -3—a, . Now 01 consists of 212 +1, and b of n. the least value of a is 102“, and the greatest value of b is 10"—1 ; -. the greatest possible error will be when a and b have the above values; (10"-1)2 (10"-1)3 102". 3 . 10412 =10"-1 10"-1< 10X +3.10" 3.10") . 1 1 1 1... <1- -— —— am ls< 10" 1 10" l+3.10" ’ ie <(il— l > " ' 10" 3.10" 8.10"" ’ which is evidently < 1. i. e. the greatest error = The error therefore is always <1, i.e. the 12 last digits can be correctly obtained by ordinary division. . From this it will be seen that as in square root, it is only at the first (llVlSlOIl that too large a quotient can be obtained for the next digit in the root. 6—2 84 THEORY or rumors. THEORY OF INDIGES. 162. The subject of I ndices deserves a separate and distinct conside- ration. It is proposed to bring together here all that has been defined or proved with respect to them in the preceding pages—to shew that the several Definitions are not in practice inconsistent with each other—and to supply the proofs still wanting in order that the Rules may be extended to all possible casesWhich can occur. (1) The primary Definition was given in Art. 63, whereby we agreed to represent a.a.a. &c. to n factors by a", where n expresses the number of factors, and therefore can only be a positive integer. (2) The next Definition was given in Art. 65, whereby we agreed to I . . represent a; by a'”; but at that stage we could only consider it as a short . . 1 . . . . way of writing a”, smce a negative quantity can 1n no sense express a number of factors. (3) The last Definition was given in Art. 70, whereby we agreed to 1 represent the ninn root of a by a", and the nth root of the mth power of a by a5. Here again we felt the restriction that a fraction can in no sense express a number of factors multiplied together, that is, a power of a, in the true sense of the word. (4) From (1) it is strictly proved in Arts. 91, 95, 138, and 144, that amxan= am—i-n) n m n 1 ' am+a =a - , or 553;, according as m> or P=l/'£i'2 (9) LEMMA III. To shew that Z/T‘fi="Z/Z (fir-=85, a={<:/y—;,m}.=(yg;m, 8:75:17; ~ COR. 1. ’LFL'f/ggg/ggjyg, Con. II . Z/(fiyz =MW= nz/Fél/fiy, which shews that the order in which the operations of involution and evolution are performed upon any quantity is immaterial. ' (10) LEMMA IV. To shew that WJl/W. W=t/W)Q=JW, (Lemma 11. Cor.) =Zl/fi, by Lemma III. Con. ylzez/a'iéyaea/Fqm, by Lemma 1. (11) T 0 shew that the Rules for the multiplication and division of powers hold true when the indices are fractional. m r. _ ______ lst, a">:/(§/a"‘)1’=~/ ,"lamp, (8) 41/255, (9), =a_’;5=a"'§. 2nd, (€/Z§)%=aE=:/a7', or U,(_i,~/:T€)P=" a”, (iJZé)P:=(:/Z; ski/2175, (8), Cor. w=ill7arfi=w dm=h§i=a§+h 1 1 r" agr-g). __ .. “if: ,1, by lst case, -a mp_ l (an)q a uq 3rd ’ (a5)— 3: -2 ’2- _z c - 4th, Let alla'ba', then .r 4=a", f/w' =:/ams _‘ 1| 1 L _— 1 .r-P= (ya’")4=, lam, = am“, 1” =3? 9 P w.- 1 ._"'___q ".14. .J; 0.0 ) ’Jamq Us” Jam” of? _: _e 9 f -e _ q 1 1 1 5th: (a a) g: (a n) P=\/ _5' = m :3” --mr a( a)? a El -2 , _E g 61"” y 1 i —— mp mp _ . a (I. 4’ __'il 1‘ .J.’ .."l 1 1 2 E 6th, Z/a_n=,r, w Q=a n, —E=-—m-:, xqzzafl’ at a," a P 3r?! 2.1.: __'_~ _-_ __e yin—5:4? ,Z/a"=a"'4 by 2nd case, =a( ); which proves the Rules for all possible cases of fractional I ndices. SURDS. 87 Thus we have proved that the Definitions and Fundamental Rules are perfectly compatible, and that no error can arise from giving to the Rules the most general application. It is not meant that negative and fractional indices can really repre- sent the powers of any quantity, but simply that they may be treated as such in all algebraic operations without error. NOTE 3. REDUCTION OF SURDS. 163. A rational quantity may be reduced to the form of a given surd by raising it to the power whose root the surd ewpresses and qfliwing the radical sign. Thus a= Mbf= 8:0. and a+w = (a+w);= \m/ (a +.v)"‘. In the same manner, the form of any surd may be altered; thus 3 c n a (a+m)i=(a+a=)i= (a,+.'v)*“5 &c. The quantltles are here ralsed to certain powers, and the roots of those powers are again taken; therefore the values of the quantities are not altered. 164. The coefiicient of a surd may be introduced under the radical sign by first reducing it to the form of the surd, by the last Art., and then multiplying according to Art. 149. Exs. a“; =\/h§><\/.-b= \/ airs; ayi = (a2y3)é; .v\/2a — .v =\/2aa'2— m3; a><(a f.v)%= {a2x(a — 5; 4\/2=\/16><2 =\/3__2. 165. Conversely, any quantity may be made the coefficient of a surd, if every part under the sign be divided by this quantity raised to the power whose root the sign expresses. Thus \/a2—a.v=ai\/a-w; \/a3—a‘~’1v=a\/a-w; l E m2 L __ .._.___ _._ (a2—.v2)"=a"(1-——¢, ”; \/60=\/4><15=2\/15; a 1 1 ‘ 1\/ 52 (7*? =6 I-j. 88 scans. ADDITION AND SUBTRA‘GTION 0F SURDS. 166. When surds have the same irrational part, their sum or difierence is found by afirving to that irrational part the sum or diference of their coejicients. Thus a\/¢_viib\/hi=(a=l=b)\/iii; 10\/8=1=5\/8—=15 3, or 5M8. If the proposed surds have not the same irrational part, they may sometimes be reduced to others which have, by Art. 165. Thus, Ex. 1. Let the sum of fled—b and J37) be required. Since W=Jdix 3b==a,\/8b_, N/3h’b+,;/3b=a~/3b+\/_3_b=(a+1)~/8b. Ex. 2. Find the sum of sci/W, bi/W, and {/5521 Here 4af/W= 4111/6363. ,i/ b: 4a2bf/b, bf/8hfb=bi/g.i/b=2agbi/b, ——i/T25—a‘bz=-i/T2_5F_Gb3.,f/b=—5aibj/b; the sum required=a2bi/b_. If the proposed surds cannot be reduced to others which have the same irrational part, then they must be connected together merely by the signs + and -. [Exercises MULTIPLIOATION OF SURDS. 167. If two surds have the same indem, their product is found by taking the product of the quantities under the signs and retaining the common indew. - - . _ 1 1 1. ,, _. Thus \VaM/b=d”>< b\/y= ab\/.by. 168. If the indices of two surds have a common denominator, let the quantities be raised to the powers eavpressed by their respective numerators, and their product may be found as before. swans. 89 Ex. 25x3%=(23)%x3$=8%x3%=(24)§; also (a + w)§><(a - w)§= { (a + w) (a — .193} 5. 169. If the indices have not a common denominator, they may be transformed to others of the same value with a common denomi- nator, and their product found as in Art. 168. Ex. (a”— m2)i><(a — .v)§= (a2— .v2)i><(a — w)§, = {(a2—w2)><(a — mfg; again 2%Xs%=2%xs%=(s><9)%=(72)%. 170. If two surds have the same rational quantity under the radical signs, their product is found by making the sum of the indices the indew of that quantity. 1) m-l-n 1 1 m Thus \/a><\7a = anx am: amnx amn= a "m ; (see Art. 162). Ex. MEX/E = 2%x2%=2%+%= 2%. [Exercises L] DIVISION OF SURDS. 171. If the indices of two quantities have a common deno- minator, the quotient of one divided by the other is obtained by raising them respectively to the powers earpressed by the numerators of their indices, and emtracting that root of the quotient which is expressed by the common denominator. 1 :2 1_ , n a a" am 1* am a For ZI= ; and —E = ( )L= (Art. 150). 6" bn I 2 I 4 5 1 p 5' r 7’" ps2 '71 r .5;- %= _ =__- - _'_ - = —— . 172. If the indices have not a common denominator, reduce them to others of the same value with a common denominator, and proceed as before. EX_ (£12.. 372)% + (a3- a3)?i= (a2— £172)? -2—- (a3— m3)§, ‘90 sunns. 173. If two surds have the same rational quantity under the radical signs, their quotient is obtained by making the diference of the indices the index of that quantity. ' 1 1 7” A Thus .\/a-:-\/a, or am divided by am, or am“ divided by am”, 7n . We: m—n ¢ that is 7 , is equal to am; because these quantities, raised to the mn a m a power mn, produce equal results — and am'". an Ex. 25+2i=2§+2§= gig [Exercises M.] INVOLUTION AND EVOLUTION OF SURDS. 174. Any power of a surd is found by multiplying the fractional index of the surd by the number which expresses the power. ;1;+1;+...to mtcrma_ '3 For (Ud)m=(a7‘)”‘=a -a" 175. Any root of a surd is found by dividing the fractional index of the surd by the number which expresses the root. 1 1 Thus fl(:/a)=’lfli;=am“; because each of these quantities raised to 1 the 112‘“ power will produce a”. It will be seen that the rules hitherto required for the management of surds are simply those which apply to quantities raised to powers expressed by Fractional Indices. [Exercises N TRANSFORMATION OF SURDS. 176. Having given a quantity containing quadratic surds, to find another quantity which, multiplied into the former, shall produce a rational result. 1. If the given quantity be a simple surd, as 3J5, the multiplier required is J5, which gives the product 3a, a rational quantity. 2. If the given quantity be a binomial surd, as Jain/(i, then the multiplier required is Jd—Jli, and the product is a—b. 3. If the quantity be a trinomial, as Jd+_\/b+,~/d, first multiply by JE+JILJQ which gives (Jd+Jl;>2—(JE)2, or a+b‘c+2,\/iib_. Next multiply by a+b-c-2Jdb, and the product is (a+b-c)”-4ab. Therefore the multiplier required is d+Jh—J;)x(a+b-c— QJcTb). SURDS. 91 The use of' this proposition is to enable us without much labour to find the values of fractions which have irrational denominators. Thus, 1 . . suppose the actual value of T were required to 7 places of deer-- 3+J2 mals; if we were to proceed to extract the square roots of 2 and 3, and divide 1 by the sum of' those roots, the operation would be long and troublesome. But if we first multiply the numerator and denominator by fl-Jé, the fraction becomes or fi—JEZ, and its value is not altered; we have simply then to extract the square roots of' 3 and 2 and subtract the one root from the other, by which the long division is en- tirely avoided. each of which fractions is thus much simplified for purposes of calcula- tion. 3/5—i/5 _ 1 . w-a ” ,3/r”+:/a—w+:/a2 ’ simpler form than the latter. Again, but the former quantity is in a More generally, 177. T 0, find the multiplier which will rationalize any binomial, having one or both of its terms irrational. 1st. Let .v+ y represent the binomial, a: and y being, one or both, irrational, and let m be such a number that a!" and y’" are both rational, that is, let an be the Least Com. Mult. of the denominators of the fractional indices of the binomial; then since .v’"¢y”‘= (.v + y) . (.v’""—.v”“”y + . . . $ .rqy”“2=!=y""1 ), where the upper or lower sign is to be taken, according as m is odd or even, the rationalizing multiplier required is .vm'l—x'may + . . . $xy"‘*2='=y’"". 2nd. Let .v—y be the binomial, and m as before, then since .v’"—y"‘= (.r—y).(x’""1+a'"‘“”y+...+.vy”""+y’“"1), (Art. 99. Ex. 6.) the rationalizing multiplier is .rm‘1+.t”“2y IF. . . +ay"‘“'"’+y"‘“. Ex. Find the multiplier which will rationalize Js-fi/s, or 5i-6i. Here m=6, the Least Com. Mult. of 2 and 3, 92 scans. mult’. reqd. ==(,,/5)‘+ b)‘ xii/6' + §)3x(\76)2+ bfx 6)” +~/5><(3/‘6'r+(3/'6>~". =25J5+25§76+ 5J5 >< (f/6)2+ so+6f5 >< f/‘6+6(3/6)2. [Exercises 0.] 178. The square root of a quantity cannot be partly rational and partly a quadratic surd. If possible, let \/n=a+\/m; then, by squaring these equal quantities, n = a2+ 2a\/n—i+ m, (Art. 81); and 2a\/m = n — ag—m, n — a2— m . 7, (Art. 82), a rational quan- tity, which is contrary to the supposition. (Art. 80); therefore Vii;- 179. If any two quantities, partly rational and partly qua- dratic surds, be equal to one another, the rational parts of the two are equal, and also the irrational parts. Let ,v+\/37=a+x/b, then m=a, and for if a: be not equal to a, let .v=a+m; then a+m+\/y=a+\/b, or m+\/_y—=\/b—; that is, is partly rational and partly a qua- dratic surd, which is impossible (Art. 178); therefore .v=a, and consequently also \/§= \/b. 180. If two quadratic surds and My cannot be reduced to others which have the same irrational part, their product is irrational. If possible, let Mhy=nm where r is a whole number or a fraction. Then wy=r2.v2 (Art. 81), and y=r°.v (Art. 82); there- fore \/y=r\/w, that is, \/y and may be so reduced as to have the same irrational part, which is contrary to the supposition. 181. £716 quad-ratio surd, \/;, cannot be made up of .two others, \/m and \/n, which have not the same irrational part. If possible, let \/a'=\/m_+ then by squaring these equal quantitieS, w = m + n + 2\/mn, and e: — m - n = 2 \/mn, a rational quantity equal to an irrational one; which is absurd. scans. 93 182. The square root of a binomial, one of whose terms is a quadratic surd, and the other rational, may sometimes be expressed by a binomial, one or both of whose terms are quadratic surds. Since (,j§i~/y)2=.v+y*2~/@, N/.r+yi2,/.r—y_=~/.;=1=~/y; hence i_f_any proposed binomial surd can be put under the form (x+y)i2~/xy its square root is at once found by inspection to be N/d=!=~/y. Now, to pro- ceed with any proposed case, take the term which contains the surd, and if it can be put into factors of the form in one or more ways, take that pair of factors for which the sum of .r and y is equal to the whole of the term in the proposed binomial which is rational. Having thus found a: and y, the square root required is + or — according as the sign of the surd in the proposed binomial is + or -, Ex. 1. Required the square root of 3+2,\/§. Here 2J5=2J§><~/—_a does not follow the rule given in Art. 167, but that it is -a, because it is that quantity whose square root_is Also _similarly ~/-_a><,,/:b is not J-ax—b, or ,jtTb, but it is ~/a><,\/-1 x,‘/b xJ-l, or JabxQ/I-IY, that is, -—,\/db'*. __To avoid mistakes in operating upon imaginary quantities, as JIb, ~/— 6, &c., __it w_il_l be_best in all cases to substitute for them thei_r equi- valgnts Jul/:1, &c., and to bear in mind that (,/—1)”=—1, (J—1)3=-~/—l, (J—1)‘=+1, &c. Ex. (w—a+J-TZTQ)(x—a—J:§)=(x-a+bN/3)(x—a—b~/j), =(x_a)2—(b\/;_1)2’ = .vz—Qax + a2+ b2. EQUATIONS. 184:. If one quantity be equal to another, or to nothing, and this equality be expressed algebraically, it constitutes an Equation. Thus as - a=b--.v is an Equation, of which w —a forms one side, and b -.v the other. In this equation it is asserted, that a certain unknown quantity is so connected with two known quantities (a and b), that it exceeds the one (a) by as much as it falls short of the other Again, agar—20:0 is an Equation, which asserts that a certain un- known number added to its square is equal to 20. An equality which admits of no question, as x+a=x+a, or 2x+3x=5m, is not an “Equation” strictly speaking, but is called an “Identity.” An * The student is recommended to have as little as possible to do with imaginary quantities, that is, with quantities .which have no meaning either as to number or magnitude. He need not wonder lat difficulties are likely to be introduced by the use of them, when he considers that \/—a signifies an operation to be performed which is absolutely impos- sible. Any discussion upon the interpretation which may be given to such symbols, and the uses to which they may be applied, would be quite out of place in an Elementary Treatise like the present. ED. ~ SIMPLE EQUATIONS. 95 Identity is therefore satisfied by any value whatever of the unknown quans tity; whereas in “Equations” the unknown quantities have particular values, which alone, and none other, will permit the expressed equality to subsist—To find these values is to “Solve” the Equations, and forms an important part of the business of Algebra. These values are sometimes called the “Roots” of the Equations, and are said to satisfy them. Thus, if Qw=6 be the Equation, .v=3, and can be nothing else; and .v=3 is called its solution. Again, if x2=4, we know that x=2, or —2; and 2, —2, are called the Roots of the equation .29: 4. 185. When an equation is cleared qf fractions and surds, if it contain the first power only of an unknown quantity, it is called a Simple Equation, or an equation of one dimension; if the square of the unknown quantity be in any term, (and there be no higher power,) it is called a Quadratic, or an equation of two dimensions; if the Cube of the unknown quantity appear, (and no higher power,) it is called a Cubic Equation; if the fourth power, a Biquadratic; and in general, if the index of the highest power of the unknown quantity be n, it is called an equation of n dimensions. SIMPLE EQUATIONS. 186. RULE. I. In any equation quantities may be transposed from one side to the other, if their signs be changed, and the two sides will still be equal. For let .v+10=15; then by subtracting 10 from each side, (Art. 80), x+10—10=15—-10, or w==15-10. Let err—4:6; by adding 4 to each side, (Art. 79), .v—Lt+4~=6+4.<, or .v=6+4~. If w—a+b=y; adding a—b to each side, .v-a+b+a-b=y+a—b; or .v=y+a-b. 18']. Con. Hence, if the signs of all the terms on each side he changed, the two sides will still be equal. Let .v—a=b—2.v; by transposition, -b+2.v=—.v+a; or. a—az=2.v-b. 96 SIMPLE EQUATIONS. 188. RULE II. If every term on each side be multiplied by the same quantity, the results will be equal (Art. 81). 189. COR. An equation may be cleared of fractions, by multiplying every term successively by the denominators of those fractions. 5m Let a+1—=34; multiplying by 4, 12m+5w=136. An equation may be cleared of fractions at once, by multiplying both sides by the product of all the denominators, or by any quantity which is a multiple of them all. (I? m :0 Let — + — + 2 3 multiplying by 2x3x4, 3x4xw + 2x4>
<5 = 300 —- 5.2? = sum forfeited ; 15w - 900 + 5n = 240, by the question, 20.2? = 240 + 300 = 540; '. a'=27, the number of days he worked, :60—.n=33, the number of days he played. PROB. 5. How much rye, at four shillings and Sixpence a bushel, must be mixed with 50 bushels of wheat, at six shillings a bushel, that the mixture may be worth five shillings a bushel? Let a' be the number of bushels required; then 9.2:: the price of the rye in sixpences, 600==the price of the wheat in sixpences, (50+d)10= the price of the mixture............ 92 + 600 = 500 +10.'c, 100=a', the number of bushels required. PROB. 6. A and B engage together in play; in the first game A “wins as much as he had and four shillings more, and finds he has twice as much as B; in the second game B wins half as much as he had at first and one shilling more, and then it appears that he has three times as much as A; what sum had each at first? ~ Let :0 be what A had, in shillings, y what B had. Then 2.2." + 4 = what A has after the first game; y—w—4=what B has; by the question 2m + 4 = 2y - 2w - 8, or 2y — 4n = 12, y - 2.2? = 6. 114 PROBLEMS. Also y‘—.n -4 + g+1= what B has after the second game; 2m+4— g—1=what A has; . 3 by the question y-.e—4+%+1=6a'+12— Eli—3, 3 or y+—‘Z-+ g—w—6a'=12—3+4—1, or 3y -7.n=12,} also y — 2.22 = 6, 3y — 6.19 =18,} and 3y - 7m =12,' by subtraction m =6, what A bad at first ; and y—2.n=6, or y-12=6; y =18, what B had. PROB. 7. A smuggler had a quantity of brandy which he expected would raise £9. 18s.; after he had sold 10 gallons, a reve- ’ nue officer seized one third of the remainder, in consequence of which he makes only £8. 2s.; required the number of gallons he ' had, and the price per gallon. ' Let a' be the number of gallons; then 392 is the price per gallon, in shillings, a' .n - 10 . . the quantity selzed, 1 8 0 o 0 xi the value of the quantity selzed, WhlCh a: and appears by the question to be 36 I shillings ; -10 l 8 ' w x ~2— = 36, 3 a: (w — 10)><66 = 36m, 660; - 660 = 36.22, PROBLEMS. 1.15 w=22, the number of gallons; 1 8 . and {-93 == 59; =9 shillings, the price per gallon. re ~ PROP. 8. A and B play at bowls, and A bets B three shil~ lings to two upon every game; after a certain number of games it appears that A has won three shillings; but had he ventured to bet five shillings to two, and lost one game more out of the same number, he would have lost thirty shillings: how many games did they play? Let a;- be the number of games A won, y the number B won, then 2:0 is what A won of B, in shillings, and 3y what B won of A; 2w—3y==3, by the question. Also (m~1)><2 is what A would win on the 2m1 supposition, and (y+1)><5 what B would Win .... 5y+5-2a:+2=30, by the question, or 5y-2a'=30-5-2=23; .0 and 2p - 3y = 3, by addition, 5y~3y=26, 2y=26; .'. y=13. And 2m=3+3y=3+39=42; a==21; and m+y=34, the number of games required. PROD. 9. A sum of money was divided equally amongst a certain number of persons; had there been three more, each would have received one shilling less, and had there been two fewer, each would have received one shilling more, than he did: required the number of persons, and what each received. 8....» ~ 116 QUADRATIC EQUATIONS. Let a: be the number of persons, 3] the sum each received, in shillings; then any is the sum divided, and (m + 3)><(y -1) = my also (w—2)><(y +1) =zry .'. a'y—a'+3y-3=a'y, or —.'v+3y=3; } by the question; and wy+zc-2y-2 =a'y, or ar—2y=2; y = 5 shillings, the sum received by each. And x—2y =w-10=2, a: = 12, the number of persons. [Exercises U.] QUADRATIG EQUATIONS. 200. When the terms of an equation involve the square of the unknown quantity, but the first power does not appear, the value of the square is obtained by the preceding rules*; and by extracting the square root on both sides, the quantity itself is found. Ex. 1. 5:02-45 =0; to find a'. By transp. 5w2=45, w2=9; (Art. 191), m=\/§= £3. The signs + and - are both prefixed to the root, because the square root of a quantity may be either positive or negative (Art. 147). The sign of w may also be negative; but still a: will be either equal to +3 or -3'l'. * It is obvious that the rules proved in Arts. 1136-192, apply to all equations, qua- dratic, cubic, 8w. as well as simple, because they are founded simply upon the Axioms (Arts. 79-82). ED. 1- This may be shewn as follows :-suppose w2=a2, then extracting the square root of , both sides, since Vdg=iag and \/a_2_=¢a, we have +a=+a .... ..(l), +w=-a .... ..(2), -a’=+a .... -.:r=-a .... ..(4). But it is evident that (l) and (4) are in fact the same equation and also (2) and (3); so that a==a includes all the four equations. ED. QUADRATIG EQUATIONS. 117 Ex. 2. am2= bcd; to find a. 2 bcd_ w=-_, a cd ‘.I O a 201. If both the first and seéond powers of the unknown quantity be found in an equation, arrange the terms according to the dimensions of the unknown quantity, beginning with the highest, and transpose the known quantities to the other side; then, if the square of the unknown quantity be affected with a coefficient, divide all the terms by this coefficient, and if its sign be negative, change the signs of' all the terms (Art. 187), that the equation may be reduced to this form, a2=pm= =q. Then add to both sides the square of half the coefficient of the first power of the unknown quantity, by which means the first side of the equation is made a complete square, (Art. 152), and the other consists of known quan-.. tities; and by extracting the square root of both sides, a simple equation is obtained, from which the value of the unknown quantity may be found. 2 Ex. 1. Let .n2+pa'=q; now we know that .n2+p.n+% is 2 the square of (0+;Z (Art. 152); add thereforeé- to both sides, and we have P2 r2 (02+ m — = + -—- ' P + 4 q 4‘ a then by extracting the square root of both sides, P \/ P2 a? + — = =l= —- ' 2 q + 4 ’ . . p P2 and by transposmon, m = — 5)- =l= q + z— . In the same manner, if a'2—pw==q, 1am <’U= ——. 4.- zol‘B '118 ,QUAnBATIc EQUATIONS." Ex. 2. w2—12a:+35=0; to find a. o I . By transposrtlon w2—12a'= -35 ; and addmg the square of a, or 6, to both sides of the equation, m2-12w+36=36-35=1; then extracting the square root of both sides, . i a' - 6 = =1; w=6=1=7, or 5; either of which, substituted for a' in the ori- ginal equation, answers the condition, that is, makes the whole equal to nothing. 2 Ex. 3. +-=3; to find a. a+1 a' 2w+2 ' 6+ =3a+3, {U 6m+2w+2=3.r2+3w, 3m2-5.n=2, 2 5.2—2 w____s 3 3 2 5w (5)2 2+2; a-- - =- - 3+ 6 3 86’ 24+25 '49 —36 86—36’ 5 7 5=1=7 . m: 6 , 1 =2, '__o 3 202. Bit. 4. m+\/5a'+10=8; to find a. By transp. \/5a' +10 = 8 -a', squaring, 5m + 10 = 64 -16.'u + m2, w2-‘21a: =_ - 54; QUADRATIC EQUATIONS. 119 2 4141 44:1 4! 4s 225 =“_; 21 15 r'U——-= ; 2 2 21*15 0.0 (1”: =18, 30 By this process two values of x are found; but on trial it appears, that 18 does not answer the conditions of the equation, if We suppose that \/5x+10 represents the positive square root of 5x+10. The reason is, that 5x+10 is the square of -\/5x +10 as well as of + \/5x+10; thus by squaring both sides of the equation \/5x+10=8- x, a new condition is introduced, and a new value of the unknown quantity corresponding to it, which had no place before. Here 18 is the value which corresponds to the supposition that x-\/5x+10=8. It should be particularly observed, that since +x>< +y is equal to -x><—y, in the multiplication and involution of quantities new values are always introduced, which, if not again excluded by the?“ nature of the question, will appear in the final equation. [Exercises V.] 203. If a quadratic equation appear under any of the forms included in ax2=bx==c, the left hand side may be made a complete square, without fractions, and the equation solved, by another method, as follows :—-- Multiply the whole equation by 4a, that is, four times the coefficient of x2, then we have 4a2x’=4abx==4ac; add b2, the square of the coefficient of x, then 4a2x2= 4abx+b2=b’=4ac, extract the square root, 2ax=b==Jb”=4ac; _ x_=l=~/b2i4ac =5 6 0 I 120 QUADRATIC EQUATIONS. Ex.1. —6—+-2-=3;tofind.r. x—l a: +2 6+2x =3x+3, 6x+ 2x+ 2 = 3a’+ 3a, 3x2-5x=2. Multiplying by 4x3, or 12, 36x2-60x=24, adding 5”, or 25, 36x’— 60x+25=94=+25=49; 6x—5=*7, 6x=5i7=12, or —.2; 1 '. a=2, or—g. Ex. 2. acxz—bcx + adx = M,- to find a. Here acx’- (bc— ad).r= bd ; multiply by 4ac, éagczag- 4ac(bc — ad )a: = 4abcd, add (60- ad)’, 4a’02x2— 4ac(bc - ad)x + (be — ad)2 = (be — ad)2+ 4abcd, = (60 + ad)2. Extract square root, 2aca! — (be—ad) = =i=(bc + ad) ; Qacx = ba—adi (be + ad), =Qbc, or —~ Qad; d b x=-, or —--—. a c 204. A quadratic equation has no more than two distinct values qf the unknown quantity which will satz'qfy it. For, if possible, let the equation axg+bx+c=0 have three distinct values of x, viz. a, ,8, 7. Then aa2+ ba+c=O. . . (1), afi2+b,B+c=O. . .(2), a72+57+0=0. . Subtracting (Q) from (1), a(a2-,82)+b(a_’8)=(); '. a(a+[a’)+’b=0 . . . . . . . . . . . .(i.) Subtracting (3) from (1), a(a2-7’)+b(a-¢y)=0; ‘. a(a+7)+b=O . . . . . . . . . . . .(ii.) " Subtracting (ii) from (L), aQ8-7)=O _ _ . , _ , . . . . . .(iii.) QUADRATIG EQUATIONS. 121 But a is not equal to 0, for otherwise the proposed equation would not be a quadratic equation; ,3—7=O, or fl=7. Hence a quadratic equation has not three distinct values of a, but it may have two. If however it be known that ,8 is not equal to 7, that is, that the given equality is satisfied by more than two values of a, it appears from (iii.) that a=0; therefore by or (ii), 6:0,- and by (I), (2), or c=0; that is, if a quadratic equation be known to be satisfied by more than two values of the unknown quantity, the coefficients of the square, and of the first power of the unknown quantity, and the term which does not involve it, are separately equal to 0, and the equation becomes an identity, being satisfied by any values whatever of the unknown quantity. ‘ 205. In any quadratic equation of the form x2+px+q=0, —p=i/ze sum of the two values of x, and q=tlzeir product. Let a, B, be the two values of a, then a2+pa+q=0, and fi2+p16+q=0; a2—,82+p(a-[a’)=0; .'. a+,6+p=0, 0r —p=a+}8 . . . . Again, g=—Pa—a2, =(a+fl)a—a’, =a,8 . . ... . ..(2)*. Con. 1. If the equation be of the form ax”+b.r+c=0, then a”+%x+%=0. Therefore, by what has been proved, b -5 =the sum of the two values of a, and g=the product . . . . . . . . . . . . . . . . . COR. 2. Hence also, if a, ,8, be the roots of the equation a2+px+q=0, it is proved that a2+px+q=x2— (a+,8)a:+a[3=0, = (x—aXx— = 0; from which it appears that if one ‘root’ of the equation be known, the other may be found by division; for, a being known, a2+px+q .r-a =.¢v-,B=O, which gives x=fl. * The above proof is open to objection in the case where a=B ; but the relations in ques- tion can then be proved to be true by actually solving the equation, as is done in Art. 207. 122 ' .QUADRATIO “EQUATIONS. Again, conversely, if the roots of a quadratic equation be given, the equation can be found. For, if a, B be the roots, the equation must be (x_a)('r _fl)=09 or .r’—(a+B)x+aB=O. Ex. Required the equation whose roots are 2 and 3. The equation is (x-2)(.:c-3)=O, or x2—5w+6=0. Con. 3. If a, B be the roots of the equation x2+px+q=0, then w’+px+q=(a'—a)(a'—B), whatever be the value of a. For it has been proved that p=—(a+ B), and q=aB; therefore, what- ever be the value of a, a’+pa'+q=x2- (a+B)x+aB=(.r—a)(x-B)*. 206. The results proved in the last Art. shewing the relation between the values of a and the coefficients in the equation, are of use in several ways; first, in enabling us to verify the solution of any quadratic equa- tion; secondly, in determining the values of the unknown quantities when an equation is proposed in which certain relations are already known respecting those values; and lastly, in solving various problems, reduced to quadratic equations, of which it is necesaary for our purpose to know no more than the sum or the product of the values of the un- known quantity. Thus, Ex. 1. flag—5.10:2. The values of a- (See Art. 201, EX. 8) are 2, 1 d. "_'_’0 an 3 Now, without repeating the work, to see if these values are cor- . 5 2 . rect, we put the equation under the form a2—§x—§=O, and smce Q+(_. -;-)=g, and 2x<—%>=—%, we conclude, at once, that these and no other are the Values required. Ex. 2. .r”—21a+54.~=0; required the values of a, it being known that one of them is six times as great as the other. Suppose a to represent one of the values, then 6a is the other, and their sum, or 7a=21, (Art. 205) a=3, and the values required are 3 and 18. a It must be borne in mind that we are here concerned not with the equation a2+pw+q=0, but with the expression x2+ px+q, irrespectively of anything that it may be equal to; but if any difficulty should arise, we might state the proposition thus: If a, B be the roots of the equation w2+pr+q=0, then yg+py+q=(y—a)(y -,8) whatever he the value of y. ED. QUADRATIC EQUATIONS. 123 ' Ex. 3.? If a and B be the values of x in the equation ax2+bx+c=0, find the value of ~1- -1-1 a B o . . . . b D1V1d1ng the equatlon by .922, a+5+55=0s c b 1 or :r—2+5+a=0. Assume y=5, then cy2+by+a=0, b a 2 _ __= , or y + cy-i-c O, 1 . 1 , 3, since y=;, I Ila 3"- all- and the values of' y are GIG" (Art. 205.) 207. Since every quadratic equation may be reduced to the form ag+pa+q=0, in which p and q may be positive or negative, we assume this as the general equation including every other. Th en, since 002+ pa: =- q, 2 2 2 2__P~4’q a+px+4__-T , a+g=il,/p2—4q, “file and x=— é ='= % ,/p2-4q, which are the only two values of a that will satisfy the equation. Now, from this result, it follows, 1st. That there is no possible value of a, if p2< 4g. 2ndly. That the values of a: are equal to each other, and each equal to ~15, if p2=4q. Srdly. That there are two unequal values of a, whose sum is --p, ifp2> Again, it appears that 1st. If q be negative, since JPQ—aq will then be greater than 12 and. always possible, there can be but one positive value of a. 2ndly. If p be negative, q positive, and 122:» 4sq, there will be two positive values of .r. The last two conclusions may also be deduced from Art. 205. 124 QUADRATIG EQUATIONS. \ Con. 1. Similar conclusions may be drawn with regard to the equation aa’+bx+c=0, by substituting% for p, and; for q. Thus, if a proposed equation be of the form ba-ax2= c, in which a, h, c, are positive quantities, there will be two positive values of a, when 4acbz. Con. 2. Hence also, if the roots of aa2+hx+c=O be equal, 6 2 <—) =42, or bg=4ac; a a b b2 b = 2 b + = 2 - = —- ' and aa+ x c a(w+ax+4a,> a(x+2a), and is, therefore, a perfect square for all values of w. This might have been easily deduced from Cor. 3 of Art. 205; for ax’+bx+c would =a(a'-a)(a:—-B) =a(ar - (2)2, a=B. 208. If an equation appear under the form (a+a)X= O, in which X represents an expression involving a, the unknown quantity; it is evident that either a:+a=0, or X=0, that is, a=—-a is one solution of the equation, as well as those which are found by proceeding with X=0. So that, whenever an equation is simplified by division, or the omission qf a factor, the divisor or factor contain the unknown quantity, one solution at least of the equation will be found by putting that divisor or factor equal to 0. Thus, let x2+3x=7x; the whole equation is divisible by .11, therefore .r=0 is one solution. Again, let x’-- 5a+ 6 =0. This may be put under the form (x— 2)(a'—3)=0; -. a—2=0, or x=2 and a, 3 O or a, 3 }; which are the only values of .v. ._ ___ , ___ By this method, therefore, the necessity for solving a quadratic in the usual way may sometimes be superseded. ' Ex. Given (away/gli— (a —b),fc§= 0; required .2). Here aJa—b~c~/ah—a~/J+ (“Eh—=0; MNZE +J5>—~/E(~/25+J176)=0’ or (J5—J55)(~/55+~/b_c)=0; QUADRATIC EQUATIONS. 125 .'. Jhz-JtTC=O, and ~/a_x_+~/h?=0; bx=ac, and ax=bc, ac be w==—, and .r=—- - b a 209. Every equation, where the unknown quantity is found in two terms, and its index in one is twice as great as in the other, may be resolved in the same manner as a quadratic. Ex. 1. z+42i=213 required 2'. e+4z%+4=21+4=25, zi+2=i5, z%==l=5-2=s, or -7; .’. %=9, 01' 49. Ex. 2. w'l+w'i=6; required a". _1 _% 1 6+1 25 w +0,” d+4_ 4 4, t 1 =l=5 a'" -= —- t _+2 2: ....1=l=5 .v 2= 2 =2, 01‘ —3, l l 1 .v2-----— or --' 2’ 3’ 1 1 0. m=_, 01"". 4 9 Ex. 3. 314- 6y2—27=0; required y. 3’4" 6y2=27’ 31*— 6y2+9=27+9=86, f—3=*Q y2=3=h6=9, or —3; y=£3, or 126 QUADRATIO EQUATIONS. 3 I Ex. 4. y6+ry3+-g—7-'=O; required a'. I 3 6+7” 3=--g—- y y 27:» 0,2 7.2 6 ,8 ___________, y+y+i a 27 , r Vrz Q3 +--=:l= -—-——, y 2 4 27 2 4: 2 8 . 2 s = JAM/LL 2 41 27 210. Some other equations may be conveniently solved as quadratics, that is, by completing the square, when they can be made to assume the form . X2+pX=q, X representing a compound expression involving the unknown quantity. Ex. 1. aa°+ W: bx ; required a. By transp. axz— be: +Jm=m adding 0, axg—bx+c+~/M=c, completing the square, '3 (ax’-6x+c) +~/ax”-bx+c+% =0 + %, ————— 1 Jaw2—bw+c+—2- = *\/0+ 1 4‘! Jm=_____*~/41+1"1 z 7 _ a axg-bx+c={i_il'£é_____m}; the equation is thus reduced to a common quadratic, from which a: may be found by the usual method. ‘ __.________ I i ’ Ex. 2. .v’-a+5J2.r’- 5x+ 6 = 5 (3x+33) ; required .12. Here 2a2-2x—l-10J2x2—5x+6=3.r+33, . QUADRATIC EQUATIONS. 127 2x2-5x+6+10J2w2-5x+6=39, (2x2-5x+6)+10~/2.r2— 5x+ 6 +25=25+39 =64, W+ 5 ==1= 8, Jm==ss-5=s, or ---13, 2a2—5x+6 = 9, 01‘ 169; which leaves two common quadratics for solution. [Exercises 211. When there are more equations and unknown quantities than one, a single equation involving only one of the unknown quantities may sometimes be obtained by the rules laid down for the solution of simple equations; and one of the unknown quantities being discovered, the others may be obtained by substituting its value in the other equations. Ex. {0- mgy =44,1 tofinda’andy. _m+3y=1 y n+2 ’i From the first equation, 2w—m+y=8, . w+y=& a'=8—y. From the 2nd equation, my + 2y - a - 3y = w + Q, or wy—2a’-y=2, by substitution, (8 - y)y - 2(8 - y) - y = 2, 8v—v2—16+2@/—y=2, 9y-y2=16+2=18, y2—9y=—18, 81 81 9 2- —-=——-—13=-, y 9y+4 4t 4 9 s —---=:E—-; y 2 2 . 9&3 .'. =———=6 or 3. 1'! 2 : And a=8-y=2, or 5. 128 QUADEATIO EQUATIONS. The solution of equations will Often be rendered more simple by particular artifices, the proper application of which is best learned by experienceii. 2 2— EX' 1' w +3} —65’} to find a' and y. reg/=28, From the second equation, 2wy=56, adding this to the 1“, {02+ 2mg + y2=121, subtracting it, :02— 2.vy + y2 = 9, extracting the square roots, .1: + y= * 11, and a: - y = i 3, adding, 20: = =I= IA, subtracting, 2y = i 8, w=7, or —7, and y=4, or -4. a2 y2 E . 2. - -=18, X y +w to find a and y. x+y=12, from second, a3+ya+ 3.1:3/(a: +y) = 1 72 8, by substitution, 18xy+36my=1728, 54wy=1728g '. ay=32. Also w2+ 2xy +y2= 14A,} and Amy = 128, From first equation, a3+y3=18xy,} '. x2—2a'y +32: 16, .v—y=*44,} and x+y= 12, 2x=16, or 8, and 2y; 8, or 10, .3 w=8, or 4', and y=4, or 8. " Many of these artifigces are pointed out, in the Appendix. En. ~ QUADEATIc EQUATIONS. 129 212. It may sometimes be of use to substitute for one of the unknown quantities the product of the other and a third unknown quantity. This substitution may be successfully applied whenever the sum. of the dimensions (Art. 63) of the unknown quantities in every term of each equation is the same. i E . 2 =12, X w +mz } to find a and y. say—23} =1, ’ Let vy==av, then v2y2+ vyg = 12, and vy2 -- 292: 1, 12 ‘ from the former y2== 2 , v +v from the latter y2= ; v -- 2 12 1 C . = , vi+v v~2 or v2+ v =12v —24, v2—11v=—24, 2 121 121 25 v —11U+“'_"==‘—4T"' =“_, 11 5 '0— --—=::l:-; 2 2 11=E=5 p 'U= =8} 1 1 And 2= =-— orl y v_ 69 a J...— 1 =l=1 00 y: ~---, I \/6 130 QUAEEATIO EQUATIONS. 213. The operation maysometimes be facilitated by substitut— ing for the unknown quantities the sum and difference of two others. This artifice may be used, when the unknown quantities in each equation are similarly involved. Ex. ‘ a+y=4, (a 2‘+_y’)(a~“+ya) = 280, Assume a=z + v, } to find a' and y. and y=z~v, then x+y=2z=4g 2:2. Also ads-y“: (2 + v)2+ (2 - v)”, =8+2v2, and a3+ y3= (2 + v)3+(2 —v)3, = 8+12v+ 6v2+ v3+ 8 —12v+ GvQ—vs, =16+12v’; (8+2v2)(16+12v2)=280, or (4+v2)(4i+3v2)=35, 16+16v’+3v‘=35, v4+156v2=132, v*+1g§v2+%=-?s%=%l, v’+-g-=*13,1-; . @2=i1;_8=l, 01' ~25; v=l; o'- and y=z-v=1. (has. In algebraical analysis it is frequently useful to observe whether the algebraical expressions under consideration are homogeneous or not, that is, whether the ‘dimensions’ of every term be the same or not; for, if this homogeneity be found at first, no legitimate operation can destroy it; or, if it be not found at first, it cannot be introduced; and thus an easy test is afforded, to a certain extent, of the accuracy of each succeeding step in the analysis. PROBLEMS PRODUCING QUADRATIC EQUATIONS. 131 For example, if the equation aa2+ 62.1: + cs: 0, _ be proposed for solution, in which every term is of three dimensions, that Is, which 1s homogeneous, every step in the process will present an homo- geneous equation, if it be correct. As a simple case it may be well to observe that, if the proposed equation be homogeneous, the final result must be so. A proper atten- tion to this observation Wlll frequently detect an error in the process of solvmg an equation. PROBLEMS PRODUCING QUADRATIG EQUATIONS. 214. PROB. I.‘ A person bought a certain number of oxen for 80 guineas, and if he had bought 4 more for the same sum, they would have cost a guinea a piece less; required the number of oxen and price of each. Let a' be the number of oxen, 80 . . ~ . . then —— Is the price of each, In gumeas, .v and 4 the price of each on the second supposition; .v+ 80 80 . '. —— = -— —1, by the question, at + 4 a: 80s: + 320 so = -——-—- - a' - 4, .12 80.1: = 80.12 + 320 - .502- 4a, .v2+ 4:0 = 320, a'2+ 4m + 4 = 324, a' + 2 = =18; .v= =18 - 2 =16, or -20, the number of oxen; 80 80 . . and — = 1—6 = 5 gumeas, the prlce of each. :0 In this, and in many other cases, especially in the solution of philosophical questions, we deduce from the algebraical process answers which do not correspond with the conditions. The reason seems to be, that the algebraical expression is more general than 9—2 132 PROBLEMS PRODUCING the common language; and the equation, which is a proper repre- sentation of the conditions, will also express other conditions, and answer other suppositions. In the foregoing instance a: may either represent a positive or a negative quantity, and cannot in the opera- tion represent a positive quantity alone (Art. 202); and the equation 80 80 ___,_____ a'+4_a: ’ when a" is negative, or represents the diminution of stock, will be a proper expression for the solution of the following problem: A persOn sells a certain number of oxen for 80 guineas; and had he sold 4 fewer for the same sum, he would have received a guinea a piece more for them; required the number sold. 215. PEOE. II. To divide a line of 20 inches into two such parts, that the rectangIeKunder the whole and one part may be equal to the square of the other part. Let .v be the greater part, then will 20 - .v be the less, and a}? = (20 - m)><20 = 400 - 20w, by the question, 4,222+ 20m = 400, 002+ 20.v+100 = 400 +100 = 500, .v+10==l=\/500; .v=\/500—10, or -\/5—05-10. The observation contained in the preceding article may be applied here; and it is to be remarked, that the negative values thus deduced are not insignificant, or useless. Here the negative value shews, that if the line he produced +10 inches, the square of the part produced is equal to the rectangle under the line given and the line made up of the whole and part produced. ' 215*. In order to ascertain the problem, the solution of which is given by the negative roots, take the equations which constitute the algebraical interpretation of the problem: write therein —.r for a', &c.: then the new equations will suggest the required question. This of course must be‘done by considering from what conditions of the given problem the several terms of the equations arise, and making the necessary alterations in agreement with the remarks of Art. 463. Thus, if we take . O the first of the preceding problems, we see that S;— represents the price QUADRATIC EQUATIONS. 133. of each ox, the price of each if 4.. more were bought, and the -1 arises from the relation between those prices given by the question. Now, . . . . . 0 80 if we write 1n the equatlon of Prob. I. —.r for x, we obtain 2+ 4! =7—15“ . 80 . 2.6. x 4|=§§+L and here ~52 represents the prlce of each of a: oxen; 80 4‘ , the price of each if there were 4 less, and the +1 gives a relation between those prices. But as a: is negative, it represents the selling of oxen (Art. 463), whence the problem above stated may evidently be seen to be that of which the altered equation is the algebraical interpretation. From this example it can easily be perceived how the interpretation of the negative result is to be made. 216. PROB. III. To \find two numbers, whose sum, product, and the sum of whose squares, are equal to each other. Let ar+y and w—y be the numbers, their sum is 260, their product w2—y2, the sum of their squares 2m2+2y2, and, by the question, 260 = 2m2+2y2, or w=a22+yh Also 2.10 = wg— y‘z, adding, 3w=2w2; 3 o 87=—. 2 2m=m2-y', or -9 4. y’ 2 9 9-12 -3 “4.. " a T’ i :5 y- 2 ' 3+ :5 Hence az+y= g/ , 3- Tie and w—y= 134 ‘ QUADRATIC EQUATIONS. both of which are “impossible” quantities (Art. 183), a conclusion which shews that there are no such numbers as the question'supposes. [A collection of Problems with their Solutions will be found in the Appendix] - [Exercises SGHOLIUM. By the method of solution pursued in Art. 202 it is clear that both the resulting values of the unknown quantity may be those of a different equation and not of the proposed one; for if the proposed equation be of the form ax +Jbx+c=d, the solution effected may be that of the equation ax—Jbx+c=d, and it is impossible to say, without trial, to which equation either of the resulting values of a: belongs. That there is no value of x which will satisfy both equations (except in a particular case) is easily proved. For, if possible, let there be such value; then, for that value, c , I,/b.r+c=O, or a:=-Z-, a value of a: which will satisfy neither equation, except in the particular 0 (1 case when -----=-(;, or ac+bd=0. 6 Hence it appears, that after solving an equation of the above form by the usual method, it still remains doubtful whether either of the values of the unknown quantity obtained, will satisfy the equation; and if one of the two be the value sought, it remains doubtful which it is. Thus from the equation 3x+ 30x—71=5, the values of a: obtained are 4, and 2%, neither of which will satisfy the equation. . And from the equation 3x+J2w-2=7, \ the values of .2: obtained are 3, and 1;, of which only the fractional, and not the integral, value will satisfy the equation. The fact .is, that in the former instance both values of a: are the values belonging to the equation 3x—~/3_Ox_—7T=5; QUADEATIc EQUATIONS. 135 in the latter x=3 belongs to the equation 3x-J2x~2=7, and the other solution x=1§ to the equation as proposed. Since, then, the values of a: in 3x+,\/ 3012—7 1 =5, found by the common method of solution, do not belong to the equation at all, as is also the case in many others of like form, where, it may be asked, lies the fallacy in the process whereby we obtain a false result? It is here. We assume, as an axiom, that if the same root of equal quantities be extracted, those roots are equal to each other in all cases; whereas we know, that they may be unequal. For instance, retracing the steps in the following operation, aw—Jbx +c=zl, ax—d=, /b.v+c, (aw— d)”: bar + c, we assume, that the same value of x which satisfies the last of these equa. tions must also satisfy the preceding one; but this may not be the case, since it may be the value which satisfies —(ax—d)=m, ‘ or an +JlT+c=d, instead of any—W: d, the equation we commonly assume to be satisfied by that value, when it is the proposed equation, whose solution is required. The fact is, that the equation really solved is not the proposed one ax-Jbx+c=cl, but (ax—d+,/bx+c)(ax—d—,/bx+c)=0, or (ax- d)2— (bx+c)=0; and it is quite a chance whether both or either of the values of .1: obtained belong to the proposed equation. It is certain, however, that the values obtained belong to one or other of the two distinct equations ax+,\/bx+c=d, aw—Jbx+c=d. Also, when there are two values of x which will satisfy one of this pair of equations, there is no value of a: which will satisfy the other; because, if there could be such a value, then the quadratic equation (ax-clf— (bx+c) = 0, would have more than two distinct values of x, which was shewn to be impossible in Art. 204. From what has been said it appears, that the results which are com-_ monly obtained as solutions of quadratic equations, when those equations. 136 INEQUALITIES. are given in an irrational form, require to be verified, before they can be depended upon. Hence also, when a proposed problem depends upon the solution of an irrational quadratic equation, the problem may, or may not, have those salu- tions which appear as solutions of the quadratic equation. This conclus10n, it must be evident, is of too important a nature, to be safely overlooked. INEQUALITIES. If one quantity be greater or less than another, or than nothing, and this be expressed algebraically, it is called an Inequality. Thus, x—a>b-a' is an Inequality, of which ar—a forms one side, and b—a the other. 217. Any quantity may be added to, or subtracted from, each sideof an inequality, and the sign of inequality remain as before. Thus, if a>b, aia>bia3 for if a>b, it is evident that a+x>b+x, Similarly, if ab, it is evident that a—a'>b—a’, as long as a: is not greater than a or b. If a>b, but not greater than a, then it is evident that a—a->b—a:, for a positive quantity must be greater than a negative one. If a>a and >b, then both sides a—a, and b-a, are negative; but a is nearer to .2: than b is to .v, therefore a—ab-a'. Similarly, if a2ab+c2, a2+ b’-— Qab > Qab — Qab + c’, or (a—b)2>02. 218. If a>b, and c>d, and e> f, &c., then it is evident that a+c+e+&c.>b+d+f+&c.; But if a>b, and o>d, it does not always follow that a—c>b-d; for a~may be more nearly equal to 0, than 6 is to d. Thus 9>6, and 7 >2, but 9-7, or 2, is not greater than 6—2, or 4. 219. every term on each side of an inequality be multiplied or divided by any positive quantity, the sign of inequality will remain as before. Thus, if a>b, 2a>2b, 6a>66, &c.; or, if—a>—b, -2a>-Qb, —6a>-6b; &c. as is sufficiently manifest. ‘ . COR. Hence an inequality may be cleared of fractions by multiplying both sides by the product of the denominators of all the fractions, or by INEQUALITIES. 137 the least common multiple of them all; provided the multiplier is a positive quantity. E abl] 1.1. ,2 l.h. . x. If Z?+6?,>5 +5, mu tIp yIng by a b , wnc 1s necessarlly positive, a3+b3>ab2+a26. But, if all the terms of an inequality be multiplied or divided by a negative quantity, the sign of inequality is reversed, that is, > is changed into <, 01‘ < into >. Thus, for example, 6>4, but 6x—2 or —12<4><-—2' or —8. Also —-g— or -3<:L-; or —2. Hence, if we multiply or divide the terms of an inequality by any algebraical quantity, it will be requisite to know Whether the quantity is positive or negative. COR. Also, if the signs of all the terms of an inequality be changed, the sign of inequality is reversed, for this is equivalent to multiplying each side by —1. 220. Both sides qf an inequality may be raised to any power or any root of them extracted, and the sign qf inequality remain as before, provided that each side is a positive quantity. Thus, 7>5 and '72 or 49>52 or 25, 73 or 343>53 or 125, and so on. But, if either side be negative, then no general conclusion can be stated as to the resulting inequality. For —3<4, and (—3)2 or 9<42 or 16. Also —2>-3; but (—2)2 or 4<(--.‘5’)2 or 9. Similarly 16<25, and Jfd or 453 ; and so also in any similar case. 3 6 Hence it appears, from this and the preceding articles, that the rules which belong to Equations must not be inconsiderately applied to in- equalities, since these latter have distinct rules of their own materially differing from those of the former. Ex. 1. Shew that a2+ b2+02>ab+ac+bc, unless a=b=c. Since every quantity, when squared, is positive, and therefore greater than 0, we have (a—b)2>0, unless a=b, 138 INEQUALITIES. or a2—2ab+b2>0, - or a2+ b2>2ab, (Art. 217. Cor.) Similarly, a2+ 02>2ac, unless a=c, and b9+ 02>2bc, unless b=c ,' by addition 2a2+2b2+202>2ab+2ac+2bc, (Art. 218.) or a2+b2+02>ab+ac+ba (Art. 219.) Ex. 2. Shew that a5+y5>x4y+y4m war-eameme-nuns). = (rave-i)- Now, whether a: > or < y, the two factors .v“—y‘, and .v— y, have the same sign, and therefore their product is always positive; hence a5+y5- (m‘y+y“.v) is always positive, '. a5+y5 > a‘y +y4a’. . 2 —4 1 Ex. 3. Gwen that x+ +£fi +-1-, find a. 3 2 2 3 x+2+fx+1+1 ' 4 3 2 ’ 2 3’ mult. by 12, 3x+6+4x<6x—24+36, and >6a~+6+4, (Art. 219.) or 7a+6<6x+12, and >6x+10, a<6, and >4, (Art. 217). Hence a: is any number between 4 and 6; and, if it be a whole num- ber, a=5. - Ex. 4. Which is greater or fié+J§? JE+JT> or or <22+2J5i squaring, by Art. 220, ............. .. 2,/‘z_0> or <5+2.f5—7, (Art. 217), ............. .. 280> or <253+20J57, (Art. 220), , ............. .. 27> or <20J35. Now 27 is clearly less than 20,/5_'i, therefore N/TQ +J3 is the greater of the two proposed quantities. / a+b+c+d P+q+r+s > the least and < the greatest of the Ex. 5. Shew that . a c d . . . , fractlons ;, ;, E , each letter representmg a posmve quantity. 'é’ INEQUALITIES. 139 Of the fractions g the least Then at“ 3 3 a a d I; , suppose-j; to be the greatest (G), and g = , - £1 - P>g, g g) 7‘. g, 8_g' '. a=pG, bpg, b>qg, c>rg, d=sg, a+b+c+d<(p+q+r+s)G, and a+b+c+d>(p+q+r+s)g, a+b+c+d p+q+r+s This proposition may without difficulty be extended to more than four fractions. And it will be equally true whether a, b, c, d, be positive or negative, provided that p, q, r, s, ..., the denominators, be all positive; of negative quantities we suppose that one to be the greatest which is numerically the least. Moreover, if the denominators be all negative, —~p, } Art. 219.. } Art. 218. g. (Art. 219.) Q.E.D. -q, &c....,' since i =2, &c., the above proposition will still hold. _ Therefore, if a number of fractions have their denominators all of the same . sum of numerators SIgn, , . sum of denominators greatest of the fractions. \ - is intermediate in magnitude to the least and Ex. 6. Shew that if a,, a2, a3, a, be any positive quantities, a +a +...+a,, .. --—_— J—2;-—>,,/a,.a,.a3. . .a,,, unless the quantities are all equal. The left hand member of the above inequality remains unaltered, however a,, a2, may change in magnitude, provided only that the sum of them all is not changed. It will therefore be sufficient to prove that, if the sum of a,, (1,, an be given, their product is the greatest possible when they are all equal. For if not, let it be the greatest possible for certain unequal values of the symbols, and let up, a, be any two of them that are not equal. Then if for each of up, a, we substi- a +a . . tute J—Q—q, (which we are perfectly at llberty to do, as the sum of all the quantities is not thereby altered) the product becomes " a+a a+a P ‘1 P 9 ~/a‘oa2-.. 2 I 2 onoan, 0 11 Instead of ~/a1.a,...a,,.a,... an. a+a a+a . P 9! P '1 T . —— Is >a,,.aq, (ap+aq)2) Which is equal to a,2+af+2a,,a,, is >4apaq. (See Ex. 1, above). But 140 INEQUALITIES. Therefore the new product we have obtained is greater than the former one ; i. e. our supposition of the former being the greatest possible is not tenable. Therefore the product is the greatest possible when they are all equal; but then it is equal to al or a2 or &c., and therefore it is equal to the left- hand member of the proposed inequality. Consequently, in all other cases, it is less, 2'. e. a +a +. . .+a ——-——— _'l__2____lz >Z/aloa2oooan, n unless the given quantities are equal. This ‘is often expressed by stating that the arithmetical mean between any number of positive quantities is greater than their geometrical mean. Another proof—By the Binomial Theorem, (Art. 308) we have 2 o-u-a .Z' "_ D a (1+72>-1+a:+1'2 a+ 1.2“, a+&c., _ 1 (1_ 1 <1__ 2 n—l 2 n—1 n—1 1.2 “7+ 1.2.3 n-1 and (I + =1 +x+ 023+ &c., n being a positive integer. \ N 1 2 & , . 1 2 ow g, 7—2, c. are Iespectlvely less than n—_—1, 7-Z—_—1, &c., and, therefore, after the first two terms of each series, which are equal, each term in the first series is greater than the corresponding term in the second. 12 n—l (I + >(1 +i) , n n—l N ow let al , a2, an be n quantities that are not all equal, a+a +...+a, " a+a+...+a-na " then 2 ) =a1"(1+ 1 2 " n , nal . a +a +...+a —na "" and 1s >a1" 1+ 1 2 " ‘ , . a+a+...+a "'1 i.e. , . . . a+a +...+a "-2 and 1s, a fortzorz, >a,.a,(—3—42_—") , n..- . a,+...+a,, "'3 I O 7:16))— 9 &c. >511 'a2'a3"'an; a1+a2+ a,+...+a,, n " >VaI-a2-aa"'an- n [Exercises Y5] RATIOS. 141 RATIOS 222. Ratio is the relation which one quantity bears to another in respect of magnitude, the comparison being made by considering what multiple, part, or parts, one is of the other*. Thus, in comparing 6 with 3, we observe that it has a certain magnitude with respect to 3, which it contains twice; again, in comparing it with 2, we see that it has a different relative mag- nitude, for it contains 2 three times, or it is greater when compared with 2 than it is when compared with 3. The ratio of a to b is usually expressed by two points placed between them thus, a : b; and the former, a,_ is called the antecedent of the ratio, the latter, b, the consequent. ‘ Since in the ratio a : b the comparison is made in regard to quantu- plicity, (Km-d wnAmo’Tm-a), the ratio must evidently be measured by what- ever expresses the degree of that quantuplicity, i. e. by what is necessary to multiply b by to obtain a. But this multiplier is the fraction ff; the fraction therefore is the measure of the ratio a : b. 223. COR. 1. When one antecedent is the same multiple, part, or parts, of its consequent, that another antecedent is of its con- sequent, the ratios are equal. Thus, the ratio of 4 : 6 is equal to the ratio of 2 : 3, that is, 4 has the same magnitude when com- pared with 6, that 2 has when compared with 3, since 2: The . . . . a C a ratlo of a : b Is equal to the ratlo of c : d, If 5 =g, because 5 and c . I . 8 represent the multiple, part, or parts, that a Is of b, and c of d. 224. COR. 2. If the terms of a ratio be multiplied or divided by the same quantity, the value of the ratio is not altered. a ma For —-=——~ (Art. 101); a: b=ma : mb. b mb 225. COR. 3. That ratio is greater than another, whose ante- cedent is the greater multiple, part, or parts, of its consequent. * Ao'yos éorrl 660 “67691511 b/toyevoiiv 1i Ka'ra‘ nnkucécrn'ra 7rpds' o’z'hhnka nouz‘ eXe'o-ts. (EUCLID, Book v. Def. III.) 142 RATIOS. . . i . 7 Thus the who 7 : 4 1s greater than the who 8 : 5; because 1, or 4 . 8 32 . . é—d, Is greater than -5-, or These concluSIons follow Imme- diately from our idea of ratio. \ v Ex. Which is greater a+a: : a—x or a"+a:2 : ag—ai? a+.v : a—.x> or or < a— a2+ 22+ 20x a2+ a” > or < L062; a2—a2 . . . 2am and smce the former Is the greater by the quantity W; '. a+a : a-.r>a2+ar2 I az—wz. 226. DEF. A ratio is called a ratio of greater inequality, of less inequality, or of equality, according as the antecedent is greater than, less than, or equal to, the consequent. 227. A ratio of greater inequality is diminished, and of less inequality increased, by adding any quantity to both its terms. If 1 be added to the terms of the ratio 7 : 4, it becomes the ratio 8 : 5, which is less than.the former, (Art. 225). And in general, let a be added to the terms of the ratio a : b, and it be- comes a+.v : b+a', which is greater or less than the former, ('1? a u >or<5, or, by reducmg them to a. common a d, a + accor In as g b + ab + bar ab + an denominator, according as W> or oror or < the ratio a: b, according as as: a . . b > or <5 (Art. 225); that Is, accordlng as .v>or : a; ai5a' : a; 8:0. are nearly equal respectively to the ratios (at=4=.v)3 : 03, (a £4104 : a4, (a=1=.v)" : a5, 820. Also ails-w : a, aiia: : a, 8:0. are nearly equal to the ratios) 3 3 '— 4 4 _ c . Jaix ; Ja; Jain : Ja; &c. Ifa' be small when compared WIth a. For (airy : aa=asi3aia+Bataan:3 ; a3, 3a: 3.222 x3 :14 2+? 5:5,, : 1, (Art. 224). Now if a: be small when compared with a, g is a small fraction; 2 2 a: 3x a: . .v . . 3.1: _ 3...; _ =- 1 1, therefore slnce — 1s small compared WIth 1, -—- 1S a2 a a a a2 and . 3a: . .a:3 . . small compared WIth —a—; a fortzorz I, Is very small compared WIth 1 or . 3x WIth—. a 1 l d hi h C Hence, neg ectIng —a; an 55, w 10 are very small fractlons, 1*??? : 1, or a=¥=3a : a, is a near approximation to (aim)3 ; as, if a be ' small when compared with a. Similarly it may be shewn that ailla : a; ai5a' : a; ‘&c. are approxi- mations respectively to (a=l=.v)‘ : a4, (air)5 : a”, &c_ PROPORTION. 145 Again, since ,laasai : J3: lag : 1, (Art. 224). a: a2 _ 1 O Q by the same reasoning this ratio is reduced to a: 14%;: 1, or aigla: : (I. Also Unix : 1/a_=ad=-,l;.v : a, nearly; and so on. The utility of the rules here proved will be sufficiently manifest from the following Examples, when it is observed by what a troublesome pro— cess the several proposed ratios would be found Without the rules. EX. 1. (1'5241)4 : (1‘524)4=1'5240+4><0‘0001 : 1524 nearly = 1'5244 : 1524 nearly. EX. 2. {77—2—93 yiés=72sg : 72s nearly. Ex. 3. 3/2-134 : i/2131=2131g = 2131 nearly. [Exercises Z.] PROPORTION. 236. DEF. Four quantities are said to be proportionals, when the first is the same multiple, part, or parts, of the second, that the . . _ a 0 third Is of the fourth; that Is, when i=8, a, b, c, d, are called proportionals. This is usually expressed by saying a is to b as c is to d, and is thus represented, a : b :: c : d; or sometimes, a : I): c : d. The terms a and d are called the ea’tremes, and b and c the means. the four quantities 237. When four quantities are proportionals, the product of the extremes is equal to the product of the means. Let a, b, c, d, be the four quantities; then, since they are pro- portionals, g=g (Art. 236); and by multiplying both sides of the equation by bd, ad=bc. 238. COR. 1. If the' first be to the second as the second to the third, the product of the extremes is equal to the square of the mean. 10 146 PROPORTION. 239. COR. 2. Any three terms in a proportion being given, the fourth may be determined from the equation ad=bc. For d=bb£, 0:95, béi-gé, a=%c. Hence we have the Single Rule of Three in Arithmetic.‘ 240. If the product of two quantities be equal to the product qf two others, the four are proportionals, making the factors of one product the means, and the factors of the other the eartremes. . . . b Let ay=ab, then diVIdlng by ay, 2=_, a y or .v : a :: b : y. (Art. 236.) 241. Ifa:b::c:d,andc:d::e:f,thenals0a:b::e:f. (EUCLID, B. v. Prop. xr.) a Because 2)— therefore = o , we \lm or a:b::e:f. 242. 11" four quantities be proportionals, they are also pro- portionals when taken inversely. (EUCLID, B. v. Prop. B.) If a : b :: c : d, then b : a :: d : c. For , and dividing c _d @la unity by each of these equal quantities, b d —-=—; that is, bza :: dzc. a c 243. [f four quantities be proportionals, they are propor- tionals when taken alternately. (EUCLID, B. v. Prop.' xv1.) Ifazbzzczd,thena:c::b:d. B h . . . 1 a 0 d . . ecause t e quantities are proportlona s, E = a , an multlplylng Unless the four quantities are of the same kind, the alternation cannot take place; because this operation supposes the first to be some multiple, part, or parts, of the third. PROPORTTON. 147 One line may have to another line the same ratio that one weight has to another weight, but a line has no relation in respect of mag- nitude to a weight. In cases of this kind, if the four quantities be represented by numbers, or other quantities which are similar, the alternation may take place, and the conclusions drawn from it will be just. 244. When four quantities are proportionals, the first together with the second is to the second, as the third together with the fourth is to the fourth. This operation is called componendo. (EUCLID, B. v. Prop. XVIII.) Ifazbzz cad, then also a+b:b::c+d:d. a c . . Because 5 = C—i, by adding 1 to each s1de, a 1 c 1 —+ =—+; b d _ a+b c+d that 1s, = “d 3 or a+b:b:: c+dzd. 245. Also, dividendo, the earcess of the first above the second is to the second, as the eacess of the third above the fourth is to the fourth. (EUCLID, B. v. Prop. XVII.) Because 5= 6%, by subtracting 1 from each side, a 1 c l b Ti ’ d t, a-b c—d Ia Is, b - d , or a—b:b::c-d:d. 246. Again, convertendo, the first is to its excess above the second, as the third is to its access above the fourth. (EUCLID, B. V. Prop. 10-—2 148 PROPORTION. a-b c-d b d u 9 By the last Article, b and — = i (Art. 242); a c _a-b ' b c-d a—b 0—d- d c that is, a—b : a :: c—di 6'; b ><-= a and inversely, a : a—b :: c : c—d. 247. When four quantities are proportionals, the sum of the first and second is to their difi'erence, as the sum of the third and fourth is to their diference. If a : b :: c : d; then a+b : a—b :: c+d : c-d. a+b c+d . 4 --== - By Art 24, b d , a—b c-d and by Art. 245, T= d ; -b -d . a+b+a =c+d+c “(11.82); b b d d a+b_c+d_ a—b—c—d’ that is, a+b : a-b :: c+d: c—d. 248. When any number of quantities are proportionals, as one antecedent is to its consequent, so is the sum of all the antecedents to the sum of all the consequents. (EUCLID, B. v. PROP. XII.) Ifa:b::c:d::e:f, &c. then a :b :: a+c+e+&c. : b+d+f+&c. a c _ Because i=8, ad=bc; 1n the same manner, af=be; also ab=ba; hence ab+ad+af=ba+bc+be’ .. or a(b+d+f)=b(a+c+e); by Art. 240, a : b :: a+c+e : b+d+f; and similarly when more quantities are taken. PROPORTION. 149 249. When four quantities are proportionals, if the first and second be multiplied, or divided, by any quantity, as also the third and fourth, the resulting quantities will be proportionals. Ifa:b::c~:d,thenma:mb::2:— n n 1 a 0 ma 5'0 F‘—-=— ,',—-=_A.t101, m b a’ mb 1 (r ) -od n , c d orma:mb::--:—. n n 250. If the first and third be multiplied, or divided, by any quantity, and also the second and fourth, the resulting quantities will be proportionals. F a 6 ma m6 dma 77710 or -~=-—-' ——=_-' _=_ b d’ b d’ 1 1 ’ '_b ‘_0d 771 n d ormaz— mc:—. 72 251. COR. Hence, in any proportion, if instead of the second and fourth terms quantities proportional to them be substituted, we 0 0 b d Q 0 have still a proportion. For — and 77/ are In the same proportlon n with b and d (Art. 249). 252. In two ranks of proportionals, if the corresponding terms be multiplied together, the products will be proportionals. Ifazb ::c :d, and e:f ::g :h, then also ae:bf::cg:dh. Because 65 = = g; therefore %2 that is, ae : bf :: cg : dh. 150 PROPORTION. This is called compounding the proportions. The proposition is true if applied to any number of propor- tions. 253. If four quantities be proportionals, the like powers, or roots, of these quantities, will be proportionals. a e a" c" If a:b::o:cl then—=—, and—=-‘ or a":b” ::o”:d"' ’ b d b" d", ’ where n is whole or fractional. 254. If a : b 1: c : cl, to prove that mainb : paiqb :: mciod : pciqd. §_§ b_d’ 7112712 b“d’ ma mo 75:71:72”, ma=l=nb_mcind b _ d Similarly Pazqb = WT",- ma=l=nb___paiqb_mcind _ pciqd b ' b _ d * d > mad: nb med: ml Paiqb =PC*qd’ or mainb : paiqb :: meind : pciqd. 255. If three quantities a, b, c be in continued proportion, that is, a: b.::'b : c, then a: c z: a.2 : b2, that is, the first has to the third the duplicate ratio of that which it has to the second. (EUCLID, B. V. Def. For =—b— ; PROPORTION. 151' 256. If four quantities are in continued proportion, that is, a : b :: b : c z: c: (1, then a : d :: a3 : b3, that is, the first has to the fourth the triplicate ratio of that which it has to the second. (EUCLID, B. v. Def. XI.) a a b c a b C- a_ll a a If??? a3 :-b—3-, 257. The Definition of Proportion here used will not serve for a definition in Geometry, because there is no Geometrical method of repre- senting the quotient of a+b, a and b being any Geometrical magnitudes whatever of the same kind. But such magnitudes may always be multi- plied geometrically; that is, a line may be produced till it becomes n times its original length—an area, or a solid, may be doubled, trebled, &c.-- geometrically. Hence the strictness of Geometry requires such a definition as that which is the foundation of Euclid’s 5th Book, and which may easily be shewn to follow from the Algebraic Definition. For suppose a, b, c, d to represent four quantities in proportion, according to the Algebraical definition; then a c b_d' cmd m a m c n'b_n'd’ 1 ma mo 0‘ — ==—- nb nd’ from which it follows, by the nature of fractions, that if ma>nb, then mc>nd; if ma=nb, mc=nd,- if' ma nb when mc> nd7 and ma=nb when mc=nd, and ma< 2, =1+98=99. o 281. If the two extremes, and the number of terms, in an Arithmetical Progression be given, the means, that is, ‘the intervening terms, may be found. Let a and l he the extremes, that is, the first and last terms, of an Arithmetical Progression, n the number of terms, and b the unknown common difference; then the progression is a, a+b, a+2b, a+n—l.b; and since l is the last term, a+n—1.b=l; l -a _ b=-- n—l , and b being thus found, all the means a+b, a+2b, &c. are known. Ex. There are four means, or intervening terms, in Arithmetical Progression between 1 and 36; find them. Here a=l, l=36, n;6,_ and the means are 8, 15, 22, 29. Con. A single Arithmetical mean between any two quantities a and b, which is called the Arithmetical mean, will be a+gill or fig—b: but this may be shewn more simply thus, Let a, x, b be in Arithmetical Progression, then by Def. x—a=b—x; x=%(a+b). 282. The sum of a series of quantities in Arithmetical Pro- gression may be found by multiplying the sum of the first and last terms by half the number of terms*. Let a be the first term, b the common ditferencef, n the num- ber of terms, l the last term, and s the sum of the series: * By this rule we are enabled to find the sum of any number of terms in Arithmetical Progression, without the trouble of adding them all together. ED. 1- The Common Difl'erence in any proposed case is obviously found by subtracting any term from the term next following. ED. ARITHMETIOAI. PROGRESSION. 163 Then, a+(a+b)+(a+2b)+ ....... ..+l==s; also, reversing the order of the terms, l+(l—b)+(l—2b)+ ....... ..+a=s; adding, (a+l)+(a+l)+(a+l)+...tonterms=2s, that is, n times a+ l, or (a+ l)n =2s; n '. s = (a + l); . COR. 1. Since l=a+(n—l)b, we have also s= {2a + (n-l)b};l. COR. 2. Any three of the quantities s, a, n, b, being given, the fourth may be found from the equation s= {2a+(n-1)b}g-. Ex. 1. To find the sum of 14 terms of the series 1, 3, 5, 7, &c. Here a=1, b=2, n=14; -. s=(2+26)><7=196. Ex. 2. Required the sum of 9 terms of the series 11, 9, 7, 5, &c, In this case a=11, b= —2, n=9; ’. 8=(22—16)><—9—=6x-9— =27. 2 2 Ex. 3. If the first term of an arithmetical progression be 14, and the sum of 8 terms be 28, what is the common difference, and the series? _ Since {2a + (n — 1)b} xg=s, 2s 2a+(n—1)b=-—, n 2s 2s—2an (n—1)b=;z-—2a=—————; 2s—2an _n(n-1) ' 11—2 164 ARITHMETICAL PROGRESSION. In the case proposed s=28, a=14, n= 8; b 56—224 7 ~28 _ 8X7 _ 7 _ Hence the series is 14, 11, 8, 5, &c. _3. Ex. 4. If the first term of an arithmetical progression be 3%, the common difference 1%, and the sum 22, find the number of terms. Generally s=(2a {ii-:1 . b); . Here s=22, a=3g, b=1%~, 20 —- 13 72 0.: __ _ ._ ,_ ~52 {3+n 1.9}2, or 396=4712+131223 and by solving this quadratic equation it is found that n, the number of terms, is 4. In this and similar examples the negative value of n is excluded by the supposition under which n originally enters the equation, since a num- ber of terms can only be expressed by a positive integer. A meaning can however be obtained for the negative value of n. For in the equation s={2a+(n—1).b}g , write — n for n, and it becomes n s=—{2a-(n+1).b}-2- , or s={(n+l).b—2a}g, or‘ {2.(b-a)+(n_1),b}2; and if we compare this with the original equation, we see that the right- hand member is the sum of an arithmetical progression whose first term is b—a, and common difference 6: this series therefore is indicated by the negative _ result. 283. The sum of any two terms tahen equidistant from the beginning and end cf an arithmetical progression is equal to the sum of the first and last terms. This is proved in the demonstration of Art. 282; or it may be shewn independently, thusz—Let a, b, c, d, e, j, g be in arithmetical progression, then, by Definition, b—a=g—j, or b+f=a+g. Again, c--b=f-e; c+e=b+f=a+g. And so on, whatever the number of terms in the series may be. 284. f the number of terms qf an arithmetical Zprogression be odd, twice the middle term is equal to the sum g“ the first and ast terms. GEOMETRICAL PROGRESSION. 165 Let q represent the middleterm, p the preceding, and r the succeed- ing term, then q—p=r—q, by Definition; - '. 2q=p+r, and p and r are equidistant terms from the beginning and end; p+r=a+l, by last Art. and 2q=a+l. 285. By the last two articles the summation of series in arithmetical progression is sometimes capable of being facilitated. Thus, if the num- ber of terms whose sum is required be even, and more than half that num- ber of terms are given, it is sometimes easier to add together two terms equidistant from the beginning and end of the series, than to find either the “common difference” or the last term. If the number of terms be odd, and the middle term is one of those which are given, then, from what has been proved, the middle term multi- plied by the whole number cy“ terms will be the sum required. . 2 7 4r 1 Ex. 1. RequIred the sum of §+15+fi+1—5-+&c. to 7 terms. Here i is the middle term; sum required=-1->< 7 =1. 15 15 15 . 5 Ex. 2. Required the sum of 1+ g+2+§+&c. to 6 terms. Here the 3rd and 4th terms are equidistant from the beginning and end, 5 27 sum =(2+ §)><3=§ =13%. [Exercises Zea] GEOMETRIOAL PROGRESSION. 286. DEF. Quantities are said to be in Geometrical Progression, or continual proportion, when the first is to the second as the second to the third, and as the third to the fourth, &c. that is, when every succeeding term is a certain multiple, or part, of the preceding term. If a be the first term, and ar the second, the series will be a, or, arz, ar“, ar‘, &c., the nth term being urn“. For a : ar :: ar: ar2 :: ar2 : ar", 8:0. 287. The constant multiplier, by which any term is derived from the preceding, is called the Common Ratio, and it may be 166 GEOMETRICAL PROGRESSION. found by dividing the second term by the first, or any other term by that which precedes it. Ex. 1. 1, 3, 9, 27, &c. are in Geometrical Progression, find the Common Ratio. Here Common Ratio=%=3. 1 Q U . Ex. 2. 2%, -2-, 116, &c. are In Geometrical Progression, find the Common Ratio. 1 1 1 H ~ C R t'0=-—+—=—. eIe ommon a 1 10 2 5 288. If any terms be taken at equal intervals in a geometrical progression, they will be in geometrical progression. Let a, ar...ar” .... ..ari"...ar3"...&c. be the progression, then a, ar", ari", ar3", &c. are at the interval of n terms, and form a geometrical progression, whose common ratio is r". 289. If the two extremes, and the number of terms, in a geo- metrical progression be given, the means, that is, the intervening terms, may be found. Let a and the the extremes, n the number of terms, and r the common ratio; then the progression is a, ar, arg, ar3,,,a-r"-1; and since l is the last term, t ar”-1 =l, and r"-1 = —; a l __1_ ‘.I I). = a and r being thus known, all the means, or intervening terms, ar, ar’i, ari’, 8:0. are known. COR. A single Geometrical mean between a and b, called the . . i _ Geometrical mean, W111 be a or Jab; but this may be shewn more simply thus, 0 ’ [Let a, x, b be in Geometrical Progression, then by Def. talc“ QIH '. x’=ab, and x=,,/c_zb. Ex. 0 There are three means, or intervening terms, in a Geometrical Progressmn between 2 and 32; find them. GEOMETRICAL PROGRESSION. 167 Here n=5, a=2, and l=32; l _1 I r=<—) "=(16)‘=2; 4\ and the means are 4, 8, 16. 290. To find the sum of a series of quantities in Geometrical Progression. Let a be the first term, r the common ratio, n the number of terms, and s the sum of the series: Then s = a + ar + ar2+ + ar"'2+ ar"", and rs= ar + ar2+ .. . + ar”'2+ ar""1+ ar". Subtracting rs — s = ar"— a; arn—a r”—1 s- _a. . r—l r—l COR. 1. If t be the last term, l=ar"'1, rl — a* 3 r — 1 from which equation, any three of the quantities s, r, l, a, being given, the fourth may be found. COR. 2. When r is a proper fraction, as n increases, the value of r", or of ar", decreases, and when n is increased without limit, ar" becomes less, with respect to a, than any magnitude that can be assigned-f. Hence the sum of the series, which in general is equal ar"- a ar" a _ _ , - a _ to or - ~-—— , Is reduced In this case, to ——, that Is, r —1 r — 1 r -1 - 1 a 1—7' *‘ The following is another method of arriving at the same result equally simple :— Let a, b, c, d, &c., h, k, I, be the series, s the sum, and r the common ratio; then, by definition, b=ar, c=br, d=cr,...k=hr, l=kr, '. b+c+d+...+k+t=(a+b+c+...+h+k)r, or s—a=(s—l)r=rs--rl; rl—a s: r—l ' En. + Thus, if r=-]3—0 or 0-3, r2=0'09, r3=0'0027; and so on, shewing that as n increases r" decreases, and that such a power of r may be taken as to produce a quantity less than any number which shall be named, however small. ED. 168 GEOMETRICAL PROGRESSION. a This quantity , which we call the sum of the series, is the 1—7‘ limit to which the sum of the terms approaches, but never actually attains*; it is however the true representative of the series con- tinued sine fine, for this series arises from the division of a by 1 —r; a . . . and therefore T— may mthout error be substItuted for It. Ex. 1‘. To find the sum of 20 terms of the series, 1, 2, 4, 8, &0. Here a=1, r=2, n=20; 1X22°-1 = '2_-1_ =2‘~*°-1. Ex. 2. Required the sum of 12 terms of the series, 64, 16, a, &c. 1 Here a=64, r=l, n=12; 1 -,,-1 4.“ 43 412-1 1 492-1 0'. =__|—"— = ‘_‘8'0 1 1 a“ 4-1 4 3 4 Ex. 3. Required the sum of 12 terms of the series 1, -8, 9, -—27, 8226. Here a=1, r=—3, n=12; . 1 1 . Ex. 4. To find the sum of the series l—;+;—g+ 8:0. in iiy‘initum. 1 Here a=1, r=—-2—; s-==--—= 1+ @3116 NIH [Exercises Zd.] *‘ That is, although no definite number of terms will amount to fi , yet, by taking a suflicient number, the sum will reach as near as we please to it; and, whatever number be taken, their sum will not exceed it. ED. GEOMETRIGAL PROGRESSION. 169 291. Recurring decimals are made up of quantities in geo- . . '1 l 1 . . metrical progress10n, where —, —~, ~—— , &c. Is the common ratlo, 10 '100 1000 according as one, two, three, &c. figures recur; and the vulgar fraction corresponding to such a decimal is found by summing the series. Ex. 1. Required the vulgar fraction which is equivalent to the decimal 123123123 &c. l23+l23+lg3 103 106 109 . . . . . 123 Here the series 1s + . . . . . .m znfinitum; a=fi5, and 123 iii 123_ 4.1 103 ()r, as follows :— Let '123123123 &c.=s; then, as in Art. 290, multiply both sides by 1000; and 123123123123 &c.=1000s, and by subtracting the former equation from the latter, 123 =999s; Ex. 2. Required the vulgar fraction which is equivalent to 'PPP &c. where P contains p digits recurring in inf: Let s='PPP &c. then 10Ps=P'PPP &c. (10P—1)s=P, P and 8:55 . Ex. 3. Required the vulgar fraction equivalent to 'PQQQ &c., where P contains p digits, and Q contains q digits recurring in inf. Let s='PQQQ &c., then 10P+qs=PQQQ &c., and 101’s: P'QQ &c.; (101”‘1—1 OP)s=PQ-P, ‘ PQ- P and s=1—Op(—1(-)q—_1—). 170 HARMONIGAL PROGRESSION. Both these results may be easily verified by expressing the proposed quantities in geometrical progressions—the .former being i + 3+ i + in inf 10,, 10,, 103,, . . . . . the latter being P Q Q Q -_+_—___. _ 101’ 10M + 101323 +1 W + . . . . . which may be summed by the rule in Art. 290, Cor. 2. 292. In a Geometrical series continued in inf. any term >, =, or <, the . . 1 sum of all that follow, according as the common ratio <, =, or >, g . Let a+ar+ar2+. . .ar’t‘l+ar"+ . . . . . .be the series. Then the sum of the series after n terms is the sum of , l ar"+ar”+1+ .... ..wh1ch=ar".—l—7_ (Art. 290, Cor. 1 and ar""1>, = or <, ar".——-, l—r according as 1>, =, or <, {1.7, (Art.219.) . . . . . . . . . 1— r>, =, or <, r, (Art. 219.) . . . . . . . .. 1>, =, or <, 2r, (Art. 217.) . . . . . . . . . 7'<, =, 01' >, (Art. 219.) HARMONIOAL PROGRESSION. DEF. Any magnitudes a, b, c, d, e, _&c. are said to be in Harmonical Progression, if a : c :: a—b : b—c; b : d :: b—c :c—d; c : e :: c—d : d—e; &c. 293. The reciprocals of quantities in H armonical Progression are in Arithmetical Progression. Let a, b, c, &c. be in Harmonical Progression; then by Def. a: c::a—b : b-c; ab-ac=ac—bc, (Art. 237), and dividing both sides by abc, HARMONIOAI. PROGRESSION. 171 Again, b : d :: b—c : c—d; bc - bd = bd —- do, and dividing by bad, 1 1 l 1 . and —c- — 5 has been proved equal to Z - ; therefore the quanti- a _ 1 1 1 l . . . t1es-, —, — —, have a common difference; that is, they are in a b c’ d Arithmetical Progression. And the same proof may be extended to ' any number of terms. Hence every series of quantities inlHarr—nonical Progression may be easily converted into an Arithmetical Progression, and then the rules of Arithmetical Progression may be applied to it. Thus, Ex. Given a and b the first two terms of an Harmonic Series, to find the nth term. . 1 1 . . . . Since a , — are contiguous terms of an Arithmetic Series, 6 the Common Difference of this Series=l— b 7 QIH . 1 ' and its a“ term=E+(n—1)<%-Zll-) , (Art. 280.) _ 1 a—b _(n-1)a—(n—2)b _ -t+<"_1)w - ab ’ ab (n—l)a—(n—2)b ' the nth term required: 294. The two extremes, and the number of terms, in an Harmonical Progression being given, the means, or intervening terms, may be found. Let a and t be the extremes, and n the number of terms; then since the reciprocals of the terms are in Arithmetic Progression, let 6 be their . l . . . . common difference, and 2- being the nth term of the Arithmetic Progression, wehave 1"Lt—Jib (Art 280 l-_a n ., . 1 1 —~—— b= a—l = (n -1)al; 172 SUMMATION or SERIES. hence the Arithmetic Means are 1 a-l 1 2(a—l) 1 3(a—l) a+(n-l)al’ Z+(n—1)al’ Zi+(n-l)al’ &c' r a+(n- 2)! 2a+(n—3)l 3a+(n--4)l & (n—1)al ’ (n-l)al ’ (n—1)al ’ the Harmonic Means are the reciprocals of these quantities, viz. (n—l)a.l (n-1)al (n—l)al &c a+(n—2)l’ 2a+(n—3)l’ 3a+(n—4)l’ ' COR. A single Harmonic mean between a and I), called the Harmonic Mean, will be (3_1)ab gab m, or m; but thls may be shewn more Simply thus, Let a, .r, b be in Harmonic Progression, then by Def. a: 6:: a—m:.r—b; '. ax-ab=ab-bw, (Art. 237.) (a+.b)x=2ab, -26 —a+b' ()Bs. There is no general expression for the Sum of an Harmonic Series, since the sum of any number of quantities is not deducible from the sum of their reciprocals. [Exercises Ze.] 295. Series are sometimes proposed for summation which are not actually composed of terms in Arithmetical or Geometrical Progression, but which may be made to depend upon the rules of one or both by arrangement or artifice. Thus, Ex. 1. Let the sum of 1: terms of the following series be required, 1+5+13+29+61+&c. The series is equivalent to 4——_3+8-3+16—-3+32—3+&c. =4+8+16+&c. to n terms—3X12, 2L1 2 42"1 2_1— n- ( — )-3n. ~ ' Ex. 2. To find the sum of n terms of the series 1 + 2w+3x2+ 4x3+ &c. =4, SUMMATION OF SERIES. 17 3 Let S be the sum required; then S=1+ 2x+3x2+ . . . . . . . . . . +nx"‘l, ', x8: x+2x2+ . . . . . .+(n--1).v"“+nx", subtracting, S- xS=1+ar+zr2+ . . . . . .+x"‘1-na:", l—w" or S(]-a:)= 1mm -12.r", (Art. 290) . _ l-x" 12.x" ' ' ~(1—ar)2 l—x ’ _ nx"+1-—(n +1)a:"+1 _ (I—wr Ex. 3. To find the sum of n terms of the series -1 - -e L.x+ZZ—-2..r2+ n .x3+ &c. n n . 0: Sum requ1red=w+x2+x3+ . . . . . .+x"—Z{l+2.r+3x"+. . .+1z.r”“‘}, 1-—.x" a: nx”+1—(n+1)a:"+1 =x.-—- --——. EX. . l-x n (l—arr)2 ’ ( 2 ) __ (n—1)x- nx2+w”+1 n(1—-.1t’)2 Ex. 4. To find the sum of n terms of the series a2+ (a+b)2+ (a +26)2+ &c. Let A1, A2, A3,...An represent the several terms in order of the series a, a+b, a+26, &c. and S the sum required; then AZ—AfL—(Al-i- b)”- -A '1': 3A§b+ 3A162+ 63, A §—A§=(A2+ b)3—A§= 31426 + 3A262+ 63, AZ—A§=(A3+b)3-A3=3A§6+3A362+ ba, I l 0 0 Q I Q ~—'loil‘s...IOtoop—QQQQQQOQQcocoon-.000 3,1-A3=(A,,+ b)3-A,}=3Aib+ 3A,,62+123 ; by addition, A3+,-A§=3(Ai +AZ+. . .+A§)b +3(A1+A2+. . .+A,,)b’+ 1263, 2 or (a+nb 3--a3=3bIS+3{Qa+(n--1)b}% +1263, ‘ 2 3 b3 {3na’b+ 3ngab2+ nsba- 3nabz— can; + 73-5} , 12362 112])2 162 =na2+n2ab+ 3 ~nab— 2 +z—6—, 1 .'. _______ 2 =na(a+n--1.b) +1262(2n2— 8n +1). 174 PERMUTATIONS AND COMBINATIONS. COR. If a=b=1, then 12” 722 n 2 2 2 2=__ _+_;’ 1+2+3+...+n 3+2 6 2n3+3n2+n 6 , _(2n+1)n2+ (2n+1)n _ 6 , +1)(2n +1). If a=1, b=2, then 413 2n 19+32+52+&c. to n terms=2ng—n+—3L—2n”+—§, (2n—1)2n(2n +1) _ 1.2.3 , and so on, when any other values are given to a and b. = g-(4n2—1)= PERMUTATIONS AND GOMBINATIONS. 296. DEF. The difi’erent orders in which any quantities can be arranged are called their Permutations. Thus the permutations of a, b, 0, taken two and two together, are ab, ba, ac, ca, be, ob; taken three and three together are abe, aeb, bae, baa, cab, cba. Some writers on Algebra call them Permutations only when the quan- tities are taken all together; and in all other cases Variations. 297. DEF. The Combinations of quantities are the different collections that can be formed out of them, without regarding the order in which the quantities are placed. Thus ab, ac, be, are the combinations of the quantities, a, b, 0, taken two and two; ab and ba, though different permutations, forming the same combination. 298. The number of permutations that can be formed out of n quantities, taken twp and two together, is n(n-1); taken three and three together, is n(n -1) (n —2). In n things, a, b: e, d, &c. a may be placed before each of the rest, and thus form n—l permutations; in the same manner, there are n—l permutations in which b stands first; and so of the rest; PERMUTATIONS AND COMBINATIONS. 175 therefore there are, upon the Whole, n(n—1) permutations of this kind, ab, ba, ac, ea, &c. - Again, of iii—l things b, e, d, &c. taken two and two together, there are (n-1)(n-2) permutations, by the former part of the article, and by prefixing a to each of these, there are (n-1)(n-2) permutations, taken three and three, in which a stands first; the same may be said of b, e, d, &c. therefore there are, upon the whole, n(n—1) (n -2) such permutations. Ex. The number, of Permutations of 7 things taken three together =7><6><5=210. 299. To find the number of permutations of 11 things taken r together. By Art. 298, the number taken two together =n(n—~1), . . . . . . . . . . . . . . . . . . . . . . . . . . . . . three . . . .. . =n(n—-1)(n—2). Similarly, . . . . . . . . . . . . . . . . . . . . . . . . four . . . . . . . =n(n-—1)(n—2)(n—3). Now, suppose the law, which is here perceived, to hold generally, that is, let the number of permutations of n things a, b, c, d, &c. taken $3 together be n(n -1)(n—2) .... ..(n——r+2). Then omitting a, it is equally true that the number of' permutations of n~1 things b, c, d, &c. taken r—l together is, (putting n—1 for n), (n—1)(n-2) .... ..(n-r+1). Prefix a to each of these last permutations, and there will be a set of permutations of n things taken r together in which a stands first in every permutation, the number of them being (n—1)(n-2) .... ..(n—r+1). The same number may be made of similar permutations in which b stands first ,- and so for each of the n quantites a, b, c, d, &c. Hence the whole number of permutations which can be made of' 12 things taken r together is n(n—1)(n—2) .... —r+1), if it be true that the number of permutations of n things taken r-l together is n(n—-1)(n—2) .... . .(n—r+2). That is, if the assumed law be true for any value of r it is proved true for the next higher value. But it has been shewn to hold (Art. 298) when r=2, and 3; therefore, it is true when r=4; and, if for 4, for 5 ,' if for 5, for 6; and so on generally for any number. 176 PERMUTATIONS AND COMBINATIONS. COR. The number of permutations of n things taken all together is n(n—1)(n—2) .... ..(n-n+1), =n(n—-1)(n—2) .... ..1, or Ex. 1. Required the number of difi'erent ways in which 6 persons can be arranged at a dinner table. Number required =number of' permutations of 6 things taken all together, = 6x\5><4.‘><3><2><1 =720. Ex. 2. Required the number of' changes which can be rung upon 12 bells. Number required=12><11x10><9><8><7x6><5><4><3><2x1, =479001600. 300. The number of Combinations that can be formed out of n —1 ; taken three and n things, taken two and two together, is n u—l n-2 3 three together, the number is n. The number of permutations in the first case is n(n —1), (Art.298) but each combination, ab, admits of two permutations, ab, ba; there- fore there are twice as many permutations as combinations, or the - , _ , 7’1. -1 number of combinations 1s n . Again, there are n(n—1)(n- 2) permutations in n things, taken three and three together; and each combination of three things ad- mits of 3.2.1 permutations (Art. 299. Com); therefore they are 3.2.1 times as many permutations as combinations, and consequently the number of combinations is n (n —1)(n - 2) 1.2.3 ' 301. T 0 find the number of combinations of n things taken r together. Number of permutations of n things taken r together =n(n—1)(n-2) .... ..(n—r-i-l). (Art. 299). But every combination of r things will make 1.2.3...r permutations r together (Art. 299. Con); and no two of these can be the same; therefore there are 1 .2. 3...r times as many permutations as combinations: and con- sequently the number of combinations is n(n—1)(n—2) .... ..(n—r+1) l .2 . 3 .......... .. r ' PERMUTATIONS AND COMBINATIONS. . 177 This may be expressed in a very convenient.form: for by multiplying the numerator and denominator of the above fraction by 1 .2.3...(n-—r), it becomes n(n-1)(n-2)...(n—r+1)(n—r)(n—r-l). 1 1.2 . 3 ............. ..rx1.2.3 ........... ..(n—r)’ L11. or _--——. Lt-ltl" Ex. Required the number of combinations of 24 different letters taken four and four together. Here n =24, r=4, 24x23x22X21 1x2x3x4’ =23><22><21=10626. number required = 902. T 0 find when the number of combinations of a given number qf things is the greatest. Since the number of' combinations of n things taken rtogether is equal n—r+1 to the number taken Z3 together multiplied by , the number - -r + n 1 first of combinations will go on increasing as r increases, until = or <1. If r be such, that n_::+1=l, that is, if r=g(n+1), and if n be odd, so that -Q1~(n+1) is a whole number, then this value of r will give the greatest number of combinations. If r be the least number which n~r+l makes less than 1, that is, if r be the first number greater than 1§(1z+1), and n be even, so that -Z,-(n+1) is a fraction, then the whole num- ber next less than -%-(n+l), that is, $42, will be the value of r, which gives the greatest number of combinations. Ex. Of six things how many must be taken together that the number of combinations may be the greatest possible? Here 12:6, an even number; the number to be taken together 6x5x4 m , or 20, combinations. X =%n=3,' which will give 303. The number of combinations qf 11 things taken r together is the same as the number of combinations of 11 things tahen n—r together. Number of combinations of n things taken r together is u . [:0 . and writing n—r for r, which may be done, because it is true for all 12 17 8 PERMUTATIONS AND COMBINATIONS. values of' r less than n, we have Number of combinations of n things taken n—r together = Le ln—r. IZ-(n—r) la 112:1- Lr’ the same as the number taken r together. The truth of this proposition will also appear from a very simple consideration, viz. that of n things if r be taken, n—r things will always be left; and for every different parcel containing r things, there will be a different one left containing n—r; therefore the number of the former parcels must be equal to that of the latter. Hence in finding the number of combinations taken r together in certain cases, that is, when r>-21~n, it will be a shorter operation to find the number taken iii—r— together. Ex. Required the number of combinations of 20 things taken 18 together. Number required = number taken 2 together, ~20><19 _ 1X2 =10><19=190. 304. Tofind the number of permutations of n things taken all together, when the quantities recur. Let a recur p times, b recur q times, 0 recur r times, 8:0. And let P represent the number of permutations required. Then if all the a’s be changed into different letters, these alone will form 1.2.3...p permuta- tions instead of one, and out of each of the P permutations we should form 1.2.3...p permutations; therefore the whole number would be P .1.2.3...p. If again, all the b’s be changed to different letters, in the same manner, the b’s would of themselves form 1.2.3...q permutations, and the whole number of permutations would be increased to P.1.2.3...p.1.2.3...q. And so on, till all the quantities are different. But, when all are different, the number of permutations is (Art. 299. Cor.) ; P.1.2.3...p.1.2.3...q.1.2.8...r.&c.=1.2.3...n; and P: 1.2.3...n . 1.2.e...p.1.2.e...q.1.2.s...1~.&c.’ L72 , or P=———~———--—~ Ex. Required the number of permutations that can be formed out of the letters of the word “Mississippi.” PERMUTATIONS AND COMBINATIONS. 179 Here the whole number of letters is 11. i recurs 4 times p .... .. 2 .... .. P _ 1.2.s.4.5.6.7.s.9.10.11 _ 123.412.32.12 = 5.7.9.10.11 =34650. ) 305. T 0 find the number of combinations of n sets 9)” things, containing respectively p, q, r, &c. things, one being taken out of each set for each combination. 1. First, suppose there are two sets of things, containing p and (1 things respectively; then the number of combinations made by taking one out of each set is clearly p taken q times, or pg. 2. Next, let there be another set introduced containing 1' things; then each one of these being combined with pq combinations, there will be pqr combinations of three sets of things, one being taken out of each set. And so on: the number of combinations required being the continued product of the numbers which express the number of things in each set. COR. If there be the same number in each set, or p=q=r=&c., then the number of combinations is p". Ex. There are 7 men, 5 women, and 3 boys; required the number of ways in which they can be taken, so as always to have one and no more out of each set. The number required = 7x5><3 = 105. 306. To find the number qf combinations of two sets of things, con- taining respectively p and q things, m being taken out qf one set and 11 out (if the other for each combination. The number of combinations of the first set taken m together is 77(7) _1)" '(P _m+1); and the number of combinations of the second set, 1 . 2 .......... ..m . q(q—1). . . (q—n +1) taken n together is 1 . 2 ........ “n . And, to form the combinations required, each of the latter must be com- bined with each of the former; therefore the number of them will be the product of these two quantities, or r(r—1)(r—2)--~(r—m+1)Xq(q—1)-~(q—n+1) _ 1.2 . 3 .......... ..m 1.2 .......... ..n And similarly if there be more than two sets of things, always taking the continued product of the respective numbers of combinations in each set. Ex. Out of 10 consonants and 3 vowels how many different collec- tions of letters may be made with 4 consonants and 2 vowels in each? 1 2--2 180 BINOMIAL THEOREM. Here 12:10, q=3, m=4, n=2; 10x9x8x7 3X2 .2 th ' =---_—— -~= = . enumber required 1><2><8X4x1x2 10x9x7 630 [Exercises THE BINOMIAL THEOREM. 307. The method ‘of raising a binomial to any power by re- peated multiplication has been before laid down in Art. 140. The same thing may be done much more expeditiously by the following general rule, which is called the Binomial Theorem. Let ac+ a be the binomial; its nth power is n—1 n-1 n-2 w"+na.v"'1+n.-—~—a2a2""2+n.————.———a3¢v""3+&c. 2 2 3 Where the index of a', beginning from n, is diminished by unity, and the index of a, beginning from 0, is increased by unity, in every succeeding term. Also the coefficient of each term is found by multiplying the coefficient of the preceding term by the index of a: in that term, and dividing by the index of a increased by unity. 6.5.4 6.5.4.3 6x5 Thus a' a 6= a)“ + 6a.v5+ —— aha:4 ~— a3w3+ a"m2 ( -+ ) 1x2 +1L2£l 1L2f24 6.5.4.22 5 +6.5.4..a.2.1 ad? “a. 1.2.3.4.5 1.2.3.455 ’ = as + 6am5 +15a2a’4 + 20a3w3+ 15a4a2+ 6a5a' + as. 308. To investigate the Binomial Theorem for a positive integral index. By actual multiplication it appears that (a: + a)(.r + b) =x2+ (a + b)c: +ab, (a: +a)(a: + b)(a'+ c) =a'3+(a + b + o).r2+(ab + ac+ bc).v +abc, (.r + a)(ar+b)(.r+ c)(x+d)= a4+(a + b + c +d).r3 +(ab +ac + be +ad + bd+ cd).r2 + (abc+ acd+ bcd+abd).v+abcd; and the same law of formation of the continued product is observed to hold whatever be the number of binomial factors, x+a, x+b, x+c, &c., actually multiplied together, viz. that it is composed of a descending series of powers of x, the index of the highest being the number of factors, BINOMIAL THEOREM. 181 and the other indices decreasing by 1 in each succeeding term. Also the coefficient of the first term is 1; of the second the sum of the quantities a, b, c, &c.; of the third the sum of" the products of every two; of the fourth the sum of the products of every three; and so on; of the last the product of all the n quantities a, b, c, &c. Suppose, then, this law to hold for n binomial factors, a+a, .r+ b, x+c, .... .., x+b; so that (w+a)(.r+ b)(x+c) . . . (a+/e)=x"+A.r"“+Bx""2+C.r"-3+ . . . +K, where A=a+b+c+...+h, B=ab+ac+bc+. . .\ C=abc+acd+. . . &c.=&c. K=abcd. . .k; introducing a new factor, x+l, we have (x+a)(.r+b)(x+c) . . . (x+k)(.r+l)=.r"+'+(/1 +l).r"+ (B +ZA).1; -‘+. . ,+Kl_ Hence A+l=a+b+c+. . .+h+l, B+lA=ab+ac+bc+. . .+al+bl+. . .+hl, &c.=&c. Kl=abcd. . .hl; so that, if the law above described holds when n binomial factors are mul- tiplied together, the same law is proved to hold for n+1 factors. But it has been shewn to hold up to 4 factors, therefore it is true for 5; and, if for 5, then also for 6 ; and so on, generally, for any number whatever. Now, let a=b=c=&c., then A=a+a+a+&c. to n terms=na. B=a2+a2+ &c. to as many terms as is equal to the No. of combinations of n things taken two together, =72. n—l 2 as’ C=a’+a3+&c. to as many terms as is equal to the No. (n—1)(n-2) , of combinations of n things :72 a ’ taken three together. &c.=&c. K=a.a.a...to n factors =a". Also (x+a)(a+b)(.r+c)...(.r+h) becomes (x+a)”; "(n—'1) 2 n-2 n(n_1)(n_2) 3 "—3 1.2 am +1.2 .3 a” n(n—1) ., n(n—l)(n—2) —-.v‘+——-——-_ 8 . 1 . 1 .2 . a “3+ (a: + a)”=:c"+ nax"'1+ +. . . +0.". Con. (l+.r)"=1+na:+ . .+.r". 182 BINOMIAL THEOREM. _ 309. Having given the Binomial Theorem for a positive integral index, to prove it, when the index is either fractional or negative. [EULER’s Pnoorj . m m—l Let the series 1+ma+-1—(——§—)-.r2+ &c. be represented for all values of m, whether positive, or negative, integral, or fractional, by the symbol f(m) ; then it has been shewn that, when m is a positive integer, f(m)=(1+w)"'- It remains to prove that this equation is also true when m is either frac- tional or negative. By the notation assumed —l 1+nx+ 3(2—2—2x2+&c.=f(n); 1 therefore, by multiplication, f(m) .f(n)=the product of the two series, which will evidently be a series of the form , 1+ax+ba2+ca23+ &c., ascending regularly by the integral powers of a', the coefficients a, b, c, &c. being different combinations of m and n. Now, although by changing the values of m and n the values of a, b, c, &c. are altered, yet their jbrms, that is, the manner in which m and n enter the series will remain the same. ‘Whatever, therefore, be the forms of a, b, c, &c. when m and n are positive integers, the same will they be when m and n are fractional or negative. But in the former case f(m)=(1+w)’". f(")=(1+~v)"; .f(n)=(1+ar)m+"=1+ (m+n).:r+ (?7z+n)(m+n_l) 1 . 2 ' m+n is a positive integer. x2+ &c. These then are the forms in which m and n appear in the product when they are positive integers; and therefore they appear in these same forms, whatever be their values: i.e. Whether m and n be positive or negative, integral or fractional, the multiplication of by f(n) always produces the series (m+n)(m+n-1) 2 which by the notation is represented by f(m+n) ; universally, f(n)=f(m+n). Since, then, this last equation is true whatever he the values of m and n, for n write n+p, and we have f(m+n+1>)=f(m)-f("+r)=f(m)finite); and proceeding similarly, we have generally f(m+n+p+&c.)=f(m) &c. to any number of terms. 1+ (m + n)x + a2+ &c. BINOMIAL THEOREM. 183 Now, let m=n=p=&c.=7]§, h and I: being positive integers; then asralatlftlft)r- and, if the number of terms be h', f (%+%+&c. to h terms)=f(llg . &c. to h factors, nv={f(§§)}”; but f (h)==(l+.r)", because h is a positive integer: (1+.v)”={f(7]:_)}k; an, _,(.,._..>, M, . . (1+.v) =f Z =1+kx+ 1 . 2 a +... y t e notation, which proves the Theorem for a fractional positive index. Again, . for all values of m and 11., let n=—-m, then f(m)-f(-m)=f(m—m)=f(0). =1, the assumed series becomes 1 when m=0 ; 1 1 f<"">=m>=- -. ..,=<._..1).,_,. e, .... .. (1%!) ,- anew-In. e, W); .;. ae,=(e-2)fl.; and so on: e, .... .. B(B:_):_':_(_fl__"+1),- rfi,==(/€—r+1),6,_,. l. .r * This proof, unlike most others, is dependent only upon the most simple elementary rules of Algebra. ED. 184 BINOMIAL THEOREM. N ow the product of the two series I+ax+a2w2+a3x3+...+a,x'+ ...... ......(1) 1+,3m+,32x2+,33.v3+. . . +[J’,a:’+. . . . . . . . . . . . by actual multiplication, is 1+(a +,6).v + terms of :02, x3, . . . a’, . . . the coeff". of at being a,+a,._1,8+a,._2162+a,_3[9’3+ . . . +afi,_1+}8,, and ............... . . xf‘l . . . a,_1+ a,_2fl -l- a,_3,82+ 0,453+ . . . + aB,_2+)8,._1. But ra,.= . . . . . . . . . . . . . . . . . . . . . . ..=(a—-’!'+l)a,._1, ra,,__“8 =(1’ —-1 )a,_16 +‘ ahlfl = (a —- r + Q)a,._._,fi +Ba,_, , Tar-ales: (7.“ 2)ar—2182+ 2ar-2162=(a "' r + 3)ar—362+ -l (Ir—2 , Tar-3183: (r- 3)ar—3/83+ 3ar-3183= (a _r + 4‘>ar-4/83+ _2)182ar—3 9 7.098151: “fir-1+ (7‘ _1>a18r-1= “Br—1+ "r +2)181—2a: rfl, = ........................................... ..=(/3-r+1)6,._1; adding, and observing that there is a pair of similar terms in every two lines, we have rxcoeff'“. 0f x'-=(a+)6—r~1)(a,_1+a,52,8+a,1_3)82+ . . . + ,.__1); + - --1 coefi". of ski—53$! >< coefi"- of w“, =a+'8-(T_l).a+/8_(r_2)><(l—1) .., " n(n-1) <1+1>, n—l x4 "av-“1+ —a2"(1+n'f—g+n *' 'a2'~a1.2.3a n-l _ _ =a2n+na2n-2x2+n. 2 .a2n-4m4+ .a2n-6x6+ &c. l 1 1 % 1 1 91 Ex. 2. J: n. 2 1 n—l (n—l)(2n-1) 8 --1+;z-.x:-----——2722 w2+——————2 . 3 . n3 .2, —&c. -‘- 1 +1 +1 2 1 EX. 3. (1+0?) "=1— Zar+L x2- Ml 2n2 2 . 3 . n3 x8+&c' % 4 Ex. 4. (a%+b%)*=(a%)*{1+%} , 6% 4x3 6% 8 4x3><2 bfi “ 4x3><2><1 5% * =a21+41+—1 <— +-—— — +——-——- -— , a2 X2 aé 1x2><3 aé 1x2x3x4 afi = a9+ 4agbé + Gab + 4aébg+ b”. 1 EX. 5. m =(1+x2)'3, 3><—4m4 —3><-4><—5 1X2 1x2x3 =1—3x’+ 6x4—10x6+. .. . . . . .. =1—3x2+ _ x6+ .... .. Ex. 6. (ax+by)§=(ax)é.{l+by}éa ax l§_~)/by3 =(ax)~{1+§.z-a-;+-——-1 . 2 (~>+ 1 . 2 . 3 \55>+...}, ax = M 1 by _1_’ _1__a_u_ (a ) +21”)é 8 (ax)%+ 16 (am); 313. If either term of the binomial be negative, every odd power of that term will be negative; and consequently the signs of the terms, in which those odd powers are found, will be changed. EX. 1. (1-x)”=1-nx+n£2:£x9_ w8+ &c. n —1 Ex. 2. (a’— x’)"= a”"- na’”'2w’+ n . —2- a’"".x‘-- &c. BINOMIAL THEOREM. 187 314. If the index of the power, to which a binomial is to be raised, be negative, and the second term of the binomial be negative, then every sign in the series is positive. _n n(n +1) 2 n(n +1)(n+2) 8 EX.1- —l+fl$+T—i—2—$+i—TTZB+ . . . . .. The particular form of this result is worth remembering. 3x4 2+3x4x5 1 X2 ‘23 l><2><3 =1+3w+6x2+10x3+ ................ .. O I I I I . Ex. 2. (1—w)_3=1+3a:+ 1(1 1) I _a_1{1+1 w2+2 2 w4+2 _ 2.a2 1X2 a4 1 2 . 05+ u v u o -.}, __1+1f+3x4+5 we ‘_"a 2.618 80a5 I o a n o a a u. 315. To find the general term of the expansion of (x+a)“. The 1st term is x", 2nd . . . . . . na1.r"‘1, -1 3rd 0 0 o o o a no fg—a2337h2’ n—1 n—2 3 H 2 ‘ a a a: ’ in which we observe that the coefficient of any term is Formed of the pro- n n—1 n-2 duct of the factors I, -_2- , __3._ ber which expresses the position of the term; therefore the coeflicient of the 1““ term will be 4th , &c. in number one less than the num- n n-1 n—2 n—r—Qa 1 —-——2 —3 . . . . .. T_1 , or n(n—1)(n—2) . . . . ..(n-r-Q) 1 . 2 . (r—l) ' Also the index of a is always the same as the denominator of the last factor of the coefficient; and the index of a: is the difference between n and the index of‘ a 1‘ ,' therefore the whole 1*“ term is n(n—1)(n—Q) . . . . . .(n--r+2)arulwmw1 1 o 2 Q 3 I o I a o o o o 0 ‘ The learner often finds a difficulty in putting down correctly the terminal factor in a proposed case. Let him write the denominator first, and after that the numerator, bearing in mind that their sum is always n+1. 1' Observe the sum of the indices ofa and .z' is always n. 188 BINOMIAL THEOREM. By substituting in this expression any proposed number for r any single term of the expansion may be found independently of the rest. Ex. Required the fifth term of (a2- b2)”. Here r=5, and 22:12; 0 . _12.11.10.9 _ 2, ,8 . . term requ1red-—————1 .2 .3 A.( b) .(a), =495amb8. 316. If the index of the binomial be a positive integer, every coefiieient in the expansion, formed from the index, is a positive integer. For the coefficient of the (r+ 1)th term is n(n—l)(n—2) . . .(n—r+ 1) . 1 . 2 . 3 . . . . . . . . .r ’ and this (by Art. 801) is the same as the number of combinations of n things taken r together. Now this latter number, by the nature of the thing, must be a Whole number, if n and r be positive whole numbers; therefore also . n(n—1)("-Q)- ' '(n_r+1) is a whole number. 1 . 2 . 3 . . . . . . . ..r Another Method. If all the coefficients of the expansion of (1+x)” are positive integers for any one value of n, it is obvious that they will be also for the next higher value, since no fractional quantities can be introduced by merely multiplying by 1+x. Now we know, that the coefficients are whole numbers in (1+x)2, and therefore it follows that they must be so in (1+x)3; then again in (1+x)4 ; and so on generally in (l+x)". 317. T 0 find the number of terms in the expansion of a binomial. The (r+1)th term of (x+a)" is n(n—1)(n—2) . . . (n -r+1) afwmr _ 1 . 2 . 3 . . . . . . . . .r ’ and if r be such that n—r+1 is equal to O, that is, if n be a positive whole number and r=n+1, there is no term after the 1"“, or the number of terms is n+1, that is, greater by 1 than the index. If n be negative or fractional, since r must necessarily be a positive integer, no value of r can make n—r+1 equal to 0, and therefore in these cases the number of terms is unlimited. Thus the number of terms in the expansions of (x+a)8, (x+a)’, is 4 and 8 respectively; but the number for (x+a)‘2, or (x+a)é is unlimited, or indefinitely great. 318. To prove that, in an expanded binomial, when the index is a positive integer, the coefiicients, formed from the index, of any two terms taken equidistant from the beginning and end, are the same. Since the number of terms is n+1, the (r+1)"h term from the end, having r terms after it, is the (n+1--r)th or (n--r+l)th term from the BINOMIAL THEOREM. 189 beginning; and its coefficient (Art. 315, putting n—r+1 for r,) is n(n—~1)(n—2). . . (n—n-r-I-l) 1.2. 3 . . . . ..(n—-r) ’ =32(n;1)(n;2)"'€;i:;x;(r;1)'nd'2'i, and divg. num‘. and denom'. by 1.2.3. . . (n—r), _n(n—1)(n-2). . .(n-r+1) _1 . 2 . 3 . . . . . . . . .r ’ =coefficient of the (r+1)th term from the beginning. This result is shewn at once by Art. 311. Or it may be obtained by writing a for x, and x for a, in the expansion of (x+a)", so as to deduce that of (a+x)", and then equating the coefficients of similar terms in the two expansions, which are clearly equal to each other. COR. Hence in expanding a binomial, with the index a positive integer, the latter half of the expansion may be taken from the first half. Ex. 1. Required to expand (a+b)7. Here the number of terms is 8; and it will be necessary to calculate the coefficients up to the 4.“ term only; 7.6 5 7.6.5 __ [,2 __ 1.2a +1.23 = 7'+ 7 asb +21 a5b2+35 a4b3+ 35 a3b4+2 1 a2b5+ 7 ab6+ b7. Ex. 2. Required the 11th term of (a+x)13. Since the number of terms is 14, the 11th is the 4th from the end, and its coefficient the same as that of the 4th from the beginning, 13.12. 1 l term required = m— a3x‘°=286a3x‘°. (a + b)"'= a7+7 asb + a‘b3+ &c. 319. Tojind the greatest term in the expansion of (a+b)“. n(n-1). . .(n—r+ 1) 2 The (r+1)‘°h term of the expansion is T n(n—1). . .(n-r+2).a an-r br th . ' n-r+l r-l and the 1' term 1s1 . 2 . . _ . . .(7__1) b . Therefore the (r+1)th term is obtained from the r"11 by multiplying n—r+1 b the latter by .,—z; and as r takes its successive values, the numerator of this fraction continually diminishes, and the denominator increases. Hence the rth term will be the greatest when n_:+1.Z-first < 1, 01- (n—r+1)b(n+l)5, (Al't- 217), or r>(n+1)&—I£Z. (Art. 219.) 190 BINOMIAL THEOREM. Take r, therefore, the first whole number greater than (n+1)z€—l—), and the rth term will be the greatest. If (n+1)i be a whole number, when r is e ual to it we shall have a+b q — 1 b . . 9L1:- .-(;=1, and then two terms, the rth and the (r+1)th Will be equal, and each of them greater than any of the other terms. If the index be negative, -n, the rth term will be the greatest when - — +1 6 . . . . . +r—1 6- _n_f__,zl- IS first < 1, 1rrespect1vely of Sign; z. e. when n .5 is first 7" b . . . - < 1, or r first >(n—1) m. And if this is a whole number, two terms Will be equal, and each greater than any of the others. COR. By thus ascertaining the greatest term we determine the point from which the terms of a series become less and less, or, as it is usually stated, the point at which the series begins to converge. Ex. Required to find which is the greatest term in the expansion of 1 (3 + 5x)”, when x: Q . _5_ b 2 Here "(n+1)a+b=(8+1).—=='-5 , 3+- 2 5 45 __ __.___... _1_. 9 11 11 411 The first whole number greater than 4% is 5; therefore the term required is the 5th. 320. Tojind the sum of all the coeficients of an expanded binomial. n (n— 1) 1 . 2 x’+&c., for any value whatever of x, let Since (1+ x)”=1 + nx + x=1, then n(n-1) n(n-1)(n-2) 1 n:- & o (1+)1+n+1.2 +1.2 . 3 + c or 2"==the sum of the coefficients. Ex. (x+a)5=x5+ 5ax4+10a2x3+10a3x2+5a4x+a5, and the sum of the coefficients =1+5+10+10+5+ 1 =32=25. Also by putting x=-l, we have n(n-1) n(n—-1)(n-2) 1 . 2 _ 1 . 2 . a or 0=sum of the odd coefficientsethe sum of the even ones. the sums of the odd and even coefficients are equal, and consequently each = %2"=2"“. (1—1 )"~=1 -- n + + &c. BINOMIAL THEOREM. 191 321. To find the approximate roots of numbers by the Binomial Theo- rem. The theorem being proved for a fractional index, we have 1 n - 1 1 73-1 lix= lea: "=li- -- .-— “a... ( ) nx n 2n m Now, if N represent a proposed number whose nth root is required, take p such that p" is the nearest perfect ntltl power to N, so that N = p" =l=q, q being small compared with p", and + or - according as N > or < p"; 1 then JN=p (121%); ; and writing 57, for x, =p{]=h71;.-q;,--1-_ni(-q;)g=e...}, p n 2n 1) of which series a few terms only will give the required root to a consider- able degree of accuracy. Ex. Required the approximate cube root of 128. 8 __ Here N:7128=,f/53+3=5\/1+-1—Z—5, 5{1.1i_11(3)*+15(3” } 3'125 3'3 125 32'9125) "" =5+l_l+l.l_...=5+3i_i°+13- .... .. 52 55 3 57 102 10‘s 3 107 = 5 +O‘O4—O'OOO82 + 0'00000412— .... .. =5'0396842. 322. A trinomial, a+b+c, may be raised to any power by consider- ing two terms as one, and making use of the Binomial Theorem. Thus, (a+b+c)"=(a+b+c)", n(n-1) 1 . 2 in which the several powers of 51-7; may be replaced by their expan- sions found by the Binomial Theorem. = (a + b)“+ n (a + b)""'c + (a + b)” 09+ &c- Ex. Required the cube of 1 + x + x”. (1 +x + x“)"’= (15+ .122)”, =(1+x)3+ 3 (1+x)’x’+ 3 (1 +x) x‘+x°, =1 + 3x+ 8x’+ x3+ 8x2+ 6x3+ 3x4+ 3x‘+ 3x5+ x“, =1 + 3x + 6x9+ 7 x8+ 6x4+ 3025+ x“. Similarly (a+b+ c+d)" may be expanded by considering a+b as one term and c+d as another; and any multinomial may be expanded in a similar manner by dividing the whole into two terms and considering it as a binomial. 192 BINOMIAL THEOREM. Also any particular term of an expanded trinomial may be easily found :--thus To find the term involving x‘ in the expansion of (1+ x+x2)3. (lTai+x2)3= (1+x)8 +3(1+x)2x"+ 3 (1+x)x‘+x6, and without further expansion it is seen that the term required = 3x2>< x2 + 3x4= 6x4. Again, the number of terms in the expansion of (a+b+c)" may be found; for it will obviously be the aggregate number of the terms in the expansions of the several powers of m, from (a+b)“ down to (a+b)”, that is, (Art. 317) = (n+l)+ n+(n—l) + &c. to iz—+_l terms, ={2n+2_n}n_f_1(Art,282)=-_W , 2 1 .2 [Exercises Zg.] 323. Required to find the Remainder after taking r terms of the expansion of (1—x)_2. By the Binomial Theorem, (1—x)-9=1+2x+3x’+...+rx"‘+ R, R representing the remainder after r terms ; 1 —————,=1+2x+3x’+...+rx*"‘+R; 1—2x+x J1+2x+3f+ ....... " +mfi4+12 o 1: —2x— 4x’+. . .——2(r- 1)x"1— 2rx’— 2Rx i + x” +. . .+ (r-2)x'-1+ (r-1)x"+ rx'+1+ Hz“, or 1 = 1 -(r+1)x’+ rx’+1+ R(1—x)”; . R: (r+1)x"—r.'i"’r1 ' ' (1 --x)2 In the same manner may be found the remainder after taking r terms of the expansions of (1 -x)"3, (l—x)-“, &c. 324. To find the number of homogeneous products of r dimensions which can be made of n things a, b, c, d, &c. and their powers. By common division, or the Binomial Theorem, 1 1 —ax 1 —bx = 1 + ax+ 6i2x9+a3x3+ . . . . . . =1 +bx+b’x’+b8x3+ . . . . . . =1 +cx+c’x’+cax3+ . . . ... 1 1 —cx &c. = &c. EXPONENTIAL THEOREM. 193 . 1 1 1 ' l—ax' l—bx'l—cx . &c.=1 +(a +b+ c+ &c.).r +(a”+ ab + b’+ ac + be +c’+ &c.).x‘ +(a3+ a’b+ ab’+ 63+ a’c+b’c +ac’+ c’+ &c.).r” . + &c. &c. the coefficient of x’ being the sum of the homogeneous products of n things a, b, c, d, &c. of r dimensions. Now to obtain the number of these products, let a=b=c=d=&c.=l, then the coefficient of x’ will give the number required. But on this supposition, the left-hand side of the equation becomes (1-x)"‘, which by the Binomial Theorem is n+1 , n+1 n+2 8 n(n+1)(n+2)...(n+r-1) l—___ o-_o-'_ +&C...-+ ’ 0... 1+nx+n 2m+n 2 3x 1.2 . 3 r x+ , _ . __n(n+1)(n+2) . . . . ..(n+r—1) " number reqmredfi . 2 . a .... .. r ' Con. Hence also the number of terms in the expansion of any multi- nomial, as (a,+ 02+ a,+...a,)", is known ; for it is obviously the same as the number of homogeneous products of r things taken 12 together, that is, r(r+1)(r+2) . . . . . . (r+n—1) 1 . 2 . 3 . . . . . . n ' If r=2, that is for a binomial, (a+b)", the expression becomes 2.3.4. . . . . . (n+1) If r=3, the number is, for (a+b+c)", 3.4.5... (n+1)(n+2) n ’ °r_i . 2 ' If r=4, or the quantity to be expanded be (a+b+c+d)", the number of terms is , or n+1. 4.5.6. . . . . . (n+3) (n+l)(n+2)(n+3) : or ' 5 1.203000... 1 O 2 Q 3 and so on for any value of r. THE EXPONENTIAL THEOREM. 325. To expand ax in a series of powers of x. a‘={(1+a_:l)"};; and expanding by the Binomial Theorem, — —l —2 e. ={1+n(a-—1)+n. n—2—1(a—1)’+ n. (a-1)3+ &c.}", ={1+[(a_1)_(“;1> fl“? &c.]n+Bn=+cn8+...}, B, c, &c. containing powers of a-l only; 13 194 MULTINOMIAL THEOREM. ={1 +An+Bn2+ On3+&c.}’7, if a—l -%(a—1)2+ SIC-=14, J! ___1 7 0 =1 Q Q p Q Q. x;n(A+Bn+...)9+ . . . . .. =1 +x(A+Bn+. . .)+x. Now, since a” is clearly independent of n, n must entirely disappear from the above series, which will therefore consist only of those terms in which n is not found ; Azxfa A3013 . ‘= ' +8: . “ 1+A“+ 1.2 +1.2.3 c COR. If 6 be that value of a which makes A equal to 1, then 6“—1+x+ x2 + $3 + _' 1 1.2 1.2.3 "" " Hence, making x=1, 1 l l = _ _' m 000: . o E o 1 g e 1+1+L2+L23+ 271828, (Art 57, x ) Another method of expanding a“ will be found in Art. 342, Ex. 9. THE MULTINOMIAL THEOREM. 326. The Multinomial Theorem is a rule or formula for expanding any power of a quantity which consists of more than two terms. The expansion of a multinomial may frequently be effected by the Binomial Theorem, as is done for a trinomial in Art. 322; for (a+b+c+d+&c.)’" may be expanded as a binomial by considering any number of terms as one term, and the remainder as another term. But a more general method is to find the general term, and to deduce the whole expansion from that term as follows :— 327. To find the general term of the expansion of (a+b+c+d+ &c.)m. Let b+c+d+ &c.=z, then (a+b+c+d+ &c.)’”=(a+z)"‘, of which the general term, expressed by the a+lth, is m(m—-1) . . . . . . (tn—a+1) 1 . 2 . . . . .. a apZa’ Where P+°‘=m’ or m(m —1) . . . . . p +1).a1’. i; , where a is a positive integer. Again, if c+d+ &c.:y, then 2a=(y+b)<1, of which the general term, expressed by the q+1‘h, is MULTINQMIAL THEOREM. 195 a<( car-l- E + +...+ [22 +...) x &c ...................................... . . Now, as this operation merely exhibits the same quantity expanded in two different ways by the same theorem, the corresponding terms, that is, the terms involving the same powers of a: will be equal to each other; therefore equating the coefficient of a’" on the one side with the coefficient of :c’" on the other, and observing that each separate term on this side of the equation which involves x’" will be the product of as many terms as there are series to be multiplied, one of which is taken out of each series, and will therefore be of the form aPmP Was" 0112’ apbqc'. &c. -_.--.-—.&c. or -_-~——~ where p+q+r+&c.=m, we have (a +b+c+d+&c.)'" apbqc'.&c. * =z__—___ |ln E.&c. ’ ll." 0.. __——_—_— C012. 1. If q+r+s+&c.=vr, then p=m-'1r, and the general term be- comes m(m-1)...(m—vr+ 1) "’"Trbq ".& . “ 1.2...q.1.2...r.&c.a c c which form is sometimes found more convenient. COR. 2. If it be required to expand (a0+a,x+a2x2+a3m8+&c.)’", the general term may be obtained from that of (a+b+c+d+&c.)"‘ by writing a0, a,.r, agxz, ass“, &c. in place of a, b, c, d, &c. respectively, by which it becomes L. m L12. &c. and all the terms of the expansion may be found as before, by giving p, q, r, s, &c. all possible values which the condition p+q+r+s+&c.=m admits of. Also any particular term involving a proposed power of x, as x”, will be found by taking the sum of the values of this general term, when p, q, r, s, &c. are made to assume all the values, which satisfy the two equations p+q+r+s+&c.=m, and q+2r+3s+&c.=n. +r+88 2. $94“? , . aopafaz'. &c. . w9+2’+“+&°' ; " ‘2 Stands for the expression “the sum of all the quantities of the form of.” 1- The proof here given of the Multinomial Theorem extends only to the case of positive integral indices, for by the Exponential Theorem m cannot be any thing but a positive integer. But if the Multinomial be deduced from the Binomial Theorem, (as in Art. 327) then since the latter is proved for fractional and negative indices, the former is also proved to hold for such indices. MULTINOMIAL THEOREM. 197 Con. 3. Assuming the Theorem for a positive integral index, it may be proved for a fractional or negative index thus :-— Let b+c+d+&c.=a:, then (a+b+c+d+&c.)m=(a+.r)"‘, of which the general term, expressed by the a+ 1th, is m(m-—1)...(p+l) T 2 ... a and since a is a positive integer, by what has been proved the general term of an, or (b+c+d+&c.)s, is LC: . &c. \1 aPx“, where p + a = m ; . '. a __ P bi 0—,. . . term requued m(m 1). . .(p+1).a . Lq . l£1510. Ex. 1. Required the term in the expansion of (a—b—c)’ which involves a’b’c”. Here m=7, m—vr=2, q=3, r=2, r ____-'17: g: ‘3’: 23 . a2. (- b)“. (—c)’, = — 210a2b302. Ex. 2. Required the term in the expansion of (a+bx+c.r’+dx3)‘ which involves x“. The required term =2 ——-—E—-—— . apbqc’d‘as, E- li- if. ii where ])+q+r+s=4, and q+2r+3s=8; and it remains to find all the values of p, q, r, s which satisfy these equations. To do this, it is most convenient to take in order, beginning with the highest or the lowest, the several values of q, r, and s, which satisfy the latter, and reject those which are inconsistent with the former, equation. And it is most advan- tageous to begin by assigning values to that quantity which has the greatest coefficient, 2'. e. in this case 8, and to take them in descending order of magnitude. Thus we see from the second equation that .9 cannot be greater than 2: i. e. it may have the values 2, I, 0: let s=2; then q+2r=2z whence r=1 or O, and q=0 or 2. Next, let s=1; then q+2r=5; and r=2, 1, or 0; q=1, 3, or 5. Lastly, let s=0; then q+2r=82 and r=4, 3, 2, 1, or O; q=0, 2, 4, 6, 8. Of these values, only so many must be taken as will satisfy the first equation, when we shall have to reject all except the underwritten:— S 7' q P '. the term required = 2 1 O l 2 O 2 O 1 2 1 O 0 4! O 0 where the simultaneous values of the quantities are written from left to right. 198 MULTINOMIAL THEOREM. Hence the required term is acd2 b2d2 602d 0‘ 8 H (F+EI§+T?+E)”” or (l2acdz+fib’d’+l2bc2d+c‘)a:8. Ex. 3. Find the coefficient of x8 in the expansion of (a— m+n-2:2)". Here + =12 . . . P +q r } are the equations of condltion: q+2r= 8 P (I 7‘ 4- _—8—_ O 5 6 1 6 4 2 7 2 3 8 O 4 coefi". required=u2_ ' { [24.11; + + €6§€2Lg+ £v§gsl§+ }, = 495a468+ 5 544a5bsc +1 3860a66‘02+ 7 920a7b203+ 1495:2804. Ex. 4. Required the term involving .22“ in the expansion of (aar— bw3+ cx5— &c.)“. Here (ax—bx3+ caza— &c.)‘°= x‘°(a—b.v2+ cm‘—&c.)‘° ; therefore it will only be necessary to find the term involving .1.“ in (a—bx’+c.r‘—&c.)‘°. Hence the equations of condition are 71+ q+ r+&c.=10,} 2q+4r+&c.= 4, P 9 7' 8 2 O 9 0 1 and by virtue of the second equation all the quantities after 1* must separately = O. 862 age the term re uired= 10. i— + 1‘ q L— is-La ’ [Q -_= (45aab'+1 0a°c).r“. EVOLUTION OF snaps. 199 Ex. 5. Required the coefficient of m3 in (a+ bw+cx2+dx3)%. 1 Here + +r+s=- . . . p q 2} are the equations of condition: q+2r+3$=3 p q i r ‘ s 5 —223 0 0 3110 2 1 ---2—0 0 1 . . 1 l l {563 1 1 _ 2, 1 -1 .. =- __ __ - --_ — _. coefficient required 2<2 1>< 2) +2(2 l)a bc+2a Q E [)3 be d ="—:_@ " "‘— +-"_1 - lfm2 411% 2a2 EVOLUTION OF SURDS. A practical method of finding the square root of a binomial surd was given in Art. 182; the following is the one more usually adopted;— 329. To extract the square root qf a quantity which is under the form a+,,/’b. Assume A/'.ri:+~/3;=I~/a+~/5, then squaring, x+y+2~/a.r§=a+~/55 '. w+,7/=6b }(Art. and 2~/.izy=~/5, from these two equations we find a: and y thus:— x2+ 2mg +ye= a’,} 4ary=b, . x2~2xy +yg=ag- b, ar—_y=,/a2—b. And_x+_y'=a,- - 2x=a+,/a’- b, 200 EVOLUTION OF scans. and 2y==a-~/a_5:h, I ...-l 2— and y a: w 2 .i fi+fl=, /€ti_____._ From this conclusion it appears that the square root of (Nu/Z; can only be expressed by a binomial, one or both of whose terms are quadratic surds, when a’-b is a perfect square. If the proposed surd be of the form tux/(7, then we assume JE—Jg =~/a-~/h, and proceed as before. 330. It must be observed that this method applies only to cases in which one of the terms of the binomial is a quadratic surd, and the other rational. If, however, a binomial is proposed which can be put under the form l,/a’c+,\/l_2ii, or ~/E(a-i-,~/h-), its square root may be foqnd, by finding the square root of a+~/h, and multiplying the result by ,1/0. 3 Ex. 1. Required the square root of -2-+J2. Assume J5+~/E/.=\/g— +J2; then x+y+2~/;y==2+~/2; 3 .. arty—'2', }(Art. and 2JE=J2> 4xy=2, . .r’- 2.13; +y’= - , m—y= , +y= 2 r 0.0 “i=1, EVOLUTION OF scans. 201 0 S zen-— and 2y=1, or y= _ 1 Jx+J§=l+gJ2, the root required. Ex. 2. Required the_square root of 21:. Here J2_7+~/24=~/§<3+,/8x3=,/3(3+~/§); +,f§; and applying the method of Art. 329, J3+J§ is found to be l+~/2_; (or see Art. 182. Ex. 1): therefore the root required=,1/3(l+,,/2), or 331. LEMMA. If jaiIJE=X+J§Z then also Um=x-,/§. For if W=x+Ji a +Jb= xa+ 3x2,/,i/_+ 3mg by]; (Art. 179) a=m3+3x_q, and J3= 3w’~/§+y~/i. hence a—~/b—=.r3—3x’~/5+3xy-y,/ij, 176362.145. 332. T 0 find the cube root of a- binomial Surd of the form a+,,/b, when it can be expressed by a binomial of the same description. Assume x+.,/;='J8a+~/b_, then a-J;=:/dljb, (Art. 33]); '. a’—y=:/m Now, if a’—b be a perfect cube, let it be equal to c’; then x’—-y=c, but x’+3a:y=a, (Art. 331); -. xa+3x(x’—c)=a, or 4x’—3cx=a. From this equation a: must be found by trial, and then 3 is known from the equation y=w’-c; thus x+,,/_y is known, which is the required root. It appears from the operation that the cube root of a+,,/b can only be expressed by a binomial of the same form when a’—b is a perfect cube. 202 EVOLUTION OF snaps. This test, therefore, ought to be applied to every proposed case in the first instance. Ex. Required the cube root of lO+,,/108. Here a”—b=100-108=—8=(-2)8; therefore the method of Art. 332 may be applied. Let a: + ij=m, '- $-~/§=:/l—(—)_:7_1__6§ ; w”—_1/= :/1_OOTOS=— 2. Also x3+3xy =10; x3+ 3x(.r2+2)=10, or 4x3+ 6.2::10, an equation which is satisfied by rs=] ; therefore y=3 ,- and a:+,,/§=l+,] If therefore the c_i_ibe root of 10+./108 can be expressed in the pro- posed form, it is 1+ ~/ 3 ; which on trial is found to succeed. [Exercises Zh.] 333. LEMMA. If n be an odd number, a and b one or both quadratic surds, and x and y involve the same surds that a and b -1 L do respectively, and also (a + b)“ = X + y, then (a _ by. ____ x __ y. By involution a + b = + y)”, n - 1 n — 2 1 (pa—Z! 2+n._____ . .___m"'-3 3+ &c. J 2 3 y where the odd terms involve the same surd that a; does, because 97, is an odd number, and the even terms the same surd that y does; and since no part of a can consist of y, or its multiples or parts (Art. 181.), n- or a+b=w”+nar"”1y+n. n—l a = w"+ .n"‘2y2+ &c. n—1 n-2 and b=nw"-1y+n.-—2-. 3 w"'3y3+&c. hence, n-1 72-1 02-2 a-b:-=w"-nw""y+n.—é—-m"‘2y2-n.—-é—. 3 w”“3y3+&c. =(w-3l)”; 1 (a - b);= w — y. EVOLUTION OF scans. 203 334. The 11th root of a binomial, one or both of whose terms are possible quadratic surds, may sometimes be eapressed by a binomial of that description. Let A+B be the given binomial surd, in which both terms are possible; the quantities under the radical signs whole numbers; A greater than B; and n an odd number. Assume \7(A +B)><\/_Q=a' + y, then \7(A - B)XME= a: - y, (Art. ass.) by mult. 6172—99; let Q be so assumed that (Az—BQVQ may be a Perfed hth Power, as p", then a2—y2= p. Again, by squaring both sides of the first two equations, we have Vmwgwww 312, VW= ag— 2ay + yg, hence + 2wg+ Qy", which is always a whole number, when the root is a binomial of the supposed form. Take therefore 8 and t the nearest integer values of \/ (A +B)”><&c. which is a perfect nth power, if a', y, z, &c. be so assumed that a+a', [3+ y, 7+z, &c. are respectively equal to n, or some mul- tiple of n. Thus, to find a number which multiplied by 180 will produce a perfect cube, divide 180 as often as possible by 2, 3, 5, &c. and it appears that 2.2.3.3.5=180; if, therefore, it be multiplied by 2.3.5.5, it becomes 28.33.53, or (2.35)", which is a perfect cube. 338. If A and B be divided by their greatest common measure, either integer or quadratic surd, in all cases where the nth root can be obtained by this method, Q will either be unity, or some power of 2, less than 2". See Dr. Waring’s Med. Alg. Chap. v9“ * The proof here omitted, and the reference to it in Dr. Waring’s work, may both be lafely neglected, as it is of no practical use whatever. En. 206 INDETERMINATE COEFFICIENTS. 339. The square root of a multinomial, as a+,,/b+ Jg+ of which one term is rational, and the rest quadratic surds, may sometimes be found by assuming Ja+Jb+JE+Jd=JE+ and proceeding to find a', y, and z, as in Art. 329. Ex. Required the square root of 21 +6Jg+ Let ,/§+~/y+~/§=~/21+6~/§+6\/77_+2~/3§, then a+y+z+2~/@+2\/ds+2~/fi=21+6~/§+6,/§+2,/3§; x+y+z=21, 2‘ “=6 5, 1 J2 J__ to find a, y, and z. 2,~/xz =6,,/7, 2 yz=2,\/3—5, Now 2@x2~/;z=4x~/y—z, or 6,,/fl5_><6,,/§=4.v,\/%, a=9, and J;=3. Also 2,,/.;y_x2,~/yg=4y~/;2, or 6J3x2,/33=12y~/'l, y=5, and Again, x+y+s=21, or 9+5+z=21, z=7, and Hence N/d+~/y+~/g=3+g/3+fi, the root required. INDETERMIN ATE GOEFFIGIEN TS. 340. If A+Bx+Cx2+&c.=a+bx+cx2+&c. be an identical equation, that is, it hold for all values whatever qf' x, then the coefiicients of like powers of x are equal to each other, that is, A=a, B=b, C:c, &c.* For if A+Ba=a+bx, then ' A—a+(B-b).r=0, an equation which admits of one value of a: only (Art. 193), unless B—b=0, or B=b, and therefore'also A—a=0, or A=a. * In the proof which is usually given a: is assumed equal to 0, and afterwards the equal quantities are divided by .12, whereas it is not proved that we may divide any quantity by a' when :1: stands for 0, in the same manner as when it stands for a finite magnitude ; and that such a proceeding will in certain cases lead to erroneous results is well known. .INDETERMINATE COEFFICIENTS. 207 Again, if A +Bx+Cx“=a+b.r+cx’, then A—a+(B-b).x+(C'-c) x2: 0, a quadratic equation with respect to a which admits of no more than two distinct values of .2: (Art. 204), unless C—c=0, or C=c, and B—b=0, or B=b, and therefore also A—a=0, or A=a. Similarly, if any number of terms be taken, or (A—a)+(B— b)x+(C— c) .v”+&c.=0, there are certain values of a', and none other, which will satisfy the equa- tion as long as it remains an equation with respect to 41:. But, by the supposition, the equation must be true for any value whatever which we may please to give to a', and consequently for any number of values of x; and this, therefore, can only be attained by that which is apparently an equation with respect to a ceasing to be such, that is, by the coefiicienls of the powers of a: being separately equal to 0; that is, we must have A—a=0, B-b=0, C—c=0, &c. or A=a, B=b, C=c, &c. COR. If there be found any power of x on one side of the proposed equation, and no corresponding one on the other, then the whole coefficient of that power is of itself equal to 0. Thus, if A+Bx+Cx2+&c.=0, for all values whatever ofa, then A=0, B=0, 0:0, &c. This may also be arrived at in the following manner. It has been already proved (Arts. 193, 204) that if a simple or a quad- ratic equation be known to be satisfied by more different values of the un- known quantity than the dimensions of the equation, the coefficients of the several powers of the unknown, and the term independent of it are separately equal to 0. This proposition will be now extended. If ac+ a,a'+a,.r2+. . .+a,,_,.v“-1+a,,.v"=0 be satisfied by n+1 different values of .r, then no, a,, a,,...a,, are each equal to 0. Let the n +1 values ofa: be .v,, .r,, a3,....:r,,+,. Then we have ao+ a,a,,+,+ a,.rf+,+. . .+ a,,_1 mfg} +ana2+1= 0, a0+a,a:' +£12.12" +...+a,,_,a”'"+a,,a’“ =0, where a' may be equal to any of the n quantities an a,. . . ay,. By subtraction we have a,(x'— ay,+1 )+ a,(x'2— a§+,)+. . . +a,,_,(.r"‘"—.v,’f;})+ a,,(.r"‘— a:,,)= O, and as x’-.v,,+, is not equal to 0, dividing by it we obtain al‘i' ag(.r’+ J'H‘)+. . . “fag-l lr'fl_2+' ' - an(xln-l+ ' ' '+ O; 208 \ INDETERMINATE COEFFICIENTS. or, arranging in powers. of x’, (a,+a,a:,,,,+. . .) + (a,+. .. )az’+. . .+a,,a""*‘= 0, where a' may take any of the n values a3, .v,,...a',,. Comparing this with the original equation, we find that it is of one dimen- sion less, and that the coefficient of the highest power of the unknown is the same as before, viz. a,,. By repeating this process we shall have an equation of n-2 dimen- sions, the coefficient of the highest power of the unknown being still an, and we shall know that it is satisfied by n-1 values of the unknown, viz. .rl, a,,....v,,_,; and so on. At last we shall arrive at an equation of one dimension, where the coefficient of the unknown will be a,” and this will be satisfied by two values, .13, a,. 0.0 an=00 And our original equation will then become 620+ arr +a2w2+ - 0 -+a,,_l$n-l= O, satisfied by more than n—l values of .v; as before, a,_,=0, so an_,= 0, &c. This, however, cannot satisfactorily be applied to the case of an infinite series. 341. Otherwise. Let A+Br+Ca2+...=a+b.r+c.v’+... be an identical equation, that is, such as will hold for any value whatever of .v ; then A ~a +(B~b).v+ (C~c).v’+...=0, and, if A~a is not equal to 0, let it be equal to some quantity p; then we have (B~b)x+(C~c)x’+&c.=-p. And since A, a, are invariable quantities, their difference p, and —p, must be invariable; but —p=(B~b).r+(C~c).v’+... a quantity which may have various values by the variation of a; that is, we have the same quan- tity (p) proved to be both fixed and variable, which is absurd. Therefore there is no quantity (p) which can express the difference A~a, or, in other words, . A~a=0, and A=a. Also (B~b).v+(C~c).v’+...=0, or B~b+(C'~c).v+...=0. (Art. 82.) Therefore, by what has been proved, B==b. And so on, for the remaining coefficients of like powers of a. This is open to the same objection as the first proof, in which it is assumed that every equation has only a certain number of values of a: which will satisfy it: for as long as .1: takes the value of any of the roots INDETERMINATE COEFFICIENTs. 209 of the equation her‘e written, the equality will hold: and as the right-hand side may be an infinite series, there may be an infinite number of values of a: that will satisfy it, considered as an equation: the conclusion that p is variable as well as fixed is therefore hardly correct. The same result may also be deduced from the corollary to Art. 392: for by virtue of it we can make (B~b).r+(C~c).v"+&c. less than —-p+lc’ coefficient in the series involving a. But this cannot be unless 12:0; A=a. as before, B=b, C:c, &c. -—p, by putting for a: any quantity less than where h is the greatest This however supposes, a priori, that the coefficients do not increase without limit. It will be seen that none of these proofs are altogether satisfactory: the reason of this is, that an identity is treated in them in all respects as an equation. The fact of the matter is, that we predicate of .v properties quite distinct from those that belong to the symbols we have hitherto had to deal with. These symbols standing for abstract or concrete quantities considered with reference to number, are obtained by the repetition of certain units which may be as small as ever we please. By that repetition therefore, all numbers, and all quantities that can be represented under any circumstances by numbers generally, are essentially discontinuous. Now in the identity We are dealing with We predicate of a: continuity, making it thereby a symbol of quantity in the most general and un- restricted sense it can possibly be conceived in. Now an equation in which .2: appears in an infinite series, certainly has an infinite number of roots: but these roots are symbols of quantity not unrestricted: and the equation is satisfied only when .2: takes the value of one or another of these roots, and not under any other circumstances. N OW, however near these roots may lie to each other, as .2: passes from one to another of them, it changes discontinuously. But supposing a: to possess continuity, which it does in the identity here discussed, it changes by insensible degrees from any one value to any other however near to the former, and consequently the infinitude of the number of values which .1: takes is of an infinitely higher order than the infinitude of the number of roots of an equation in the form of an infinite series. On this principle, therefore, and not otherwise, in the first proof, we can say that the infinite equality is satis- fied by more values ofa: than the number of its dimensions; and in the second proof we can conclude that p is variable as well as fixed, and by this manner we overcome the objections to those proofs. Or we may proceed to reason as follows: When we say a+bx+cxi+&c.=0 is an identity, for all values of .v, we thereby assign to .r the property of continuity, and make the series continuous likewise, and therefore capable of taking all manner of different values by the variation of .27. These may or may not lie between certain limits, but that is quite another question: what we say is merely that the series does depend upon .2: for its value, and that therefore it can change 14: 210 INDETERMINATE COEFFICIENTS. in value, and moreover that if a: is unrestricted, it does change, however much 1t may not when w is restricted. But by its continual equality to zero, we see that it is incapable of taking different values, during the variation of a: which is unrestricted. This cannot be unless it only involves .2: apparently and not really, 2. e. unless a: disappears. Now the only way in which a: can disappear is by each of the coefficients vanishing separately, therefore we have a=0, b=0, 0:0, &c. And by the same reasoning we shall have, if A +Bx+Cw2+&c.=a + bx+cx2+&c., A =a, B=b, =0, &c. 342. If A+Bw+ Cx2+... a+bw+ex2+... +A’y+B’a'_y+... = +a'y+b'a’y+... +A"_y2+... +a”_zf+... for all values whatever of x and y, then the coefiicz'ents of like quantities are equal to each other, that is, A=a, B=b, C:c, A’=a’, B’=b', A”=a”, &c. Since a: may receive any value whatever, suppose it to have some fixed value while 3 is variable, then the equation may be put under the form A,+Bl_z/+O,y2+...=a,+b,y+c,y2+... where A 1, B,, C,, &c. a,, 6,, 0,, &c. are invariable coefficients, and AI=A +Bx + Caz +... B,=A’+B’.r+0"x2+... &c.=&c. a,=a +6.2: + 0.222 +... h,=a'+b'.:v+c'w2+... &c.=&c. Now, by Art. 340, 11,: a,, B,= 6,, C,=c,, &c. ; therefore A + bx + 000’ +...=a + 6.1: + 0x” +... A’ +B'..'::+C"a'2 +...=a'+ b'w+c'w2+... A"+B"x+C”.:c’+...==a”+b"a:+c"x2+... &c. = &c. Then by Art. 340 again, since a may have any value whatever, A=a, B=h, C:c, A’=a', B’=b’, A”: a”, &c. The application of the preceding Theory will 'be shewn in the follow- ing Examples :— Ex. 1 . Expand o 0 o a+cw to four terms, that Is, d1V1de a—ha' by a+ca'. INDETERMINATE COEFFICIENTS. 211 Let Zl::=A+Bx+Cx’+Dx3+...; in which the coefficients A, B, C, D, &c. remain to be determined, a - bx =Aa +Baa: + Cax2+Dax8+ . . . +Acx+Bcxg+ Ccx3+... =A a +(Ba +A c)w + ( Ca +Bc)a:2+(Da +Cc)a:a+. . . ; and equating coefficients of like powers of x, Aa=a, or A=1. Ba +Ac=- b, Ba=—(b+c), or B=- 6:0; 6+ 6 Ca+Bc=O, Ca=—£. c, or C= +,c . c,- a a b b Da+Cc=O, Da=- 4.20.0”, or D=-—~i;-E.c’; a a a-ba: b+c b+c +0 , , =1-——~ -——.,—cx—- 8 cw+ .... .. a+cx a Ex. 2. Extract the square root of 1+:c. Let ,/1+x=1+A.r+Bw2+ Ca3+ ---- -- assuming 1 for the first term, since that is the root when x=0,' Squaring, 1+x=1 +2Ax+2Bx9+20x3+. . . . . . . . . +A2x2+2ABx3+ . . . . . . . . . (Art. 142). +13%“ +. . . Hence, equating coefficients of like powers of a, 1 214 = 1, 0.0 A: A2 1 2B A2: 0.0 =_"—' =__ . + 0, B 2 8 , 2C+2AB=0, C=-AB=+iI-6-,- and so on: . “"L 1 1 2 1 a ..,,/1+x-1+2a: 8x+-i—6a:+...... Ex. 3. Let ys- 3y+w=0;- required the value of 'y in a series of ascending powers of w. Assume y = Ar + Bats-i- Cw5+Dw7+ &c.*, ' The even powers of a- are omitted because, from the given equation, it appears that the relation betwixt a: and y is such that, if --a' be written for a', and - y for y, the equation is not altered. If the even powers were retained, their coefficients would be found separately equal to 0. ED. 14—2 212 INDETERMINATE COEFFICIENTS. ' then y3 = A3m3+ 8AQBw5+ 3A2 (2227+ &c. + 3AB2w7+ &c. __ - 8y: - 3Aw - 3Bm3— 3Ca25- 3Da'7— &c. "' + w = + m and supposing the coefficient of each poWer of a: to be equal to 0, (Art. 340. Cor.) 1 A3 1 —3A+1=0, or A=gs A3-3B=0, mB:-g =5,; 1 01' C==AQB= -'-6'; &c. 3 31423 -- 3C = O, at w” m5 ' y=§+§1+§ +&C. Ex. 4. Let a' = ay + by2+ 0313+ &c. required the value of y in terms of an Assume y = Aw+Bw2+ Cw3+ &c. then ay = aAw + aBw2+ aCw3+ &c. by2 = bA2av2+ 2bABac3+ &c. ey8 = 0A3w3+ &c. = 0; &C. = &c. -— a? = -— d? I 1 hence aA -~1=0, or A = -; a - bA2 - b a __ -2bAB 0A3 262- cu; _ a _ a,” ’ c a0 + 2bAB + 0A3 = 0, or a; be: (262— ache3 y=—- -—;§-+~——~—;,-——~—+ &c. a Ex. 5. Let w=y-ay3+by5- &c. required the value of y in terms of w. INDETERMINATE COEFFICIENTS. 213 Assume y= Am+ Bm3+ Cm5+ &c. (See Note, Ex. 7.) then y = Aw +Bw3+ Cw5+ &c. - ay3 = - aAsara- 3aA2Bm5 - &c.l + 63/5 = + bA5m5+ &c. = 0; &c. = &c. —m=—w hence A—1=O, or A=1; B—aA3=O, or B=0; C-3aAzB+bA5=—-O; or C=3a2—b; &c. y = w + aw3+ (3a2+ b)m5+ &c. [Exercises Zi.] Ex. 6. Tojind the sum ofthe series 12+ 22+ 32+...+n2 by the method of Indeterminate Coeficz'ents. Assume 12+22+32+m+n2=A+Bn+Cn2+Dns+ &c. A, B, O, &c. being unknown coefficients independent of n ; then 12+22+. . . + (n +1)2=A+B(n +1) + C(n +1)2+D(n +1)3+ &c. by subtraction, n2+2n+1=B+2Cn+ C + 3Dn2+3Dn+D, the following terms being omitted, because their coefficients are separately equal to 0, being the coefficients in the last equation of n3, n“, &c. Hence, equating coefficients of like powers of 12 (Art. 340), a _ __1_ oD—l, 01' D-3 , 2C+3D=2, or C:%, B+C+D=1, or 13%,- . n n2 n3 sum requ1red=6 + E + 73- +A, = é—nQz +1)(2n+1) +A. To find A. It will be observed that all this investigation will equally hold good if we had begun at any other term of' the proposed series, instead of 12; therefore the value of A depends upon the term of the series from which we start. Putting then a=1, which we can do, because we do not thereby alter the value of' A, which is independent of 12, our original equation becomes 12=A+B+C+D ,' but we already have obtained B+0+D=1, 11:0, and the sum required 12 1 723 n2 = 6n(n+1)(2n+1), or g +5 + 55, which is the most convenient form for recollection. 214c INDETERMINATE COEFFICIENTS. Ex. 7. To find the sum of the series 13+23+33+ . . . +n3. Assume 13+ 23+ 33+ . . . + n3 =A+Bn + 0n2+Dn8+En4+ &c. 13+ 23+ . . . +(n +1)3=A+B(n +1)+C(n +1)’+D(n +1)’+E(n +1)‘+ . . . and by subtraction, 228+ 3722+ 3n +1=B + 2 Cn+ C+ EDnQ-t-BDn" +D+4iEna+ 6En2+ 4.~En +E, omitting the remaining terms for the reason assigned in the preceding Example. Hence ~ 413:1, or E=%, sD+6E=a, or 1):; QC+3D+4E=3, or 0%, B+C+D+E=1, or B=0; 2 3 4 '. sum required=% + Z;- + ’7;- +A=in2(n2+2n+1)+,4, n(n.+1)}2 _l1 . 2 7"‘4' As in the last example, to find A we put n=1, when we obtain by a similar process A=O, and the sum required = n(n+1) l. . 2 13+23+33+. . .+n3=(1+2+3+. . .+n)3. COR. Since 1+2+3+. . .+n , therefore Ex. 8. To solve the equation w4+1= 0. Assume {04+ 1 = (xQ—mn +1) (mg—nae + 1), then a>4+1= :04- (m +n).n3+ (mm. + 2).n2— (m + n)a' +1; and equating coefficients of like powers of .n, m+n=0, and mn+2=0, n= -m, and m2= 2, or m= i\/§; w4+1= (w2+ \/_2_..n+1)(m2— +1) = 0, or w2+\/§.a: +1: 0, and w”--- \/§..n+1=0, from which two quadratics we obtain the four roots, -1a\/-.-_i 1=l=\/:T as we ' , and INDETERMINATE COEFFICIENTS. 215 Ex. 9. T o expand ax in a series g” powers of x. Since a.”-=(l+ a—1)”=1+x(a—1)+a:.fl(a-1)2+x. fig—1 . x—;%(a-l)3+ &c. =1+{a—1-%(a-1)’+. +Bx2+ Cx3+. ..* assume a"=1 +Ax +Bx2+ Cxs-I- .... .. then ay=l +A y+By2+ 038+ .... . ., and rim/=1 +A(.r +y)+B(a: +y)’+ C(a' +y)3+ .... .. Multiply the first two series together, and equate coefficients ofy with the last, and the result will be , A+2Bx+3Cx2+4Dx3+...=A+A’x+ABa:’+AClr’+... from which, by equating coefficients of like powers of x, we have A2 A=A; QB=A2, B=—2—; C AB C ‘43 1) AC A4 d 3 == ,.. =m, 4 = ,..D=1.2.3A!;an soon. ,, Ax 112.222 A‘°’.r3 Air" _ Hence a -1+—1—+ 1.2 1.2.3+1_203A‘+ . . . . .. where A Is equal to a—1--§~(a-1)2+;};(a—-1)3— . . . . .. 1 . . . . Ex. 10. Resolve (x+a)(x+b)(x+c) Into Its partial fractions. 1 A B C L et (x+a)(x+b)(x+c) = m+n +331?) + m; 1=A(.r+b)(a:+c)+B(.r+a)(x+c)+C(w+a)(x+h). Now, since this equation (by the theory) is true for any value of .r, let ar=-a, that is, x+a=0, ___. __ 1 then 1=A(a-—b.a~c); 01 A=W; let a=—b, that is, x+h=O, _.___ __ 1 then 1=-B(a-b.b—c), 01 B=—(a'TbXT;)~, let w=-c, that is, x+c=0, __ __ 1 then l=C((L-C- b—C), 01 C— , 1 __ 1 1 (x+a)(x+ b)(x+c) _ (a—hXa— c)(a:+a) (a- b)(h-c)(x+b) 1 + . (a- c)(b—c)(x+c) “ This transformation is made to shew that the series will involve no other than positive integral powers of a. Q 216 INDETERMINATE COEFFICIENTS. 2 Ex. 11. Resolve (1+a‘:;(_fg;i€+cx) into its partial fractions. A + Bar + 0.12 P Q R Let (1+aw)(1+b.r)(1+c.r)=l+a.n+ 1+bar+ 1+cw’ then A +Bx +Cx2=P(l+ bacX 1+ ca)+Q(1+ax)(1+cx)+R(1 + ax)(1+ bx), l . let rs=-5, that Is, 1+ax=0, ><1_%>; or P=Aa -Ba+0. B 0 b then A—z—+—a—.,=P<1— m: E 1 . let a=-—Z, that Is, 1+hx=O, B 0 la 0 Ab’—Bb+C, then A--b—+Z-2'=Q, 01‘ Q=-(—a__b)(b——:c—), l . let x=—-C-, that Is, 1+cx=O, 2— then A’§+'CE=R(1_£\l<1-2>; or R=AC Bel—0 0' C c/ c (a—cXb—c); _ A+B.r+C.r"’ _ Aa2—Ba+ C _ Ab”-—Bh+C ' ' (1+ax)(1 + ba')(1+ ex) _ (a—bXa — c)(1+ax) (a—b)(b—c)(1+bx) A02-Bc+C' + (a—c)(b—c)(1+cx)' Ex. 12. Resolve into its partial fractions. 8.1: -1 Fg Let ___3x___1_ =£+§+i +_.‘_Z_)___ a:2(a'+1)2 a: x2 a+1 (a+1) many fractions to include every possible case which can produce the proposed fraction, although either A, or C, or both, may possibly be equal to 0,) 2, (it is necessary to assume so then 3m —1=A.r(.r +1)2+B(x +1)2+0a:2(x+1)+Dx2; let sir-:0, then —1=B, let x=-1, that is, x+1=0, then —¢1 =D, let a=1, then 2=4~A+4$B+2C+D, or 4A+2C=10, . . . . . . . .. (1) let x=2, then 5=18A+9B+12C+4D, or 18A+12C=30, or 3A+2C= 5, . . . . . . . .. subtracting from (1), A=5, and C=5——2A =-5, az-I 5 1 5 a _____ -__-_ w2(x+1)a=x a“ a+1 (x+1)2' INDETERMINATE COEFFICIENTS. 217 Ex. 13. Resolve into its partial fractions. .r 2 .r(.:z:+1)3 .x(a:+l)"_ a: a+1 (n+1)2 (a:+1)3’ then 3x+ 2 =A (a:+1)3+Ba'(:r +1)2+C.r(.r+1)+Dx , let a:=0, then 2=A, Let let a:=—l, then -—1=—D, or D=l, let a=1, then 5=8A+4B+2C+D, or 4B+2C=—12, (1) let x=2, then 8=27A+18B+6C+2D, or 18B+6C=-48, or GB+2C=-—16, ......... (2) subtracting (1) from (2), 2B=—4<, or B=—2,' and O'=—-2B—6'=—2. 3x+2 _g 2 2 + 1 ' a:(a:+1)8_a: a+1 (a+1)2 (x+1)3' 342 i“. The four preceding are examples of the resolution of a “ rational fraction” into its partial fractions, of which we shall now proceed to describe the general method. By a rational fraction is meant one whose numerator and denominator, when arranged according to the powers of a variable quantity .r, contain only positive integral powers of w; and furthermore it is supposed that the numerator is of lower dimensions in .2: than the denominator: if however this be not the case, we can perform the division as far as it will go, thus Obtaining an integral quotient and a fraction with its numerator of lower dimensions than its denominator: to this therefore we can confine our attention. Let %:be such a fraction ; then V can always be resolved into factors of one or two dimensions, which may or may not be repeated, 5. e. V: ..(x—b)’ . . . . . .(x2+ax+,8)...(a;2+ 7.2+ BY... a—a being one of the linear factors not repeated, .r—h one repeated r times, w2+ uni-,8 a quadratic factor that cannot be resolved any further, and m2+ yx+3 a quadratic factor repeated s times. The following assumptions must then be made: for every factor such as x-a, assume a partial fraction m: for every factor repeated as (.r—b)’, assume a series of partial fractions B. B. ., a, (.r—b)’+(x—b)"‘ "..n-b- for every quadratic factor xz+ax+fi assume a partial fraction Ca2+D _ x2+aw+fl ’ 218 INDETERMINATE COEFFICIENTS. and for the other, assume a series of fractions Elm +11"l E,a:+F’, '_2__——;+000000 ‘2———:;' ; (a:+7x+3) x+7x+o A, B &c. being independent of .r. It will be obvious on consideration that these assumptions must be made, because all that we know is that the numerator of the proposed fraction is of lower dimensions than the denominator. (1) To find A. Multiply by a:--a, the denominator corresponding to A, and suppose V=Q.(.r—a), then g Now neither Q nor any one of the denominators of the other frac- tions contains .r—a; and if we put x=a, the value of A is not changed, =A+ (the other fractions) ><(.r- a). . U . . . and neither a nor any of the other fractlons becomes Infinite, and the other fractions are all multiplied by zero; A=y when a:=a. Q In this manner all the other corresponding partial fractions can be determined. (2) To find B1, B2, &c. Multiply by (x-b)’, and suppose V= Q(.z'— b)’, then =B,+B,(a:—-b)+. . .+ (the other fractions) ><(.r- b)’ ; Bl=g when w=b. Q By transposing this fraction we have U Bl B2 Q(x—b)' _ (IE—by = (a-b)’-1 +"' And if the left-hand member be reduced to a single fraction, x—b will be a factor of the numerator, which if cancelled will make the denominator involve (.r—b)""1: B2 can then be determined in the same way as B,; and so on, if there be more. (3) To find 0' and D. Multiply by x2+ aa'+,8, and suppose V = Q(x2+ax+,8), then g=Cx+D+(the other fractions)x(a:"'+ax+,8). If then we put w2+ax+,8=0, we have g— Cx+D. Q __ We then must substitute for .r’ continually -(a.r+;6’), and after multi- plying out, we shall have at last a simple equation in a, whose terms involve C and D. Now we know that this is satisfied whenever w’+ ax+,8=0, INDETERMINATE COEFFICIENTS. 219 i. e. it is satisfied by two values of av, therefore the two terms are each =0, thus giving two equations to find C and D. (4) To find E, F,, &c. Multiply by (x”+7.r+3)', and we can deter- mine El and I?"1 as in (3): let this first fraction be transposed as in (2), then we shall have w9+ 'yar-l-3 to be cancelled, and we can then find E2, F2; and so on, for any others. _1 U I O . Ex. Resolve -,-f,_-- Into 1ts partial fractions. . x (x +.r+1) . D Let It =‘1 + B Um- x“ E + x2+x+1 ' Then by (2) multiplying by :02 and putting a=0, we have A=—1. Transposing this first fraction, w~l 1 __B C'x-l-D _ m’(a:’+a:+1) x2_ x x2+x+1 ’ n+2 ___B+ Car+D . ' a:(a:2+.r+1)— x x2+x+1 ’ therefore by a similar process B=2. Applying we have Cm +D= at; when x2+x +1= 0, or .r’=—.r-1; x—I —x+1 C D=——=——; x+ —x—1 a+1 ‘. —x+1=(x+1)(C.r+D)=Cx2+(C+D)a'+D = C(—-a:—1)+(C'+D).r+D; O=(D+1)x- C+D-1, an equation which is satisfied by the two roots of x2+x+1=0 ; D+1=0, and -C+D-1=O; '. D=-—1, C'=—2,' therefore the proposed fraction 1 2 2w+1 a-— +-- -——- . a” a: x2+x+1 220 CONTINUED FRACTIONS. CONTINUED FRACTIONS. a O U I 343. To represent B 27?. a continued fractzon*. Let b be contained p times in a, with a re- b) aflp mainder 0; again, let 0 be contained q times in b, c)b(q with a remainder d, and so on; then we have d) 0(1- a=pb+c, b=qc+d, c=rd+e, &C- e&c. a c c 1 01‘ 5=P+5=P+W=P+7 Z 1 I =p+ d =P+ a 830‘ q+rd+e (1+ e T-l-E r __ +s+&c. 344. C011. 1. An approximation may thus be made to the value of a fraction whose numerator and denominator are in too high terms; and the farther the division is continued, the nearer will the approximation be to the true value. 345. COR, 2. This approximation is alternately. less and greater than the true value. Thus p is less than g; and P+_1_ q is greater, because a part of the denominator of the fraction is c 1 0 0 omitted: q+ — Is too great for the denominator, therefore p+ r . a q + Q Is less than I; ; and so on, * Although a ‘continued fraction’ _(see Def. Art. 8) may be of the form p +q__:-B I ' 1 l &c. the term is commonly restricted to those of the form p+ -—- or —-1- q +" P + *—* 8: . 1 ?+&a ED. CONTINUED FRACTIONS. 221 O I a O 0 It is obvmus that, when 7)- is a proper fraction, p=(), and c=a, so . . a . . . . . that the operation of converting E into a continued fraction in this case commences with dividing b by a. DEF. The quantities p, q, r, &c., which are always positive inte- gers, are called the Partial Quotients; and 2%, p+l, p+ , when 9+; reduced to simple fractions, are called the Converging Fractions, or Convergents, to (71)- . Ex. To find a fraction which shall be nearly equal to w , and in lower terms. 100000 100000 ) 314159 (s 300000 14159) 100000 (7 99113 887) 14.159 (15' 887 5289 4435 "ii—El) 887 (i 854 E8zc. Here 12:3, q=7, r=15, s=1, &c. therefore 314159 1 100000 = 3 1 +______ 15 + &c. The first approximation is 3, which is too little, the next is 3+-;-=?7E, too great; the next is 3+ 1 = +1-1-0—56=?§-Z—, tOO little; and so on. \ 7+ '1—5— The proposed fraction expresses nearly the circumference of a circle whose diameter is 1; therefore the circumference is greater _ 22 . ass . than 3 diameters, less than -7— diameters, and greater than To? dia- meters, &c. 222 CONTINUED FRACTIONS. f 346. To convert any continued fraction into a series of converging ractzons. Let the continued fraction be (Art. 343) then the quotients are p, q, r, s, &c. and the converging fractions are respectively equal to 1 1 1 1,1" P'l‘ga P+_'i a 17+ -"'_l'> &c- qi“; (7+ 1 r+- 8 mp) pq+1 pqr-l-p-l—r pqrs+ps+_rs+pq+l’ &c. 1 q ’ qr+1 ’ qrs+q+s p pq-i-l (pq+1)r+p (pq+1.r+p)s+pq+l or —, ——-—, —————~-, , &c. 1 q qr+1 (qr+1)s+q in which the law of formation is observed to be as follows :— Write down in one line the quotients p, q, r, s, &c., and the first and second converging fractions at sight; then the other fractions may be obtained thus :- For the 3rd, num’. =3rd quot:>< num‘. of 2nd fract.+ num’. of 1st fract. denom'.=3rd quot.xdenom‘. of 2nd fract.+ denom'. of 1st fract. For the 4th, numr. =4th quot.>< num’. of 3rd fract.+ num“. of 2nd fract. denom'.=4th quot.> the other, that is, lies between them, (Art. 345) ; therefore the diffe- a . . rence between -— and elther of them is less than 6 ll N ll, < N11),~N,Dl 1 D, D, DlD, < 0,12; l x 0 o a N 1 Or, since D,>.-D,, a fortlorl, 7)- ~ 3' < 5-2 . l I 22 Thus in Ex. Art. 345, 7 differs from the value of 314459 by a ntit 1 ss thin I or --1-— ' and :33? d'ff f h qua y e a 7X106 742, 106 1 ers rom t e true value 1 ba ntitlsthn—l— r——-—- y qua y es a (106)20 11236' COR. 1. Those convergents, which immediately precede large quotients, approximate especially near to the true value of the continued fraction. N N N a F r 'f —1 i ——3 b th on ' _ .. o , 1 D! , D2, D3 e ree c secutlve convergents to b , corre sponding to quotients q,, q2, q3, then 525 u a M 131’ D3’ 5’ I); O are in order of magnitude: and 1:73 N1 q3N2+N1 N1__Qa(NsD1~N1D2)_ gs ~_-——._—— D3 DI:QsD2+D1~T)_1—_ D1((IsD2+D1) “DIDZ also &~%——i- 1), D,_D,D, And if q, be especially large, since D,b; . I then It appears that xn ; so that :c=n+?—, where y>1, and 1 l 1 ”+-- - - b y . . a y=b, or a".ay=b, and a”=;1—,,, or =a. Agam, Since $91, and p, so that y=p+g; and x=n+ EX. Required the value of a: in 10”=2. By substituting O and 1 for as, it appears. that x>0, and <1; let 1 a=1, then 103:2, or 2”=10. Again, it appears that y>3, and <4,- .9 1 1 3+;— 3 .2: . i; . - . z ' et y=3+;, then 2 =2 .2 =10, 01 2 =§=125, . . (1 25) =2. Again, 1 it appears that z>3, and <4; let z=3+£, then (1'25)8+“=(1'25)3.(1'25)“=2, l '_ 2 o (1'25)“=6?5)3=1'0241, (1'0241)"=1'25, and by trial u>9 and <10. Hence by successive substitution x_ l — 1 3+-—*-—-1- 3+§+&0 230 INDETERMINATE EQUATIONS INDETERMINATE EQUATIONS AND UNLIMITED PROBLEMS. 354. When there are more unknown quantities than inde- pendent equations, the number of corresponding values which those quantities admit is indefinite (Art. 198). This number may be lessened by rejecting all the values which are not integers; it may be farther lessened by rejecting all the negative values; and still farther, by rejecting all values which are not square or cube num- bers; &c. By restrictions of this kind the number of answers may be confined within definite limits; and problems are not wanting, in which such restrictions must be made. 355. If a simple equation express the relation of two unknown quantities, and their corresponding integral values be required, divide the whole equation by the coeflicient which is the less of the two, and suppose that part of the result, which is in a fractional form, equal to some whole number; thus a new simple equation is obtained, with which we may proceed as before; let the operation be repeated, till the coefiicient of one of the unknown quantities is 1, and the coeflicient of the other a whole number; then an integral value of the former may be obtained by substituting O, or any whole number, for the other; and from the preceding equations integral values of the original unknown quantities may be found. Ex. 1. Let 5w+7y=29; to find the corresponding integral values of m and y. Dividing the whole equation by 5, the less coefficient, 2y 5+4 m+ _— _- y+5 — 5, or .v=5-y+ y, a whole number; 4,...0 ~’l is a whole number. Assume AND UNLIMITED PROBLEMS. 231 y=2—2p-g, a whole number. And 5 is a whole number. Let g=8, or p=2s, then y=2—5s, and w=5—y+p=8+5s+2s=3+7s*. If s=0, then .v==3 and y=2, the only positive whole numbers which answer the conditions of the equation; for, ifs: and y are positive integers, 5s cannot be greater than 2, that is, s cannot be greater 2 Q . than 5, and s cannot be negative, for then a would be negative. If negative values of x and y are not excluded, then an indefinite number of such solutions may be found by putting 1, 2, 3, &c. -1, --2, ~3, &c. for s. Ex. 2. To find a number which being divided by 3, 4, 5, gives the remainders 2, 3, 4, respectively. Let a be the number, i d?— 2 then = p, a whole number, or a’ = 3p + 2 ; also, from the second condition, (0—3 3p-1 »———, or = a whole number 4‘ 4‘ q, a , q+1 that Is, 3p-1=4q, or p=q+—~3;--; +1 let g-é—=r, orq=3r--l, then p=4r—-1, and a=3p+2=12r—l; -4 12r-5 again, from the third condition, or ————5——~— is a whole num- . 27‘ . her, that 1s, 2r + ---5— -1 1s a whole number; * This operation might have been abridged thus, 2 (2-21), 5 i then x=5-y+2s=3+7s. ED. x=5-y+ let ggl=s, or y=2-5s, 232 INDETERMINATE EQUATIONS 2r , '. ~5- IS a whole number; 2r let -5— = 2m; then r = 5m, and w=12r—1=60m —1. If m=1, rs= 59; if m= 2, m=119; &c. The artifices employed in the two following examples are deserving of notice. \ Ex. 3. Let 11.n-17y=.5 ; to find the integral values ofx and y. Here $=17y+5= 22y—5y+5=2y_ sail-1) = a whole number. 1 l l 1 Let 'y—1_—11=p, or y—1=11p; y=11‘p+l. And w=2y—5p=22p+2—5p=17p+2- If p=0, r=2, and y=L; if p=1, x=19, and y=12; &c. EX. 41. Let 11w—18y=63 ,- to find the integral values of .r and y. Since 18 and 63 are divisible by 9, let a=92, then 11x92—18y=63; and dividing by 9, 1153—2y=7; - 2—1 . . y=5z— 3+ 7 . Let 224:1), or z—l=2p; z=2p+l. Hence x=18p+9; and y=10p+5-3+p=11p+2; and by giving to p the values 0, 1, 2, 3, &c. the integral values of x and y are determined. 356. If the simple equation contain more unknown quanti- ties, their corresponding integral values may be found in the same manner. ‘ Ex. Let 4.v+3y+10=5z; to find corresponding integral values of .v; y and a. Dividing the whole equation by 3, the least coefficient, (n+1 2.2: w+y+3+ 3 =z+—é-, y=z-.v-3+ ——-3————, a whole number. AND UNLIMITED PROBLEMS. 233 22—m-1 Assume -——§——=p, or 22—w-l=3p, then .v=2z-3p—l, and y=z—2z+3p+l—3+p=4~p-z—2; and for p and z substituting O, or any whole numbers, integral values of .v and y are obtained. If 2 =3, and p=1, then az=2 and y= -1; if z=41, and p=O, then w=7 and y= -6; &c. 357. If the number of independent equations be less by one than the number of unknown quantities, the equations may be reduced by elimination to one equation only betwixt two unknown quantities, and then the preceding method of solution may be applied. Ex. Let 2x+5y+3z=108, and 3x—2y+7.2=95, From 1st equu 6x+15y+92=324,} 2nd 6x— 4y+14~z=190, '. 19y—52=134.<; from which equation the values of y and 2: may be found by the usual method, and then the values of u may be deduced from either of the original equations. } to find the integral values of a, y, z. [Exercises Zl.] A more complete Theory of Indeterminate Equations, distinct from the preceding, is contained in the following Articles of this Section :— 358. To shew that, a is prime to b, the equation ax+by=c has one integral solution at least. . c—by Since ax=c—by, x=T,- now give to y the several successive values 0, 1, 2, 3, air], and since a is prime to b, the several values of c-by, divided by a, will all leave different remainders. [F or, if not, let y] and y, be two of the values of y, which make c—by, divided by a, if possible, to leave the same remainder r, ql and q2 being the quotients, then c-by,=aq,+ r, and c—by,=aq,+r, 5(91-y2)=a(q2—q1), or b(y,—y,) is divisible by a without remainder; but 6 is prime to a, there- fore y,—y2 must be divisible by a, which is impossible, since yl -y, or <5 , that is, accord- ing as? occupies an even or an odd place in the series ;) either a.cq+b.(—cp)=c, or a.(- cq)+b.cp=c, a(cq— bt) + b(at- cp) = c, or a(- cq —bt)+ b(cp+at)= c. . a But ax+by=c, .'., according as —> or <20 b "' 9 a:=cq—bt,} or x=—cq—bt,} y=at—cp, y= cp +at. Similarly, if the proposed equation be aJc—by=c, the general solution . . a p Wlll be, according as Z > or < z]- , x=cq+bt,} x=—cq+bt,} or y=cp+at, y=-cp+at. N.B. Of course, in any proposed case, Where we can see at once the values of a: and y which will make ax—by=1, or —1, there will be no need to form the series of convergents. Ex. 1. 21a+40y= 4000. l 2 19 4.0 The convergents are T , —1-, T6, QT, 21x19—40x10=—1, 21x19x4000—40x10><4000=—4000, 21(—76000 — 40t) + 40(40000+ 2 1 t) = 4000, .v=—76000—40t, and y=110000+21t. Ex. 2. 5x +7y=29. Here we can see.that a:=3, y=2, will satisfy the equation 5x—7y=1, that is, 5x3—7x2=1, 5x3x29+7x(-2x29)=29, or 5x87+7(-58)=29, 5(87—7l)+7(—58+5l)=29, '. a=87-7t, and y=-58 +5t. 236 ' INDETERMINATE EQUATIONS COR. 1. If only positive integral values of a: and y are required, t will be restricted within certain limits dependent upon the particular Example. Thus, in Ex. 2 above, if a: and y must both be positive, 7t<87, and 5t>58, that is, t<12%, and >11%, t can only be 12; and the only positive solution is a=3, y=2. COR. 2. Since a=a—bt, and y=fi+at, is the general solution from which all the values of a: and y are found by substituting for t, 0, ='=1, =2, =3, &c., it follows that the values of a and y, taken in order, form two arithmetic progressions, the one decreasing, and the other increasing; so that when two contiguous values of each are obtained, the rest may be assumed. N .B. If an equation of the form ar=by=c be proposed for solution, in which (when reduced as low as possible) a and b are not prime to each other, it may at once be affirmed that it admits of no solution in whole numbers. For, if a: and y be whole numbers, dividing by the common measure of a and b, which is not a measure of c by supposition, we have a whole number equal to a fraction in lowest terms, which is impossible. 361. To shew that the number of positive integral solutions is limited for ax+by=c, but unlimited for ax—by=c, a being prime to b. 1st. It has been shewn (Art. 359) that all the solutions of ax+by=c are deducible from a:=a-bt, y=[a’+at, where ar=a, y=fi, is one solution, and t any integer, positive or negative. But, if a: and y must both be positive, then it is plain that bt must be less than a, that is,t is restricted to the integral values which are less than 2 . Hence the number of positive solutions is limited. b ar— 5 values of a: greater than 2 and divisible by b, the number of which values 0, which is positive and integral for all 2nd. If ax—by=c, y= is unlimited, that is, the number of positive solutions is unlimited. 362. Tofind the number of solutions in positive integers of the equa- tion ax +by = c, where a is prime to b. Let the series of fractions converging to g- be found, and let ‘3 be that which immediately precedes 2, then (Art. 349) either aq—bp=1, or aq—bp=-1, according as 25> or <75 . a I. If -b->§, then a.cq—b.cp=c, . a (cq—bt)+b(at— cp)=c; and comparing .this with the original equation, ax.+ by=c,_ a=cq— bt, and y=at—cp; AND UNLIMITED PROBLEMS. 237 and the several solutions will _be found by giving t such values, that cq-bt and at—cp may be positive integral quantities, that is, t may be . . . c any positive integer less than 7? and greater than 223, but no other num- ber, (supposing 0 to be excluded as a value of either x or y.) Hence the number of solutions will be the number of integers which lie between ('11- and 2P- . b a II. If %<£, then aq—bp=—1, and it may be shewn, as before, that the number of solutions is the number of integers which lie between 0 1'1 and Cl, a b Com 1. If a]: , and gbq— are both whole numbers, the number of solu- tions will be Pfl~ gq—l *. a b If either 25, or 961 , be a whole number, and the other fractional, then the number of solutions will be the greatest integer contained in exp ~ 0% 'I‘. If 92?- and g—q be both fractional, reducing each to a mixed number, the number of solutions will be the greatest integer contained in 9a£~ Egg ,1, _ . . c or 0.2212 ~€ljl+1§, according as the greater of the two quantltles, 7?, 21:! , has the greater or smaller fractional excess. . c c b ~a c . Con. 2. SIIICG-al) ~—bg=c. pub q=E, (Art. 3419), the greatest Inte- . . c . ger contained In a—b- cannot differ by more than 1 from the number of solu- tions; so that the nuqmber of solutions may be determined within this error (which may be either in excess or defect) without solving the equa- tion at all. And hence also, if c c it is impossible to have a positive solution; for a=1, y=1, are the smallest positive values which a: and y admit of, and for these ax+by>c, therefore dfortiori for any other positive values ax+by>c. Ex. In how many different ways may £1000. be paid in crowns and guineas P * Ex. 7a+2y=42. + Ex. 2la+5y=400. i Ex. 9w+7y=110. § Ex. 3r+5y=26. 238 INDETERMINATE EQUATIONS Let x be the number of guineas, and y the number of crowns, then, reducing all to shillings, 21w+5y=20000; and it is required to find how many solutions this equation admits of in positive integers. The fractions are %, p=4.~, q=1; 5 ’ Ci 8‘ _c_p__ 20000 ~ 20000><4t hence b a — 5 21 ’ 11 =4000—38092—i, 10 =1902-15 . I ' c c O and Since one of the quantities -a£ , 7;]— , 1s a whole number, the number of solutions required is 190. Or, if x=0 be admitted as a solution, that is, if the payment may be made in crowns only, then the number required is 191. 363. (Another bfetlzod*). To find the number of the positive integral solutions (f ax+by=c. Let .r=a, y=/3, be one of them, then all the others are found from x=a—-bt, y=/a’+at, subject to the condition that t —-'§-; a b and there will be as many positive solutions as there are integral values of t between these limits, (reckoning t=0 for one of them to include the solution x=a, y=fi, which is positive by the supposition). But aa+b/J’=c, g-=a£b._§, ‘. t<—c--é, and >--[3:. ab a a Also c>b/3, aa is positive i>é. ’ ab a Hence the number of negative values of t will be the greatest inte- ger less than £- , and the number of positive values the greatest integer less than big—g, that is, the total number will be (excluding zero values 0 of a: and y) the greatest integer either in (Lil—7, or {lib-1, 01- ab +1, ac- . . c cording to the particular forms of a; and g. * This method may be employed when one solution is either given, or easily discovered by trial. AND UNLIMITED PROBLEMS. 239 lst. If a—Cb- andg be both whole numbers, the number of solutions will be i- -1 *. ab 2nd. If one of the two he a whole number, and the other fractional, c ab'l'. the number of solutions will be the greatest integer in 0 El 0 O Q o c I bers, the number of solutions Wlll be the greatest Integer in —+1 1, or In ab 0 o I O c O —5§, according as the fractional excess in E is greater or not greater than (1 3rd. If both and? be fractional, reducing them to mixed num- that in a Ex. Given .v=4~, y=9, one solution of 2a+3y=35, find the total number of solutions in positive integers. Here Ila—b: ?%).é=5§—, and §=g=4-é—=4%, both fractional, and the frac- . . c . . . tional excess in 55> that in g, therefore the number of solutions required 0 . is the greatest integer in ab Oi' thus, with less burden on the memory :—Since x=4~, y=9, is one solution, x=4—3t, and y=9+2t, (Art. 359) is the general solution; and in order that both a: and y may be positive integers, t will be restricted, +1, Viz. 6. .\ 4, _ so as to be less than &- or 1%,, and greater than g or —4%. Hence the values of t will be -4, -3, —2, —l, 0, 1, or the number of positive solu- tions will be 6, as before. 364. To find the number of solutions in positive integers of the equation ax+by+cz=d, each term being positivell. Transposing one of the terms, cz, the equation becomes ax+by=d-cz, ’ where 2 may have any integral value greater than 0 and less than zero values of the unknown quantities being supposed to be excluded. 0 Hence, by Art. 362, the number of solutions will be the number of integers which lie between gala—)1? , and @, where p and q are those quantities which satisfy the equation aq~bp=1. * Ex. 3w+4y=24. 1' Ex. 3w+4y=39. I Ex. 2.v+7y=125, § Ex. 11w+7y=108. 11 If any term be negative, as by, then the equation being of the form aa*—by=(’, the number of solutions is at once declared infinite. (Art. 361). 240 INDETERMINATE EQUATIONS Thus a certain number of solutions is determined for each value of z, and the sum of these numbers will be the number required. 0%. The application of this rule is attended with considerable difficulty (see Barlow’s Theory of Numbers, Art. 162, or Peacoch’s Algebra, Art. 506); but the following Example will shew the student how he may determine the number of solutions in some cases without much trouble. Ex. Required the number of positive integral solutions of 17w+19y+212=400. . . . 2y 4k . _9_. Divuling by 17, a+y+ 1—7-+Z+1—7—Q5+ 17 a __ 2y+42-9 a+y+z-23- 17 - 2y+42-9 Assume =t, or 2y+42-9=17t; 17 then y+2z=a+st+t-g_1; t+1 Assume —g—=s; t=2s—1. Hence it appears, (1) that t must be an odd number; (2) that it must be less than 23; also since a+y+2 cannot be less than 3, (zero values being excluded), t cannot be greater than 20, that is, its greatest value is 19; and therefore all the different values of t are I, 3, 5: 7: 9, I]: 13a 152 17: 19; and the number of solutions required is 10. 365. The theory for the solution of indeterminate equations of more than one dimension is too difficult to be admitted into an elementary work, like the present. The reader is referred for further information to Barlow’s Theory of Numbers, Chaps. III. and Iv. But there are two classes of such equations, which admit of easy solution: 1st. Such as do not involve the square of either of the unknown quantities; and 2nd. Such as involve the square of one of them only. Ex. 1. Required the positive integral solutions of 3xy—4y+3x=14,_ Here 3r(y+1)=14+4y; _14+4y_4(y+1) + 10 _ y+1 _ y+1 y+1 ’ =4+-l)—; y+1 AND UNLIMITED PROBLEMS. 241 hence y+1 must be either 10, or a divisor of 10; that is, it must be either 1, 2, 5, or 10; and therefore the values of y can be only 0, or 1, or 4, or 9; of which the first and last make a: fractional. x=3, 2 y=1, 4 Ex. 2. Required the positive integral solutions of 2xy—3x’+ y=1. _ 3x2+1 3x a 7 _.2_x;_i =—2--1+m), by diViSion; } are the solutions required. Here y 4y = 6x— 3 + a whole number. 7 2x+1’ Hence, 7 being a prime number, 2x+1 can be no other number but 7, or 1' , 2x=6, or x=3, and y=4; or 2x=0, i.e. x=0, and y=1; which are the only solutions in positive integers. [Exercises Zm.] 366. In the solution of different kinds of unlimited problems different expedients must'be made use of, which expedients, and their application, are chiefly to be learned by practice. Ex. 1. To find a “ perfect number”, that is, one which is equal to the sum of all the numbers which divide it without remainder. Suppose y"x to be a “ perfect number”; its divisors are 1, y, y2...y", x, xy, infamy/"'1; y"x=1+y+y2...... +y"+x+xy+xy2...... +xy""1. n+1—1 l+y+y2.....l I, y—l n —1 and x+xy+xy2......+xy”"‘=::/ 1xx; . .,___r+1-1+(r—1)><(2”+1—1) is a “perfect number”. Thus, if n=l, the 16 242 - SCALES or NOTAITON. number is 2x3 or 6, which is equal to 1+2 + 3, the sum of its divil sors: If n=2, the number is 22x(23—1)=4¢x7=28. Ex. 2. To find two square numbers, whose sum is a square. Let are and y2 be the two square‘numbers; assume 222+ y2= (nw - y)2= ngwg— Qnwy + ye, then w2=n2w2— Qnxy, w=n2x—2ny; hence (n2—l).r=°)ny, Qny 01‘ £17= 2 . 92—1 And if n and y be assumed at pleasure, such a value of w is obtained, that wU-y‘3 is a square number. But if it be required to find integers of this description, let y=n2—1, then m=2n, and n being taken at pleasure, integral yalues of a? and y, and consequently of a)“ and 312, will be found. Thus, if n=2, then y=3, and 00:4, and the two squares are 9 and 16, whose sum is 25, a square number. ~ Ex. 3. To find two square numbers, whose difference is a square. Let :02 and ye be the two squares; assume avg—f: (w - rag/)2, = we— Qna'y + ngy‘i. Then y2=2nwy-ngy2, or 2nm=(n2+1)y; n2+1 0.; (U= .2. 2n “I And if y=2n, then m=n2+l. Thus, if n=2, then y=4.~, and .z'=5; hence we-y2=25—16=9. SCALES OF NOTATION. 367. To explain the dzf'erent systems of notation. DEF. In the common system of notation each figure of any number’f‘ increases its value in a tenfold proportion in proceeding from right to left. Thus 3256 may be expressed by 6+50+200+3000, or 6+5><10+2x102+3><103. 1 4' In this Section and in the following one by number a whole number is always meant. . 2 .. SCALES OF NOTATION. 243 The figures 3, 2, 5, 6, by which the number is formed, are called its digits, and the number 10, according to whose powers their values proceed, is called the radix of the scale. It is purely conventional that 10 should be the radix; and therefore there may be any number of different scales, each of which has its own radix. When the radix is 2, the scale is called Binary; when 3, T ernary ,' when 10, Denary, or Decimal; when 12, Duodenary, or Duodecimal; &c. H 3256 expressed a number in a scale whose radix is 7, that number might be expressed thus, 6+5><7+2><79+3><73. And generally, if the digits of a number he a0, a“ a2, as, &c., reckoning from right to left, and the radix r, the number will be properly repre- sented by ac+ alr + a2r2+ a3r3+ &c. ‘ Or, if there be n digits, the number will be (reversing the order of the terms) a,,_1 r"“1+ an_2r”-2+ a,,_3r”‘3+ . . . + a,r +ao. OBs. In any scale of notation every digit is necessarily less than r, and the number of them, including 0, is equal to r. Also in any number the highest power of r is less by 1 than the number of digits. 368. To express a given number in any proposed scale. Let N be the number, and r the radix of the proposed scale. Then if a0, a,, a,, &c. be the unknown digits, N = ao+alr + a2r2+ a3r3+ &c. ; and if N be divided by r, the remainder is do. If the quotient be again divided by r, the remainder is a,. If . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . is a,; and so on, until there is no further quotient. Therefore all the digits a0, a,, a2, as, &c. are found by these repeated divisions ; and consequently the number in the proposed scale. ,_ , Ex. To express 1820, written according to the denary scale, in a scale whose radix is 6. 6 1820 6 303, 2 1st rem’. (10:2, 6 50, 3 2d rem’. a,==3, 6 8: 2 3d rem’. (12:2, 6 1, 2 4th rem“. a,=2, 0, 1 5th rem’. a,=1; the number required is 12232. 16_2 244 SOALES OF NOTATION. This may be easily verified. Thus if the result be correct, 2+ 3><6+2X62+2><63+1X6l must amount to 1820 ; which upon trial it is found to do. By the same method a number may be transformed from any given scale to any other of which the radix is given. It is only necessary to bear in mind throughout the process that the radix is not 10, as'usual, but some other number. Or the same thing may be done by first expressing the given number in the denary scale, and then proceeding as in the last example. Ex. 1. Transform 12232 from a scale whose radix is 6 to a scale Whose radix is 4. - (Observe, 1st, that in proceeding below to find how often 4 is con- tained in 12, 12 does not mean twelve, but 1x6+2, or 8. So also 23 is fifteen, 32 is twenty, and so on.) 4 12232 4 i 2035, 0 1st rem”. a,=0, 4 305, 3 2d rem’. a,=3, 4 44, 1 3d rem”. a,=1, 4 11, 0 4th rem’. a3= 0, 4 1, 3 5th rem’. a4: 3, O, 1 6th rem’. a5=1, the number required is 130130. This number expressed in the denary scale is 1x45+3><44+0x43+1x42+3><4+0, or 1820, which proves the result correct according to the preceding example. iEx. 2. Transform 3256 from a scale whose radix is 7 to a scale whose radix is twelve. Bearing in mind that the digits in 3256 increase from right to left in a sevenfold proportion, the division by twelve will be performed thus, twelve 3256 twelve 166, 4 lst rem”. ao=4., twelve 11, 1 2d remr, a,=1, 0, 8 3d rem’. a,=8; the number required is. 814. .SCALES OF NOTATION. 245 369. The most useful scale of notation, after the common denary scale, is the' one which has twelve for its radix, called Duodecimal. Here it is necessary to have two new symbols, in addition to 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, for the purpose of expressing ten and eleven; since 10, and 11, in the duodenary scale, signify twelve and thirteen respectively. These new symbols may be any thing distinct from the other digits, and are usually two letters, t for ten, and e for eleven. The common pence-table will be found of considerable service in the use of the duodecimal notation. Ex. 1. Multiply 25 ft. 7in. by 7ft. 10in. The lengths are here expressed in the denary scale. In the duodenary the question is, what is the product of 21'7 by 7 't? 21"? 7'1t 193$ 12e1 148'4l the product is 148ft. 4’. t” in the duodenary notation, or 8+4><12+1><122ft. 4'.10” in the denary; that is, 200ft. 4’. 10” . . . . . . Ex. 2. A floor in the form of a rectangular parallelogram contains 1532 square feet, 9’.9”, and is 81ft. 9’ long; required the breadth. The whole area-:-the length-=the breadth; and in the duodenary scale the given quantities are t7 8'99 and 699 respectively. 699 ) t7 8'99 Q16'9ft. in the duodenary scale, 669 39e9 34156 5139 5139 * * in the common scale the breadth of the floor is 18ft. gin. [Exercises Zn.] 370. To find the greatest and least numbers with a given number of digits in any proposed scale. ’ Let r be the radix of the scale, and n the number of digits; then it is evident that the number will be greatest, when every digit is as great as it can be, that is, r—l ,- in which case the number will be r—1+(r-l)r+(r—l)r2+ .... ..+(r—1)r"", or (r—1)(1+r+r2+ .... ..+r"-‘); 246 SCALES 0F NOTATION. r"-1 r-l’ _ Again, the number will be least, when the extreme digit to the left 1s 1, and every other is 0 ; in which case it will be equal to r"“. Ex. Thus in the common, or denary, scale, the greatest number having 4 digits is 104-1, or 9999; and the least is 10", or 1000. or (r—l) or r"—-1. 371. Tojind the number of digits in the product or quotient of two given numbers. I. Let P and Q represent the numbers, having p and q digits respectively; and r the radix of the scale; then P=a0+a1r+ .... ..+a,,_,rP-‘, Q=b0+b1r+ .... ..+b,,_,rq“; PQ=a series of terms in which the highest power of r is found in arIbHrM-z, the greatest value of which is (r—l)2.r1’+q"2, and the least rm“. Therefore the highest power of r in PQ is less than r2. r1394, or rm, and the lowest power is rm”. Hence the number of digits in PQ cannot be greater than p+q, nor less than p+q-1. II. To find the number in P-r-Q, when P is exactly divisible by Q; let P+Q=R, then P=QR, and QR contains p digits. Suppose R to contain x digits; then, by former case, p cannot be greater than 9+4", 1101' 1688 than q+x—1; x, the number required, cannot be less than p—q, nor greater than p—q+1. Con. Hence the number of digits in P” is either 2p or 2p—1; in Pa either 3p, or 3p—1, or 3p~2; and so on. Also, if JP have x digits, then JFX J13, or P, has either 2x, or 2x—1, that is, p==2x, or 2x~1, and x+ép, or -é-(p+1), Whichever is integral. Again, if Z/F have x digits, then P has 3.1:, or 3x—1, 0r 3x—2, that is, p=3x, or 3x—-1, or 3x—2, and x=-,l;p, or ;.l,-(p+1), or §(p+2). 372. In every system of notation, of which the radix is r, the sum of the digits of any number dividegt~ by '13 will leave the same remainder as the whole number divided by r—l. Let a0+a1r+a2r2+a3r3+ .... ..+a,,r” be the number (N), then N=ao+a,+a2+...+a,,+a,(r—1)+a2(r2—1)+ .... ..+a,,(r"-1); N a +a +...+a r"—1 o o o I In+ano r 1 O ' ' r—l r—l But rn—l r— be; (Art. 99, Ex. 6) ; _ N ==PLa,I +a,+--.+a,, C. | is a whole number, whatever positive whole number u may , P being a whole number, SCALES 0F NOTATION. 247 or the number divided by 17:1 leaves the same remainder as the sum of the digits divided by Fl—l. COR. Hence, if the sum of the digitspf any number he divisible by 1?:1, the number itself is divisible by r—l. Similarly, it may be shewn that, if the difference between the sum of the digits in the odd places and the sum of the digits in the even- places be divisible by It, the number is divisible by r_+_1. PROB. To find what numbers are divisible by 3 and 9 without remainders. Let a, b, c, d, &c. be the digits, 01" figures in the units”, tens", hundreds’, thousands’, &c. place of any number, then the number is a+10b+1000+1000d+&c. this divided by 3 is a b c d —+3b+—+330+-.—+333d+— +&c. 3 s 3 3 a+b+c+d+&a + 3b + 330 '+ seed + &c. a+b+c+&d which is a whole number when is a whole number; that is, any number is a multiple of 3 if the sum Of its digits be a multiple of 3. Thus 111, 252, 7851, &c. are multiples of 3. In the same manner, any number is a multiple of 9 if the sum of its digits be a multiple of 9. a +10b+1000+1000d +&c. 9 For b d =E+b+~+uc+ g+111d+—+&c. 9 9 9 a +b+c+d+&a = 9 +b+llc+111d+&c. . . b . . which 1s a whole number when a+ + 09+d+ &C 1s a whole number. Thus 684, 6588, &c. are multiples of 9. 248 SCALES 0F NOTATION. COR. 1. Hence, if any number, and the sum of its digits be respectively divided by 9, the remainders are equal. COR. 2. From this property of 9 may be deduced a rule which will sometimes detect an error in the multiplication of two numbers, called casting out the Nines. Let the multiplicand be 9a+x, that is, let it consist of a nines, with a remainder x; and let 9b+y be the multiplier; then 81ab + 9bx + 9ay +xy is the product; and if the sum Of the digits in the multiplicand be divided by 9, the remainder is x; if the sum of the digits in the multiplier be divided by 9, the remainder is y; and if the sum of the digits in the product be divided by 9, the remainder is the same as when the sum of the digits in xy is divided by 9, if there be no mistake in the operation*. 373. Thus far we have treated of whole numbers only; but fractions also may be expressed in different scales of notation. Thus 23°21, radix 10, signifies 2x10+3+—-2—+-L,, . 10 10 d' 6 2 6 2 l , ,,,, ,,, r3, 1x , . . . . . . . .. X +3+6+-6—,, , 2 I , ,,,,, ,_ radlx 4, ....... .. 2X4+3+Z+Zp ...... .. radix r, 2xr+3+g+ r r And generally, if there be u digits 'after the point which separates the integral from the fractional part of a number, the fractional part will be expressed by + + + +a'" r O O 0 l I O I Tn ’ or a_,r"l +a_,r-2+ a._,r’3+ . . . . . . . + a_,,r*". It is to be understood that as any vulgar fraction in the common scale may be converted into a decimal fraction, the same may be done in any other scale, and in the same way, bearing in mind only the difference of radix. * It will be easily seen that this method fails to detect an error in any of the following cases :--(I) when one or more cyphers have been omitted in the product; (2) when any of its digits are misplaced ,- and (3) when the error is equal to 9 or any multiple of 9. En. SCALES 0F NOTATION. 249 374- To transform a fraction from one scale to another.' Let N be the given fraction, r the radix of the new scale, and a__I , a_,, a.,, &c. the unknown digits; then N = a_,r"1 + a_,r_2 + a_,r‘3+ &c. rN= a_,+a_,r"‘+ a_,r'2+&c., the integral part of which is a_., the first digit required, leaving the fractional part a_,,r“1 + a_,r“’ + &c ; multiply this by r, and the result is a_,+a_3r“1+&c., the integral part of which is a_3, the 2nd digit required; and so on, continually multi- plying by r, and separating the integral part for a new digit after each operation. If the proposed fraction be not a proper fraction, the integral part must be dealt with separately according to the former rules for whole numbers. Ex. 1. Transform g from the denary to the ternary scale. 4: 12 . . 3x-9-=—9—-=1-13—, 3x-,l,-=1, the No. required 1s 0'11. Ex. 2. Transform 43"] or 43,76 from the denary to the senary scale. 6 43 43:11 1 in the senary scale. 7 42 6 6 7, 1 And 'i—OX &c. 6 1’ 1 the fractional part is '41111, &c., 0, 2 and the No. required is 111'411] .... .. Ex. Transform 23%- from the senary to the septenary scale. Here 23};=2x6+3+-f,-=15% in the denary scale; 715 7 21 7 Q, 1 TiX7=E=ITEU %X7=—4-=5715, &c. 0, 2 the No. required is 211515 .... .. Or thus, without introducing the denary scale at all, but bearing in mind throughout that the radix is 6, 2371:2313, 7 23 '13 7 2,1 __7_ 0, 2 1'43 7 $15 No. required is 2l'1515...... .250 PROPERTIES OF NUMBERS. Ex. 4. Transform 4.56 '16 from the duodenary to the ternary scale. 3 . 456 0-16 3 Tbs—,3 e 3 56,1 5Z3 3 1e, 2 3 , a 7, 2 I713 3 2, 1 _ . A 0, 2 No. required is 2122100101 .... .. [Exercises Zo.'_'| PROPERTIES OF NUMBERS. 375. The product of any trvo consecutive numbers is divisible by 1x2. Of the two numbers one must evidently be even, that is, divisible by v2, therefore their product is divisible by 2. 376. The product of any three consecutive numbers is divisible by 1x2x3, 0r 6. Every number must be either of the form 3m, or 3mil, or 311242, since it must be either divisible by 3 without remainder, or have a remainder 1 or 2,- therefore the product of any three consecutive numbers may be represented by one of the forms 3m(3m+1)(3m;l-2), (3m—1)3m(3m+1), —1)3m. Now, since by last Art. both (3m-r-1)(3m+2) and (3m-2)(3m—1) are divisible by 1x2, and 3m is a multiple of 3, therefore the first and third forms are divisible by 1x2\.><3. Also, if m be an even number, that is, divisible by 2, it is clear that 3m, and therefore the second form is divisible by 1>£2x3 ; and, if m be an .Odd number, 3m is odd, and 3m+1 even; consequently the second form is divisible by 1x2x3. Hence, in all cases, the product of three consecutive numbers is divisible by 1x2x3. 377. The continued product cy" any 1' consecutive numbers is divisible by 1.2.3.. .r. [This has been already shewn indirectly in Art. 316; but the follow- ing is a more independent proof] - PROPERTIES OF NUMBERS. 251 +r—l) 1 . 2 . 3 r h be represented by "P, for all values of n and 1'. Then nPr=n(n+1)...(n+r—2).n+p—1 =nPH.<_n;1+1 , 1 . 2 ....... ..(r-1) 1' r Let n be the least of the numbers, and let __(n—-l)n(n+1). ..(n+r—-2) P _ 1.2.3. . . . . . . ..r“ +“ "4’ or nPr=n-1Pr+nPr—1' Now assume that the product of any r—l consecutive integers is divisible by 1.2.3...(r—1), that is, suppose ,,P,._l is an integer, then‘ ,P,=,,_,P,.+ an Integer, for all values of n and r, write, n—l for n, ,_,P,=,,_,P,+ an Integer, .... .. n-2 ,_,P,=,,_,P,+ an Integer, .......... .. &c. = &c. ....... .. 3P,=,P,+ an Integer, ....... .. ,P,=,P,+ an Intege1'=1—2—3——:-+’an Integer, =1+ an Integer=Integer, adding and cancelling, _ nPr=the sum of Integers=an Integer; which proves that, if ,P,_, be an Integer, then also is ,P,. But we knew that “P, is an integer, (by the last Art.), therefore also is ,P,; and if 7,P,, therefore also ,P5; and so on generally for ,P,.; that is n(n +1)(n+2).....'. (n+r—1) is divisible by 1.2.3...r. Ex. 1. If n be any whole number, then will n(n2—1)(n’—4) be divi- sible by 120. n(n’—1)(n2-— 4) = n(n—1)(n +1)(n —2)(n +2) =(n—2)(n—-l)n(n+1)(n+2), which is the product of 5 Consecutive numbers, and is therefore divisible by 1.2.3.4.5, or 120. Ex. 2. If n be any even number, na-J- 2.0n is divisible by 48. Let n=2m, since it is an even number, then n3+20n=8m3+ 40m, =8m(m2+ 5), =8m(m2.-l)+48m, 8(m—1)m(m+1)+48m. _ _ New ‘(m—1}m(m +1), being the product of three consecutive numbers, 1s dIVISIbIe by 1.2.3 or 6; therefore n3+20n is divisible by .48. 252 PROPERTIES OF NUMBERS. 378. Every number which is a perfect square is of one of the forms 5m or 5m_=1=1. v. For every number is of one of the forms 5m, 5m+1, 5m+2, 5m.+3, 5m+4; all Of which are included in the forms 5m, 5mil, 5m*2, Slnce 5m+3=5(m+1)-2=5m'—2, and 5m+4=5(m+1)-1=5m’—1. But (5m)2=5(5m2)=5m', which is of the form 5m; (5m*1)’=25m2i10m+]=5(5m’i2m)+1, which is of the form 5m+1; (5mi2)2=25m"’i20m+4=5(5m’i4m+1)—1, which is of the form 5m-1; every square is of one of the forms 5m, 5m+1, 5m—l. 379. Every “prime number” greater than 2 is of one of the forms 4mil. For every number is of one of the forms 4m, 4m +1, 4m+2, 4m+3; but neither 4m, nor 4m+2, can represent prime numbers, since each is divisible by 2; therefore all prime numbers greater than 2 are represented by 4m+1 and 4m+3. But 4m+3=4(m+1)—1=4m’—1; therefore the two forms for prime numbers are 4m='=1. COR. Since m may be odd or even, that is, of the form 2n +1, or 222, all prime numbers are represented by 8n=bl, or 8n=*=3. 380. Every prime number greater than 3 is of one of the forms 6m41. For every number is of one of the forms 6m, 6m+1, 6m+2, 6m+3, 6m+4, 6m+ 5, of which the 1st, 3d, 4th, and 5th obviously cannot re- present prime numbers; and therefore, all prime numbers greater than 3 are represented by 6m+1 and 6m+5. But 6m+5=6(m+1)—1=6m’-1. Therefore, 6m=i=1 will include all prime numbers greater than 3. Con. Since m may be odd or even, that is, of the form 2n+1, or 2n, all prime numbers greater than 3, will be included in l2nil, or ‘12ni5. [Exercises Zp.:| 381. N0 Algebraical formula can represent prime numbers only. Let p+qx+rx2+&c. be a general algebraical formula; and let it be a prime number when x=m; therefore (P) the prime number in that case 1s p + qm+ rm2+ &c. Now let x=m+nP; then p+q(m+nP)+r(m+nP)2+&c. is the number; and is equal to p +qm +rm2+ &c.+MP, (M signifying “some multiple of,”)=P+PM, which is divisible by P, .and therefore not a prime; consequently the formula does not represent prime numbers only. i PROPERTIES OF NUMBERS. 253 382. The number of primes is indefinitely great. For, if not, let there be a fixed number of them, and let p be the greatest: then 1.2.3.5.7.11 . . . . ..p is divisible by each of them, and 1.2.3.5.7.11 . . . . . .p +1 . . . . . . . . . . . .not one of them. If this latter number, then, be divisible by a prime number, it must be one greater than p; if not, it is itself a prime, (since every number is either a prime, or capable of being resolved into factors which are prime) and is greater than p. Therefore, in either case, there is a prime greater than p; that is, we may not assume any prime to be the greatest; or, the number of primes is indefinitely great. 383. T 0 determine whether a proposed number be a prime or not. It is obvious that this may be done by dividing the proposed number by every number less than itself, beginning with 2, until we have either proved it to be divisible by some one of them without remainder, or that it is a prime, from not being divisible by any one of them. But there is no necessity to proceed so far, as may thus be shewn. If the proposed num- ber (p) be not a prime, then p=ab, the product of two other numbers. If then a>~/p, bJp. Hence in both cases p is divisible by a number less than the square root of itself. Or it may be that the proposed number is an exact square, in which case it is di- visible by its square root. If, therefore, a proposed number be not divisible by some number not greater than the square root of itself, it must be a prime. 384. To find the number of divisors of a given number. Let a, b, c, &c. represent the prime factors of which the given num- ber is composed; and let a be repeated p times; b, q times; 0, r times; &c.; so that the number=aPb’c’, &c.; then it is evident that it is divisible by each of the quantities ‘ 1, a, a”, a3, . . . . .. aP, p+1 in number. 1, b, b”, b“, . . . . .. bq, q+l l, c, 0’, ca, c’, r+1 &c. &c. &c. and also by the product of any two or more of them, that is, by every term of the continued product of (1+ a+a2+. . .+aP)(1-i-b+bg+. . .+bq)(1+c+c"’+ . . .+c’).&c.; for (l+a+a’+. . .+aP)(1+b+b2+. . .+bq)=1+a+a’+ . . . . . .+aP +b+ab+ . . . . .. +a1’b +b2+ab2+...... +a1’b +-......-............ +b1+ab9+ . . . . . . +al’bg, which are all “the different divisors (including 1) of aPb’. 254 PROPERTIES OF NUMBERS.‘ Similarly, if this result be multiplied by (1 +c+ 09+. . .+c'), the ‘pro- duct will consist of all the different divisors of aPbqc’; and so on, if there be more factors of the given number. Now the number of these divisors is Obviously p+1 in aP; in aI’bq the number is p+1 taken as many times as there are terms in the second series, thatis, (p +1)(q+1); in apbic" it is (p+1)(q+1) taken r+1 times, or (p+i)(q+1)(r+1); and generally the number of divisors in aPb’c'.&c.=(p+1)(q+1)(r+1).&c. including 1 and the number itself. Ex. Find the number of divisors of 2160. Here 2160=2 ><1080=22 X 54O=23 >< 270:24 X 135=24X 3 X45=24X 32 X15 =2“ 2332' 5. Therefore the number of divisors =(4+1)(3+1)(1+1) or 40. OOR. 1. The number of divisors will always be even unless each of the quantities p, q, r, &c. be even, that is, unless the proposed number be a Perfect Square. COR. 2. It is also obviOus from what has been said above that the sum of all the different divisOrs of aPb'qc’.&c. is the sum of all the terms in the continued product of (1+a+a2+...+ap)(l+b+b”+...+bq)(1+c+c“ +...+c’) &c. aim—1 bin—1 c'+'—l r . I Q 0 a-1 b—1 c—l _ 385. If m be any prime number, and N a number not divisible by m, then Nm‘I—l is divisible by m. (FERMAT’S THEOREM). For, it is easily seen by the Binomial Theorem, that in the expansion of (a+b+c+&c.)"‘ m is a factor of every term except a’", b’”, c’", &c.; and that the coefficients are always whole numbers. But m, being a prime number, will not be divisible by any factor in the denominator of ,a coefficient, and will therefore remain as a factor of each term, when the coefficients are reduced. Therefore we may assume, (m being a prime number), (a+ b+ c+&c.)’"=a’”+ b’"+ cm+ &c.+ mP. h Let, then, a=b=c'=&c.=1, and the number of them' be N; and we ave N’i‘=N+mP, ’"—N=mP, ’ _or lV(N”"—1-1)=mP. But, 'by supposition, N is not divisible by m, Nm‘¥=-1 is a multiple of 172, Or is divisible by m. PROPERTIES OF NUMBERS. 255: 386. T o prove that, for any positive integral value of n, n"- n(n-1)"+ 7;(7l;1)(n—2)"-&c.=1.2.3 . . .n. By the Binomial Theorem, (s-1)"=e"-ns—r+ s-r- &c.. . .. . . .(1). Also by the Exponential Theorem, r n $2 n mg n (e -1) -(1+x+-L—2 +.. .-1) -(w+-1-_-2-+--~)s =x"+Ax"+1+Bx"+2+. . . by the' Binomial Théor. . . . . (2 15 O I. 0 72 Now, the'coefficient of x" m 6"” 1s ' 1.2..'.n ’ e(fl—l)2 . (n _1 ).n 00000000000000000000000' "1.2...7z) . I O I Q I O O O O 000 one. 1.2..n) and equating coefficients of x" in (l) and (2), , n" n(n-1)n 201—1) ("-2)n_&c _1 1.2...2 1.2...n 1 . 2 ' 1.2...n ' ’ n(n —1 and n"--713(n-1)”+ 1 2 )(n—2)”——&c.=1-.2.3...n. COR. Let p be a prime number, and p-1‘=n, then __ - -1 -2 1.2.3...p—1=(p—1)P‘1--PTl(p—2Y‘1+W(p—3)P1-&c. = .p+1-’-’-]—’~&c» by .,_ _, Fermat’s Théo’rem; =p{M,-*(p=-1)M,+ (P--—:ll—X-'P—2>'~'—-)M,—&c.} +1 P-I _,_ (P-1XP_2)-&c. to pTl- terms, 1 1 . 2 §pQ=l-(1=*1)P"1-‘—‘1 ;' 1.2.3...p-1+1=19Q, or is divisible by p; which is Wilsqn’s Theorem for determining whether any proposed number be a prime or not. 256 . VANISHING FRACTIONS. VANISHING FRACTION S. 387. To find the value of a fraction when the numerator and denominator are evanescent. Since the value of a fraction depends, not upon the absolute, but the relative, magnitude of the numerator and denominator, if in their evanescent state they have a finite ratio, the value of the fraction will be finite. To determine this value, substitute for the variable quantity its magnitude, when the numerator and denomi- nator vanish, increased by another variable quantity; then, after reduction, suppose this latter to decrease without limit, and the value of the proposed fraction will be known. ail-a2 Ex. 1. Required the value of , when x=a. 617—0; Let x=a+h, and the fraction becomes a2+ 2ah + h'Z— a2 2ah + h2 a+h-a _ h = 2a + h, and when h=0, or x= a, its value is 2a. 1 — (n + 1)x"+ nx”+1 (be; Let x = 1 +h, and the fraction becomes 1 - (n +])(1 +h)"+ n(1 + h)"‘H h2 i ’ Ex. 2. Required the value of , when x=1. n-1 n-1 n—2 , 1 - (n+])(1+nh+ n.—-2—-h2+n.——2——.—-3-—h3+ &c.) h2 —1 n{1+ (n+1)h+(n+1)%h"+ (n +1)g-.%-h3+ &c.} 'l’ hr ’ n +1 n 72 4?, -' 1 o . 0 - n. + ( ) ( ) h + &c. the remaining terms being 2 3 multiples of h; \ , n + l and when h= O, or x =1, the fraction becomes n . VANISHING FRACTIONS. 1257 388. It is clear that every fraction, whose terms are made evanescent for a particular value (a) of some quantity (:2) contained in them, is capable . m o v o of being reduced to the form 296—25,;- ; and, upon d1V1d1ng the numerator and denominator by their greatest common factor, the fraction will no longer have both its terms evanescent, when w=a. Hence by whatever process this common factor can be discovered, and divided out, the value of the fraction will be found. A simple algebraical reduction is frequently sufficient for this pur- pose, as will be shewn in the following Examples. __ 3 Ex. 1. Required the value of }—(31—x%-7?-, when x=l. _ 2 8 __ 2__ 2 _ Here 1 3.2: +3.1: :1 a: 2x (21 .21)‘1 o-w) o-o _ l+ar—2a:2 __ 1-x2+x(1-x) _ l-a: _ l—a: ’ =1+m+x, =3, when x=1. x—a + Qax—Qa’ Ex. 2. Required the value of , when x=a. ,Jaf—az Here .r—a+ 2aw—2a2_x—a+./2a.~/x—a Jwg—ag Jm../x_a ’ =,./.z'-a+ 2a .512 ’ =~/_2_‘f=1, when x=a. ,Jfla Ex. 3. Required the value of W, when k=0. (“mi-$3 (w+h)T*-afi Here k _ (a:+lz)-x ’ 1-( ) ’i'-1 x+b =(J;+h)n .I—I—T , —x+h 17 “.258 . INFINITE SERIES. 1 321 1-2’" .Z’ " = in .__.__ 'f‘ (m = (x+) 1—2”’ 1 x+b.) z . ._ (Peso-2) ‘ ‘.<‘”+") 'o-rf-JZYIEY 3‘4 1+ z+z’+23+... z”°"1 v 1 +2+22+z3+ z""’ ‘ ~ 3-1 1+1+1 +...to m terms a =wn a ’ 1 ‘ 1+1+1+...to n terms ~_~ ..- 772 2‘71 Eur" , when 12:0. [Exercises Zq.] INFINITE SERIES. 389. DEF. An iwgfinite series is a series of terms proceeding accord- ing to some law, and continued without limit. Thus the series treated of in Art. 290, Cor. 2, is an “ infinite series”. DEF. The sum of an infinite series is the limit to which we approach more nearly by adding more terms, (as in Art. 290, Cor. 2,) but cannot be exceeded by adding any number of terms whatever. DEF. A.converg‘ent series is one which has a sum or limit, as here defined. A divergent series is one which has no such sum or limit. Hence every infinite series in Geometrical progression, in which the common ratio is less than 1, is convergent. 390. To determine generally in what cases a series is convergent or divergent. _ I. Let a,+a,+a,+a,+ .... .. be the series, in which all the terms are posmve. Then the sum is equal to a a a a,{1+-5 + i +~5+ .... ..}, a1 a: a1 a a a a a a or a,{1+—3+—§.—5+-—4.~3.-3+ .... i v o (I “a a . And If each of the quantities 3, 3, J- &c. be less than some quantity ’ a1 ‘72 as, - . p, the whole series is less than a,{1+p +pi+pa+ .... INFINITE SERIES. 259 But if p be less than 1, the sum or limit of this latter series is al ; therefore the proposed series also has a sum or limit less than this quan- tity. Hence an infinite series of positive terms is always convergent, the ratio of each term to the preceding term is less than some assignable quantity which is itself less than 1. II. Next, let al—a2+a,-a4+ .... .. be the series, in which the terms are alternately positive and negative, and go on decreasing without limit. Then, since the series may be written in the two following ways, a1—a2-i-a3—a4-l-a5—(l6'l- u u s o 00 al—a2_ a5_ a . . o o a a on and since a,—a,, as—as, &c., a2—a3, a4—a5, &c., are severally positive, it is evident that the sum or limit of the series is greater than al—ag and less than 11,, that is, the series has a sum or limit; consequently it is convergent :-—or, every series, in which the 'terms continually decrease and are alternately positive and negative, is convergent. 5%, &c. be less than p, the Con. If each of the quantities 9, g, ’ a b c . . . 1 series a+bx+cx”+d.v3... 1s convergent whenever a: Is less than And &c. then the series QIRI if p be less than each of the quantities %, %, 7 . . 1 a-i-bx+c.v2+ .. . 1s divergent for every value of a: greater than 1 1 1 1 . o _ _ _ +... bea EX 1 To determine Whether 1+ 1+1.2 +1.2_3+ 1.2.3.4, convergent series. a2 a3 1 a, Here —=1, —=— , — a1 a2 2 a3 0 I O i 1 . O I each of whlch ratlos, after the second, 1s less than 5, which Is Itself less than 1. Therefore the series is convergent. . l 1 1 1 Ex. 2. To determine whether 1+—2-+§+-4-+ . . . . .. be a convergent or divergent series. a2 1 a, 2 or4 3 ---—:,-—- -—-=_ _=_ &c. Here a1 2, a2 3’ a3 4: and, as these ratios continually increase, and approach towards i, no quantity can be named which is greater than each of them, and whlch 1s Itself less than 1. Hence the proposed series is divergent. ' ' 17-_2 260 INFINITE SERIES. 391. Another method of determining the convergency or divergency of series is to find the limit of the sum of the series after the first n terms; which also determines the limits of the error arising from taking any number of terms instead of the whole series. Thus, 1 l 1 . . . ' 1 — -~ —— h m fthe series Ex 1 In the serles +1+L2+L2.3+ t e su o afternterms 1 l 1 =-+ - . - - . . . . - . - . . .- Ln n+1 [3+2 1 1 l =-. l+-——+-——-—-+...... n { n+1 (n+l)(n+2) }’ 1 l l l < a o u u n}, o 1 1 <'-7Zo_1, 1.__ n < l 1 or < l l lli—l'n—l’ 1.2.s...(n-I)'n-1' But this quantity decreases as n increases; and, by increasing n without limit, it may be made less than any assignable quantity. Therefore the series is convergent. And if n terms be taken for the whole series, the error is less than 1 l 1.2.s...(n-I)' n—l' . 1 . Ex. 2. In the series 1—-%+%—z+ .... .. the sum of the series after in terms a 1 1 1 1 +1) '{m'm+(;;§5n+i)+ "" _ ' n+1 n+2 n+3 n+4 n+5 ’ and, since the quantities within the small brackets are all positive, this sum ,, 1 1 ' ><-1) '{Zli'iT-Tei’ ..l. and <(--1)“. n+1 , both which quantities are diminished without limit as n is increased. INFINITE SERIES. 261 Hence the series is convergent; and if n terms be taken for the whole series, the error is less than the n+1th term. ' . 1 1 . Ex. 3. In the serles l+é+g+z+mm the sum of the series after it terms - 1 + I + 1 + —n+1 n+2 n+3 ——-1—-+-~——1 + 1 + + I _n+1 n+2 n+3 "" " 2n 1 I 1 1 +2n+1 + 2n+2 + 2n+3+°"+4_n I O O I 0 i 0 0 I O O 0 0 0 00. I O O 0 l I 0 O .0. I I I I Q Q I 00 > 1-+ —1—+&c. to n terms +i+ 1- +&c. to flu terms 2n 2n 4n 4n 1 l 1 > "0% +'2n.E;z +47Z-é-7-i-Fun..." > 1 + 1 +l+ 2 2 2 c Q 0 0 0 I 0 0 0 0 0 0000000000000... > any assignable quantity. Hence the series is divergent. 392. In the series a,x+a,x2+a,x3+...... in inf. such a value may be given to x, that the value of the whole series shall be less than any pro- posed quantity p. Let h be the greatest of the coefficients a1, a,, a,, .... .., then the. whole series is less than hx+hx2+hx3+...... in inf. 1, the series is infinite; yet we know that it arises from the division of 1 by (1-a')2, and the sum of n terms may be determined.‘ The series after the first it terms becomes (n 1).n”+ (n + 2).v"+1+ (n + 3)a~"+2+ &c. in which the scale of relation, as before, is 2 -1; and therefore the series arises from the fraction (n +]).v”+(n + 2).v"+‘— 2(n + l)n" 'H 1 —2.v + m2 9 (n +1).v"- new“ (1-"4'19" ’ 1+2a’ + 3.v2+ &c. to n terms = 01' 1 — (n +1).v"+na""H (1 —a;')2 _ Con. If the sign of a- be changed, 1-2x+3.v2— &c. to n terms 1:1: (n +1)m°’¥ nut"+1 " (1+wr 2 where the upper or lower Sign is to be used, according as n is an even or odd number. Ex. 3. To find the sum ofn terms of the series * 1 + 3m + 5a2+ 7w3+ &c. Suppose f+ g to be the scale of relation; then 3f+g=5, and §f+3g=7; hence f=2, and g=-1; 264 ' RECURRING SERIES. and, by trial, it appears that the scale of relation is properly determined; ‘ 1 -|- 3m - 2w 1 + a: hence S = = , . 1—2a' +.v2 (l-a'y After it terms, the series becomes (2n + 1).v"+ (2n + 3).v"+1+ (2n + 5).v"+2+ &c. which arises from the fraction (2n + 1).n”+(2n + 8).v”+1— 2 (2n + 1).v""'l _ 1-2w+a'2 ’ (2n+1).v"- (2n —1).v”+‘_ (l-w)2 ’ hence I + 3n + 5a2+ 7w3+ &c. to n terms, 1 +x - (2n +1).v"+ (2n —1).v"+1 " (1-wr ' Ex. 4. To find the sum of 1+2m+3m2+ 5w3+ 8w4+ &c. in inf. when the series converges. In this case the scale of relation is 1+ 1, and consequently the sum 1s 1 + 2n: — .v I + m 1—.v-.v2 = l—w-aii' If .22 becomes negative, l-ar 1-200+3.v2- 5w3+ &c. in = ——————~. 1+.v--.v2 Ex. 5. To find the sum of n terms of the series (n -1)a' + (n - 2).v2+ (n — 3).v3+ &c. The scale of relation is 2—1; therefore the sum in inf. is (n -1).v+ (n— 2).v2— Q (n --l).v2 (n -1)sc “nae2 _. 2 ’ °“ ——_—2—' (1 a') (1 a) After it terms, the series becomes -.v"+1-2.v"+2- &c. the sum of _, n+1 which is found in the same manner to be M; RECURRING SERIES. 265 (n -l).v + (n — 2)a'2+:(n - 3)a*3+ &c. to n terms (n —1)a' - nm2+ an“ (1-wr (n —1).v +J(n - 2).v2 CoR. Hence +&c. to n terms (n - l)n — nm2+ x”+1 — 72(1—w)2 Ex. 6. To find the sum of n terms of the series 12+ 22w +32.v'"’+ 42w3+ &c. Ilet the scale of relation be f + g + h; then 9f+ 4g + h =16, 16f+ 9g+ 4h =25, 25f+ 16g + 9h =36. From these equations we obtain f=3, g=-3, h=l, which values, when substituted, produce the successive terms of the pro- posed series; therefore 1+4m+9w2- 3.v—12.v2+.‘5.v2 1+.v _ , =_- = , the sum of the series in 1 _ 3(1) + 3t'b'2— LII/,3 —’ (0)3 when a' is less than 1. S After the first n terms the series becomes (n +1)2.v"+ (n + 2)2a'"+1+ (n + 3)9.v"+2+ &c. of which the sum is (n + 1)%"— (2922+ Qn— 1)a""+1+ 92%” +2 _ (1 - (’03 ’ and consequently the sum of n terms of the series is 1 + a' -(n +1)2a'"+ (2412+ 2n ._1)a,n+1_ n2mn+2 (1 — (0)3 On this subject the reader may consult De Moivre’s Misc. Analyt. p. 72; and Euler’s Analys. Infinit. C. XIII. 266 LOGARITHMS. LOGARITHMS. 394. DEF. If there be a series of magnitudes a0, a1, a", a3,...a’; a“, it”, a'3,...a'-’/, the indices, 0, 1, 2, 3,...a’; -—1, =2, —3,...-y, are called the measures of the ratios of those magnitudes to 1, or the Logarithms of the-magnitudes, for the reason assigned in Art. 230. Thus a', the Logarithm of any number n, is such a quantity, that a“=n. Here a may be assumed at pleasure, and is called the base; and for every different value so assumed a different system of logarithms will be formed. In the common Tabular logarithms a is 10, and consequently O, 1, 2, 3,...a', are the logarithms of 1, 10, 100, 1000, ".10: in that system. 395. COR. 1. Since the tabular logarithm of 10 is 1, the logarithm of a number between 1 and 10 is less than 1; and, in the same manner, the logarithm of a number between 10 and 100 is between 1 and 2; of a number between 100 and 1000 is between 2 and 8; &c. These logarithms are also real quantities, to which approxima- tion; sufficiently accurate for all practical purposes, may be made. Thus, if as be the logarithm of 5, then 10’=5; let :3: be sub- stituted for .v, and 10% is found to be less than 5, therefore #3- is less than the logarithm of 5; but 10% is greater than 5, or g is greater than the logarithm of 5; thus it appears that there is a value of as between % and %, such that 10”=5; the value set down in the Tables* is 069897, and 100‘69897= 5, nearly. 396. Con. 2. Since a°=1, b°=1, &c. in any system the logarithm of 1 is 0. Also since a‘=a, the logarithm of the base is always 1. The method of finding the logarithms of the natural numbers, or forming a Table*, is explained in Treatises on Trigonometry. ‘ DEF. If n be any number, logan signifies the logarithm of n to base a ; and logn the logarithm of n to any base. * Tables of Logarithms have been published in a very cheap and convenient form by Taylor and Walton, London, under the superintendence of the Society for the Diffu- sion of Useful Knowledge. ED. - - - i LOGARITHMS. 267 397. In the same system the sum of the logarithms of two numbers is the logarithm of their product; and the difl'erence of the logarithms is the logarithm qf their quotient. ' Let w=logan, and y=logan'; then a'”=n, and ay=n'; hence n 0 I - n at+y= nn', and a "t/= a, ; or .v+y 1s logann, and .v -y 1s log, 7?; I I I n I that Is, logann = logan + logan ; and log, a, = logan —- logan . Ex. 1. Log3_x—'I=10g3+10g7. Ex. Log'p;r=logpg+logr=bgp+logq+logr. 9 Ex. 3. Log§=log5=log 7. 3° Ex. 4.. Log,,0'6=log,oi%=10g106-10g1010=0'77815—-1. Ex. 5. L0g100'006=10g106—10g10103=0'77815—3. The last two results are usually written T'77815, 3'77 815. 398. If the logarithm of a number be multiplied by n, the product is the logarithm of that number raised to the nth power. Let N be the number whose logarithm is a', or a‘”=N; then a”“"’=N"; that is, run is the log. of N ”, or logaN”= n.10gaN. Exs. Log(13)5= 5><1'0767=5-3835=5£. 7s. 8d., the amount required. 404. When compound interest is named, it is usually meant that interest is payable only at the end of each year; but there may be cases in which the interest is due half-yearly, quarterly, &c. ; and then the amount found in the last Article will be altered. Thus, if r be the interest of Li? paid at the end of the year, it has been shewn that the amount of P£ at the end of n years=P(1+r)". 272 COMPOUND INTEREST. But if 22: be the interest of 1£ paid at the end of each half-year, then 1+-;-= the amount of 1.15 for half a year, 7.2 (1+§)=......... . . . . . . . . . ..1 year, 7.3 (1+-é): . . . . . ...............1%years, r4 (1+é)=.. . . . . . . ..Qyears; 0.7.0.2; O C t O O O O O 0 O I I I 0. (1+5) = . . . . . . . . . . . . . . . . . . ..n years; . 2n amount of P£=P<1+é~> . Similarly, if be the interest of 1£ paid at the end of each quarter, 4n amount of P£ in u years=P<1+2> . And, generally, if interest be considered due q times a year, at equal intervals, each payment for 1£ being 2 , "9 amount of P£ in n years=P(1+:1;) . Ex. Find the amount of 10043 in 1 year at 5 per cent. per annum, when the interest is due, and converted into principal, at the end of each half-year. Here P=100, r=0'05, n=1, Amount required=100><(1+0-025)==105-0625£, =105£. 1s. 3d. 405. Required the amount of a given sum at compound interest, the interest being supposed due every zustaut. The interest being paid q times per annum, by last Art., the amount “it M=P(1+%) , __ Z1 m1(m1—1) rj ~P{1+nq.q+—-—————-1 . 2 .q,+...}, 1 u(n-- -.=P{1+nr+-----2-1 2 r”+. . .}. Let g be indefinitely great, that is, the intervals between the payments indefinitely small, then, neglecting % and its powers, COMPOUND INTEREST. '273 an” nsr8 T5 +__1.2.3+"'}’ =Pe’", e being 2'7182818. (Art. 325.) M=P{l+nr+ 406. To determine the advantage, when compound interest is reckoned, of having interest paid half-yearly, quarterly, do. instead of yearly. It appears from Art. 404 that the advantage per 1£ for a year, when interest is paid half-yearly, and the halfl-yearly payment is half the yearly one, 7. 2 r9 =(1+-2->-~(1+r)=1+r+ Z— (1+ r), el‘i. In the case of quarterly payments of interest, the advantage =(1+-2>4— (1+r), 3r 1'3 r4 -—1+7'+ “5+ E 'i' 2—5'6--(1+T), =15 nearly, r is a small fraction. And generally, when the interest is paid q times a year, the advantage 7'7 =1+—>-1+T, (q < > _ . q(q--1) if _ -1+1+1 . 2 .q,+&c. (1+r), =2;1 .r9 nearly. Hence it appears, that, for a single year, the advantage of having interest paid frequently is very small. But it increases as the number of years increases, and is expressed in n years, for every 1.1-3, (when interest is paid q times a year at equal intervals, 5 being the payment per 1.50,) by (1+ gyq— (1+ r)". 407. It must be observed always, when interest is paid q times per . r annum, each payment belng - for every 159, that the true annual rate of . . r q . . . interest is not r, but (l+-> —l, smce this expresses the value of the inte- rest for 1£ for a year. 18 274 COMPOUND INTEREST. PROB. In What time will any'sum of money double itself, at any given rate of interest simple or compound? I. In the case of simple interest M=P+Pnr, here 2P=P+Pnr, I . or 72:; , the number of years required. II. In the case of compound interest paid yearly M =P(1+r)", here 2P=P(l+r)", or (1+r)"=2 ,- log 2 __1 , the number of years required. 0g 1+r The following Table, calculated from these two results, and shewing the several times in which any sum will double itself at the rates of interest there given, is taken from Baily’s Doctrine of Interest and Annuities, and will furnish good practice to the learner, who will verify it by means of the Logarithmic Tables. Rate No. of Years at of IntereSt' Simple Interest. Compound Interest. 2 50- 8500278878 2%- 40- 2807108458 8 2844977225 3% 2857142857 2014879168 4 25- 1707298769 4%; 2222222222 1574780184 5 20- 1420669908 6 1666666667 11 '89566105 7 1428571429 1024476835 8 12-5 900646884 9 11-11111111 804828178 10 10- 727254090 In general practice, compound interest is only reckoned for an integral number of years, so that if there be any fractional part of a year remaining, for this simple interest is taken. DISQOUNT. \ ' 27 5 DISCOUNT. 408. Discount is defined to be “the abatement made for the payment of money before it becomes due”, and although in ordinary business the quantity of such abatement is generally according to private contract, there is besides a true mathematical discount which affords exact justice both to the payer and to the receiver. It is clear that if A receives from B a sum of money n years before it is due, n being whole or fractional, A is benefited by the interest of that money for the time; and therefore, in justice, B ought to receive an abatement such, that the sum thus diminished, paid to A, would, if put out to interest until the proper time of payment arrives, amount to the sum due. This sum is called the Present Value of the debt. Hence if D be the discount, and V the present value, of a debt of £P due at the expiration of 22 years, V =P—D, and V would amount at the end of n years to P, i. e. V (l+nr) =P, reckoning simple interest. Pnr 1+nr' Hence V: P , and D=P——V= 1 nr COR. 1. If n be sufficiently great that compound interest may be reckoned, then n__ . _ P _ V(1+r)_-P, i.e. V~————-U+T)n, P and COR. 2. If I be the interest on £P for the given time, we shall have P+I=P(1+nr), or P(1+r)", according as simple 01' compound interest is reckoned; P O P+I’ P _ PI P+I _ P+I ’ _1_ ._ _1_ +1 D _P I ' From this result we see that the discount on any sum is always less than the interest. ‘ in both cases V =P D=P-P 01' COR. 3. Since D=P—V, if these be put out to interest for the time in question, we shall have amount of D=amount of P— amount Of V ==amount of P-P =interest on P. 1 8-—-2 276 EQUATION OF PAYMENTS. EQUATION OF PAYMENTS. 409. When various sums of money due at different times are to be paid, it may be required to know the time at which they may all_be paid together, without injury to either debtor or creditor. T 0 determine this time, which is called the Equated Time, it is clear that we must suppose the interest of the sums paid after they are due to be together equal to the discount of the sums paid before they are due, the debtor being entitled to discount for that which is paid before, and the creditor to interest for that which is paid after, it becomes due. 410. To find the equated time of payment of two sums due at diferent limes, reckoning simple interest. Let P, p, be two sums due at the end of times T, t, respectively; r the rate of interest, and .r the equated time; then supposmg T>t, the interest of p for the time ar—t must be equal to the discount of P for the time T --.r, or P(T— x)r P(x_t)7~1+(T—a)r whence we may obtain the quadratic equation x2_pr(T+i)+P+p x+prTt+PT+pt=0 1”“ ' P" from which a: may be found by the usual method. , (Arts. 401, 408-) 1 411. Since the above method does not furnish any simple Rule, and is more complicated as the number of payments is increased, another method is generally used, although incorrect, which is founded on the supposition that the interest of the sums paid after they are due should be equal to the interest (not the discount) of those which are paid before they are due—Thus, if P and p be the sums due at the end of times T and t, and w the equated time required, p(.v—t)r==P(T-.r)r; PT+pt x=~ —— . P+p Or, more generally, let P,, P2, P,, &c. be the sums due after the equated time, at the end of times T1, T2, T,, &c. and p,, p,, 7),, &c. the sums due before, at the end of times t,, t2, is, &c. then we have p,(.r-t,)r+p,(a'—l,)r+p3(a-t,)r+&c. =P,(T,-.r)r+P,(T,~a)r+P3(T,—-x)r+&c. 2 Pl Tl +P2T2+ P3T3+ &c.+ 7211‘, +p,t2 +p3t3+&c. P,+P,+P,+&c.+p,+p,+p,+&c. ’ which furnishes a simple rule easy of application. By this rule a small advantage is given to the payer, because he reckons on his side the interest, instead of the discount, of those sums which he pays before they are due, whilst the opposite side of the account is confined to strict accuracy; and it has been shewn in Art. 408, Cor. 2. that the interest of any sum is greater than the discount. 0.. 1x ANNUITIES. 277 412. Another method of finding the Equated, time is to find the Present Value of each payment, and make the sum of them equal to the Present Value of the sum of the several payments supposed due at the Equated time. Thus, if P, p, be due at the end of times T, t, respectively, r the rate of interest, and a: be the equated time, P Present value of P=I—m, (Art. 408.) P . .............. .. f =___ O P 1+tr’ P+p l+a'r’ . . . - . | . . ¢ Q u - "Of p _P+p 1+ Tr 1+tr_ 1+arr’ and from which simple equation with respect to a: we get _PT+pt+r(P+p)Tt a:— P+p+r(Pt+pT) COR. If the quantities multiplied by r be neglected, since r is gene:- rally a very small fraction, we have _P T+ pt _ P+p ’ which is the common rule. ANN UITIES. 413. To find the Amount of an annuity or pension left unpaid any number of years, allowing simple interest upon each sum or pension from the time it becomes due. Let A be the annuity; then at the end of the first year A becomes due, and at the end of the second year the interest of the first annuity is rA (Art. 401); at the end of this year the principal becomes 2A, therefore the interest due at the end of.the third year is 2rA; in the same manner, the interest due at the end of the fourth year is 3rA; &c. Hence the whole interest at the end of it years is _- n—l q'A+2rA+3rA,,,,,,+'n-l .rA =n.—2——rA (Art. 282); and the sum of the annuities is nA, therefore the whole amount n- 1 M=nA+n. rA. 278 ANNUITIES. 414. Required the Present Value of an annuity which is to continue a certain number of years, allowing simple interest for the money. - Let P be the present value; then if P, and the annuity, at the same rate of interest, amount to the same sum, they are upon the whole of equal value. The amount of P in n years is P+Pnr (Art. 401); and the amount of the annuity in the same n —1 rA; hence time is nA+n. -1 P+Pnr=nA+n.n rA; nA+n. _ rA P== 1+nr From this equation, any three of the four quantities P, A, n, r being given, the other may be found. - A 415. COR. Let n be infinite, then P=-n? an infinite quan- tity; therefore for a finite annuity to continue for ever, an infinite sum ought, according to this calculation, to be paid; a conclusion which shews the necessity of estimating the Present Value of an annuity upon different principles. 416. To find the Amount of an annuity in any number of years, at compound interest. Let A be the annuity, or sum due at the end of the first year; then 1 : R z: A : RA, its amount at the end of the second year; therefore A+RA is the sum due at the end of the second year; in the same manner, 1 : R :: (1+R)>;-1 q 1’ COR. 4. If the annuity is to continue for ever, this Present Value becomes A 1 5' 1 ' (1+:>m-1 (I 421. The Present Value of an annuity, to commence at the ‘iexpiration of p years, and to continue q years, is the difference between its present value for p+q years, and its present value for p years, __-___- A AR‘PTQ { A AR'P} =RT1" R—l 1.3-1 _R-1 AR'P =R__-1_{1-R"q}. COR. If the annuity commences after p years, and continues for ever, the Present Value will be R—l ' RENEWAL OF \LEASES. 281 Ex. What is the Present Value of an annuity of 1£, for 14 years, to commence at the expiration of 7 years, allowing 5 per cent. per ann. compound interest? . 21_ The Present Value for 21 years= W=12'82£; (1'05)*1><0-05 . 7__ and the Present Value for 7 years = (105) 1 =5'79£; (1'05)7x0'05 hence the value of the annuity for fourteen years after the expira- tion of 7 is 7 '03£, or 7 years’ purchase, nearly. The preceding Article contains the whole Theory of the RENEWAL OF LEASES. 422. T 0 determine the fine which ought to be paid for renewing any number of years lapsed in a lease. Let p+q be the number of years for which the lease was originally granted; p the number lapsed; and A the clear annual value of the estate, after deducting reserved rent (if any), taxes, and all other fixed annual charges. ' Then it is clear that the lessee has to purchase an annuity of A11 to commence at the expiration of q years, and to continue p years, the Present Value of which is the Present Value for p+q years—Present Value for q years, A = m,{R-q— R-u’wl}. COR. The number of years’ purchase is ._l__ -q_ -(p+a) R__1 {R R Ex. In a lease of 21 years 7 years lapsed are to be renewed, the reserved rent is 10.45, and the estate is really worth 150.10 a year, what fine ought to be paid for the renewal, reckoning interest at 5 per cent? In this case the lessee has to pay for an annuity of 140£ to com- mence at the end of 14 years and to continue 7 years; therefore the fine required is %%%{1-05 I‘M-105147.43. 140 NOW 10g,011051-“=-0-29666 =P70334=10g,00'50505, log,,fi5 “lb-044499 =T-55501=10g,,0-85895. 282 SCHOLIUM. the required fine=2800x{0'50505—0'35895}, =2800X01401, =409£, very nearly. - [Exercises Zs.] SOHOLIUM. 423. The method of determining the present value of an annuity at simple interest, given in Art. 414, has been decried by several eminent Arithmeticians, and in its stead a solution of the question has been proposed upon the following principle; 5‘ If the present value of each payment be determined separately, the sum of these values must be the value of the whole annuity.” Let a; be the value or price paid down for the annuity, a the yearly payment, n the number of years for which it is to be paid, r the interest of If. for one year. The present value of the first , a payment IS I: (Art. 402); the present value of the second pay- ,. . . a ment, or of a£ to be paid at the end of two years, Is W; and + T a a a so on: therefore rs= -- +-——~ .,,+ . 1+r 1+2r 1+nr These different conclusions arise from a circumstance which the opponents seem not to have attended to. According to the former solution, no part of the interest of the price paid down is employed in paying the annuity, till the principal is exhausted. Let the annuity be always paid out of the principal a' as long as it lasts, and afterwards out of the interest which has accrued; then a", w-a, .v-Qa, a-Sa, &c. are the sums in hand, during the first, second, third, fourth, &c. years, the interest arising from which ran, ra—ra, ra—2ra, r.v-3ra, &c. that is the whole in- . n—l _ terest, 1s nra— {1+2+3...(n—1)}><(R-1) ( e6 r ) CHANGES OR PROBABILITIES. 424. Chance, or Probability, has two meanings; the one a popular meaning, without any very distinct signification; the other a mathematical meaning, pointing out a real value existing in the circumstances. DEF. Most questions of probabilities will fall under one of two classes, called direct and inverse probabilities. A question of probability is termed direct, when, certain causes being given as existent, from which a certain event may proceed, the proba- bility of that event happening is required. A question of probability is termed inverse, when, an event being given as existent, and proceeding from one of several causes, the proba- bility of one proposed cause being the true one is required. Some more complex questions may partake of the nature of both kinds of probability. 284 CHANCES. I. DIRECT PROBABILITIES. ‘ 425. If an event may take place in n diferent ways, and each of these be equally likely to happen, the probability that it 1 will take place in a specified way is properly represented by - , n certainty being represented by 1. Or, which is the same thing, if the value of certainty be 1, the value of the expectation that the event will happen in a specified way is n For the sum of all the probabilities is certainty, or 1, because the event must take place in some one of the ways; and the Q I I . 1 probabllltles are equal: therefore each of them 1s - . n 426. COR. If the value of certainty be a, the value of the expectation is 3. But in the following Articles we suppose the value of certainty to be 1. 4:27. If an event may happen in a ways, and fail in b ways, any of these being equally probable, the chance of its happening aib , and the chance of its failing is ail). The chance of its happening must, from the nature of the supposition, be to the chance of its failing, as a : b; therefore the chance of its happening : chance of its happening together with the chance of its failing :: a : a+b. And the event must either happen or fail; consequently the chance of its happening together with the chance/of its failing is certainty. Hence the chance of its happening : certainty (1) z: a : a +b; or the chance of its hap- is penmg: a + b . Also, since the chance of its happening together with the chance . - . . a of its failing 1s certamty, whlch 1s represented by 1, 1 - m, that is the chance of its failing. , b Is, a+b - CHANCES. 285 428. Ex. 1. The probability of throwing an ace with a single 1 die, in one trial, is 6; the probability of not throwing an ace is g; \i 2 the probability of throwing either an ace or a deuce is 6; &c. 429. Ex. 2. If n balls, a, b, c, d, &c. be thrown pro— miscuously into a bag, and a person draw out one of them, the 1 probability that it will be a is —; the probability that it will be n i Q 2 elther a or b 1s — . n 430. Ex. 3. The same supposition being made, if two balls be drawn out, the probability that these will be a and b is 2 n(n-1) ' For there are n. combinations of n things taken two and two together (Art. 300); and each of these is equally likely to be taken; therefore the probability that a and b will be taken is 1 2 or -——-— . n—l’ n(n—1) 2 431. Ex. 4. If 6 white and 5 black balls be thrown pro- miscuously into a bag, and a person draw out one of them, the 6 probability that this will be a white ball is H; and the probability ’ 5 ‘that it will be a black ball is 1—1 . From the Bills of Mortality in different places Tables* have been constructed which shew how many persons, upon an average, out of a certain number born, are left at the end of each year, to the extremity of life. From such Tables the probability of the continuance of a life, of any proposed age, is known. 432. Ex. 1. To find the probability that an individual of a given age will live one year. * Some of these Tables will be found at the end. of the Section, pp. 299-301. ED. 286 CHANCES. * Let A be the number, in the Tables, of the given age, B the number left at the end of the year; then 2 is the probability that the individual will live one year; and the probability that he will die in that time (Art. 427). In Dr. Halley’s Tables, out of 586 of the age of 22, 579 arrive at the age of 23; hence the 5 probability that an individual aged 22 will live one year is 3%:- or I 1 2 1 d 7 1 1 i th 0r —— OI' _- near ' an _ or —— near S e 7 1 83 y’ 586’ s y’ 1+ 579 1. 5 82+— 7 probability that he will die in that time. 433. Ex. 2. To find the probability that an individual of a given age will live any number of years. Let A be the number in the Tables of the given age; B, C, D,...X, the number left at the end of 1, 2, 3,...0), years; then 5 is the probability that the individual will live 1 year; 5 the probability that he will live 2 years; and zgvthe probability that A-B A—C A— be will live .22 years. Also A , A , AX bilities that he will die in 1, 2, :0 years, respectively. , are the proba- These conclusions follow immediately from Art. 427. _ 434. If two events be independent of each other, and the I 1 probability that one will happen be 5 , and the probability that ' l the other will happen be — , the probability that they will both n 1 . . . happen is -—; m and n being the numbers of ways in which run the events can severally happen or fail. CHANCES. 287 For each of the m ways, in, which the first can happen or fail, may be combined with each of the n ways in which the other can happen or fail, and thus form mn combinations; and there is only one in which both can happen; therefore the probability that this . , 1 Will be the case is -—- (Art. 425). mn 435. COR. 1. The probability that both do not happen is 1 mn—l 1-——, or mn mn together with the probability that they do not both happen, is certainty; therefore, if from 1 the probability that they both happen be subtracted, the remainder is the probability that they do not both happen. 436. COR. .2. The probability that they will both fail is (m—l)(n-1) For the probability that they both happen, For the probability that the first will fail is mn _1 __ m , and the probability that the second will fail is “"1; .. . . . m—l n—1 therefore the probability that they Will both fail is x , n or (m—1)(n -1) . mn 437. COR. 3. The probability that one will happen and the other fail is For the first may happen and the second fail 77272 in n-l ways; the second may happen and the first fail in m—l ways. Hence the number of favourable cases is m+n—2, whilst the number of favourable and unfavourable cases together is mn. Therefore the pro- m+n—2 bability required is 438. Con. 4. If there be any number of independent events, .. . . l 1 1 and the probabilities of their happening be —, 7;, -, &c. respec- m r . . l tively, the probability that they Will all happen is —-——— For mnr &c. 1 the probability that the first two will happen is E; , and the pro- 288 CHANCES. , 1 bability that the first two and third will happen is 772-75; and the same proof may be extended to any number of events. 1 When m = n = r= &c. the probability is Ev , v being the number of events. 439. Ex. 1. Required the probability of throwing an ace and then a deuce with one die. . 1 . The chance of throwing an ace IS 6, and the chance of throWing 1 a deuce in the second trial is 5; therefore the chance of both hap- . Q 1 1n _ 0 pen g 36 440. Ex. 2. If 6 white and 5 black balls be thrown promis- cuously into a bag, What is the probability that a person will draw out first a white and then a black ball? 6 The probability of drawing a white ball first is 1—1 (Art. 431), and this being done, the probability of drawing a black ball is 5 1 . 1—0, or 5, because there are 5 white and 5 black balls left; there- . . . . 6 1 a fore the probability required is fix-Q- or i-i . Or we may reason thus; unless the person draw a white ball first, the whole is at an end; therefore the probability that he will have a chance of drawing a black ball is 1%, and when he has this . . . . , 5 1 chance, the probability of its succeeding is 1—6, or 5; therefore, the . . 6 probability that both these events Will take place 1S fix; or 1—31- . 441. EX. 3. The same supposition being made as in the last example, what is the chance of drawing a white ball and then two black balls? CHANCES. 289 The probability of drawing a white ball and then a black one is 3; (Art. 440); when these two are removed, there are 5 white and 1 4 black balls left; and the probability of drawing a black ball, out 3 4 4 _ 4 . . . . of these, is 5; therefore the probability required is fix— or E . 442. Ex. 4. Required the probability of throwing an ace,' with a single die, once at least, in two trials. The chance of failing the first time is -5—, and the chance of . . . 5 . . . failing the next is 6; therefore the chance of failing tW1ce together . 25 . . . . 25 11 is 56; and the chance of not failing both times 1S 1- -— , or - 36 26 ' 443. Ex. 5. In how many trials mayia person undertake, for an even wager, to throw an ace with a single die? Let a' be the number of trials; then, as in the last Art., the . 5 f . chance of failing .2? times together is , and this, by the ques- tion, is equal to the chance of happening, or (5 f 1. 6) ‘5’ 1 hence a'xlog 6 = log;; or (ex (log 5 --log 6) = log 1 — log 2, _log1—log2_ log2 _ - ; ' 1 1=0; log5—log6 log6—log5 Slnce 0g '. .v = 3'8, nearly; that is, a person would safely undertake to throw the ace in four trials, but not less, and then have some probability to spare in his favour. 444. Ex. 6. To find the probability that two individuals P and Q, whose ages are known, will live a year. 19 290 CHANCES. Let the probability that P will live a year, determined by Art. 1 I 1 432, be —; and the probability that Q will live a year be 9;; then m the probability that they will both be alive at the end of that time 1 1 1 IS ~X'_ '— a m ’n m”: 445. Ex. 7. To find the probability that one of them, at ~least, will be alive at the end of any number of years. —1 , and the The probability that P will die in a year is m probability that Q will die is 7fib—I; therefore the probability that n (m —1)(n — 1) 777/77, they will both die is ; and the probability that they will not both die is m—l n—l m+n—1 1_g____)_.(—)’ _—______' mn mn 1 In the same manner, if i) be the probability that P will live 1 . t years, and a the probability that Q will live the same time (Art. 483), the probability that one of them, at least, will be alive at the end of the time is —1 —1 -1 _(r )(q ), or P+q . 19‘] P9 1 446. If the probability of an event’s happening in one trial be represented by a a +b (Al't- 4127), 10 find the probability of its happening once, twice, three times, &c. exactly, in n trials. The probability of its happening in any one particular trial being 227;, the probability of its failing in all the other n—1 01—] trials is W (Arts. 427, 438); therefore the probability of its onAnoEs 291 . . . . , , . , abn-l happening in one particular trial, and failing in the rest, IS W ; a+ 7b and since there are n trials, the probability that it will happen in some one of these, and fail in the'rest, is n times as great, or nab”-1 (a+b)"' The probability of its happening in any two particular trials, . . . , (l2 7"”? 92—1 and failing in all the rest, is ——-b—)—n, and there are n. a+- ways 2 in which it may happen twice in n trials and fail in all the rest (Art. 300); therefore the probability that it Will happen twice in n n—l no '_ trials is '(a+b)" In the same manner, the probability of its happening exactly n-10n—2 a3bn_3 2 th t' ° ~ d h bb'l't f'th - ree imes is (CH—b)" , an t epro aiiyo is ap pening exactly t times is (a +1))" 447. Con. 1. The probability of the event’s failing exactly t times in n trials may be shewn, in the same way, to be n—1 n—2 n-t+1 “if. M 00.000 __—___ 2 3 t (a + b)" 448. Con. 2. The probability of the event’s happening at least t times in n trials is 92—1 a”+ nan-lb +n. T a“'2b2+... ...to (n —t+1) terms (a+b)" For, if it happen every time, or fail only once, twice, .... .. (n— t) times, it happens t times; therefore the whole probability of 1 9-2 292 CHANCES. its happening, at least t times, is the sum of the probabilities of its happening every time, of failing only once, twice, n- - t times; and the sum of these probabilities is n n-lb n a +na +n. -] a"-2b2+ .... ..to (n -t+1) terms (a+ b)n 449. Ex. 1. What is the probability of throwing an ace twice at least, in three trials, with a single die? In this case n=3, i=2, a= 1, b=5; and the probability 1+3X5 16 2 6X6 X6 _ 2T6 ' 57' ' 450. Ex. 2. What is the probability that out of five indi- viduals, of a given age, three at least will die in a given time? required is Let 1- be the probability that any one of them will die in the 7711 given time (Art. 433); then we have given the probability of an event’s happening in one instance, to find the probability of its hap— pening three times, at least, in five instances. In this case a= 1, b=m — 1, n= 5, t= 3; therefore the proba- bility required is 1 +5 (m- 1) +10(m-i)=2 m5 ' II. INVERSE PROBABILITIES. 451. AXIOM. When an event can proceed from one of a system of causes, the probabilities of these causes having produced the event are proportional to the numbers of ways in which they can severally produce the event. 452. Con. Hence the probabilities of the several causes having produced the event are proportional to the chances of the event happening on the assumption of their being severally existent. If p,, p,, p,, &c. be the probabilities of several causes from which an event may proceed; a“ a,, as, &c. the chances of that event happening on supposition of these causes severally existing; we have Bl=lle=ll§=gm a1 a2 as . P1__2(P) . a:~m. (Art. CHANCES. 293 But as some one of the system of causes is known to be the true one, 2(1)) is certainty, or 1, a1 pl=im . a, a, So also p,- 2(a), p,_2(a), &c. Ex. An urn contains 3 balls which may be white or black. A ball is drawn out and replaced three times, and in each case a white ball is drawn. What are the probabilities of the urn containing (1) Three White balls, (2) Two white and one black, One white and two black, (4) Three black? (1) Supposing the 18* state of the urn, the chance of the event hap- ' 0 o 0 3 2 . Q pening which did happen is ~x§x§ or —7, that is, certainty. 3 3 3 27 (2) Supposing the 2m1 state the chance is gxgxg or —8~ . ’ 3 3 3’ 27 (3) Supposin the 3’“ state the chance is lxlxl or -1- g ’ a s s’ 27' . . O O O (4) Supposmg the 4“ state, the chance 1S gxgx-é or 0. 27 _ 27+8+1 27 27'_~277_’°rs_6° the chance for the 1“ state of the urn is i . s 36 s Th ch nc 'th 2“dis——+——, or—. e a efor e 27 27 86 . 1 36 1 I a o a o o o o a n a u u I o p u o o a o o on 3rd "_". 1 27 27"” 36 36 th --.:___ ...................... .. 27.27,0r0 The following Proposition involves the nature of both direct and in- verse probabilities :— 453. If a,, a2, a,, &c. be the chances of an observed event on supposition of each of the system ofcanses being the true one; a,’, a2’, a3', &c. the proba- bilities of another proposed event on the same separate suppositions; the 2(aa’) 2(a) ' For, as has been seen, a,, a,, as, &c. are proportional to the chances given by the observed event of the separate causes existing. Hence the number of ways of the first cause existing and the second event happening from it is proportional to alal'; and the number of ways of that event happening from some one of the causes is proportional to 2(aa’). But since some one of the causes exists, 2(a) is in the same proportion to chance of this event happening: 294 CHANCES. the number of ways in which one of the causes may exist, and the event either happen or fail from it. Hence the chance of the event happening _ 2(aa') _ 2(4) Ex. 1. An urn contains two balls, but whether white or black, uncer- tainfwe draw one ball, and find it is white. The ball is then replaced; what is the chance of next drawing a black one? ‘ Before the drawing takes place the two states or causes may be, (1) Two white, and (2) One white and one black. . 2 Now the chance of the observed event under (1) is g- , or 1, . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . IS &- . Again, the chance of the proposed event under (1) is O, . 1 0 u o e a e a o Q 0 e a e n a s e e o a a Q e o o o e e Q e e a o a 0 o 2 '_’ 0 < >15 2 1 1 XO+ é'Xé- 1 the Chance required = 1 =6. 1+ '5 Ex. 2. Taking the case supposed in Ex. Art. 452, what is the chance of a white ball coming out at a fourth such drawing? Here before the drawing takes place the states or causes may be, (1) Three white, Two white and one black, One white and two black. Now the chance of the observed event under (1) is 1, . 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2 1 —— I 1 e o u e 0 0 0 I 0 o I a a a o o o o a o o t o I o a o o o I O o o o on 3 —. (>182, Again, the chance of the proposed event under (1) is 1, . Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 18g, . 1 s e - . e - . a Q . - . . . . . . . . . . . . . . . . . e . . - - e .- 135, 1x1+ 8><-2-+ 1XI 27 3 27 3__ 93 _49 1+_s_+_1_ Hos—54' 27 27 the Chance required= LIFE ANNUITIES. 295 SGHOLIUM. Much more might be said on a subject so extensive as the doc~ trine of Chances; the Learner will however find the principal grounds of calculation in Articles 425, 4.27, 434, 4.46, 448, 452, and 453; and if he wish for farther information, he may consult De Moivre’s work on this subject*. It may not be improper to caution him against applying principles which, on the first view, may appear self-evident; as there is no subject in which he Will be so likely to mistake as in the calculation of probabilities. A single instance will shew the danger of forming a hasty judgment, even in the most simple case. The probability of throwmg an ace With one the is -6-, and smce there is an equal probability of throwing an ace in the second trial, it might be supposed that the probability of throwing an ace in two t'l ' 2 r1a318—. 6 This is not a just conclusion (Art. 442); for, it would follow, by the same mode of reasoning, that in six trials a person could not fail to throw an ace. The error, which is not easily seen, arises from a tacit supposition that there must necessarily be a second trial, which is not the case if an ace be thrown in the first, LIFE ANN UITIES. 454. To find the Present Value of an annuity of £1, to be continued during the life of an individual of a given age, allowing compound interest for the money. Let R be the amount of £1 in one year; A the number of persons in the Tables of the given age; B, C, D, &c. the number B left at the end of 1, 2, 3, &c. years; then 2 is the value of the C D . :4 , 2 , &c. its value for 2, 3, &c. years respec- tively; and the series must be continued to the end of the Tables. Now the Present Value of £1, to be paid at the end of one year, is life for one year, 1 "R (Art. 408); but it is only to be paid on condition that the * The more modern writers on this subject are Laplace, Galloway, and De Morgan. ED.- 296 LIFE ANNUITIES. annuitant is alive at the end of the year, of which event the proba- .. . B .. . bility is 2; therefore the Present Value of the conditional annuity . B . is 2% (Art. 426); in the same manner, the Present Value of the second year’s annuity is the Present Value of the third year’s C I ARQ’ &c.; therefore the whole value required is D AR3 ’ ix<§ +1%+—g,+&c. to the end of the Tables.) 455. De Moivre supposes, that out of eighty-six persons born, one dies every year, till they are all extinct. This supposition is sufficiently exact, if our calculations be made for any age neither very young nor very old, as will appear from an inspection of the Tables; and, on this supposition, the ~11; +-}(—; + ~B{23+&c.) may be found. Let n be the number of years which any individual wants of 86; then will n be the number of persons living, of that age, out of which one dies every year; and n—1 n—2 n-3 ———, —— , —— , &c. n n n will be the probabilities of his living 1, 2, 3', &c. years; hence, the Present Value of an annuity of £1, to be paid during his life, is n—1 n—2 n—3 nR + nR2 + nR?’ The sum of the series 2 a (n—l).v+ (n—2).v + n n n annuity is , 1 sum of the series 2x( + &c. continued to n terms. +&c. to n terms is found to be (ii—l)n; —ntvg + an“ n(1_w), (See Art. 295, Ex. 3); 1 . let .v =2? , and the sum of the series n—l n—2+n—3+& t t (n—1)R-n+R'("-1) c. o n erms = nil + nR2 n183 n(R—1)2 Present Value of the annuity. , the LIFE ANNUITIES. 297 456. COR. 1. This expression for the sum is the same With nR—n R-R'ln‘ll 1 R l—R'” n(R-1)2- n(R-1)2’ (“PITT-$153?“ Let P be the Present Value of an annuity of £1, to continue 1__ —n —1 certain for n years, then P: (Art. 418); and the expression 13.10 n becomes R - 1 457. Con. 2. The Present Value of the annuity to continue RP n(R- 1) i For the whole Present Value of the annuity, to continue for for ever, from the death of the proposed individual, is ever, is (Art. 420); and if from this its value for the life of R-l the individual be taken, the remainder is the Present RP. . n(R-1) Value of the annuity, to continue for ever, from the time of his death. 458. To find the Present Value of an annuity of £1, to be paid as long as two specified individuals are both living. Find, by Art. 444, the probability that they will both be alive at the expiration of 1, 2, 3, &c. years, to the end of the Tables; call these probabilities a, b, c, &c. and R the amount of £1, in one a b c B+F+R3 required. (See Art. 454.) year; then + &c. is the Present Value of the annuity 459. To find the Present Value of an annuity of £1, to be paid as long as either of two specified individuals is living. Find, by Art. 445, the probability that they will not both be extinct in 1, 2, 3, &c. years, to the end of the Tables, and call these probabilities A, B, C, &c.; then the Present Value of the annuity is B %+ —— +"—,+&c. (See Art. 454.) 298 LIFE ANNUITIES. 460. COR. If the annuity be M43, the Present Value is M times as great as in the former case, or A \ B C MX(R R2+R3+&c.>. 461. These are the mathematical principles on which the values of annuites for lives are calculated, and the reasoning may easily be applied to every proposed case. But, in practice, these calculations, as they require the combination of every year of each life with the corresponding years of every other life concerned in the question, will be found extremely laborious, and other methods must be adopted when expedition is required. Writers on this subject are De Moivre, Maseres, Simpson, Price, Morgan, and Waring. Other writers on the subject are Milne, Baily, and De Morgan, of which the last mentioned is now most accessible. 462. T 0 find the Present V alue of the next presentation to a Living. Let i£ be the average annual net income of the living; oil the cost of a curate, that is, the money value of the work to be done; and oil the unavoidable expences of the admission of a new incumbent. Then the Present Value of the next presentation will obviously be the present value of an annuity of (i—c)£, to commence at the death of the present incum- bent, and to continue during a life then 24 years of age, deducting the present value of a£ payable on admission. Let n be the number of years which the Tables give to the present incumbent, p the number for a person ’24 years of age; then the Present Value required will be that of an annuity of (i-c)£ to commence at the expiration of n years, and to continue p years, deducting the Present Value of 0.59 to be paid after n years if, _ Q ' ~n_ 57+? __ n _R_1.(R R )m-. COR. The Present Value of an Advorvson, or perpetual nomination to a living, will be that of an Annuity of (i—c).£ commencing after n years, and continuing for ever, deducting the present value of a£, to be paid at the end of n years, and also of the same sum to be paid at intervals of p years for ever afterwards, i_ _ ‘__ . . =-'R:_01 . R_n—aR_n-aR_n+p—CbR_u+2P—- I I n o o i—c 1 = . R_"— R‘”. —— . R—l a 1~R"P * Of course this is on the supposition, that the laws, which permit such traffic in spiritual cures, remain in force, and that the values of i, c, and a remain unaltered, for n+ p years at least. LIFE ANNUITIES. 1687...1691, by Dr. Halley. TABLEiL For determining the Probabilities of the Duration of Life, from Obser- vations on the Bills of Mortality of BRESLAW, made in the years Age 1135;? .1334. Age .3338. Age 1 1000 145 81 528 8 61 2 855 57 32 515 8 62 8 798 88 88 507 s 68 4 760 28 84 499 9 64 5 782 22 85 490 9 65 6 710 18 86 481 9 66 7 692 12 37 472 9 67 8 680 10 38 463 9 68 9 670 9 89 454 9 69 10 661 8 40 445 9 70 11 658 7 41 486 9 71 12 646 6 42 427 10 72 18 640 6 48 417 10 78 14 684 6 44 407 10 74 15 628 6 45 897 10 75 16 622 6 46 887 10 76 17 616 6 47 377 10 77 18 610 6 48 867 10 78 19 604 6 49 857 11 79 20 598 6 50 846 11 80 21 592 6 51 885 11 81 22 586 7 52 824 11 82 28 579 6 58 818 11 88 24 578 6 54 302 10 84 25 567 7 l 55 292 10 85 26 560 7 56 282 10 86 27 553 7 ' 57 272 10 87 28 546 7 58 262 10 88 29 589 8 7 59 252 10 89 80 581 8 ‘ 60 242 10 90 _ Persons living. 232 222 212 202 192 182 172 162 152 142 181 120 109 98 88 78 68 58 49 41 84 28 23 19 15 11 8 5 3 1 Deer. of Life. l—ll—lh-IHHI-db-ll—ll—‘l—‘l—ll—Jh-li—lhll—Jl—l QOOOI—ll—‘l—‘I-‘OOOOQQOOO size NJCm 0019 4> s-tn o>~q @183 300 LIFE ANNUITIES. TABLE II. For determining the Probabilities of Life at NORTHAMPTON, as deduced by Dr. Price from the mortality of that town in the years 1741...1780. Age £38. Age .1338. Age it??? .1238. o 1 149 800 81 428 7 62 187 8 i 849 127 82 421 7 68 179 8 2 722 50 88 414 7 64 171 8 3 672 26 34 407 7 65 168 8 4 646 21 85 400 7 66 155 8 5 625 16 86 898 7 67 147 8 6 609 18 87 886 7 68 189 8 7 596 10 38 879 7 69 181 8 8 586 9 89 872 7 70 128 8 9 577 7 40 865 8 71 115 8 10 570 6 41 857 8 72 107 8 11 564 6 42 849 8 78 99 8 12 558 5 48 841 8 74 91 8 18 558 5 44 888 8 75 88 8 14 548 5 45 325 8 76 75 8 15 548 5 46 317 8 77 67 7 16 588 5 47 809 8 78 60 7 17 588 5 48 801 8 79 58 7 18 528 6 49 298 9 8O 46 7 19 522 7 50 284 9 81 89 7 20 515 8 51 275 8 82 82 6 21 507 8 52 267 8 88 26 5 22 499 s 58 259 8 84 21 4 28 491 8 54 251 8 85 i7 4 24 488 8 55 248 8 86 18 8 25 475 8 56 285 8 87 10 2 26 467 8 57 227 8 88 8 2 27 459 8 58 219 8 89 6 2 28 451 8 59 211 8 90 4 2 29 448 8 60 208 8 91 2 1 80 435 7 61 195 8 92 1 1 The preceding Tables require but little explanation. The former com- mences by stating that out of 1000 persons who were born at the same time and attain the age of 1 year, 145 die before they attain the age of 2 years, LIFE ANNUITIES. 501 n Consequently at 2 years of age there are left 855 out of 1000. Of these 57 die between 2 and 3 years of age; and so on. Thus, of 1000 persons who attain the age of one year, the Table indicates that 346 live to be 50 years of age; &c. The latter Table commences a year earlier by taking 1149 persons born together, that is, at the age 0 ; and then proceeds in the same manner as the former. Thus, we have given by this Table, that after 50 years, out of 1149 persons born together, 284 are then alive, and that of these 9 die before attaining the age of 51 ; and so on. It has been objected to both the preceding Tables, although the latter is very generally used by the Assurance offices, that they make no distinc- tion between male and female life, and yet that a very material distinction can be proved to exist. TABLE III. Shewing the Expectation of Life, as deduced by Professor De Morgan from the Statistical Returns of the whole of BELGIUM made by M. Quetelet and Smits. Towns. Both. Country. Age. Age. Males. Females. Both. Males. Females. 0 29-2 88-8 322 320 329 0 5 450 47-1 45-7 46-1 44-8 5 10 42-9 450 439 44-4 429 10 15 39°C 41-8 405 41-2 400 15 20 354 88-0 373 88-1 370 20 25 88-1 350 84-7 357 842 25 80 304 321 320 88-0 315 80 85 27-5 29-2 28'9 29-7 287 85 40 24-4 265 258 26-0 259 40 45 21-5 28-8 227 22-5 232 45 50 18-8 201 19-5 19-1 20-0 50 55 15-5 17-1 164 162 16-9 55 60 128 14-0 134 133 13'7 60 65 10-4 112 108 106 109 65 70 8'2 8'6 8'4 8‘2 8'5 70 75 6'3 66 6'4 6'3 65 75 80 4'8 51 5'0 5'0 5'1 80 85 3'7 40 3'8 3'8 88 85 90 29 3'0 3'1 81 3'2 90 95 1-8 2-0 21 2-2 1-9 95 100 0-0 05 1'3 05 0-5 100 302 DI CUSSION AND INTERPRETATION The extent of the error which arises from not distinguishing between the sexes may be seen in Table III. constructed by Professor De Morgan from the statistical returns of the whole tyf Belgium for three successive years, as given by M. Quetelet and Smits, in the Recherches sur la Reproduction, &c. Brussels, 1832. This Table is calculated to .shew the “ expectation of life,” that is, the average number of years remaining to any individual, at intervals of five years, from the age of O to 100. It distinguishes not only between male and female, but between town life and rural life; and the middle column gives the general average for the whole kingdom, male and female, town and country. DISCUSSION AND INTERPRETATION OF ANOMALOUS RESULTS. 463. Negative Results. It often happens, that the result of our operations for the solution of a proposed question or problem appears in a negative form, although strictly speaking, there can be no such thing existent as an essentially negative quantity. But it will always be found, when such a result occurs, that there is something in the nature of the question which will either dispose of, or supply a meaning to, the negative result. Thus, to take a simple example; suppose a man wishes to ascer- tain the amount of his property—he puts down What he has, together with what is due to him, as positive, and all his debts with a negative sign. If then he finds that by taking the sum of both positive and negative quanti- ties, the result is negative, its meaning Will be sufficiently obvious, viz. that his property is so much less than nothing, that is, he is so much in debt. See Scholium, p. 44,- and Art. 214. Also, see Art. 282. Ex. 4. In this and like cases it is true that two solutions may be found for the equation, that is, two values of n ; but when either of those values is fractional or negative, it is clearly inadmissible as a solution of the question proposed. It may be observed also here generally, that when in solving a problem, expressed algebraically, we find it necessary, as in the above Example, to extract the square root of a quantity, the double sign 4, (that is, + or —), need not to be prefixed to the root, at least for the object before us, if we have sufficient data beforehand for determining which sign the problem requires. Is it to be wondered at, that we produce an anomalous and unintelligible result, if we wilfully make a quantity negative which we know to be positive, or vice versa' .9 Oftentimes, however, the negative solution, whether it results from carelessness or necessity, will satisfy another problem cognate with the proposed one; which may be determined by substituting the negative quantity for the positive in that step of the process which most clearly expresses the conditions of the question; and then interpreting the result- ing equation with the assistance of the given problem. This has been done in the cases above referred to. - or ANOMALOUS RESULTS. 303 464. Interminahle Qaotients. Strictly speaking a Quotient can only exist when after the division by which it is determined there is no Remainder; but the term is applied to those cases also where a remainder is left which cannot be got rid of. Thus we say generally, that the quotient of 1+1—a: is 1+x+x2+x3+ .... ..; whereas the true quotient is w? 1+a+a2+a~3+ .... .. +.22"1+ 1 Thus, whatever be the value ofa', 1 2 3 'r-l fr ————=1+x+a: +a: + ---- .-+.r , l—x _ 2 —1+ x or 1+x+ m or 1+x+w2+ _ l—x’ l—a" l—x N.B. By taking the Remainder into account no unintelligible result can arise From substituting any particular value for .r. ; and so on. COR. 1.- If'x <1, then the Remainder maybe neglected, if' a sufficient number of terms of the series are taken. (Art. 290, Cor. If' a=1, then %=1+1+1+1+&c. in. inf.=an infinitely great number. . 1 . . . Gen. 2. It‘s: be neoative, we have —-—-=1-a:+x’—x3+&c. in inf; in a b 1+.n \ Which if we put 1 for a', we get %=1—1+1—1+&c.=0, 01‘ 1, according as an even or an odd number of terms is taken; both of which results are obviously impossible. N ow, taking the Remainder into account, we have 1 (—x)"“ ————= — dag—$3 ...H. —- \r ————-— 1+1“ 1 x r + < x) + 1+.” 7 and l——1—1+1—1-{- =’=-1—--1- as it on ht 2- . . . . . . . .. 2-2, g . Again, as it was shewn in Art. 323, I (r+l ) .r’-— ram ————.=1+2a:+3x2+ . . . . ..+rx’“1+———————— (l-az)‘ (Ina)£3 ’ without which fractional Remainder no arithmetical equality subsists be»- tween the series and And it may be observed generally, that no 1 <1-wr' equality subsists, for purposes of calculation, betwixt any infinite series without the “Remainder”, and the primitive quantity from which it was derived, unless the series is convergent, so that we can make the Remainder after 2' terms, by increasing r, as small as We please. 465. To explain generally the results which assume the forms axO, O 0 a a 3 a0) 00, '6 , 6 a 304 DISCUSSION AND INTERPRETATION (I). Since axb signifies a taken 6 times, it is clear that, if 0 .is to follow the same rule as other multipliers, axO signifies a taken 0 tunes, and is therefore equal to O. . b ' . . . . 0 (92). Since a expresses the number of times a is contained 111 b, a will signify the number of times a is contained in O, that is, 0 times, 01‘ 9:0. a m-n c By the general Rule of Indices, amxa"=am+", and a’"-:-a"=a Now in the first of these let m=0, then a°.a"=a”, if the same rule holds when one of the indices is 0; therefore a°, as far as regards the rule for multiplication of powers, is equivalent to 1, or a°=1. m . am . . . . Also, smce a-=a’""’", 1=a°, 1f the rule for d1V1s10n of powers holds when the powers are equal; therefore it also accords with the general rule for division that a°=1. Hence a°, 6°, 0°, &c. are separately equal to 1, if 0 be admitted as an index subject to the same rule as other indices. (4). Since it has been already explained, that any quantity raised to a power represented by 0 may be safely expressed by 1, it follows that (a—b)°=l, whatever a and b may be. If then a=b, We have 0°=1. a (5). When an algebraical quantity is made to assume the form 6, it is said to be infinitely great, and its value is expressed by the sym- bol 00. All that is meant is, that if the denominator be made less than any assignable or appreciable quantity the fraction becomes greater than any assignable quantity. This is easily shewn by taking any fraction, as 66%, which=10a: for if the denominator be successively diminished one-tenth, we obtain the series of quantities 100a, 1Q00a, IOOOOa, &c., proving that as the denominator of the fraction is diminished, the value of the fraction is increased, and without limit. . . . O (6). Suppose that an expresslon mvolvmg .2: assumes the form 6 when some particular value (a) is substituted for a', then it is clear that the . . , .r—a m expressmn 1s capable of being reduced to the form w, where p and q have no factor m—a in them ; and by dividing numerator and deno- minator by their highest common factor, the value of the fraction may be found when x=a. (See Art. 387.) . , . 0 Thus it appears that a quantity WhlCh assumes the form — may have a O determinate value. And, conversely, since p is equal to p.x_a, what— x-a . O . ever be the values of x and a; 1f x=a, 12:6 , that 1s, any quantity may be OF ANOMALOUS RESULTS. 305 “made to assume the form But this is, in fact, multiplying and dividing by 0, on the supposition that the rules which apply to finite quantities apply also to O as a multiplier. It may be said, generally, that to speak of absolute nothing as the subject of mathematical calculation is absurd ,- and that it can only become so when it is taken to represent some finite quantity in that state when by indefinite diminution it has become less than any appreciable quantity. The mathematical symbol 0 has, then, always reference to some other quantity from which it is derived; and it is this primitive quantity which must be the subject of our investigations when by becoming 0 it produces an anomalous result that requires to be explained. That mistakes will constantly arise from considering 0 as an absolute quantity is easily seen: Thus, it has been shewn that a°=1=b°, therefore we might say, if 0 is a quantity, that a=b; or, since 2°=4°, that 2:4, both of which conclusions are manifestly wrong. Again, if .r—1=0, then a(x-1)=a'><0=0, also a=1, and x2=1, or .v2—1=O; a(x—1)=O, and (a:+1)(x-1)=0, '.x=x+L or 1:0, which is absurd. This amounts to saying that, because ax0=0, and bx0=O, therefore a=b, which is obviously incorrect. 466. Given ax+b=cx+d, and x=:::, to explain the result (1) when a=c, and when a=c, and b=d. (1) When a=c, aid—32:00. In this case the proposed equation becomes ax+b=ax+d, which can only hold true on the supposition of a: being such, that can is not affected to any appreciable amount whichsoever of the two different quantities b or d be added to it, that is, a: must be immeasurably great, agreeing with the result already found. b—b 0 . . . (2) When a=c, and b=d, arm =6. In this case the original equation becomes ax+b=ax+b, an identity which is clearly satisfied by . . . O . . any value whatever of x ,- and this is the meaning of 5 in this case. b'c—bc' ac'—a'c ab’—a’b’ y ab'—~a'b’ - 467. Given ax+by=c}, so that X: and ax+b’y=c’ explain the results when a’=ma, b’=mb, c’= me. a' b’ ’ , 0 Here Z=m=Z=%, or ab=ab’, ac’=a’c, b’e=bc'; a=6 , . . . o’ ’ 6’ From the original equations we have ax+by=c=E=§nx+3Zy=ax+b$ O and y=5. 20 306 DISCUSSION AND INTERPRETATION or ANOMALOUS RESULTS. an identity which is satisfied by any values whatever of a' and y,- agree- ing with the results before obtained. . _ a ‘2- . . 468. Given ax2+bx+c=0, and X: “2%; emplam this result when a=0. - b When a=0, one value of .1: becomes 9, the other -%— or 00 . From . . . . 1 the original equation, putting g for .v, we have a+by+cy2=O, and ifa=0, hy+cy2=0, or y(b+c_y)=0, 51:0, or b+cy=0, 1 0 —=0, a a' c . . x=m; or w=--Z-, which is the value of (6) in this case, as may be easily shewn. Thus - b+Jbz—4ac_ 4ac __ 20 _ 0 when a__0 2“ 2a(—b—Jbz—4ac) —b—~/62—4ac b _ . 469. A and B are travelling along the same road, and in the same direction, at a uniform pace of a miles and 6 miles per hour respectively. Given that at a known time B is d miles before A, find the time when they are together; and explain the result (1) when a=b, and (2) when ab. (a -—li )a: a2x2+ 62 (a2- b2).x a2x2+ 62: (a2— bz)yx, agag— (a2— bays =— I)“, 2 a2__62 a2__ 62 2 a2___ 62 2 b2 x_( a, ),,+( M, ),2=( 4a,) ytgg, 2_ 2 9.2-??- . y=h J(a2- Wy- use. Let =p, then 0.. x= Now, that a may have a real value, (ag— b’Yy’ must not be less than 4a’bg, but it may be equal to it, or y=5j—aébé, which is therefore its minimum, or least value. Hence also in this case, a-W.W-5 . 471. If the quantity under the radical sign remains positive what- ever value be given to y, then the proposed quantity will admit of neither a maximum nor a 20-2 308 MAXIMA AND MINIMA. . . 4 2 4x—3 . . Ex. Required to determine whether admits of either a 6(2x-i-1) maximum or a 4 2 4‘ —3 Let M x =3], then _6(2x +1) 4x2— 4(3y—1)x= 69 +3, and x= S'yg’l i -12-, /Qy2+4. Now, whatever value may be given to y, the quantity under the root will always be positive; therefore the proposed expression does not admit of any maximum or minimum. 472. If the quantity under the root be of the form my2+ ny +p, then by solving the equation myz+ n_2/+p=0, we can find the greatest or least value of g which will permit ,/my2+ny+p to be real, and therefore the required maximum or minimum. Ex. 1. Let a and b be two numbers of which a>b, required the dim—w. (a+b); and 00:36-13. x 4-ab a—b greatest value Ans. lllaximum= Ex. 2. Required the smallest value of aim—gig?) . Ans. lllinimum=(\/a_+~/h)2,- and x=~/E. 472*. Sometimes the introduction of another symbol may be dispensed with: thus, to find the minimum value of x”+px+q, we see that it may be 2 2 ' , 2 written as (n+8) +q—Zl— , and therefore it has its least value when 2 2 is the smallest possible. But this is when =0, 2'. e. when x=-g, 2 and then the given expression becomes g—% , which is consequently the minimum sought. Problems of this kind, however, usually require for their solution the aid of the Differential Calculus. APPLICATION OF ALGEBRA TO GEOMETRY. 473. The signs made use of in algebraical calculations being general, the conclusions obtained by their assistance may, with great ease and convenience, be transferred from abstract magnitudes to every class of particular quantities; thus, the relations of lines, surfaces, or solids, may generally be deduced from the principles of APPLICATION or ALGEBRA TO GEOMETRY. 309 Algebra, and many properties of these quantities discovered, which could not have been derived from principles purely geometrical. 474. Simple algebraical quantities may be represented by lines. Any line, AB, may be taken at pleasure to represent one quantity a; but if we have a second quantity, b, to represent, we must take a line which has to the former line the same ratio that b has to a. Instead of saying AB represents a, we may say AB=a, sup- posmg AB to contain as many linear units as a contains numeral ones. 475. When a series of algebraical quantities is to be repre- sented on one line, and each of them measured from the same point, the positiiie quantities being represented by lines taken in one direction, the negative quantities must be represented by lines taken in the opposite direction. Let a be the greatest of these quantities, then a—x may, by the variation of :0, become equal to each of them in succession. Let AB be the given line, and A the point from which the quan- tities are to be measured; take AB equal to a; and since a-x must be measured from A, D' A D B BD must be taken in the contrary direction equal to x, then AD=a—a'; and that a-x may successively coincide with each quantity in the series, beginning with the greatest positive quan- tity, a' must increase; therefore BD, which is equal to .22, must increase; and when .n is greater than a, BB is greater than AB, and AD', which represent the negative quantity a—x, lies in the opposite direction from A. Con. 1. If the algebraical value of a line he found to be , negative, the line must be measured in a direction opposite to that which, in the investigation, we supposed to be positive. COR. 2. If quantities be measured upon a line from its inter- section with another, the positive quantities being taken in one direction, the negative quantities must be taken in the other. 310 APPLICATION or 476. If a fourth proportional to lines representing p, q, r, be taken, it will represent g; and if p=1, it will represent gr; if also q and r be equal, it will represent qg. If a mean proportional between lines representing a and b be taken, it will represent \/ ab, which, when a =1, becomes \/ b. Hence it appears that any possible algebraical quantities may be represented by lines; and conversely, lines may be expressed algebraically; and if the relations of the algebraical quantities be known, the relations of the lines are known. 478. The area included within a rectangular parallelogram may be measured by the product of the two numbers which measure two adjacent sides. Let the sides AB, AC of the rectangular parallelogram AD be mea- sured by the lineal quantities a, b, respectively; then axb will express the number of superficial units in the area, that is, the number of squares it contains, each of which is described upon a lineal unit. For instance, if the lineal unit be a foot, of which AB contains a, and AC contains b, the parallelogram AD contains axb square feet. A B C - D For, 1st, if AB, AC be divided into lineal units, and straight lines be drawn through the points of division parallel to the sides, the whole figure is made up of squares, which are equal to each other, and to the square upon the lineal unit; and the number of them is evidently a taken b times, or axb. .‘Zndly. If AB and AG do not contain the lineal unit an exact num- ber of times, that is, if a and b be fractional, let a=a+-1m—, and b==fi+Ilz . Then take another lineal unit which is fli—nth part of the former; and by what has been shewn the square described upon the larger unit con- tains mnxmn of that described upon the smaller. Again, the sides AB, AC respectively contain mna+n, mnfi+m lineal units of the smaller ALGEBRA T0 GEOMETRY. 311 kind, and therefore, by the first case, the whole figure contains (mna+ n)>< (mu/8+???) square units of the smaller kind; that is, the area =m2n2{a+7%}.{,3+%} of the smaller units, 1 1 . =(a+ 77,-)(18+ 7-2) of the laiger units, =ab, as before. 3dly. If the sides AB, AC be incommensurable with the lineal unit, a unit may be found which is commensurable with certain lines that ap~ proach as near as we please to AB, and A0, and therefore the product of such lines will represent the area of a rectangle differing from the rectangle AD by a quantity less than any that can be assigned, that is, we may, in this case also, without error express AD by ABxAC. (See Art. 260.) 479. Con. 1. Since, by Euclid, Book I. Prop. 35, the area of an oblique-angled parallelogram is equal to that of the rectangular parallelo- gram upon the same base and between the same parallels, therefore area of any parallelogram=basexaltitucle. 480. COR. 2. Also, since by Euclid, Book 1. Prop. 41, the area of a triangle is half that of the parallelogram upon the same base and between the same parallels, therefore . l . area of any triangle:éxbasexaltitude. 481. COR. 3. Since any rectilineal figure may be divided into triangles, its area may be found by taking the sum of all the triangles. 482. The solid content or volume included within a rectangular paral- lelopiped may be measured by the product of the three numbers which measure its length, breadth, and height. Let the base of the parallelopiped be divided into its component squares, as in the preceding Proposition, and through each of the parallels suppose planes drawn at right angles to the base; and let the same thing be done with one of the faces adjacent to the base. Then it is evident that the whole figure is divided into a certain number of equal cubes, each cube having for its face one of the squares described upon the lineal unit ; (that is, if the lineal unit be a foot, each of these cubes will have its length, breadth,-and height equal to a foot, and is called a cubic foot). Now the number of these cubes is manifestly equal to the number of squares in the base taken as many times as there are lineal units in the height; therefore content or volume=base><39, ... =51+(19-17)><22=95, Art. 195, EX. 5. a+x+J2ax+x2= a+x—,/2ax+a:’ a+a: __b’+l ./2ax+x2_ 52-1 ’ a’+2a.r+a:2_ a2 +1_ P+I)” 2a.z.'+.'z:2 —2aa:+x2 — 62—1 ’ [)2 ; find .13. By Art. 195, 316. EQUATIONS. a” <62+1>2 452 2ax+x2_ “(1.2-1)“ 2ar,.z'+a:2 __ (bk-1)8 a“ _ 462 ’ (its): (If-1)”, l_(b’+1)’ a _ 462 + _ 4b“ ’ w_bz+l a _ 26 ’ £_bz+l—Qb_(b—l)2. a_ 26 _ 26 ’ a x=2—b(b—1)2. __ _ 2 Ex. 6. a “£23” “ =5; find .12. J2ax—w’ 1+ —-—-—-—=b a—x , flour-~11:2 _______ = -1 2 a2- 2ax+x’ (b ) ’ (pay-141,4)”, (ct—it”); 1+(b-1)”, air =,,/1+(b—l)2, a-x_ 1 a J1+(b-1)2’ 1__1____. ,Jl+(b—1)“” .'. x=a—-——__g———_: . ,jl+(b—1)" EX. 7. W=(4J5—J§Yg find at. 81J§.(J§+4J.;§ = 4 x_ _2 <1./weavers (I J3” slJ's' - 2 W=(4‘~/x_~/§)a EQUATIONS. '317 (4~~/§-/75)3=81~/§, 4JE-J§=3J§, Mat/s, w=3. Ex. 8. f/m+f/e:r=bg find an. (.3/a+x+:/Z—-—a_')3=2a+3:/a2-x”.(:/m+:/a—w), and f/a+x+,f/;:._r by the supposition is equal to b; cubing both sides of the proposed equation, 2a+3,3/ ail-.1:2 . 6 =63, Soyuz—.122: ba—Qa, 2 2a flag-$2: E —-g—b , 62 2a 8 9_ 2: _____ a x (a sb)’ 2 2 2a 3 $2“ "=1 3 3 a: 3 ’ 2 2 w+§=O, or x=—§.. . . . . . . . . Ex. 12. (m+a+Jma+(x+a—Jm’=141(w+a)3. Observing that (A+B)3+(A-B)’=2A8+6AB’, we have 2(x+a)3+6(x+a)(x’+2aa:+b’)=14(m+a)“; .'. m+a=0, or a:=—a. (Art. 208.)... ... Also 6(w’+2aw+b”)=12(w+a)’, .z’+2aw+b’=2.v’+4ax+2a’, EQUATIONS. 319 .z”+2ax+a’=6’- a”, m+a=*~/bg-a2; :r==1=,(/b’—a2—a. . . . . . . .(11). Ex. 13. (1+d+d=)9=g.(1+w*+r); find .2. ]+x+x,_a+1 l+x2+x‘ —a—I ' 1+x-1-a:2 ’ _a+l 1+x+w“-.r(1-x") a—l ' 1+a¢+zv2 ’ a+1 -—- =fi (l—x. l—x); _ 1+x+ar’_a+1 ' 1—a:+.r’_a-1 ’ 1+x2_a a: 1” 2 2 x’—ax+Z=z-1; ' x—Ei a2 1 I _2 4 I _ _ 2 Ex. 14. fi(5_l’+1)(“ 2Z’+1)=-‘-‘--d,- find 2. 11' (I! a: 6(a-5 +2) (a—Qb+x)=a”x—a.z”, b (a—b) (a -Qb)+b (2a—8b)a: +bx’=a2w—aa:"’, (a+b)x"'- (ag—Qab+Sbz)x+b(a—b) (a-26)=0, (a +6) x2—{(a--b) (a-26)+ b(a+b)}.r+b (a—b)(a—26)=O, wiw—(a-ab>}-b{} =0, (a+b)x—(a—b)(a—26)=O, (Art. 208.) a-b .'. w=m(a—Zb), ....... Also w=b ........................... ..(n). x+a é a—b ‘ EX- 15. -2—(-;—-+c)—1, find J7. x+a__ a—b+ (a—b)” x+b~ x+c 4(x+c)” 320 EQUATIONS. fig—1 or gz__—b=€l—_'--b-+———-(m--b)2 x+b x+b x+c 4(w+c)” 1 1 0—6 a—b 53—5-2. or (x+b)(.v+c)=4~(x+c 9 ’ glé_ a-b x+b—- 4(w+c) ’ {a—b—4(c—b)}w=4c(c- b)-b(a-b); x: 4c(c-b)-b(a—b) . a+3b—4c 2 Ex.16. EL. w ;find A. x+b 2x+b+c a—b a—b )2 l+g+__ 1+2x+b+c ’ 2(a-b) (ti-b)” _1+2x+b+c+(2x+b+a)2’ 1 _ 2 __ a—b x+b Qw+b+c_(2x+b+c)2’ c—b a—b x+b = 2x+b +0 ’ Qcx— Zia-62+ cg=ax- bx+ab— 62, (a+b—Qc)x=cg—ab; _ cs—ab — a+ b—Qc ' Ex. 17. I/(1+x)2—',:/(1-w)2=r/1—x2; find 2:. Converting the roots into fractional indices, the equation is (1 +x)§‘— (1—w)_’2;=(1—-x2);"£=(1+w)’%(1--m)%. 2 Dividing by (l-xf, (fl>m—1=<1+x)m l-a: 1T3: ’ 3 l. 1+.r’" 1+x’" 1 1 5 (1-2 "(3 +z-“1I-I’ I! l. (BYJAAAE l—x 2 2 ’ EQUATIONS. 321 131=(1*~/3)"‘ . l—a: 2m ’ =E m_ m w=Mll. (Art. 195). (1*~fl§m+2m 4 Ex. 18. i=a; find as. (l+x) 1+w‘=a(l+a:)‘, =a(1+4x+6w2+423+a24), (1—a)(1+x4)=4a(x+x8)+6ax2, . . . , ‘ ,“1 _ 1 dlwdmg by x, (1—a)<.:c +F,)-41a(m+w>+6a, x2+1 ' 4a(x+1 _ 6a a" l—a x _1-a (w+1>’ 4a +1>__ 6a +2_2+4a‘ x 1—a2 2+4a 41a2 2(1+a) l-a a: l-a l—a + (1—a)2= (1—a)” , 1 2a J2(1+a) 0.0 ____, a: l—a l—a 3 1 2a *J2(1+ a) w+—=———————— x l—a =2p, suppose; '. x”-2px=—-1, x2—2px+p2=p2—l, w—p=i,/p2—1, x=p¢./p2—1. Ex. 19. “Mei/553%,. find 2. Assume w+jd=ffiyi then cubing these equals ar+a+ +_Z/Z)=y; '. x+a+3W=y. N ow, comparing this with the original equation; it appears that y=6 ; W+UZ=~7R f/5=i/5—i/5, w=(i/g—i/d)3. — 21 322 EQUATIONS. Ex. 20. x’—I=0; find all the values of .r. x°-1=(x—1)(x’+x+1)=0, .-. x—1=O, or w=1 . . . . . . . . . . . . . . Also x’+x+1=0, 1 3 :c’+.r+Z-;—1_-Z 1 -—3 z+§==1=’\L2_, x=-_—1—*—2—"/—:§.........(11). Ex. 21. x‘+1=0; find all the values of x. Dividing by x’, w”+xl,=0, 1 In a similar manner to that of the last two Examples may the roots of w‘-1=O, m°+1=0, x6+1=0, and like equations, be easily found. Ex. 22. (x-a)fi—(w+a)fi=b(fl—Jg); find x. a: x-aJ5-xJ5-afi=b(~/_-fi), a(J5+~/5)=(w- b)(J5—J5)=(~/5+~/5)(J5-JF)’. fish-=0, J5=-JZ, and 2:6 ............. ..(1). Also fl—Jfi=fi, fl=JZi+~/5, w=(~/Z+JZ)’ ......... ..(II). EQUATIONS. 323 Ex. 23. a+w+J2ax+x”=Jax~w2+J§a2—ax—w2; find .r. a +x—Jaw—x2=~/2a”— ax-fwz-Jm, #275. {J53 -fi}. a’+ 3a.22-2(a + x)~/?z_.r':a.7’= (2a + x) {an—QM , 2am: ag—Qam, 2 ax—zc”=a-2x, 4am-4w’: a’— 4am+4a’, Ex. 24. a+ (b+J5)J5=(b-J§)Jm; find as. a +w +Jm= “Mg—fl), (a +x+WY= 62(2a + 2x—2JW), ' =262(a+x—m , (a+x+m)3=259.{a +;]’—2ax+ .112}, =2a262, a+w+ 2ax+x2= 8 241262, 2dx+afi=aba dab-22/2dfib'§(d+d)+d*+2ad+w’, —23 2a”? 2 ab, 8 _ x_g a ab\8/_2_:_a ' -2\/ 2W2 ab - Ex. 25. L— ‘mg=a+b; find .2. 1—:z:+,./1+a:2 2a 1+x”=(a+b)(1—x)+(a+b),/1+.r’, 21—2 324 EQUATIONS. (a—b) 1+a:2=(a+b)(1-a:), 1"” a—b JfiF=bTlf 1+x2—2x _ ($19)“ 1+2:2 _ a+b ’ 2x (a—b)’ 4a!) 1+.2‘-'= _ (a+b)”= (a+b)“ (1+2)2 __ (a+ b)”+ 4ab _ (a—b)2+8ab (1—x)’_(a+b)’—4ab _ (a-b)’ ’ l-l-.1r_\/1+ 8ab ' 1_x_ (a_b)2: / 8ab 1+-———-2--1 _-—--_- _ ((5-5) _(a—5)2'{\/1 8ab 1F _ 8ab _ Sab ' +(Zz-—b_)3_ Ex. 26. fii———— JT;=Q/x’-ag.{Jxg+ax—JmQ—am}; find .2. Squaring, Mil = , /.2:2- a”.2.x{.x—JM}, x + JxsL as (L— ‘W = 2xJx2— a”, 2.1:”- 412- 222/.122— a2= 2a2xJx2— a2, 2022— a2= 2x(a2+ 1). /x”- a”, film‘— 4a”w”+ a4: 4a‘x‘+ 8a2a:‘+ 411:“— 41a6w2— 8a‘w2— 412222, a4: 4a4w4+ 8a2:c‘- 4a622- 8414.29, a2: 4(a2+ 2 )24— 4a2(a2+ 2)::2’, 2 a 4 22 m—ax=———- 4(a2+2)’ 4 2 2+a‘ a2 a4+2a2+1 .r—aa: —=—.———————— 41 41 a2+2 ’ w2=g{a* a’+l }, m 2 ./a’+2 ggg 'SNOIIvnoa .<_I_-v.6.>./;=, I-Pg 129—1 17-1 ) C = x _ + (Fog) “ (8 r—vsI v—I _ P—I _ fi‘ 22-[ 77—I_ ‘7 =g+zw~V/~il'+8+8x5 file/‘Hgwfijifii/Wwwzfl ‘p_~[= __ 5 g +3217 fa." + [+8205 ‘29 game/“Menard _ x pug . --8_I_%M7/~lx_l_In!”g 8% XH Z I I II. =x ... 0 WW '(H)"'L/~'I7=F6=1’ ‘7? =g—w _ I g ‘Lf$=F=6—$ ._ ‘1 = g w ‘zII=Is+w8P-.w 21 .., a "' ‘5g+291=1+a:'5-Z:v ‘5+~v=[+w5-$x’ weal-Ar 1m {Aw—=11» w iéi'f/Ffiffl'li’F Pug fo=wfl+rsf f{Zfi'£/?+L'3'/”}'fi£/?=€fi/”-I—w ‘m/Z'I-Wr- ‘¥(F5+w)'fi(I+wE—zw)=g+m 5-1-2' '2' pug fz+wg-gx/:+I=§fi/*5—x '15 'x5[ , J’fQ-isfl W” _ _ . 5333/” .i 27 z(5+zv/~=F 77) . 22%” 6212+ 222/Ws fil‘fi‘ 'SNOIINIIOU 93g -(I)..--... v C I. O l C O I l I I l O U O C I O I O ... -‘0 =(I—.w.v) 1211/“(I—v)+.(I—w)v—(1—.w.v)(I—.v) ‘0= Meow (1— 1’) +.(I—w)v— (I—.v)- ( [-z”)axz” ‘zxzmg +zvg= (I-axzv) M (1—29) +z([- am) 29 ~52? + are,” +I+ 5x529 ‘axzvg +6v5=([~zwav)§vi/” (1-29) + z( [-x29)29—(629+1) (Ina-3'10) 211.1917 +2917 = ( has) 'wi/‘(I- 2) fish—1'”) 195'”(2”+1X1+8%”) % ‘zx'zm'z +aviv= (619+ 1)([+amg +8252) + (l-awzv) 2111 /‘ (I— 1% +.( I—”)( I+wz—.=v.v) ‘MvZ=fil/”(I+wv)+ (I—vXI—wv) ‘wwv—zwfl 22+ (xv—I)v=£il_/“'v-—zv +I +zvv-I zv+I/“:v-zx+I/"+xv—1 . _x zm+I/*v—£wfi-_]/*+wv-[ 08 H 'a’ pug f v: I I v _x O O afl—I /”v=I=I d; _' ' flan—{form} 322-1 /"=F :2 [—5295 /"+82)) ‘80—I/“; =I—,v7¢/“.r+:vzv+v (1—812) .10 ‘av—I/“GF =',Inousp _8('lmousp) I_5v .{(aw+ a”) —a(x +v)av}_ = ‘8(w+ 0);) -8w +av= ‘a_(x+ v)sv—xvg—z(x+ n) =xvz; -8(:v+ v) (522 —[) osIV {gird-av) Qz(x+v)zD}(I-av) 10 ‘iaw "'29 _sxzlv+xs'vz +vv}([_zv) 10 (K [_87))sx + $806+ (I_zv)zv}([_z77) ‘10 2017.”); + ( I—zp)xspz +.%( I *2”) 1° ‘81. +awavz*sxvv+(t_sv) xspg+aflz(l_av) ‘IO ‘( [7475)?” “six 817+ ”( 1'17)» 51- sousH ssmossq IomIsumu sq; ‘xomnnnousp sq; Kq 110902.13 sq; 30 .Io;'euunousp pue .Io;e.IsIunu sq; fiuyildmnm [—198] x+xzv+v(I-zv) _ . w Pay fwZ_‘(x+vX“n_U=I—,va wsmsv+v(I—,v) 65 x3 1.68 'SNOIIvnOa .§q+1/~=F89’ ¥=w ". frq+I/~¥zq=sw ‘79 + 1:99 +8x895_fx ‘ 1:51.59 z_ix usq; f ssoddns ‘zqg={,v-I /" i I} “62 - = ins ¢_z'1____ £23 A“; 112.7 fil ‘7 r .(z)+(l 3)T+.17 A_zm 322 +u —u a: v v I I % s _a__+.[+_z£ x' v . ’1' +8275” —§fl ‘ [—u u xv vx'l'axz” +137 [+u [_u5 _z(zx _ap) _ (vm'i'zwzv _zpXI'i' u) I .x Pu c(Zw—Zv '88 'X y '. w) I <.w+.w.v+.v>o—u> H c 8 = o. ‘8__ 8 = 0 aI/‘zqvfi I 9*? ( g 8 ‘8 g =——|r —- =——{l.' gl/‘ss r1 9F 2 6 6 8 8 6 8 8 ‘-—=;7+—= —>+--—a: .10 ‘ -=<->+--w as 91.17 ws“ H9181? ws“ r a s g— .10 ‘ -—3—- =-—— I I sF as ‘r/ (2:? =2 5“ /\ £._ 6 1+5 I+ws “r +219 “w c =5“ A _9__ 8 :08 ,w +308 ,2 ‘ s s —-:v a: - —-a;' = (8 >/‘ (. )a 6 - 8 (6 8 —+w —+ — —+——-w .rzg_ mg a“)? __6 s , _ ____z_‘_8 _“W{Gar/“2221282) *5 r + <§—:r)5 I = i - (g-w) ~22 'Xu '2’ u 5 pg 25 I (E- W 8 6 $+ax8 'A{(I+.v—I/”)(I—.v+I/‘)}€-=F=w ‘IIOBQHPQI Kq ‘10 ‘{(rv_I/‘*I)5/~¥vD_I/~¥I_/~}%=F=x .'. ‘{v”"'I/ £17132: v77 _I/ zivv-z}!izz+_[=rq+[ z'l) ‘{.v—I/**I}—I-- =zq WEI 688 'SNOIdvnos ‘0flew—17)+(,w-1)(.q—.v)/~(:w*v)wz—(.w+1).(w-v)+(dv—I)zq— (.w-I).w.v w—I ar—I w—I .0 =z(T9_p)L—+5(L_> + WW (x0 _v)x5 _aq _zwzv '98 'XH 818+]: wo—v- ' fll¥= fl ¥=Ir 81'. :91 = i = M 99 81x9 6+5”? : 85 = 88 x ___ <6_zwq7) .(8I)><% a(st) 09 8 6+,m £217.21: 6+5w17 09 K 6‘5”? (6 +.wr)% +(6+,wr)z ‘6-53'%__8_Ix +8+$25+8—1‘5=(6"zm7)<fl) ‘Sutqng 6+2»? 8 s-wz s+w8 8 6+2“? 8 8 ' .61.”? 8t 8+wz g—wz .w u 0—.~= __ Q I P H 6%” 8 gf 178 xg “ '{5—2 +Z+u Z .10 I I/ }22 c{I:_ZL/\’+ I_'u ..- [+u g-ug v _— ¢5_u Jo ¢fl_5_z— -” ‘0=Z-(“i11>”/"wz—2+ “’ w __l_____ 8w_I ‘0=2+5-G+1)fi 2+1 g-{5-xifl-fi}(x-1) ‘0=(w—I)z+ m /\ Agra-m-{d-(§-+1)v}.v}(8w-[) wrisr—Z—GH)» New '2 PHH fI—wa/‘HH =Z-(€-+I)”/\(w-[) '92 221 .27¥9 __— _- 9 *1? f q im=w(v¥o) ‘g¥=:vv=+=wo-v ‘sq =8zvzv+ (xo—v)mvg=l=a(xo—v) ‘(xo— 2?) mug; (ax—1);? =s(zro -- 22) +39 -z22 ‘(.r—I)(w—v)wvzq=.(.w—I).v= <.w-I>(.w—v)+ (.r—IXA —.v) ‘(zw-I)(wa- 22)xvz¥,(,x—1),v +,(:m-n)5:r =z(.ra - v)+ (zw- [)(zq 122) ‘(z'T—I)v¥= (x0 _v)x"z(x9 —v)+ (sx_l)(zq —3lv)/~ ‘z(zx_I)zv=z(wa _ n)ax+ z(x0_v)+ (sw_I)(zq _zv)/~(m0_v)w5 “hero-D) + (zm_I)(zq is”) ‘O=s(wo _v) + (zx_[)(zq_aD)/~(x9_v)w5 '- zwz(x9 _ 27) +a(w9 "' 27) + (5x _ 1X83? 2” Ta”) _ (ax _I) (69 _av) EQUATIONS. 331 _‘__-___—_- 0—2 =1= _—1—1 *,~/a+1-.12 =1= a+1_1 2 w2+20_____~/5:_2—— .x+( _2 ) =< ) * a+1—1 a+1-1 xu'w—a-Q _ a-2 X(*‘Va_1)’ i _1 ' w= LXGhA/a—l—l), (Zr-2 w.(i,/a-1-1), ~ = (a—1)-1 _ (*0 a+1-1X*~/a——1—1)_ * a+1-1 _(*~/_a—1+1)(*./fi-1)1./E+1 ' Ex. 37. wig/=8, }; find as and y. ar"—_y‘=14.~560, ASS‘LIZS (Art.213), then x—y=2'v=8; 11:4. Also w2+yg= (z + 41)’+(2—4~)’= 222+ 32, and w”— 512: (2 + 4)”- (z — 4)”=1 62, (x’+y2) (mg—y”) = .24— 31‘: (222+ 32)1 62:1 4560 ; °. 28+162=455, 2‘+1622=4552=65 x 72, 24+ 6522+ ((2%); 4929+ 65 x 72 + (%)2, 22+-6—5-=72+6~5 , 2 2 22:72; z==7. Hence x=z+4»=ll,} and y=2—4,=3. ‘ . . a: bc—a: = .2: —ac 1 ' Ex 38xy(a_y+bx(-xy)Zlbiify—c;."(2%; find x and 'y' From (1) c(bw+a_y)=xy(x+y), C(bfv+lly—wy)=xy(x+y—c) .... _ Q__qgg . . . . c - my , dIVIdmg (2) by (3), 01‘ ($y)g=abcs, OI' xy=c.,‘/a_b_; 332‘. EQUATIONS. _x_ _ dry—ac __ cJeTlI—ac .9 _bv—xy _bc—m/Efi’ = JAE-£195 MS—fi) b-M ./b‘- <fi-fi) ’ and from (1) '. x’=wy.§=c,/ab.,\/2—l=ac, '. 2=*Ja_c} and y=iJZE Ex. 39. w3+y(xy-1)=O...(1) ys—w(xy+1)=0...(2) From (1) w‘+ m°y2—-x_y=O. From y4-x2y2—xy=0 ; ar‘—_y‘+2.r2_2/2=0, w‘+2x2y2+y4=2y4, m”+y2=J2.y2, w2=(~/§-1).z/2> Q49 }; find a: andy. ;—:=~/§-1, 0 Q s Q on (a) .'. ;=JJ§—1. . w3_y l-wfil, AgaIn from (1) and ?—5-l+xy, __ 4 .. =Z;=3-2~/§, from (a) 2 2—2 1 by Art. 195, wy=Zg—~/—§=—~/—.2_; . .£_ __1_ _f. .. 2yxy_22_J§JJ2 1, 3”: '}//2{1=\/ iHé—l) And = i = 4 1 . 9 J2“ ~/2(~/'2-1) Ex. 40. xy=a(a:+y). . . (1) xz=b(x+2) . . find 11;, y, z. yz=c(y+ z) . . .(3) EQUATIONS. 333 From (1) w=-I-, my a 1 1 1 01° “+'_'=—o a: y a 1 1 1 From (2) 5+E-—b-, _1__1_1__1. ' y z_a 6' From l+-1-=-1-; y z c _2._l+l_..1_ y_a c b’ 2 1 1 1 and 2:??? 2 1 1 1 also ;-Z+-5——c-, ‘ x_ 2a60 " _ac+bc—ab _ 2abc y_ab+bc—ac ' 2abc km Ex. 41. a:(.r+y+z)=a2, y(x+y+z)=b”, z(x+y+z)=c’; find .31, y, 2.? Adding all together, (a: +y+ 2)(x+y+z)=a’+ b’+c”; ' '. x+y+2=*~/m; a2 ./a2+62+02’ 62 Q ~ - o a. 2 =:l:-_.____, () y ./a2+b’+c’ 02 JW’ '. from (1) w=i .... .. 2==1= Ex. 42. x(y+z)”=1+a’; w+y=g+z; and yz=136;‘find 2,9, 2. 1+2a _x 3 From (1) (y+z)2= . __ Llifit __ 0. Z)— x 334 EQUATIONS. But from (2) y—g=g_w’ 1+a8 3 3 9 9 2 - x —Z-(§—x>-Z—3x+x, 1+ a8 1+a3= xa- 3x2+ 3.x, a”: w3~ 3w2+ 8x—1 , a=w—1, -. x=a+1. Hence from (1) (a+1)(y+z)’=1+a3, (y+z)”==a2—a+1, y+z=*~/E’:ETI~=* JW, 1 also y—z=-2--a; 2y=-;-—a=l=~/m Elx.43. m(y+z)=a, y(.r+z)=b, z(a:+y)=c; find m,y, z. ‘ my+xz=a, wy+yz=b, xz+yz==c; .'. 2wy_+a:z+y2=a+b; 2xy=a+b-c. Similarly 2yz=b+c—a, and 2w2==a+c-b; 2xy.2mz=2x2= (a+b-c).(a+c— b) , 2yz b+c-a 2xy.2_yz __QyL (a+b—c).(b+c*,a) 2m: _ — a+c—b . ’ 2yz.2wz=222: (b+c—a).(a+c~b) 2mg _ a+b-c ; EQUATIONS. 335 (a+b-c).(a+c—b) 2(b+c-a) ’ _* (a+b--c).(b+c—a) y_ \/ 2(a+c—b) ’ (610“a).(a +c—b) 2(a+b—0) ° '. w=:l: 2:5: Ex. 44. wy+n(x+y)=a, wz+n(a:+z)=b, yz+n(y+z)=c; find x,_y, and 2. From (I), (w+n)(y+n)=a+n’, .... .. (2), (a:+n)(zl+n)=b+n’, .... .. (3), (y+n)(2+n)=c+n’. _ Multiply the 1st and 2"“ together, and divide by the 3’“, of these equations, then 2 2_, (y+n)(z+n) or (x+n)_ (oi-n”) ’ a x+n=*\/(a+n”)(b+n2), c+n’ (a + n2)(b +12”) and x=—n* 2 0+2: Multiply the 1"t and 3"1 together, and divide by the 2"“,- then we have (a+ n2)(c+ n”) b+n2 ’ y+72==h and y=—n* gig—3% . Multiply the 2‘“1 and 3" together, and divide by the 1“; then (b+n"’)(c+n”) a+n2 and z=._n=l= W ‘V a+n’ ' Ex. 45. x“yb=r, and w°y"=s; find a: and y. 2+72==l= , Raising the first equation to the dth power, and the second to the b“1 power, 336 EQUATIONS; ‘ w“y”=r", and xb‘ybd=s"; mad '. F=§, or a: 1 8 .a _1__ Similarly y=(;-c-)“d"’”. a M_7' , Ex. 46. flaky“, and 3“ =1”; find a: and _y. ‘ an From first equation x“ =y, .... .. second .r”'+—y=y; H-y d __ - .r+ '. x4“=.r‘+~", or ——'y=-l—; 4a x+y (x+y)2=4a’, or x+y=2a; " m=32> $22.94“; .'. y2+y=2a. From which it is found that y=%{—1*J8a+1}; and then 1 _— £U= §{4:a+1¥~/8a +1}. Ex. 47. J5?“ =Jafi+3 .... .. (1)1 and w(y+1)’=3fi (y8+ 156) . . . . From (I), multiplying the N umerator and Denominator of the first fraction by “, /y’+1—1, of the second by Jx+9—3, then multiplying the resulting equation by (l), Jy2+1+1 =./.z+9+3 _ Jy’n-l x+9—3 ’ um J3711= 3 , (Art. 195). ; find a: and y. 2+1=x+9, y 9 o .0 992: x, EQUATIONS. 332 From (2), 9y9(_92+ 2y +l)= 36y8+ 64, < 99"(312- 23 +1 ) = 64, 3.9(.y~1)==i=8: And w=9y2=g{19i~/T65}. Ex. 48. 2w+~/w2_92=5{\/'%9+\/HH ‘71 2 ; find a: and y. x+yii x~2 % and (?)+(?’)=9 j 33+ $— Assume —7'y='v, and 2'9 rd then 2x= 2(1) + w), = 71), also rug—9% 4mm, and y: 1) - w ; by substitution in the first equation 14: "v—w' 2(v+w)+2fl7v= 7 or v+w +Jvwv=m; vé— 122%: 7. But 25+ mg: 9, from the second equation; 20km", or v: 4, and 21v%=2, or 10:1; x=5, and y=3. By finding all the values of v and 12) which satisfy the equations v3-64=0, w8—1=O, by the method employed in Ex. 20, other values of x and y may be determined. 22 338 EQUATIONS. Ex. 49. y-W “Lg—1 =‘Y x+1 yz-Q .r’-1 a? and ig/Ey’x—l J ; find a: and y. From 2nd equation 9“— 4~xy2= - 4, y‘- 4.1:.y9-i- 4w”=4x’— 4, f—2a:=*2./x’—l, y8=2x*2Jx2-1; and y=*(./x+1'*~/E:i) .... ..(1). Substituting for y in the first equation, and taking the upper signs m (10’ m .53 2 J55 m 2.2 *"é?+§'-§r= .. ’ 3JFF=JTH. 9x—9=x+l, 8.2::10; '.x=18—0=1;1-. And y=Jm+¢5ii=A By taking the lower signs in (1) other values of .1: and y may be obtained. 1 E o 50: 2 é__ a... 1 X 417,? y‘r findxandy. and w’g— 3=x§y5(x§-y5) . . .(2) From (I), y(.r’+ = 4ym%+ 4, y(x°+ .r_y+ =xy9+ 4yx§+ 4 ; y’i’ x+%)=*(w5y+2). i Taking the + sign, xy%-x5y=Q_-y_2. , EQUATIONS. 839‘ ‘3 gt 0. $%yé(wé_yt)=g_%; % by (2), xg-3=2—'22— , 2th y%=10 ......... ..(3). Also from (2), 3x5y5(x%—y5)=3x§-9, “vi-yak by (3); = fi-t/t-awtfiot-tshlt extract'cube root, x§—y%=—1 ; yé=m5+l . . . . . . . Substituting in (3), 2x5+w§+1+3x1i(.r§+1)=10, 3w%+3a:+8:c5=9, xg+x+xi=3 ,- (t%-1)+(.t-1)+(t%-1)=o, or (x5-1)(w+2x5+3)=0; w‘5—1=O, and x=l, substituting in (4), y%=1+1=2 ; _y=4t.} Also m+2x5=-3; '. w+2xi+1=(a:5+1)2=-2, which is impossible; '. x=1, y=4, are the required values. Ex. 51. a(1-ty)=tJ1-_y2...(1) and J5(1—xy)=y—x. . . . . From (2), x5-w%y=y—w; }; find a: and y. wé+x _ 1+.rg , .'- xii-ta? 1—x2 l-xy=l—~ —-———- = 1+.Tg 1+.r’3‘ , .r+2.r%+a:2 and 1— 2=1--———————— y 1+2x%+x3’ _ 1—a;'--.r’+a:a __ (1—w)2(1+ .r) (natty mtg)” 22—2 340 EQUATIONS. _ 2 — _— Substituting in (l), a l—f— a: 1 xJl+w; '1+x%= . 1+w'3 a(l+x)=xJm; also 1—x=0, or w=l. . . . . . . . . . . ..(i), .. w=a.~/l_+_a_:-; also 1+x=0, or x=—l............(ii). x2=a2+a9x3 4 a4 '2 2 2. x —a.r +——=—+a 4‘ 4t ’ a2—1 2a (iv) —— , __________.—___-__,——— Q ~ 0 o o . a2+1 ~/(a2+1+aJ(t2+ 4t)2—1- 2a EX. 52. y‘=w2(ay—bw), and x2=ax—b_y; find x and 3}. Also y=1, or —1, or i 4 From first equation, %;=ay— bar, and by second equation, x2=aw—b_y; Lb y“ ay—bw am __4: z“. -1 x at?! )3] a_b:2 {I} y at—b . Assume ;=t, then at“: 5:1”, '. at‘—bt°=at—b, or at(t8—1)—b(t5-])=0, az(t*+z+1)-b(t4+t3+12+t+1)=0; t-1=0, or i=1 . . . . . . . . . . . . . Also bt‘+(b—a)13+(b-a)t’+(b—a)t+b=0, 60%;)—(a-b)(f+%>—(a—b)=0, b t+%)2--(a-b)_*— z" b ’ b - t l__a_:_l.)-L\/(—_L—b_ig+l_zfl ‘ "L?" 2b "' 46’ b ’ __a-bd=,\/a2+2ab+5122 " 26 =m, suppose; EQUATIONS. 341 Then t’—mt==-1, m m2 . =_ A: __— o o o c a 0 o 01 t 2 \/ 4‘ l If from (1) i=1, then w=y=a~b . . . . . . . . . . . . . . . . . . . . . . . 2 1;. L.__ If 1=Zg.=1=\/7—Z—-1, x=a—bt=a—é(mi~/m2—4), .(ii). and y=tx= % {a- g (m=l=~/ m2—4)}{m=l=,~/ 1922-4} . . . . . . . . . . . . . . . (iii). where m: Qlb{(.'—b=l=~/ag+2ab+5b'l}. Ex. 53. .:r"+'y5=.ary(a:+_y)3 . . . . fi d d and ’ n m an y' 'From (1), dividing by x+y, 684" way +6199”- wys+y‘= WW4 2% +.92), =x3y+ 2x2y2+ .1193, x‘- Qx‘ly — xgyz— 2xy3+y4= 0 ; . . . x2 a: ’ d1V1d1ng by xgy’, 55— 2 -1- 2 {1% == 0, 2 2 ‘35 +33%, 2 (5 +3)=1, .7 y fi.2)_t(@+g)+1=t, .1/ x y “1 +2=1i2=3, or -1. a: Q]?! Taking the first value and multiplying by 5, 2 <§>+1=3.f, .9 .9 .x 2 x 9 5, (git-5+1"? g_3=!=\/_5_ O y— 2 0 But from (2), .agy‘=(x+y)3, . M 5_(‘x )3 3_(5*~/53 a . 2 .y-y-H .y- -—2—)._y, 342 EQUATIONS. , ,_ (ma/5)“ _50e20j5' __ 50i10~/_5- "‘9 4(3*J5)_ 3.445 4 5 / 2 — U.Cy=§ _ /50e=20~/5_5\/ 22 _- and {13-— Ex. 54. (x°+1)y= (32+l)a:3...(l) _ and (y°+1)w=9(w’+l)y"...(2)} ’ find ’0 and 9' I i From (I), (wa-t- £—8)y=y’+1, 1 1 '. wa+—= +- . . . . . . . .. 3 . x. y y () 5 a l 8 1 __< _1_ _ or 5(y +§§>-3 w+x), . _1. l L a < l) l . .3(y“+y8)+y+y-a:+3 w+$ +w8, 1 1 1 a y8+55+3(y+5)=3 , From (2), y8+55 =9(.v+ 1) 1 l a _ 3 l I a -' And x+?=<\/%(b—0)4+(62+02)2x2 = (b—cf— (62+ c”)c.z:}>< d2- (6 + c)‘+ (62+ 02)’a:2; find .r. 'SNOImvnb'a 9:99 ‘(29) Aq ‘za-g-rmqy= ‘Za—(a+n)g=g-5n17 ‘00) 4q ‘(n— a)a= ‘([+ ag—n—mzg)a= ‘(I-n)(I—aa)a=s—nz any ‘(I) “IO-Ia .(n).......l=(a+n) _anz 10 c ha .10.} a pun ‘%=(Q+n)g—rmql ‘(5) moi} ueqq law .103 n QIIJM (a) """ "s=(I—tfiEXI—twa) PUm 'fi' pus w pug f I—zzv I O I . Q I 6. _ . OX (I) (. z I). $w,7_zxz+g 69 a _ Zq +80 @_ au-t-zw __ 8q—zo 22_ 2mm _ 5 (824—3214)? =09 pue ‘Q=vuzu+a~(,u +52“) ‘60 _5q = um , ', ‘0 = 3052‘52’“ +wvuw(zu +6x8(5u +82“) ‘0 = oqsvauzmg + w(ao no) v(,u -,ut) +szro 95(zu +,ut)2; ‘Oqauzugu -8w)5 — x80n(,u qua) =zxoqz(zu +zut) g + oqsniug + xzqv(,u -,zu) ‘gaovgu +5ua) (,u -,w) + 1093298246,“ -,zu.) 5 -zqgv,u(zu —zut) = 309451! +,w)z + mo 98 ova (,u +zw) z +axzqo(zu +zw) (,u - ,zu) +5 980,24 (an Wm) : zmz(z_u +32“) +zvvu = zwzoz(zu +37“) + xoqv(zu+sm)zuz _zQzvvu v(zu _Zw) w99(zu +8w)5 +zqv(zu '17“) ¢ 5xz(zu +zw) +32%“ ___ arzazgu +31“) + w0qv(zu +Em)zuz _sQzvt'u 80(1ru wt“) xoqvz(zu +aw>5 +z9av(vu _rw) ‘ zws(su +zw’) +zvvu = ax89z(au +zw) + $99245” +zw)zu5 “flap?” axs(zu +52“) +577?!“ zxaaz(au +52“) + w 9917 (zu +zw)aw5 +z9809w ‘ ,w,(,u +zw) +5152: /\ _ xo(zu+ zm) -qvzu Help 8w3(6u +Zflf'w — “'0 (Bu +52“) + qvaw ‘(zu+zw)%-=ao+zq mp 0s ‘u=0-q pue ‘ut=a+q qe'I PROBLEMS. 347 4llg—4Ul)+'l)2=l, 2u-v==*=1, -. v==2uel . . . . . Taking the upper sign, and substituting in (a), 4u2—2u—n—2u+1=1, 5 or 4ng=5u, n=2, or 0 ....... . . Taking the lower Sign, and substituting in (a), 4242+2u-u—2u—1=1, 4u2—u=2, O 1 2 1 33 4u-U+1 .. _ _ > a b ’ ab ’ and ac+ bc>ab. Similarly it may be shewn that ab+bc>ac, and ab+ac>bc. PROB. 7. The fore-wheel of a coach makes 6 revolutions more than the hind wheel in going 120 yards; but if the circumference of each wheel be increased 1 yard, the fore-wheel will make only 4 revolutions more than the hind-wheel in the same distance. Required the circumference of each wheel. Let x=circumf. of the hind-wheel, in yards; 31: ................ .. fore-wheel .......... .. 120 . . then —x-=number of revolutlons by former In 120 yards ; 5%): . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. latter . . . . . . . . . . . . . . . . . .. .9 120 120 I... __"' + ='—' , .9 or 20x—20y=xy ................ . . . 120 120 ) — 4‘:— Agam on the 2nd suppOSItlon, x+1+ y+1 , or 30(y+1)+(x+1)(‘y+1)=30(a+1); 29.27—31y=.ry+1 .... ..(2). 352 PROBLEMS. Subtr. (1) from (2), 9x—l]_y=l, 9x=11y+l ................ But from (1), 20x9x~20>§9y=9ajy, 20(11,2/+1)—180y=11‘2/2+]/, or lly’—39y=20, 7,_s22+§q ’_1521+39_2401 J 11J as (aw n'cza)“ Bra-t .2 " 22 " ’ 11’ 11y+1 and x:- ' =5, or—---: 9 9 the circumference of the wheels are 4 and 5 yards respectively. PROB. 8. Find two numbers in the ratio of m to n, whose sum is equal to their product. Let mm, and me, be the two numbers, then - mx+nx=mw.n.r, m + n = mnas, m '. m.r=--+1 n , the two numbers. n and mu: —- +1 722 P303. 9. The product of two numbers is p, and the difference of their cubes is equal to m times the cube of their difference. Find the numbers. Let x+2, and .r—y, be the two numbers, then (x+ —g/), or w2—y2r-p, (w +y)”— (4 —.1/)3= 1429?} ' From 2nd equation, 2y3+ 6x’y=8my3} .... .. 1st 6w2y—633=6py 8y3=8my3—6pg/, 4(m__l)y2-:3P> . _1 \/ M ' ' y— 2 m—l ' 1 3p (4m-1)p ] 2= 2 '.::—,—--—-- :z. _ ' A so .1: y +p 4 __1 + 4(m__n ) PROBLEMS. 353 . x__1 (Mn-4);) ' ' —2 m—l ‘ Hence the required numbers are 1,~/(4m—1)P+~/3P and _1_.~/(4m-1)P-/5’5_ 2 Jm_1 ’ 2 Jh-l P303. 10. Find two numbers whose product is equal to the difference of their squares, and the sum of their squares equal to the difference of their cubes. Let a: and my be the two numbers, then x’y=xQyQ-xg .... and w2y2+x2=w8y3—x3 .... ' From (I), y=yg-l, and y=%(~/—5+1). Q. the required numbers are é-Jg, and i6 Pnon. 11. There are four numbers in Arithmetical Progression. The sum of the two extremes is 8, and the product of the means is 15. What are the numbers? Let x—fa’y, w—y, a:+_y, x+3y, be the numbers; then by the question, x—3y+x+3_y=8, '. w=4. Also (x-y)(x+y)=15, or x2-y2=15, '. 16—y2=15, or y=1; the numbers are 1, 3, 5, 7. 23 354 PROBLEMS. PROB. 12. There are three numbers in Geometrical Progression, whose product is 64, and sum 14. What are the numbers? Let Z, w, my be the numbers; then, by the question, gammy or m8=64, x=4. 3} Also -x-+w+x ~14 or l+1+ —-1_%-Z y y“ ’ y y_x_2: 5 5 B 25 9 . 2_____ __ =_____ =___ "y 2y+(4> 16 1 16’ 5=1=3 1 y=T=Qa 01 g; the numbers are 2, 4, 8, or 8, 4, 2. PROB. 13. Two labourers A and B, whose rates of working are as 3 to 5, were employed to dig a ditch; A worked 12 hours and B 10 hours a day: B being called away, A worked one day alone in order to com- plete the work: when they were paid, B received as many pence more than A as the number of days they worked together. Now, had B been called away a day sooner, A would have received 39. 11d. more than B at the conclusion of the work. Required their respective daily wages, on supposition that the payment to each was in proportion to the work per- formed. Let a: be the number of days they worked together; 37?), and 512/, the work per hour of A and B respectively ; 36w=A’s daily work, 50w=B’s ............. .. let then 3Gy=A’s daily wages, in pence, 50y==B’s ......................... .. and we have 50yxx-Sfiy(ar+l)=x, or 14xy—36y=a: ............. Again, on the second supposition, A and B work w—l days together, the work done in that time =86w(x-l), but the whole Work =86wx+36wg work left to be done by A=122¢22g PROBLEMS. 355 1 2 2 w days, .'.A ork -1+ . w sa: 36w 7 or x+2+—1--85 days. Hence (.70 +2 + i7—8~)36y—(x—1),5Oy=47, or 136y—14xy=47.. .. . . . . . 86y _ 14y—1-’ 14><36y2 14y—1 14xl3632—136y—14x3692=47X14y-47, 14009L 794y=—47, ' ,2 794 897 2_ 91809_ J_14009 1400 “(1400)” 'From (1), x: '0 =47) _3974303_1 or 47_ ~ 1400 '2’ 700’ A’s daily wages=36><-%=18d.=1s. 6d. and B’s . . . . . . . . . . .=50x%=25cl.=2.s'.1d. PROB. 14. It was calculated that, if the gross revenue of a state were increased in the proportion of' 271-: 1, after deducting the interes of the national debt and the cost of collection (the latter of which varies as the square root of the sum collected), the available income would be increased in the proportion of 3.3% : 1. If', on the other hand, the gross revenue were diminished in the proportion of 1% : 1, the available in- come would be reduced in the proportion of' 7%: 1, and would in fact amount only to 4 millions. Find the amount of the revenue, and the interest of the debt. Let x=the gross revenue, / y=the interest on the debt, in millions of pounds slerlz'ng. z=the expense of' collection, a: , . _ ‘ x Then 9—4 =1ncreased revenue, and expense of collecting : 2 z: : MQZ; '. expense of collecting:%; 9x I o o u TEL‘dImITHSl‘IGd revenue, and expense of' collectmg== 7L- , 23—2 356 PROBLEMS. 92: 3a ... i-E—y—E-zlls - . . . . . . . . . "(1). Also x—y-z : -$19—2-—y—% :: 7g: 1 :: 31 : 4, by (1), x-y—z=31 .... ..... Again, ar—y—z : gf-y—‘iz'in 1 : 2: 3l : 109; x 32 by (2), 9Z-—_2,1—--2—=109 .... Subtracting (1) from (2) {%-§=27, 5a: a ’ ............. .. (2) from -4T—§=78, _ 5.17 7.1:__ _ . 4 ——8—-78—54-24, 3.2: —§ =24, and x=64. 7w_ T6 27=28-27=1, and 2:4. 2 Hence 1: Also y=x-z— 31=60—-31=29. PROB. 15. A steam-boat sets out from London 3 miles behind a wherry, and having got to the same distance a-head it overtakes a barge floating down the stream, and reaches Gravesend 1%— hours afterwards. Having Waited to land the passengers %th of the time of coming down, it starts to return, and meets the wherry in ~2- of an hour, the barge being then 5% miles a-head of the steam-boat, and arrives at London in the same time that the wherry was in coming down. Find the distance between London and Gravesend, and the rate of each vessel. Let x=rate of the boat, y=rate of the wherry, t=rate of the tide, that is, of the barge; then boat’s speed against the tide=wherry’s speed with it, $439.“, =‘”_2__.Z. m- 3 -- Hence :c+ or a; y=boat’s speed down, I 6 . fi=t1me before the boat overtakes the barge; PROBLEMS. 357 —6— +§=whole time of boat down, .r—y 2 ——6——+ é-=time for landing the passengers; 5(x—y) 10 b 6 3 3 3 6 l —— + -— +—+— or —_ +—=interval of time between the boat 5(x—y) 10 a 2 5(x—y) 20 . _ passing the barge, and meeting the wherry in returning, in which time . -_ __ __ ._ __ _ d h the barge moves 0\81 2 {5(a) y)+20}, Oi 5+40(x y) miles, an t e boat has come up £2?- miles, ‘. + g + 2% x—y) + 221% distance from Gravesend when boat passed the barge down)=g. avg—'1], from which equation we get 4x+y=39 ................... .1. . Again 8441 = time wherry takes to get from 3 miles behind to 5% miles a-head of the barge, 33 6 51 00 . . . . . - s - s - t ~ - ..(2), from which two equations (1) and (2), x=9, y=3. Hence also distance from London to Gravesend=time downxspeed, = 2x12 =30 miles. PROB. 16. A and B travelled on the same road and at the same rate to London. At the 50th mile-stone from London A overtook a flock of geese, which travelled at the rate of 3 miles in 2 hours; and 2 hours after- wards he met a stage-waggon which travelled at the rate of 9 miles in 4 hours. B overtook the flock of geese at the 45th mile-stone from London, and met the stage-Waggon 40 minutes before he came to the 31st mile-stone. Where was' B when A reached London? Since A and B travel in the same direction on the same road and at the same rate, the distance between them is always the same. Let x=the number of miles per hour of' A’s and B’s travelling. Then, since the places at which A and B overtake the geese are 5 O I Q 3 l 0 0 miles apart, which the geese travel over in 5+5 or —3— hours; therefore in . lO . that time A has moved forward -3-¥: miles ,' 358 PROBLEMS. 1 . . . —glr~5=d1stance in miles between A and B ............. .. .0 Again, A met the waggon 50—2.v miles from London, 2.21 .B . . . . . . . . . . . . . . . . . . . - . - . . . . . . - . . . . . . . . . . . . .. 0 o s ¢ ' distance the waggon travelled between the meetings is —é—--19 miles; and the time elapsed between A and B meeting the waggon = {gr-19) +9 or % =———— = 4. 8 64 +189 64 ’ s _ s ’ 288 4x=~—8——=36,' x=9, 10.2: N . . . . and ——3— —-o=25, the required distance in miles of B from London. PROB‘. 17. Fine gold chains are manufactured at Venice, and are sold at so much per braccio, a braccio being a measure containing about two feet English. \Vhen there are 90 links in an inch, the value of the work- manship of a braccio is equal to the whole value of a braccio when there are but 30 links in an inch ,' and the whole value of the braccio in the former case is equal to three-times the difl'erence between the cost of the material and workmanship of' a braccio in the latter, together with 4% francs. Supposing that the workmanship in each braccio varies as the number of PROBLEMS. 359 links in an inch, and the weight of metal varies inversely as the square of that number, find the values of the material and workmanship in a braccio of each of the chains. L t the value of the gold in a braccio 30 links to an inch, in e { francs, y= the value of the workmanship . . . . .. . . . . . . . . . . . . . . ._ . .. .. . Then 1 _ 1 _x _ a: {the value of gold in a braccio 90 links to an 302 ' 902- ° 5 inch, and 30 : 90:3] : 3y, the value of the workmanship; m+y=3y, or x=2y . . . . . . . . . . . .(1), and g+3y=3 (x-y) +44 9, x+27y=27w-27y+40. 54y=26x+40, 27y=18x+20, =26y+20, from (1); y=20 francs, workmanship of 80 link-chain per braccio, x=40...... gold . . . . . . . . . . . . . . . . 3:44, francs, gold in the other chain, per braccio, 3y=60 . . . . . . workmanship. . . . . . . . . . . . . . . . . . . . . . . . . . . PROB. 18. A pack of up cards is dealt regularly round to 1) persons with their faces uppermost, every card dealt to each person being placed upon that previously dealt to him; the hands are then taken up, turned so as to have their backs uppermost, and placed upon one another; that hand which contains a particular card (A) being always placed below 7' other hands. The cards are then dealt again, the hands taken up, turned, and placed upon one another as before; and so on :-—Shew that, if m and q be the whole numbers next greater than Lg 2n], and 732)— respectively, the card A will, at the end of the m“‘ and log p p—1 every succeeding operation, occupy the (1‘11 place, or be restricted to the (1‘11 and q—I‘h places from the top, according as m is indivisible or divisible by p-l. (Senate-House Prob. 1839. by Mr. Gas/tin.) Let the 1st operation be performed, and let a be the number of cards which stand before A in its own hand ; then A< n, and rn+a=N° of cards before A in the pack, whole N° next} m+a _ .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. own hand, less than... p after 2nd operation, 360 PROBLEMS. whole N ° next rn +a N° of cards before A in the pack, less than 72+ = after 2nd operation, . . . . . ................ownhand, “' ’1;- P2 = after 3d operation, rn rn+a . . . . . . . . . . . . . . . . . ..the pack, ' ' " ' ° " ' rn'l‘ _ P2 = after 3rd operation, &C. = &c. “n rn+ whole number next less than rn+rLz+2—, -'-- F12, m—l that is, the whole number next less than +-%, P P_1 P rn a rn . or i +—m—_; -—m—_,_—=number of cards before A in the pack, after the P—1 P P (P —1) at“1 operation. - rnp . . . Hence, if g be the whole number next greater than —-——1 , it is en- dent, when p—mE-l is not an integer, that the card A will be restricted to the q“’ PIaCE, PI‘OVided %_1—]7_Trnp—1) is a fraction so small that it cannot make a difference of 1 in the value of rap a rn _ + s - s Q no 1 s P—1 P ‘ P ‘(P—l) ( > Now a<12=8. (2) The rule for ‘DOUBLE POSITION’ is applied to those cases in which the required quantity is not a multiple, part, or parts, of a given quantity, but furnishes an equation of the form ax+b==ear+d. By transposition this equation becomes (a-c)x+b-d=0 . . . . . . . 362 iuLES OF SHOT 0R OANNON BALLS. Now suppose s to be the value of .22, which by substitution does not satisfy the equation, but gives (a--c)s+b—d=e. . . . . . . . . . . . (2), then subtracting (1) from (2), (a- c)(s--.:c) = e. Again, suppose s' to be the value of .r, which by substitution and subtraction, as before, gives (a —c)(s’— a) = e'. Then (a—c)(s—a:) _ e (a—c)(s'—.v)_ e’ ’ s—a: e or —— =— s'-x e" , r r __ ' 01 es—ex-es-e.r, es'—e’s ='_e___e/"> which proves the common Rule, namely, M ahe two suppositions (s and s') for the required quantity; treat each of them in the manner pointed out by the question; and note the errors (e and e'); then the required quantity will be found by dividing the diference of the products es’, e’s by the d-ifl'erence of the errors e, e'. Ex. What number is that which, upon being increased by 10, becomes three times as great as it was before? 1st. Suppose the number to be 20, (s) then 20+10=30, but 3x20=60, e=-30. 2nd. Suppose the number to be (30), (s') then 30+10=40, but 3x80=90, e’=——50. ~30><30+50><20_ ~30+50 “ Hence the true number= PILES OF SHOT OR OANN ON BALLS. Ex. 1. _IN a pyramidal pile of shot of which the base is a square, the number in one side of the base is given, find the Whole number n the pile. Lbt n be the given number in one side of the base, then u2 is the num- ber in the base; PiLEs OE SHOT 0R CANNON BALLS. 363 n—l is the number in a side of the next superincumbent square, (n—l)2 . . . . . . . . .the whole square. Similarly (n--2)2 is the number in the next square; and so on, until the squares are diminished to a single shot. Hence the whole number in the pile is equal to the sum of the series n2+(n--1)2+(n- 2)2+ &c. . . . +12, or 12+22+ 32+. . .+n2, which by Art. 295, Ex. 4, is equal to -én(n +1)(2n +1). Ex. 2. In a pyramidal pile of shot of which the base is an equilateral triangle, the number in a side of the base is given, find the whole number in the pile. Let u be the given number in a side of the base, then u—l is the number in a side of the next superincumbent triangle, oocoicuooooloioololoo s o a o co .lIoOltlllliloooilooo 721—3 000000uoolooooooocoooooooo. ooooolooaoooooooocooa and so on, until we get to a single ball; that is, there are n triangles, which, beginning at the other end of the series, may be represented thus, 0 o 0 ..... 0 ° ' ’ 0.. .0 .... 810' so that the whole number of shot in the pile, being the sum of the shot in these it triangles, will be the sum of the series 1+8+6+10+15+21+&c. to n terms. lst. Let n be even; then the series, taking pairs of terms together =4+16+36+&c. to 2 terms, 71 =4{1”+22+82+&c. t0 5 terms}, 1 1 = 4,6, 2%.,1)(,,+1)=6a(n+1)(n+2). (Art. 295). 2nd. Let n be odd; then the series may be written n+1 1+9+25+49+&c. to terms, 364 PILES OF SHOT 0R CANNON BALLS. or 1’+3’+5”+ 7’+ &c. to ZZZ—1 terms; which is equal to 1 12+] n+1 ’ _1), (Al'to EX. 4‘.) =%.(u +1)(n”+ 2n), =én (n +1 +2), as before. Ex. 3. To find the number of shot in a pile of which the base is a rectangular parallelogram, whose sides contain a given number. Let Z and 6 represent the given number of balls in the length and breadth respectively of the base ; then bl=number of balls in the base. The next layer will be a parallelo- gram whose sides are b-1, and l—l; and (b-1)(l—1) is the number of balls in this layer. In the next the number is (b —2) (l—2): and so on, until we come to (b—l/l—lfll—hj), or l><(l- 6+1),- the last parallelogram being reduced to a straight line of l—b +1 balls. Hence the whole number of balls in the pile =bl+(b-1)(l--1)+(b—2)(l—2)+ .... ..to 6 terms, =bl+ bl+bl+&c. to b terms—(1+2+3+&c.+b_—T)(b+l) +1’+2’+ 32+ &c.+b_-il’, =bsl_b_§_;1_(b+l)+%(b_l)b(gb_1), b_.1 ' b_1 1 E '_ s- _— _ __ __. _ _(b b. 2 >l b. 2 .(b 3 26 1), _b(b+1) b-l 5+1 __T'l_b"§_' _a" =%b'(b+i)(sz_b+i). OBS. To find the number of balls in an incomplete pile it is only necessary to find the number in the pile which is wanting to complete the given one, and subtract that number from the number in the given pile supposed complete. EXAMPLES. [N.B. Where App. is subjoined to an Ex. it signifies that the Solution may be found in the Appendix. Also Comp. signifies that the Solution may be found in the “ Com- panion. to Wood’s Algebra” lately pub1ished.] IN TRODUOTION . VULGAR FRACTIONS. . 602 4139 12332 45739 1. REDUOEt d _— . ' omixe numbers 11 , 15 , 1111 , 60 Ans. 54%, 275%, 111—111%, 762%. . 2 5 12 l 2 at is 5 of 9g also 9 of 25 2 (1) Ans. (2) Ans. 1%. _7_q 24.2 13% 109375 462’ 1111’ 2961’ 10000000' of g of of 41%? 3. Reduce to lowest terms An, 5 ea 11 _7_ ' 33’ 101’ 63’ 640' . l 1 1 l 1 1 4. Reduce to the least common denominator g, g, E, 3, E, Ansala 140mm 12 60 ° 420’ 420’ 420’ 420’ 420’ 420' 5. Find the least common multiple of 2%, 631;, and 5. Ans. 95. 11 and find 6. Reduce to the least common numerator 1—3, T6, 56 ; which is the greatest. I32 132 132 __ __ __ . st' (1) Ans. 429, 44,0, 482 (2) Ans The 1 . . 2 3 4 4 7. Which is the greatest 5, Z, or 5? Ans. 3. 8. Which is greater 71—3, or 7%.}; and how much? Ans. 711% by Egg. 9. What is the difference between 7%— and 7 x2? Ans. . . 191 10. What fraction of £1 is 19s. 10§d.? Ans- 366 VULGAR ERAOPIONS. 11. 12. 13. 14. 15. 16. What fraction of £5 is £3. 6s. 8d? 2 Ans. g, . _ 3 _ Which is the greater, N/g, or N/g? (Comp. p. 1.) Ans. The latter. Reduce 5 yards 2 feet to the fraction of a mile. . . l 1 Whrch is greater, T9- of £ 1, 01 56 Ans. of a guinea? 17 5280 ' Ans. -—1— of £1. 19 What is the difference between 3- of £1, and i- of a guinea? 12 Add together 100%, is, and 7%. Ans. 2d. Ans. 110%. 17. Find the number which exceeds by 148%,: the difference between 195% and 95%. 18. 19. 20. 21. 23. 24. 25. 26. 27. Multiply 45%- by 17%, and divide the product by 4%. Divide 2% of 7171 by g of i of 18%. . 1 l Flnd the value of 7;><%><é><§. 12x11x10x9x8 Ream" ' Find the Value of gxl-gxmg-z—Gg. 4%x4%x4%~1 - __.—u_-—_--—. C \ Q 1. Reduce 4%X4,J_I ( 0m!) P ) _ s 41 4. . 1 _ __ __ Reduce to Simplest f01 in 18+ 3 of 34+ 5% Reduce g-xgxl 3%+(,-19-x-3~+ 54) . ’ 2 1 a 1 1 1 Reduce (E+—3->-.-<3~§ >< 5+3). 18 641 s 1 <1 5) RGdIJCE 'i—7X{1— '1‘ HX'GX 5+ 1—2 . Ans. 247%. Ans. 190%. Ans. 3. Ans. 53g. Ans. 792. Ans. 1,1,. Ans.‘ — . VULGAR FRACTIONS. 367 28. Simplif'y the following peculiar fractional forms: 2 1 5 1 1 2 '1', "5", "T, 2%+ 1 . Ans. 3, 1%, —3—, 291%. _ __ .1_ __ a 11 “*2 38+ 421i . . 68 29. Simphf'y , and 2 (1) Ans. 2+ 3+--- 1 6 82 ———- — A . —-. 3+ 1 5+ 7 (2) ns 151 4+~ 5 30. 5 multlphed by 3 Signifies 5; taken three izmes, that 1s, §+§+§; what does 2 multiplied by g signify? (Art. 129, and Comp. 11.1.) 31. If two~thirds of an estate be worth £220, what is the value of T31- of the same? Ans. £90. 32. An article which cost 35'. 6d. is sold for 3s. log-(1.; what is that per cent. profit? (Comp. p. Ans. 10?. 33. How much per cent. is 14s. 6d. of £3. 10.9.? Ans. 20?. 3%. How much per cent. is 27%,- parts out 'of 36? Ans. 7617-5. 35. A shilling weighs 3 dwts. 15 grs. of which 3 parts out of 40 are alloy, and the rest pure silver. How much per cent. is there of alloy, and what Weight of pure silver? (Comp. p. 2.) (l) Ans. 721; per cent. ; Ans. 3 dwts. 835% grs. 1 . . . 36. The length of gé—d of the Earth’s circumference 1s 69%2— miles nearly; what is the Earth’s diameter, assuming that the diameter of a g a 7 I i (y 0 Circle 1s 55 of 1ts elrcumference? Ans. 7908131- miles. 37. There are five numbers, of which the first two are 2%, 355;,- and each number exceeds the preceding one by the same fraction ,- find the numbers, and the sum of them. (1) Ans. 2%, 3%, 4.1%, 43-3}, 5%. (2) Ans._2015¢;. 38. What is the sum 1% of which is 58. 3.1.? Ans. a... 3d. 39. Divide i- into two parts, so that one is greater than the other 4 59 19 by l—é. (Comp. p. 3.) Ans. Is—O , Tg—O . 368 DECIMAL FRACTIONS. DEOIMAL FRACTION S. 1. Reduce to vulgar fractions 0-375, 081%, 4.075, 00064.. a 13 3 4 Ans. g, 1—6, in, 2. What is 0003 of 027 of 90? Ans. 00729. 3. Divide 0'27 by 0003 ; 0'06 by 60; 600 by 0'06; and 0'006 by 600. (1) Ans. 90. (2) Ans. 0001. (a) Ans. 10000. (4) Ans. 000001. a _1_ A 3’32 25 ’ 800’ 128’ 256 ' Ans. 0'08, 0'00125, 0'0078125, 15'625. 5. Find the value in shillings and pence of 35097216. Ans. 19.9. 5%(1. 6. Reduce to vulgar fractions 08333.", 4041666", 009009009... 4. Reduce to their equivalent decimals . 10 (1) Ans. 5; (2) Ans. 4.1.; (a) Ans. m. 7. Add together 2%, 72%, 316%, and 2875. Ans. 394. 7 s 41 a 8. Add together 1%, — of -—, ~—, and 06666... Ans. 6. 3 a4. 5%.; 9. What decimal of a square mile is one acre? Ans. 0 0015625. 10. What decimal of a Year is 1 second? Ans. 00000000317. 11. Divide 04.454545... by 0'121212... Ans. 3'75. 12. Find the decimal which does not differ from by the ten- thousandth part of an unit. Ans. 314-15. 13. Shew that 3+ 1 1 =3'14159 nearly. 7+1—6 141.0 Divide 2-é~+% by 3%——;-; and express the result in a decimal form. Ans. 0790123. 15. Extract the square root of 1% to 4 places of' decimals; and the cube-root of 3% to two places. (1) Ans. 1'1726. (2) Ans. 1'56. . 1 . 5'04 16. Find ~—§ correct to 7 places of decimals,- and 0012 to two places. (Comp. p. (1) Ans. 05773503. (2) Ans. 20-49. . . l 17. Express in a dec1ma1 form 2+ £3-+ ————5 + 4“ Ans. 26057. 5 1000 20000 ' DECIMAL FRACTIONS. 369 'l 8. Reduce the following expressions to simple decimals having 7 decimal places: (1 ) -;-><{6l+22--3}. (1) Ans. 30833333. 1 1 1 1 1 1 1+...}. (2) Ans. 009314.72- (2) 2x{-8—+§x§§+5><§5+§x—37 l l l 1 1 1 1 3 _ _ _ _ _ _ _- & ¢ 0 o ' o ( ) Q><{5+3><5.+5><5.+7><5,+ c} (s) Ans 04054650 19. Express the following in decimals of 41 places: , 1 1 J 1 1 . 1 1 1 1 41 (2) 16X{3_3X53+ M5; 7X57+&c.} - 2-35. (Comp. p. 4..) (2) Ans. 31416. 20. Express a degree (69-21- miles) in metres, 32 metres being equal to 35 yards nearly. (Comp. p.74.) Ans. 11183542857. 21. The true length of a year is 36524224 days. Find what the error amounts to by the common reckoning in 41 centuries. (Comp. p. 4.) Ans. 0104 days. 2.2. A square inch plate of metal of 005 inches thickness is drawn into a Wire of uniform thickness 50 feet long: find the thickness of the wire. (Comp. p. 5.) Ans. Section of wire=0-0000833... of a square inch. 23. A quadrant of the meridian in French metres is 10000565'278, and 1 metre=39‘37079 English inches: required the length of the quad- rant in English feet. Ans. 328108462868. ‘24:. The number of degrees in an arc of a circle which is equal to the radius is 57 '2957 8 : required the number of seconds in the same. Ans. 2062648. 1 French foot . . = ' '- th 25 Gwen that 1 English foot 10607654, and that e equatoreal and polar radii of the Earth are respectively 3271953854 and 32610729 toises, each ioz'se being 6 French feet: find the Earth’s equatoreal and polar radii in English feet. Ans. 20922811 and 20853232. 26. The length of the pendulum which vibrates seconds in the lati- tude of Greenwich is 391393 inches, and the acceleration of gravity is measured by the product of (length of seconds pendulum)x(3'1415927)2: required the expression for the acceleration of gravity in feet. Ans. 321908. 27'. Two distances are measured in inches, and are known to be cor- rect Within a quarter of a hundredth of an inch each way, being 1187 and 995. How far can their product be depended upon for accuracy? (Comp) p. 5.) Ans. Only in its integral part. 24 ( 10 IL 1% 1& 1% Ii 16. 1) a+bx—cy, (1) Ans. 3. ALGEBRA. NOTATION, &c. WHAT is the difference between 3+a, and 3a, when e135? Ans. 7. What is the difference between 3a +.r, and 3am, when a=2, and arr-3.? Ans. 9. How many terms are there in 3a: and in 3+a? (1) Ans. 1. (2) Ans. 2. How many terms are there in each of the following quantities? (2) abcxy, 2a—3b-l-4axxmnp. (2) Ans. 1. (3) Ans. 3. What are the coefiicients of a and er in via +.r? a (I) Ans. n. Ans. 1. What is the coqflicz'ent of a: in any, and of a in 2am? (I) Ans. y. (2) Ans. 21'. What is the difference between 3a, and as, when (i=3? Ans. 18. What are the simple factors of Qab(a+b) P Ans. 2, a, b, a+b. What are the simple factors of m(a+b)(c—d)? Ans. m, a+b, c-d. 1 Shew that Lil-i- =x, when x=1, or 2, or 3, or any number. -+1 x . +6 1 Find the value of Z_ 6 , when a=§ , when 6:2. Ans. 9, . ax+by Find 6+“: , when a=5, b=3, x=7, y=5. Ans. 5. . 3 3 Find the value of 115+ m, when erg . Ans. 8. . ax’+ 62 Flndm, when (1:3, 6:5, 0:22, 113:6. ADS. 7. Find x8—2xy2+3/—13, when r=2, and y=-3. Ans. -44. I _1 b 2 Find the value of L + r+3—2.£i-, when .0=5. Ans. 0. a:+1 .r—3 .r—2 "‘ ADDITION AND SUBTRACTION. 371 17. 18. 19. 20. 21. 23. 5. 6. and 20?) 7. 8. 7x+5 9x—1 00—9 2x—3__ t1 __ Shew that 23 + 10 5 + 15 ~238, 1f x-19. 9 68 Find the value of get—3, when a=3, and b=~2. Ans. E . a -_b 35 . 1— Find the Value of £-,\/-tv-, when .2:=1—, and y=-1- . Ans. 0. y 1—3/ 4 fl 5 Find the value of 5J62+ 3x- %\/95§—5w, when x=6%. Ans. 4d. . 3 3 ————— 1 Find the value of Z—x+ ./2x—§J1—4x, when .r= 'l—é'o Ans. O. I 2 62 Find the value of $lj—y, when a=-4l, and b=—3. a ——2b(a -b ) ' 1 Ans. wéé . . 2 (1,9+0"2 200 a2 a—b Find the \alue of 3a 6+ +abc+——§—+Z)—3+m, when (1:4, 6:3, 0:2. Ans. 1895,79; ADDITION AND SUBTRACTION. ADD together 3.2?- 5x+1, 7.r2+2.r—4~, and —w2—4x+13. Ans. 9.129—7x+10. Add together a—3b+3c-d, and a+3b+3c+d. Ans. 2a+60. Add together a+b+c-d, a+b+fd—c, a+c+d-b, and b+c+d-a. Ans. 2a+2b+2c+2(l. Add together x3—2aa22+ agar, .t’+3aa:2, and 2a3—aas’eagx. Ans. 2.22% 203. Add together Ans. Gax- 563/. Add together Bag- 2a+ 2, 6b”- 5ab + 50’- 300, a”+2b’+a+2, + 350 + 302, Ans. gag—a + 862- 3116+ 809+ 4. Add together M’y — 4a by - 2ab’+ 263, and My + aby+ 009— b“. Ans. 502y- 3aby- ab”- 63. Add together (12+ 62+ 09+ d3, ab - 2a2+ ac— 202+ ual-2&2, 113- Sal) +63- 3a0+03-3aa', and 2ab—a+2ac—b+2ad-c. Ans. 0.3+ 63+ 03+ 62— 02— 02-— d2- a — b — c. 5ax-7by-1-cs, and ax+2by- cs. 24--2 372 ADDITION AND SUBTRACTION. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. Ans. 17002— 257-026. Add together 5aba+gagb, and ZabLAaQb, __3_ 1 1 l 1 1 1 1 1 1 together Ell-'56 +56, za—gb—é'c, and 'ga-l-Zb +56% 13 17 11 ADS. '1—2-a—'€66+%C. Add together 11"- 3ab - g b’, 26L; 03+ 0’, ab {3; 62+ 68, and 2ab-éb’. Ans. a’+b’+c"’. Add together a”‘-— b”+ 3111?, 2a”‘- 3b"-.r1’, and a’”+ 40"— x9. Ans. 40’"+ 2xP-w’. Add together 2(a+b)+c, 2(a+b+c)+d, and 2(a+b+c+d)+e. Ans. 6(a+b)+5c+3d+e. Add together (a+?) :c’— by”, and (b— 7)x’— 7f. Ans. (a +6) x2- (6 +7)y’. Add together car—by, x+y, and (a—l)x—(b+1)_y. Ans. 2a.:r—2by. Add together 2 (a + 0) xy’, 3 (a—2b).ry’, and 41(2a-b)a:_y2. Ans. (13a— 80).ry’. Add together at”+ 3bt"— 9at"+7bt". Ans. (1 06-— 8a)t". Add together 1-(1—13), 2.x—(3—5w), and 2—(—4s+5a:). Ans. 4+w. From 6a—b—c take a—b+2c. Ans. 5a—3c. From 8a+az-5b-50 take w+26-5c. Ans. 8a—7b. From 7.29—2.11: +41 take 2x2+ 3x —1 . Ans. 5x’- 5x + 5. From Sxy- 7x2—ar-y take 7mg + w2+x +y. Ans. xy- 8x2-2x-2y. From 4am+2xP—.r’ take a’”-b"+ 3x? and 2am-3b"-xP. Ans. a’"+45"-.z’. a 36 -—+-—. Ans. 2 2 From a+b take MULTIPLICATION. 373 25. 26. 27. 28. 29. 30. 31. From 9:52 take 2;; and é-(ll+b) from a—b. (1) Ans. b. Ans. g—g—g. From 2(a+b)—-3(c—d) take a+b—4(c—d). Ans. a+b+c-d. From (a+b)x+(b+c)y take (a—b)x-(b—c)y. From (a’+ be) x2— (a2— 0’) bar take bcxg-(az— 0’) bar. Ans. a’m’- (If—02) bx. Ans. 26(x +31). From :03— aw2+ bar—c take .r’—p.r’+ qx— r. Ans. (p—a)a:’- (q— b)a:+ r—0. 9» 9°?19‘9‘t1‘ From 'a-m-(x-2a)+2a-x take a—2x—(2a—x)+(x-2a). Ans. Set-3.2:. Simplify the following quantities: a—{b—(Qb+.r)}+{b—(ac-2b)}. Ans. a+4b. a-(b—c)-(a—c)+c-(a—b). Ans. 3c-a. a—{a+b-[a+6+c-(a+6+c+d)]}. Ans. -b-d. a+b-(2a-3b)—(5a+7b)-(-13a+2b). Ans. 7a—5b. MULTIPLICATION. MULTIPLY Saab by 2ac, and the product by - 50. Ans. —30aab’c. Ans. Multiply a+2.r--3a:2 by -m. 3mx2—2mx—ma. Multiply 4a2-3a0-i-2 by 5am. Ans. 20a“x—15a”cx+10a.r. Multiply 5a —2ab + 1 0 by - gab. Ans. —- 45a”) +1 8a202—90ab. Multiply 2m + 3y by 2x— 3y. Ans. 4x2— 99”. Multiply 4.02- 60 + 9 by 2a + 3. Ans. 8a3+ 27. Multiply a‘+aab +agbz+ aba+ b‘ by a- b. Ans. a5— 05. Multiply 2a + lac—262 by 2a—bc+2b’. Ans. 4a2— 6202+ 4030—464. Multiply a3+ 31126 + 3ab’+ 03 by aa— 301’?) + 3a62— 03. Ans. a6— 3a402+ 3a264— 0". Multi 1 w‘—2ax8+4na’.r2— 8a3x+16a‘ b 2a +.r. P y y Ans. 32a5+ x5. 241agbz—6a’c". Multiply 4ab-2ac by 6ab+3ac. Ans. 374 MULTIPLICATION. 16. 17. 18. 19. so. 21. 22. Multiply a—ZH-c—d by a+b-0-d. Ans. a’- 2---c'“+al‘~’-2ad+26a Multiply a+ba: by a+cx. Ans. a’+(ab+ac)x+bca:2. Multiply x3—ax2+ba:—-c by x2—p.v+q. Ans. .r‘- (a +p)x4+ (b + up +q).r“-(c+ bp + aq)x2+ (bq + cp).r-—qc. Find the product of (x-aXx-bXx—c). Ans. xs— (a + b +c)a:’+ (ab + ac + 60):: - abc. Find the product of (x-10)(x+1)(a:+4~). Ans. x3—5x2—46x—40. Find the product of (.r-5)(.r+6)(.r—7)(x+8). Ans. x‘+2x3-85.r2—86x+1680. Find the continued product of .r+1, x+2, x+3, and .r+4~. Ans. x4+10x3+ 35.11% 50.1: + 24'. Multiply (12+ ax + x2 by ag- ax +.r’. Ans. a‘+ a2x2+ x“. Find the continued product of .r- a, m+n, x2—ax+a2, and x2+ax+a2. Ans. 326—06. . 1 1 1 1 1 2 2 1 x — "_ __ "- Q 1.... —_~ _b _' 20 Multiply 1+2a+3b by 1 2634-30 Ans 1 404-3 +90 . ., 1 2 1 1 11 2 7 4. Multiply a—éawé- by g.r+2. Ans. 3.6+ 6 w+§w+-3-, Ans. x4+.r"4- x2— cc”. Multiply x3— x"3 by .rc- .171. Multiply (1 + a)a2_g +y‘“’+ (13/2 by ag—y. Ans. (1+ all/(014— 3}”). 43122134. 1}_ Multiply tL5—+agasg+arfazi+c15.z:5+.rEv by a5-x5. Ans, a_w_ Multiply x+2y§+32§ by .rv—2y‘5+32-%. Ans. w2—4y+6wg§+9g%. Multiply ag—2azb§+4a%b%—8ab+16a%6=i-826% by as. 20%. Ans. (13—6462. Ans. 9abii-25a‘303. Ans. a3mb’1’"'c’. Ans. a’m— 3amc"+ 20’". Ans. a"bg— a""'bg+a’b". Multiply 3a56%+5a§b% by eases-ssh? Multiply a""‘2"‘b"’1’*"ca by asm‘2b‘gc“. Multiply a.’"—2c" by a’"— 0". Multiply a"'”115--a""’b*'-1—ab"'1 by ab. Ans. amx”"1’+a'"+1.r“1’- am”. Multiply .r“31’+ ax‘gp- a’x‘“? by 12%”. Find the continued product of an+lb""xc’Pd">}+§(x—1)x(x+1). (Comp. p. 6.) Ans. x3. Prove that (a—b)(x—a)(x-b)+(b- c)(a'- b)(ar-c)+(c -a)(a:- c) (av-a) is equal to (a-bXb-cXa-c). (Comp. p. 6.) Find the difference between a(b+c)“+b(a+c)’+ c(a+ b)2 and @+®@—®@—®+@-®@—®@+®—@-@@—®@+®. (Comp. p. 7.) Ans. 12abc. DIVISION. DIVIDE SOasb’c by 5:150; and --ac%“ by —a.r3. (I) Ans. 601%. (‘2) Ans. 02x. Ans. 1+3c—4cd. Ans. 1+4w-9a. Ans. 3a+4b—5c. Divide Qab+6abc-8abcd by Qab; . . . . .. 5xy+20xiy~4i5axy by 5xy; . . . . .. --9a"’bc'--lQab20+15611/02 by -—3abc. Ans. (2+2. Ans. a4- asb + aibz— aba+ b‘. Ans. (12+ 2a!) + 269. Divide 2a’+a—6 by 20—3. Divide a5+b5 by a+b. Divide (1.4+4114 by a”-2ab+262. Divide x0-2a3w34-a8 by ac”— 2ax+a". Ans. .v‘+ 2axa+ 3a2x2+ 2a3a: + (2‘. Divide ass—a" by a23+ 2ax2+2a9x+a3. Ans. :03- 2ax2+2a2m_a3, Divide as+ba+ 03—3abc by a+b+0. Ans. a2+62+c’—ab-ac—bd. Divide 1222+ 2772115-712—212q+1)2~q2 by m—-1z+p-q. Ans. m+ n+p+qs Divide 1+Qx by 1-3x. Ans. 1+5m+15x8+45x3+...- 376 DIVISION. 11. 12. 13. '14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. - 31. Divide a4—81 by a—3. Divide 64—a6 by 2—a. Divide Ans. 0.8+ 3a2+ 9a + 27. Ans. 32 +16a + 8a’+ 4a8+ 2a4+a5. 5 52a1163x4y32' by 1 84a663x‘3y2. Ans. 3665.1;2 y”. Divide 1 15a"‘b"c”1’**1d‘*’_1 by —69a"b"cP“d. a + b by f/d+ 2/52 Divide wm+"+x”y"+ x’"ym+ym+" by x”+y”‘. Ans. — g- am’"c"d2'"2. DiVide Ans. (1% _ + 6%. Ans. xm-l-y". Ans. cam—20". Ans. Divide agm— 3amc"+ 20’" by a’"— 0". Divide ax3—(a2+b)a:2+ b” by awfb. x2—ax—b. Divide .r’-ap:c2+ a"p.z:-a3 by x—a. Ans. x9-apx+ax+a”. Divide x5—px‘+ qx3—qx2+px -1 by x—l. Ans. x4— (p—1)x3+ (q—p +1)x”- (p -1)x+1. Divide mpx3+ (mq- np)x2— (mr+ nq).r + nr by mar—n. Ans. px’+ qx—r. Divide .x3-2a.1:2+ (az—ab—bg)x+cizb+ab2 by .r—a—b. Ans. .r”--- (a- b)x—ab. Divide (x3—1)a3- (.23+ .r'z— 2)a2+ (4x2+3x+2)a —- 3(a: +1 ) by (x—l)a’—(x—1)a+3. Ans. (x2+x+1)a- (x+1). Divide a:(.r—-l)a3+ (.ra+ 2.17 —2)a2+ (3122- .x3)a —a:“ by a2x+ 2a —-x2. , Ans. (ar—1)a+.1:2. Divide .v‘+.z:‘4- .222— are by .r—x“. Ans. wa- .x‘s. Divide —2a“8.r5—l7a‘4x6—5x7—24a‘x8 by 2a“3x3— 3a.:v“. Ans. - a_5.'v’+7a“1.z-3+ Sa’x‘. Divide (2x —y)2a4- (a: +y)2a2.222+ (a: +_1/)2a.v4- me by (2x—y)a2— (a: +y)ax + .23. Ans. (2x—y)a2+(x +y)a.r-a:8. Divide (30—66)a2—- (cg— 462)a +03—6602+12620— 863 by 0—26. Ans. 3a”- (0 +Qb)a +cg—4ibc + 4.6”. Divide %-6a’+27a‘ by %+Qa+3a2. Ans. 1—6a+9a2. 5 11 1 1 3 s 0 4___ 3 __ 2___ 2___ . . 2__ . DIVIde a: 4.11-1- 8 a: 2.2: by x 2.1: Ans a: [ix-+1 . . 3x5 4 77023 413.1? 33.1: $2 DIVIde T—ésw +-8_- T... 2.__+g7 by E_x+3, 8 Ans. §g~—5x’+€+9. GREATEST COMMON MEASURE. 377 32. 33. 35. 36. 37. 38. N.B. . . $4 11.223 41.122 23.12 2.222 5m x2 3‘” D1V1de T "-71." +6 -3—'-6_+1- AnS.-2—~-4:'+6. “2413682823345 23421a D1V1de—5-ax 75ax+5ax+10ax a: by §a—5aa:+§w. Ans. gas-2.x”. . . 1 — 1 J— 1 1 1 I 3 D d _ §£____ 10___ 12 __ 15 __ 2__ 2o __ ’ 1V1 e 2a 10a 3a +5a +Qa 20a by J; Ja 1 13 14 Ans 2 a—3 —; a Divide w'a— % by xi-yé. Ans. x+xiyé-i-y. Divide a—b by i/a-g/E. Ans. a%+a%bi+a‘ib%+b%. D ivi d e a8m-2nb2pc_ aBm-i-n-l bl~pcn+ a-nb-l cm+ a8m—nb3p+2Cn_ a2m+2n—1 63 0211-1 + bp-H. cm+n-1 a—nb-p—1+ ban-1. Ans. a3m_n b3p+1 c _ a2m+2n-1 62 611+ bpcm. Divide mpg-1 by $4, and write down the last three terms of the quotient. Ans. x("'1l9+ w(P‘2)q+x(P“3>’+ . . .x29+xq+1. Since the Product of two quantifies-:-Multiplicand:Multiplier, or Product +Multip1ier==Multiplicand, each Example in Multiplication with its Answer suppli :s two others in Division. Also, since Dividend -:— Quotient = Divisor, or Divisor x Quotient = Dividend, each Example in Division with its Answer supplies two others, one in Divi- sion, the other in Multiplication. £95”? GREATEST COMMON MEASURE. FIND the Greatest Common Measure of 3a‘a253/ and 6aibx. Ans. 3a2x. Find the (2.0. M. of ax+x2 and abc+bca2. Ans. a+x. Find the G. 0.11. of a2+ab—12Z)2 and a2-5a6+669. Ans. a-3b. Find the G.C.M. of 6a2+7a.7z:--3.222 and 6a2+1 lax+3x2. Ans. 2a+3x. Find the G.C. M. of x4+ a2.ar."“’+a‘1 and m4+ aws- aax—a‘. Ans. Find the G. 0.14. of 3x2+16x-35 and 5.1:”+33x-14.‘. x2+ am+ a2. Ans. x+7. Find the G.C.M. of 3x4+14x3+9x+2 and 2x4+9x3+14x+3. Ans. x’+5x+1. 378 GREATEST COMMON MEASURE. 10. 11. I3. 14. 17. 18. 19. Find the G.o.M. of 20x‘+.z”-1 and 25x‘+5xa-w-1. Ans. 5.19-1. Find the G.C.M. of' 6a”-6a’_y+2ag’—2_y3 and 12a”—15a_y+3y’. Ans. a—y. Find the G.C.M. of 48x9+16x-15 and 24x3—22x’4-17x— 5. Ans. 12x—5. Find the G.C.M. of 6x5—4'x4-11x3-3x2—3x-1 and 4~a:"+ 2x3-18x2+ 3.1: - 5. Ans_ 2x3__ 4x24. 3;. 1 , Find the G.C.M. of x4+ axa— Qai’xga-l 1 a%— 4n“ and Ans. (x—a)“. x4— axa— 302w2+ 5:231: - 2a‘. Find the G.C.M. of x‘-p.z:3+(q—1).r"'+px-q and .r‘— qx3+ (p— 1).r"’+ qx - p- Ans x2—1 Find the G.c.M. of 3x'"’- (4.12 +26)x+ 2ab+a2 and .123— (2a + Z2).v’+(2ab + a2).v- agb. Ans. w - a. Find the (3.0.11. of 2403+ (2a+36)x’+ (2b+3ab).z2+3b2 and 2x“+(20+3b)x+360. Ans. 2x+3b. Find the G.C.M. of x9-2x-3, x2—7x+l2, and x9-x- 6. Ans. :c—3. Find the G.0.M. of 902+ 5.r+4, w’+2.z:-8, and x2+ 7x+12. Ans. x+4. Find the G.C.M. 0f 15a4+10a312+4a262+Gabs—-354 and 6a3+19a96+8a69-5ba. Ans. 8a2+2ab-bz. Find the G.C.M. of 61'1)+2a2—-869-11‘bc--ac-cg and Qac+2ai—5ab+4ca+ 8bc~121fi Ans. 2a+8b + c. Find the G.C.M. of q21113+3121fq2—211pq3—2nq4 and Qtnpgqg- 4m 12“— mpaq + 3mpq". Ans. p - q. Find the G.C.M. of w6+ 4.115— 3w‘—16x3+11x’+12x -9 and 6x5+ 2Ox‘-12x8—48a:2+ 221’ +12. Ans. w8+ x9— 51' + 3. Find the G..G.M. of (19+ 26%- (a + 26). /ab and as“ 59+(a—b)~/c_lz, (Comp. p. 7 Ans. Jd+./5. g “— I- Find the (MM. of '%+1%-f~/x+1—x-s1 and .tQ—g—lli. (Camp. | 8- P ) Ans. .t-éJawl. LEAST COMMON MULTIPLE. 37 9 27. 30. 31. Find the G.c.M. of (b—0).z9+(2ab-2ac)x+a’b-a’c and (ab-ac+bg--bc)x+a2c+abg- ago—abs. Ans, 6-0, Find the G.O.M'. of a2x3+ai—2abxa+bsw3+aabQ—2a‘o and 2a2x4~ 5a4x2+ 8a6— 262m4+ Sagbgwi— 3a462. (Comp. p. 8.) Ans. (a-b)(x+a). Find the G.C.M. of 2(93—2312—3/ +2)a:3+ 3(y’-1)x2—- (2 ys-g/Q—Qy +1 ) and 8(y3—4y2-l- 5y—2).r2+ 7(y2—2y +1)x-(393- 5yg+y+1). (Comp. p. 9.) Ans. xy-x—yH. Find the value of y which will make 2(y”+y)x2+(113/-2)x+4 and Q(g/3+_e/’)x3+(1ly’-2y).r2+(y’+ 5y)x+5y-l have a common mea- sure. (Comp. p. 9.) Ans. y=5. Find the G.C.M. of 6m?—6abg-54a—21w2+2162+189, 6aw5+ 60a-21w5-210, and 6abs-12a—2168-l-42. (Comp. p. 9.) Ans. Gel—21. Find the G.C.M. of 6aBx9+ 4am8—10axy - 302$.11-2d22y + 5 y“, lOagxz— 20:03]] -- Gaxf— 5ax_y + $232+ 33/3, and - 40x2— Ga’bxy + 2abzx'zy — 20 6:0 + 2mg + 3abyg— Izzy/2+ by. (Comp. p. 10.) Ans. y-2ax. If ar+a be the G.C.M. of w9+px+q, and x’+p’x+q’, shew that q-q’ a: . Com . .10. MP, ( P P ) Shew that 002+ qx+1, and a3+px9+ qx+1 have a common factor of‘ the form m+n, when (p—l)”—q(p—1)+1=O. (Comp. p. 11.) N.B. Since each proposed quantity is divisible without remainder by the G.C.M. cach Example in Greatest Common Measure with its Answer supplies two or more Exam- ples in Division. LEAST COMMON MULTIPLE. FIND the Least Common Multiple of So”, 12:23, and 20a4. Ans. 1200“. Find the L.C.M. of I-a, l+a, and 1~a9. Ans. l—a". Find the L.C.M. of (La—.13 and (1.2—.132. Ans. a4+a3.r—ax3—a:‘. Find the L.C.M. of 2x—1, Avg—1, and 4x”+1. Ans. 16.2211. of x2+ 502+ 4, w”+2x--8, and 29+ 7x+12. Ans. x4+ 6x3+ 3x2-26x - 24. Find the LAM. 0f (lg-LA”, a9+ 62, (a—~b)"’;. (a+-bf, (ta—b“, and a3+63. Ans. a‘°- aeb4- a466+ 6‘”. Find the L.C.M. 380 , FRACTIONS. 7. Find the L.C.M. of .r-l, .22’-1, .r—2, and w’—4. Ans. .r‘-5.z~’+4. Find the L.C.M. of 4-(1—x)’, 8(l—x), 8(1+x), and 4(l+.r2). Ans. 8(1—x)(1~.r‘). 9. Find the L.C.M. of Gas-11.12% 5m—3, and 9.13—9x2+5w—2. Ans. 18x4- 45w3+ 37x2- 1 9x+6. 10. Find the L.C.M. of 3.t2—]1x+6, 2x2—7ar-l-3, and 6x’—7x+2. (Comp. p. 11.) Ans. 6a3-25x2+23.r—6. 11. Find the L.C.M. of 6x’-l3x+6, 12x2— 5.2—2, and 15x2+2x—8. Ans. 1 2014-1 34x3—129x2+ 7 4x +24. 12. Find the L.C.M. of m8—3x2+3w—1, x3—x’—.x+1, m4-2m3+2x—1, and w‘-2x3+2x’—2a:+1. (Comp. p. 12.) Ans. w6—2x5+a:‘—x’+2x—]. l3. Shew that if x+c be the G.C.M. of w’+ax+b, and .r’+a'.r+b', their least com. mult. will be x3+(a+a.'—c).r2+(aa'- 2).:z:+(a-c)(av,’--c)<,-, (Comp. p. 12.) N.B. Since the Least Com. Mult. is divisible without remainder by each of the pro- posed quantities, each Example in Least Common Multiple with its Answer supplies two or more Examples in Division. FRACTIONS. 1. ADD together 6%, Z—g, and g. Ans 22% 2. Add together :1, 5;, 3—1‘r-, and Ans. 12753: . a. Add together 6%, 2%, and Ans. 135515013. 4.. Subtract 73% from $5; and -2- from Ans. ~25, 5. Subtract 25:135 from 25:18 . Ans. J—ZEZ . 6. Subtract 5%.w—% from (ix—g;- - Ans. %x—% . 7. What is the difference between 5% , and 651—962? Ans. a +a_b:b . .- b 8. What is the difference between aabb , and Ea; - 572-? Ans. 0. FRACTIONS. 381 Reduce the following fractional expressions:— 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 24. 350—1 312—5 5 ———--———— —. ' Ans. 1. 24- 24: 6 4722—372 m+3n 222 m_ -- -—-. Ans. ———- . 3(1-72) 3(1—n) + 1—1: 1—12 au: 5:0. A... 0. a6 ac 66 7æ—10__ 3æ—7 _ 2743—30. Ans. . 5 6 30 6 — — - - ' -2 3a: 4y_2a: 3/ 1+15x 4. Ans. 85x 0y. 7 3 12 84 a (ad—bc)æ a Ans bx Ë— c(c+a‘æ) 6+dæ' c+dæ a+ ———b—a {cg—l I+ba 4 1 ' — 1 Ans. b. 2 Ans. x. 1 a b—a’ ><2a:+1+2 U U '1+ba 62 %+a—153— a+b+z 6+1 b 3 ———.3- (1) Ans. ——2— (2) Ans. — . 13—14—1 a+b+ï b a b b 1 1 Ans 2 1+na: n—1—(n—1)x n+1+(n+1)æ° ' 722—1 1--a:2 ° et (cf-62).?) a(a*—b2)a:2 An a+bæ b— b3 bg(b+ax) ° b+aæ 1 3m+2n _1_ 3772—212 ns 6mn 2 o 3772—27! 2 ' 3m+2n ' ' 9m2_4m2 ELLE. wS-æ’y Ans _3_ y w+y ya—xzy' ' w+y 4_ 4 2 2 2 m y 2.5.x “t'y, Ans. æ+g— . x—2æy+y æ—y æ æ(æ+1)(æ+2) æ(æ+])(2æ+1) .r(x+1) _ o a . o A . we 3 1 . 2 . 3 (Camp p 13) us 2 1 1 æ+3 Ans x+3 æ-l 2(a:+1) 2(x2+1)' . aft-1 O 3 + 3 + 1 1—.2: Ans 1+æ+æ2 4(1---w)2 8(1—æ) 8(1+æ) 4(1+x”)' ' 1—æ—a:‘+a:5 382 FRACTIONS. 25. '26. 27. 28. 29. 30. 31. 33. 34. 35. 36. 3a 2a +32 5 Ans Qûaœ—Qfäæ’ (ct—2.217)2 (a +æ)(a—-2.r) a+w ' ' (a -i-a:)(a--2~zc)2 ' I I 1 l 4:113(a+æ) + 4a“(a-—.:v) + 2a2(a2+æ3) ° Ans' (fwæ‘ ' x3" mg}? 1 88 F1 -— m - m æn+1 o (007721), P. Ans. æ +29 a9+ 112+ ab ac” À ad ) . (002772). P. 13.) Ans. W. (as-baxæè—aæ duce tol w Æ l'îl’i _ o 1 79“" L 0 est terms 3m, and 3m_3n Ans 3, 3 111—1 2n 2 -I -‘2 -3 Reduce t0 lowest terms 52,7% , Lazy/Q , ——-——pæ l'y “Z a . a b bæïyè qx"ëy"ëg—2‘ l (1) Ans. 6"“ ‘ cm:2 79 m I 2 _‘_' Q 3 — o FRACTIONS. 383 a"(b*+260+ c2) -—a"'b(2l)2+ 360%") + ab“(b + c) ' Ans a(b—c)- b2 2a’-— 2a!) ax + 33° 1 4a’--7a b 3 . R (I .____ H“ ___ 7 e uce to lowest terms 56289506 , 36æ_cx , 10ac_5bC , 19a3x‘+2a"æ° 5a’+ 5aæ 2 a+x 7a ,‘3a‘3ac3 5a —“__2__sïî: an "'"2__2—' ": T: "_a "_—2'1 “a” ' 18ab æ+3b w a a—æ 5 312— c 50 3b aræ Reduce to lowest terms the following fractions ;-- x3+(a+c).r+ac æ+a ac+by+ay+bc 6+1, 3 a ‘— r" - . v s o ' 9 af+2bæ+2ax+lgf Ans' f+2ac ' 40 Gap-1,1060 +9aw+1 5129) A 3a+5b ' 609+9cæ—20-3æ ' us 30—1 ' 41 a9+63+ cg+2ab+2ac+flbc A a+b+c ‘ ag-bS—cL-Qbc “s 5:5:0 .:r:“+a:"2"+:1:a +2/3 æ°+z &2' Ans' 2 '12 ' '3 W "3/ (38+ abs-as!) — b“ a9+ b” ,°Zaz“+ab—bg 9a,. z, “- m- Ans. :02—1 ' 5 4 __ .3212 t Q 2 2 ,1 4‘5. 6æ3+15x_;; 4a, +100r'1/z 3 . Ans. à . 2æ+oy . 9xy—27a: yz—6x_2/2“+1 83/2 8y az—3z acæ9+ (ad+bc).r +bd cx+ d 46. ŒQŒL be . Ans. 0&4) . a3+(a+b)ax+bæ° (Ha. 4'7. a4? bzæs 0 Ans. m . 4x2—12ax+9a” n 21:— 3a 438. W“. Ans. 4æ8+6aæ+9a2 o axm___ bæm-iul æm—l 49' aFbx-b Bæa' Ans' b(a+b.r) ‘ r0 30a3m—lbrcr+î_ 6a2n—4630rdr-1 Ans gaga—1661- J 20a"6"102d"— 4az“‘°’('22d”“l ' Qd" 2 2_ a __ 2 _ 2 3 51. a (b ) ab(26 +60 c )+b (6+0) (Comp. P. 16.) ' a2(b+c)-ab'2 ' 384' FRACTIONS. 52. 53. 54. 55. 10. (c-d)a”+6(bc-bd)a+9(b%-b”d) (bc—bd + 02- cd)a +3(bgc +bc2— 62d~ bad) ' .1 +xé+ x+ mg 2x+ 2xg+ .‘5'.r’+3-l’g . §5_0q+30mp+1850+5mpq 4adq-42fg+24~ad—7fgq ‘ x— 4—3xé+ 419%— (xyY-l’ x—S -2w%+12y%-3(wy)%' (Comp. p. 16.) m” a a: MULTIPLY a—-- by __+_ _ a a: a . 1 1 Multlply a+z by a+5 . . b a 1 z + b z- + Z . . 2a —b Gan—25 Multlply -—4a—- y ————62_2a6 . . 2 1 1 Multlply a: +x+1 by 52 - 5 +1, . 1 1 Multlply x+1+ —- by x—1+ - , J? a: . a b a b Multlply m+aiz 2;; *2)??- Multiply 3wg— by xi- at" $71? . 11’ ab 6’ 3a? gab 62 Multlnly EE-ézy-q-f ?__5?y+y_2. Ans. & a: . a2 203d‘ 70’ a Mnltlply ~58 + T _. W Ans (Comp. p. 16.) a+36 Ans 6+6 . 1+.r Ans' 2x+ 3.1:2 ' 350+ 5212)) ADS. W , l. i w‘1—y?+1 Ans. MT~ . mé—SyH 2 l Ans. a’+ 3+2. a Ans. a+l+2. a b—Sa 2a!) ' Ans. x2+1+1— a: Ans. Ans. x2+1+ 1, _ a: (a2+ 62):: Ans' (a + b)(a + c)(/) + c) ' Ans. 3x’—26.ry+35_y”. in: lgaab Qia’b” Qab8 b“ 2 Qc3d4 by F"T+_' 406d“ 5.11951SI — 10.2293 +54 ' 1405114 490‘ FRACTIONS. 385 11. 12. 13. 14. 15. 16. 17. o a 3 D1V1de m m . ADS. z . . . 2x 2x a;.. 1 D1V1de .2:+—al_—_-?3 by w—EZ—é . Ans. w_5 . . . 4m(a2- x’) a’—- ax 4(a+.r) DIVIdG m m . Ans. 3(c_x) . Divide .14— l‘ by w—l. Ans. x3+ 33 +x+ 1. x x a: a: . . 1 D1V1de a6+ —1-é +a‘+ —1-‘+a”+ l, +2 by aa+ —8 + a+ Ans. (28+ 1; , a a a a a a . . a’3 2a as be 0’ a c a b c D1V1d€ ZE~F+ZE+-J2—Zé 5—2. ADS. 'E—2+-e-. . . a’a:3 _ abas’3 acx’ b’x a’a: a an: 6 , DIVIdG ——6Z2—'— EE'Q'i‘ b '8 ? _~ (—1, and verify the result by multiplication. 18. 19. 20. Divide a'+a: by a+x to 4 terms of quotient. I a a—a' a—a’ a' Ans. 5+ a- 0x_ .x2+‘__.ws_u. a;2 a3 a‘ 2 3 Ans. 1-—-2;;2+§‘Z -:1‘—'r— Divide a2 by a’+2ax+x”. a, a 3 .Q. Divide a-bx by a+cx to 4 terms. 3 2 3 Ans. 1— (b+c)-g+(b+c) Zia—(6+0) %+... 2.1: 2a: __ _-_ __ h = l, , FIND the value of x+x_3) . (a: x_3), w en .1: 59 Ans. 9 . a: 2x-3 3.2—1 _.:v—1 _ Find the value of §-{ 3 - 4 }-.- 2 when x-éé. Ans. 2%. . 0 —br ar-cp Find the value of ax+by, when ar=ag_bp, an y=aq_bp. Ans. c. 25 386 INVOLUTION AND EVOLUTION. b 4; Find the value of x+2a “‘26 4a x_2a+ 5:27) , when w=a—+—b . (Comp. p. 17.) Ans. 2. l ' . an 6n an+ b1: 5. Fllld the value of 2nan_2m+ 2nbn_2m, when w- 2 . (Comp. p. 17.) Ans. 6. If two fractions are together equal to 1, shew that their differ- ence is the same as the difference of their squares. 7. If the difference of two fractions is equal to 2, shew that p times their sum is equal to q times the difference of their squares. 8. Two fractions are together equal to g , and the one exceeds the other by g ; What are the fractions? Ans. gig-(22+ , d — . ‘ I O . . 9. D1V1de -8—§ into two parts differing by 1 2823 381 a 19 ‘ us“ 8382 ’ 3382 ' INVOLUTION AND EVOLUTION.* 1. FIND the square and the cube of 3429603. Ans. Qa‘bgcs, 27asbac". 2. Find the square and the cube of §a3.r’"+2y1"1. 44 8 l A u 6 2,72% 2P—20 o 9 3m+6 3p_3l () ns 25am 3/ Ans 125mm 3/ 3. Find the square and cube of each of the following quantities, and Verify the results by extracting their square and cube roots :- ' . 2 l 4 2 1 4:; nd th S ‘ f —- —— . , -_ 9__ _ 2 F1 e quale 0 5a 26 Ans 25a 542121-46 . . 1 2 1 8 1 2 . 11 e cu e 0 2a 8?) 4“ins 8a 276 2a b+3ab . ' 6. Find the cube of 1+2x+8x2+4x3. Ans. 1+ 6x+ 21w2+ 56x3+111x4+1 74x5+219w°+204w7+1114~x8+ 64x9. * Each Example in luvolution with its Answer supplies also an Example in Evolution, and vice versfz. INVOLUTION AND EVOLUTION. ‘ 387 10. 11. 12. 13. 14-. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. Find the 11‘“ power of a+bx. Ans. a4+ 4aabx+6a°bgxi+ 4a68x8+b4x‘. Find the cube of fi-JZ. Ans. (a+3b)JZ—(b+3a),j5. Ans- Ive—835147 “3/5)- Ans. m3— 15— 3(x— m a: Find the cube of ,3/5—3/5. Find the cube of w—-;-. 1 Find the cube of w-iu-I. Ans. m3-£§—3w2—35§+5. Find the square of rig-($44.. Ans. a%‘—2a +3a%;-2a%+1. . x a2 mg 3.10 3a“ 0‘3 Find the cube of -;——. Ans. -—6--—2+-—--—3. a a: a _a a: w Find the cube of er—e-l’. Ans. ”-e“3”-3(e“’-e"‘”). Find the cube of mpg—1. Ans. w8Pq_3x’P9+3xP‘1—1. Find the square of amb"-3l»2”“c”. Ans. Limb“—6a’"63""10”+964""202". If 00+ % =p, prove that x8+ 2% = 10"— 3p. . _ 1 2 a a 3 a 4 Find the squale of 1 2a:+3.2: 4.24-5.70 &c. Ans. 1—.:r+1—9:c2— 5.22%... 12 6 3 5 2 2 4 2 Prove that x—?—+§——... + 1——— +w—-... is equal to 1. E [2 (Comp. p. 17.) Find the square root of 9a4b’c“, and of 16x”"+2a."". Ans. 361260“, 4xm+1a'", Find the square root of 4x4—12w3+ 25w’~24x+16. Ans. 2.22”-3.r+4. Find the square root of 9a4-I2agb+344a”b’—-20ab3+25b4. Ans. 8a2—2ab+5b“. Find the square root of w6+4x5+10w4+20a3+25x2+24w+16. ‘ ' Ans. m3+2x2+ 3x+4. 3x2 a: 1 Find the square root of .r‘-2.r3+- 1 -— —~. Ans. x2—x+-, 2 2+16 4. 25—2 388 - INVOLUTION AND EVOLUTION. 25. 26. 27. 28. 29. 33. 34;. 35. 37. . 255126” c‘ 5abc’ 5ab (:2 t f _— a I _ __ 0 e square r00 0 4 + 9 3 Ans 2 3 Find the square root of are +£+§f+-2—aw--1-ab-bx. 9 4 3 3 1 1 2 3 Find the square root of 103;“: -6?x—145x +9+49.'v‘. Ans. 7w’—g+3. _3____a~/a_§'l/g. flil+L Find th s f ’ e quare root 0 a 2 2 + 16 Ans (vii/3+1 4r . a” b’ a b Fmd the square root of F+;.;.—2. Ans. 5-5. . 2 7)" 2a 2b a b h f _a_ _ _ _ . _ _ _ Fmd t e square root 0 bg+ag+ b + a +3 Ans b +a+1. Find the square root of a2m—4Iam+"+ 4a“. Ans. am—2a". 2 2 4 2 _ 2 - h is we > - Find t e square root of ye 4y2+1 + $2 xg-H +3 231’ A “' ns. é??+—;2-+1. (5+2) ,fé+2%. ya: F‘ d h t f“? 9’ In t e square r00 0 ?+a_§_ 1 E Z_.——_- 4a“ -_l- 9a‘w’+ 6a2a:‘+ x6 Ans. a2 + .2”. 402+ :0“ Find the square root of Find the cube root of a6—6a5+15a‘—20a3+15a’-6a+1. Ans. a2—2a +1. 8 2 Find the cube root of %;-6a‘+12a’ 9-86‘. Ans. %—2b%. 3 ’ 3 3 2 3ab 68 Find the cube root-of ab: .r°—-—%£.x5+ Tet-25.x”. - ac b _|x2— —J.'- b c INVOLUTION AND EVOLUTION. 389 38. Find the square root of (x+.r“‘)’—4(x—x"'). Ans. x—r‘—-2. 39. Find the square root of 49a‘“‘6—4-2a6'""+9a8’"*’. Ans. 7a2m'4_ 3a4m+lo 40. Find the square root of 9a” + 6a3m+‘+ 25cm" — 30a”‘c”"‘“ + am“ __1 Oa’2m+lcm--2o Ans. 3am__ 5cm-I+ 02mm. 4.1. Find the cube root of (a +1)6”‘.r3+ 60a1’(a +1)"".r’+ 12c’a2"(a + I)“; + 803a31’. A ns. ((2 + 1)’"‘a: + 20a”. 42. Find the 6‘Lb root of a6+ -1—6—6+15(a’+—1;)—20. a a a Ans. a-l . a 2 4 6 m 7 27m 6 27m 8 43. Find the 4‘11 root of a8-3_n_-a + 5 811" a - n a + _____ a‘o 8n” 1612.3 2561a4 2 3m." Ans. a —Hn. 4.4. Find the square root of (a-b)‘-2(ag+b*)(a—b)2+2(a‘+b‘). Ans. a’+b’. 45. Find the square root (without the aid of the common rule) of 4{(a’—b2) cd+(c’-—d2)ab}”+{(ag—bi)(cg—d’)—4abcd}2. (Comp. p. 18.) Ans. (a’+b’)(c’+ d”). 416. Find the coefficient of $8 in the square, and in the cube, of 1—2x+3x’+4x‘-x7. (1) Ans. 20. Ans. 168. 47. Find the coefficient of .22“ in the square of (.r+a)(.r+b)(.r+c). (Comp. p. 18.) Ans. agbz+azc’+b’c’+4abc(a+6+0). 1. IF two numbers differ by a unit, prove that the difference of their squares is the sum of the two numbers. 2. Shew that the sum of the cubes of any three consecutive integers is divisible by three times the second of them. (Comp. p. 18.) 3. In the extraction of the square root according to the common rule, shewr that, if the_remainder is not greater than twice the root obtained, the last digit in the root is correctly found. 4. Prove that x4+px3+ qx’+1'x+s is a perfect square, if p’s==r’ and J)” — —Z+2~/s. (Comp. p. 18.) 390 SURDS. _ 5, ax3+ bx2+cx+d is a complete cube. 6. Find the relations subsisting between a, b, c, d, when (Comp. p. 19.) Ans. aca=dba, and bg=3ac. Find the relations subsisting between a, b, c, d, e, when ax4+6x3+0x2+dx+e is a complete 4th power. (Comp. p. 19.) .10 $7! 10. p 11. 12., 144. 15. Ans. bc=6ad, and cd=66e. SUBDS. REDUCE to simplest form J15, JEFF, 3/526, Ans. 5J5, 19,15, 453, 62/15, 2J5. Reduce to simplest form U268, 5608, 5/646, I/W. Ans. 43/3, 23/19, 21/5, abw’". Reduce to simplest form $14732, Z/dsb—a, Ja4“b21’03, Jeri—ail). Ans. aQbJZ-Z', abs/4?, aQW’cN/z', Reduce to simplest form Jax’—6ax+9a, J(a:2—y2)(w+y). ~ Ans. (w—3),/Zi, Simplify the following surds :-s-- 11 — 17 —- —7-,~/72~ 5 J50. J48a62+b~/;i’5_(_l_+ h/ 3a(a - 96)“. 6yfaz+2y2d+ 3/5513. sjs— 7Ji§+5./?2—./5—O. 5 -- -— s 2\/§+~/60-~/15+A/3. Ans. 6law/s. Ans. a~/3_a. Ans. 92/271. ‘ Ans. 8J2. 28 _ A . —— 5. 115_15J1 ' - — 25 _ 8+‘443—é,\/12+4~/2_77—2 Ans. -2—J3_ 7351+33/Fifl/2—5f/f25 Ans. 8,;75. UR+~7§1“~3/“512+3 192-73/2- Ans. 10. Ans. 485/4. 184a563+N/5-0a31)3. ' N55??)+,/144416-~/289a%. Ans. (3(zQb+5ab)N/2a*b. Ans. O. SURDS. 391 16. 17. 18. 19. 20. 21. 23. 24'. 25. 26. 27. 28. 29. 30. 31. 32. 36.3/242562—7 i/Qa5bfi+ SaE/Qazbs. yn—‘mn _ 2/) 20 ' — Zi cJZF—ac c~/ab.\/Z-, andm . Ans. Arabs/2:121)”. 3 T Ans. . (1) Ans. ac. (2) Ans. \/5‘b! .' ab N/mnnz an ' A/z ms: m-‘Fl-E. (Comp. p. ABS. W. 555 I T 2 a '—-' ~/-62— +§EJa 6—416126 +4aba. Ans' éE'Jab’ 5J§><7 I Ans. 140. f/gx7i/_6><~éf/5. Ans. 731—5. (3 +~/§)><(2—~/5). Ans. 1-.,\/_5‘. § 5 - - — - - l. < 5 5+N/4). Ans. 24.4. Ans. l4~§-+l3\/g- . Ans. 30 2+3Jé. Ans. 5+3i/Ig-l-3i/T2. +2\/g>. ><<-i/Q+U§>- (3/IQ+./T§)>—. Ans. 16a4b(a+b)8. Find the square root of —2+a9\/§+ a-Wgc. Ans. a‘Æ— "Vâ. 2 Find the square root of 35E —qf+2ac , / lg. c __ Ans. a N/ähj—cfi L3. 1 l % Ans. æ“ — ëb’äx”. Find the square root of 攑+È”Ë/W— UÊ.W Find the square root of {/rcilés- + QJilim/U (comp P' 20-) Ans. 5/23; Find the square root of each of the following quantities: 18+2JŸÎ. Ans. 94"42Jä Ans. 7—3N/Ê. 28+10fi- Ans. 5+JÊ. 13+2J3T0- Ans. Jinx/'3'. 10%+2J5. Ans. JÏÎ)+-%Jë. Jr“; A Ï”; M5 a+x+ ~ax+ar . 115. N a+ë+ 2 . SURDS. 395 2 ~——- '79. Find the square root of ag + g-A/ag—c”. Ans. a2- 0”. 80. x+y+z+2Jxz+yz Ans. Jx+y+~/§. 81. mxQ/i-f/g) is a rational quantity; find 122. Ans. 82. m><(2%+55’) is a rational quantity; find m. - 0 11 Ans. 5%-25><2§+5%><2%-20+5=5><2%-23. 83. mx(.j§+,i/§) is rational; find m. Ans. 2%-4.xs%+2%xa%-6+2%> 1+x +y —x’y2 q‘ 1—-ab+,~/l-_i-_ci”--ar~/1_-U2TI __ _____. 001 . .26. 1-ab+J1+b*-6J1+a’ ( "PP ) 107. Simplify 1—a+,\/lTI-_Z" Ans. __ . 1-b+~/1+b2 108. the value Of , when n9=1_-_i-_~/_'_3_+?___ NQ'JS. n Ans. . / J2J§ l—ax 1+bx when w_1 2a 1 1+ax' l-bx’ _a 7)". ' (Co-mp. p. 27.) Ans. l. 109. Find the value of OI, 398 EQUATIONS. 111. 112. 113. 114:. 115. H 0 £99"? 10. 11. 12. Find the value of' J-Lz—“JM i , when waif—é. (Comp. p.27.) ~/a +a:-—~/a-w 5 +1 Ans. 6. Find the value of 2(uv—N/1—u2W/l—v2), when 2u=x+w"1, and 222 =y+y‘~‘. (Comp. p. 28.) Ans. wy+(xy)“. Find the value of 2a*/1+af~, when w+~/1+a:" 2 b a (Comp. p. 28.) Ana. a+b. If x: JlilQ/g—l), and y=~ 19 (Jig—+1), find the value of' .v”+y”. 2/5 2.5 A Q ' ns. . “bi-e? lag—bgp— ,- If .2:=(E___5) P, find the value of 2.»——-—ag+bg(~/x+~/x), a+b 111’ (007221), P. 29') Ans- (5:5)??- EQUATIONS. IF 20—4x+8=5.v+10, .r=2 If 2x-2L1+3w—30—6x+66=0, ar=12. If 7x+20—3a:=60+4x-50+8x, 02:17,:L If x—3%=3—é-+15x, w=-% If lag-g=2x-8-§;, a=9- If 35’ '17_“5’+1-é_‘”-=366, m 120. If g+g-g+§=7-g-, $310. 9 3 =31 If é; —4$—3 , a: o 1 1 1 7 1 If a: 23—5 19;, “"5 a 1 ~- If ’3; -%=10 “46—1, x=14. If x wig-=31 +4, m=5 EQUATIONS. 399 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. If If If If If If If If If If If 3x—2 x-2 4—x 4. 73 =2”? 7’ “2' .r+ 1 x—2+.r+3 7 -- ----=--——-- —'—-~ ’27: o 2 1.12- a ’1 ’ w-7-1- 3.2—9 27—52 , 22: 4 + a ’ “717' 10+a: .r-2 .1: £17 5 1, 37:5. 3.x—2 2.2-5 w_- 5 =3_ 3 , $249 1200-2 18-400 6 - 3 =.z'+2, 11:81}. 22—1 2+2 5.71- +1=3$+ ~5— +7, man-8. 2+6 16-3w__2_§ 8 1 12 " 6 ’ w“ ’ 7w+5 16+4w 3x4- T-T+6='—2-9’ “1' x+7 2a:+5 10—50: 2x+7 9x—8 _.'v—11 _7 7 11 " 2 ’ w“' 3.2—1 6:1: 2.x-4._ _a_:_+__2 x__5 7 4. " 12 ' 28 ’ _' 7x+9 3.1:+1__9a:—18 249—91! x_9 8 7 '" 1 11 ’ _° 8x+3 72-11 7x+1_2 W7 16 15 2o - ’ " ' a: 12-.2: .1: 8l 7w+5 9.2-1 x—9 2x—3_ 1 _ If _2—3_ 10 5 + 15 "235’ ‘7“19' ' 1102— ' 4 1—6 2.1+1_ 3.8_ __ 7 17) $272. If 29 12— 2 400 EQUATIONS. 30. 31. 32. 33. 34.. 35. 36. 37. 38. 39. 40. 4d. 42. 43. 44. 45. 46. 47. If If If If If If If If If If If If If If If If If 52—5(2.r-1 ) =27, 4(x—3) - 7(w—41)=6-.r, 20m-50(3.v - 1) = 200- 20(4x—5)-(3x -4) 50, 4(5x-3)—64(3 -x)—3(12x-41)=96, 1 1 1 4 5 1 Ew—Zx+18-—9-(41x+1), 1 1 1 5(2117—1 -' —15-'5(57-‘r): 1 1 1 11 1—2$—§(8-$)-;(5+Q7)+Z —0, 1 -1 _l L _. 13 7G” 2) 5(3 “>431” ' 1 3 1 1 31 1(4+§x>-7(2‘”-§) ~25, 1 2 1 1 — '77 (4W— = -2- (517—6), _ _§_ 7w~9_j$_< > 323 4 5- 3 —-5 6+ 9 1 3 _Z%:£_£(§_) 2(ax+4) a “2 x 1 ’ 3_!;x{28—(§ +24)}=3%x{2%+g}, w—l 3 10a:+17_12.r+2 _5x-4 18 112—8 9 ’ 6.2:+13_ 3x+5 _2.r 15 " 52-25" 5 ’ 4.2 20—42 15 5-.v .r a: \t 5x+3+2.v—3_ x—l 2.1v--2T 9. EQUATIONS. "4'01 4s. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59.. t 60. 61. 62. 63. If If If" If If If 11? If If If If 'If If ‘If If If 30+62+ 60+82_1 48 ms 2+1 2+3 _ 2+l’ 3'— 1 2 _ 1 “7 2—1 2+7_7(2—-1) _ ' 2 32 _ .___ 4 2+1 2+2_ ’ ‘1’“— 5' 62+8 22+38 -- = , w=2. 22-1-1 2+12 2 __ 2- _ 52+2 3:72 32 9, “:3. 52—41 72—10 42—17 3§—222 6< 22) A ._ = --- -— . .31 . ~r=3- 9‘ 33 w$154,(PPP 4') . 42-17 102—13 82-30 52—41 = A . .3l4. £=2l 2-4. 22-3 22-7 2-1 ’ ( P3P )- 2 2 2—9 2+1 2—8 __ . =___.__ ___.__. {0:41. 2—2+2-7 2— +2—6’ 1 72—1 8 2-»; __ =__ , 2=1 -. 3 6é—32 3 2—2 115 1 l x2+2 w_4‘% 106 I) ’2'22-2’ ’ 2 32 8122-9 3 222—1 57—32 1 __ = '__n____'___——*_""' A 0 03150 2 (32-1)(2+3) 3x 2 2+3 2 ’ (ppp ) .- 2:10. a 5_ ’ _a+b a2 02 02 ' 12+ -—+—-+ = 2+~+k . . . x: g _ 6 f g g f , a—bq a b___ 2 2 . 1 1 1 l b ab—ax+ 60—62 _ac-aa: ’ x=5(a_6+0)° __ 8 a+2+./2a2+22=b, ((1266) . 402 EQUATIONS- 64. 65. 66. 6'7. 68. 69. 70. “I 29 73. 7 4s. 75. 76. 77- 78. '79. 80. If If If If If If If If If If If If If If If If If a + 2 +W= b, J5+m=fi. a + 2 +Jm= b, M+2Ja+2=J2—2+W, 2211' = 22m 4;. ./1+2+22=a—./1—2+2’, Ja +2 +.Ja-2 =b,:/ag—22, %~/m+%./a+2=%\/2, (Comp. p. 29.) Jas+ 22+. / a”— 22= b, J2+~l2—~/13=1, 1+.‘/1+2-~/1+2+ 1—w=0, 9w!2 2+./a2+22= W. ./4a+2+ a+2=2./2-2a, M+J1+2+J1~2= 1— w, . - _ 2 J2+J2—~/2—\/2=a\/———w+~/5, ax +. /a"’22 + 62 =\/ (12+ (122/ 452+_$2, EQUATIONS. 81. 82. 83. 84'. 85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95 96. If If If If If If If If If If If If If If If If 29-2” ----————=b- 22+2’ 2+./22+22 ’ (12—52 22—6 __. = - C . . 30. J1+2+22+ 1-2+2”=m2, . /a*- 22+ 2. /az—1= nth/1:25., Va + 2 iii/22+ 8a2 + b”, 1222 ./22+ 222 i a —1 (NH-1 = + L;—- , (Comp. p. 30.) J22+ 222 +,\/22-2ax = 22—1 J(2a +2)2+ b2 +~/(2a -2)2+ 62:22, Ja+ (2125+J2—Jag— 22:72 1—22 1+62 m'dmr-l, (6077727. I). 1+2+2a __ 62 1+2 1--2-l~22 _ 63' 1—2 , (Comp. p. 31.) 2 (w) =1+§%, (Comp. p. 31.) a— 1+2a 1—28 (1+ 2)” (1—2)8 =a’ J(1+2)2—a2+J(1—2)2+ a2=2, 1 1 1 __.—__- + ="'", ,‘/1-2+1 1+2-1 w 2c2!./1+22 a+b=-—-—-—-—— 00m). .31. 2+J1+22’ ( I p ) ___—-—_, a+./a”—2‘°" is H U'IIH $2 II a PM H H- § ml 0% W 21011 EQUATIONSZ (P 100. 102. 103. 1 O4. 105. 1 O6. 107. 101;. 109. 110. 111. 112. _If_ J22+2a2+ 62+JF-2aw + 62: 101. If i If, a-|-2+./2av..z:+2’a If . /a+2+ 2—2: I+JEZI 1+2a~/2G 22 J a +,~/a"+22 , =~“/‘:2:12_1 , (Comp. p. If 222 9 22— a2 u If \/%4Ta—2+./3a2-2=-3—22\/1—4<2, If Jm+3./1—2=3/2; (See App. p. 317.) i ' 1+2 1-2 If ‘ __ + ' __=a, \ 1+2+~/1+29 ]—2+,\/1+22 If Jm+JwL9=4+ 32, If J{a222+ b~/(ab.r + 4a222+ lk/2ab2 + 5a222)} = 22+ 6, 1+.2'+\/22-1-22 —————————=1—a2 1—2+._/2.11:+2"J , If J(l+a)2+(1—a)2+ (l—a)2+(1+a)2=2a, 1+2—J22+22_a ./2+2+~/._2_ If 1+2+./22+22— 'J2T2—J2 ’ __.-.62, A . , '_ a+2—,~/222+22 ( PP p 315) If az—2+,~/2a2—22 2—2 =5, (App. p. 316.) a+~/2a2—22_ a+~/2a2—22 _ i If fi—x/a—Jag—ax _ fiM/aq/az-aar If 5, w=a_. x=J1+4(a-1)’. 2: f[ix/22+ b2. .x_i 32—1 _12'a+1 ' 2=1. 2:. /(2—a)2—1. _ (2+2)2 _4a(a+1) ' J --— 1+22 1—211‘ 4‘7 If J1+a. 1T5) + l—a.(l—+—x~) =2J1—a2, (Oomp.p. 2=—a. 2=8. 1 1 $='2~{i/Z+ fi}—1. a 2= 27 (b-1)2. a ./1+(b-1)2 ' (la/Zr _ w=a' 1+6 M11121}- EQUATIONS, 405 113. 114. 115 116 117. 11& 119 120. 12L 123 124. 125 126 ,/a2+zz'"—-a_ _QaJE —— If T .2lx2 __l—a—b, x—WJI_LB. If ~/a+x+~/m=f/a2+x2+:/az—xz, am??? 45'. If 1”” _““”__= ' =95. 5 0. Ja+~/m+Ja+~/a-x Ja’ x 2 J ’ or If __‘Vl-i-H +f/T:';+I_a ,jT-EH £54 _ ’ If x+~/(.r2_w~/l-;z'_)= 1 , w=l_6_. .x-JQzFr-ak/l-x) ,Jl-a: 25 If (l+.2:),/1+aa+(1—.cv)~/1-a=2~/1+azr’3 x=l—~/<—l-)2~1 ’ a a 4 1' 1 - If fl.1c—‘-_1+x,/x‘-1=x3, ar= §+§~/2. If fi+2~/2a—x=~/x+N/2az—3ax+x2, _ 961a w-Qa, or a, or 1025 . If2 2:; 24J220 51- “it—22¢ 315) T9-Jx+ 937+ 7 '- {13+ 27+ — w+l7a PP’P' ' w=7& 243+324J§5 -2 If 16x_3 _(4fl-Js), (App. p. 316.)? x=8. If (x+a+Jx2+2aw+52)3+(x+a—Jx2+2ax+bz)3=l4(w+a)3, (APP- p. 318.) x=—a, or iJb”-a‘-(z. __ n 8 If b ’ ( PP. p. ' “+6436 ' 406 EQUATIONS. 127' W+gii= a , a+c(a—ar) a: a~20x (Comp. p. 32.) (1 1+0 ' CC='-'a, 01‘ 61% . --a 3 w—2a+b a+b J =m, (Comp. p. w==-—-~2~—~ . 129. If x+a+sfi/Ez)‘l»é=b, (App. p. 321.) “(3/54/2131 130. 11f (w-a)J;-(¢+a)fi=b(J5-JE), (App.p.822.) w=b, or (fl+,~/Z)2. 131. If a+(b+fl)J5=(b-J5)./2a+x, (App/p323.) 3 “q 8 _— _2\/L+£é\/3_a x~2 262 2 ab ' =a+b, (App. p. __ (a—b)”{\/W8ab JI~ Bab . ‘1 l 2 .- 133. 2a: 4.1+.” 4“” +3 =a; P. az== M.___3—a-._____:__L:___— - 2w2+3+x~/4w’+3 J(1-a)(9a-1) 134 If (a2—1)a+a’m~x~/2a2-1 (a2-1)a+a2x+:r,/2a”-1 _—'§ 132, If M 1—ac+,~/1+.11:2 =(1— aBXa-i-xY—Qax, (App. p. 326.) 1 *aJl—ae =N/1— 2. ——-—~—————————— . x a a2+,,/2a2-1 If .‘l—-amr+,\/l+af-—az~/l+.1c2 135. __M -W :a 0 o 1--a:1:+,~/1-1~w”--a:,\/1-¥-a2 ’ ( 11) p ) __ __ z 2__ w=_1_’ or (a 1 Jl+a) 3a a a(8a—&) 2m+3 Y1» 2x-3 -% 8 4x2+9 136. If ) (“__. =___._______ . _ ‘ = 2x~3 + Qw+3 13 [hf-9’ (App P 329) x *1“ 137. If _ (b+c)’+ (62+ cficx}. N/£Ig (b-c)‘+ (52—1~02)2.222 b 2 “ ={% (b - c)g— (69+ 02km} . N/é' ([1 + c)"+ (69+ ceyxe, a 02—112 2 . 02+ [)9 ' (App. p. 245.) a}: EQUATIONS. 10. 11. 12. 13. IF 3x+5y=8,} {m=1, 4x+3y=7, y=1. If 9w—4g/=8, 13x+7y=101 _y=7. 1 1 1 1 y=216. If 21-9 ‘3 - 5”” 5 " 2 ’ 31g§§+2y=2x+44 If (x+5)(y+7)=(w+1)(y—9)+112,} 2x+10=3y+1, 2x+y+ 7y+6x+11_§ 5.22-17 ‘ If 9 18 _2_T’ §{5w+sy+2}=%{9y+6}, 3x—5y _ Qx—Sy—Q _ If 3 12 _ 3 2+ .2: C y 31.{7+%+1-g}=6g.{1w-Esq-21}, 10 M1_+ 1 _ 452 21:0 22y _ 462mg ’ If w+y=38%—.4x+12y ,l 2. If 4x-3y=2, 4 {33:4, 4. If“ 45x+ 83/=350, {Ln-=6, 21y-131=122, {$=14¢4¢, ' 60 y=10. } {M y=9~ {20:2, y=3. {tr-9: .y=4- {wt-5’ y==9~ {13:3, y=5n {x=72 y=4- y=11u 408 EQUATIONS. . 14. 15. 16. 17. 18. 19. 20. 21. 22. Q3. 12859-18y2+217 If 16x+6y—1- 8x_8y+2 ,1 {30:6, 10x+10_y-35_ _ 55 3-5 2x+2y+3 — - 3x+Qy—1 ’ x+ly éy- 1 2 3 5 1 1 5 ~ If §< _5y)+ 6 _§_§{ 6 '_C(w+y-C), .9=='=~/56- 4 (17:: 1 Q (Mg—1); 1’ If w8+y(xy—1)=0, 93-w(wy+1)=0,} (APP- p. 332.) __.... __ 1 _;___2 If Jw+5+~/w—5I=;(wy—y~/=v'—y)l .4/x+y+,:/._r:?=b, (Comp. p. 35.) 2 3 '— [ .-1{@_._~12£}11 W, 4. 2 b2 2 62 3 b— a ly=§"\/§+Z*2' If (w2-wy+y”)(w’+y”)=91, {5:2, or -3, (w"—wy+y”) y=46, 1 3:10. 1 2=34. y+52=341, J z+i(x+y)=61, J w+.y_4<4.~ w-y__20 w2-.y’_3520 z "17’ z _f'17’ and _ 17 ’ x=128, y=48, 2:68- 4x-5yi+mz=7x-l1y+nz=x+g+pz=8, (C'omp- P- 36-) w=2, y=1, 2:0. 63—412: 52-0: y—Qz 32—7 29-52 3y—2a: 1’ w ’ y 7’ 1 2 3 w+§(y+2)=y+§(x+3)=2+z($+y)=$+y+3-43 ' {17:10, y=6, 2:2. 1 ° ‘ ?..,._=§, 10. 113--5—+-1-=3w, 5:6 .1: y 2 x__1 .2: 3y .8 ’ _ ’ §_2=2, y=-2, l+l+g=6_7L%-, rL y=97 z 3/ 3 4.1: y z 1 3:: 1 1_41 ' 5 1 41 1 Z==—-. 35+z—3’ 5&_y_+§_12m 3 xy+86_4x—y 1 3x+2y_ 5 ’ w=6 9 x2+4~é w—z = . 1:1, a2+32 2 ’ 1 J 1 2:3. yz+4-.i__y+3z 2y+z _ 7 ’d 3 4. 1 3 1 ' 1 __--- -—= .5, w=_: .2: 5y 2 1 2 1 1 2__ l L 1 _1 y- 13—3, 4 1 4 1 ___._ -=16—1— z==—- 5a: 2y+z 10’) L 4: EQUATIONS. 411 13. 14‘. 15. 16. 17. 18. 19. .x-l-z-3 4y-4z+5 arr—4 1 If ' =2“: 14 9 3 w=25, 5w-Qy+z+3x—y+2+1= 2y+nZ~5 , {r {y=16, 7 l I a 2:60 5y—Sz+4_x—~z+b =2y+3z_2w, 12 6 J Fx__ 1 _ a-c b—c ’ If x+y+2=0, ( X ) (a+b)w+(a+c)y+(b+c)z=0,} (COmP- P- 37-) 19: m : abx+acy+bcz=l, 1 Lg: (a-bXa—c)‘ If x—ay+a’z-a8=0, x—by+bgz—ba=0,} (Comp. p. 37.) ar-cy+cQz—-03=0, x=abq {y=ab+ac+bm z=a+b+a If 55-11J§+132/5=22, FL 45: + 6,.[1] + 5:/z=31, y=4h w — J; + Z/§=2. 2:27. x__ 2abc “ac-Hm—ab ’ If xy=a(w+_y), ‘ _ Qabc wz=b(w+z). (APP-11' 332') y_Zz_l;+bc-ac ’ yz=c l 255. ' 222m 0 ab+ac-bc r ___.2.”__. —Ja2+62+69 , If ZUCJ? +31+2)=02, b2 y(a:+y+z)=b’,} (App. p. 333.) iy=WTc§ 1 z(x+y+z)=c”, 62 z: m , 1 J52+52+52 f __ ‘(a+b-c)(a+c—b) m— 2(b+c-a) ’ If w(2/+z)=a> (a+b—c)(b+c—a) y(x+z)=b,} (APP-P-334") 1’”: flaw—6) ’ z(.r+y)=c, CM (b+c—-a)(a+c—b) ,4— L 2(a+b—c) 412 EQUATIONS. 20. If 2Jw+éy+3x+3Jx+5y—z-—4Jx+y+z=10, -.§—\/4x+2y+122—J2w+lOy—22+J2x+2y+22= 3, J2x+y+62+2./x+5y‘—z-3 x+y+z= 4~_J2, 21. 23. 25. 26. 27. If x_2y"z=1%, Jx=1, 22. $492218, y=2, xy223=108, = 3. F {7/60" x: 8 , If x2y324=a, 2, a a 4 2 J abw my z :1), ) = 08 2 4 2 3__ my 2 -c, , 27 cam (2: b8 : If .r(y+z)’=1+a3; ar+y=g+z; 1a1 x=1+a, y= Z—ai §J(a—1)2+a, If x+2y+ 32+4u=27, 3x+5y+ 72+ 21:48,] 5x+8y+102—2u=65,] 7x+6y+ 5z+4u=53, If 7x—2z+3u=17, 4y—Qz+t=ll,l 531—310—221: 8, 4y—3u+2t= 9,] 32+8u=33, x=7, 3:6, 2‘=5. If x“1y_1z_1=(105)'1, x=3, wy’12-1= 3 X (3 5)'1, y = 5, xyz'1=2%, 2:7. 241. If xy2=231, x=3, wyw=420, y=7, yzw==1540, l 2:11, xzw = 660, w = 20. and yz=1—36; (App. p. SHEW that the following equations are not sufficient to determine 5:, y, z. (1) 3x—2y+ 52=14, (Art. 198.) 2x+ y —82=10, 8x—3y+22=38. (3) 2.22—y+2z=8, 6y-1-13—g=4.w—1+32, w 1 —+z-y 4 2—4. 3+6" (2) 3w-2y+52=14,‘ 6x—4g— 32:15,} 9x-6_y-—7z=20. 1 (Comp. p. 38.) EQUATICN‘S. 7. 10. 11. 12.‘ 13. 14. 15. 16. 17. 18. 19. 20. IF 5x’—12a:+2=1 1, .If 52=511+6000, If 3x”—53w+ 54=0, If 11552-255=1000, If 11052-2111+1=0, If 780x2—73x+1=0, If 125x2-7x=17%, '1 -1 If w+1 _x x—l w+1 If x—l _ar+3 33’ fix-1+ 13 __3a:+5 2.23-1-1 11_3.z'—5’ 12 8 32 If 5—x+ 41—x_x+2 ’ 1 1 1 _ __ =0 If 3+3+x 3-1—22: ’ x4+ 3x3+ 6 x2+ .r — 4.- If' 2,./3x+7=9+./2w-3, If \/(x2+-é-./x2+1664)=x+1, If ~/(1"-1)(xi23+~/(x-3X$—4)=~/§, =x2+2ar+15, a:=3, or —§ . 1 _ x=2=l=—~/10. 5 1 52-2, or 1—6 00:2, or —3, w=11, or -13. ar=5, or Q% . x=2, or 4.165. 1 _ $=3(—1i§~/Q)_ x=2, or ~2T75. x=14, or x=2, or 3. 414 EQUATIONS. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30‘ 31. 32. 33. 34. 35. 36. 37. If w”+Jw’-5=11, 6 If $+2¢ gig-=1, If w’+w+1=——§—2- , 87 +127 If 9x+2w 9.r+4=15w’-4, If 2’-w+3J22“-32+2=g+7, If 2(J5+1)2=102(2+J5)~2576, If (9+5J3)22-(15+7J3)2+6=0, If 11./624 3'5--;~./952-'5'2 =11, 3J5-fi __ 13+3J2-22 2+2 — 2fl_3 ’ 1 1 1 If 22+11x-8 + xg+2x-8 + x“—-13w-8 2==1=3. .z=1, or -4. 2:2, or --3. 2:14, or --§, or 810. x=2, or --%. x=49, or 64, or &c. a:= 3-/3, _or Q-fi. =0, (Comp. p. 38.) 00-148, or *1. N If we _2 x a +6—2- @219 Q 2.1:(a-w) _Z If 312—23 4’ If nx-tbznatb, J. '2.“ 12a If ./a+a:+ a—x: 5Ja+m’ If ma” + “'3 =2JJ, Jan—m Jm If ax+2./n”x+nax’=(3x-1).n, If x+~/1172~— (lit-1'62: 7;- +6, 9W 212011211311 . 2 LU” I =F=x ‘32 +.-P33= +;.('6+.“’)QI Jl 3-—v/" _, . .(r—I) .(x‘H) _ v= + .11 3/‘~E£/‘* 8m ="""+I . 17 (v) 3 v} . 83-1 3"“ ___. _ =F-_u.- :3] p: a 3/ I 2 43") 4”“) H .g/éaax ‘%1=.w—I 23—3-33 .11 .~ I a 1 1w 3 (I—.v)§/”2329—.29 ‘ —Z-+v/V;zq)-sv/\¥=w . .2 _ .§___ __2 1 _____. v I a” /”.-,=.Io q ~.c "111+ 23/“43H23 =w ‘rcv =32? --'[/”'517—I/\’+1-251’3/v JI ‘2=£:3/?+£t£’/? JI ‘(qz+w)—q=(w+v)wq+(w+ 11).“: 51 ‘5/‘+.v= (1)—@312 +.(v—w) .11 ._9_ :8 ._ ‘M _9_.='[—1l?2; ~32 -- -- g .10 T5-zv wot /'8 /~ /‘ I wZ/~+a.~/" _11 w -_C_l-_ .10 (27:3; ‘O=clu-wbd+xuw—axbm 31 “(n—q); .10 ‘g-za+qv'5/"=x' o I 0 ‘ 5% §--— i...- (017 (I dump) (q+v)fl_a(qv x) _H ‘ .w+wvz/W mpg—axe =aw+xvzf+x+v 31 8 9 '22 — - .10 ‘—-=x 413/1.) .0. -uv{§/”343}=w ‘xéwg = gw-g(m+x) JI . . . .§_I’_ i’§./\_i ”_9 (Ggdclwog) v 8063%“. -x5+8w31 --u =1“ .10 ‘§-;Z_/\'¥=3' 1+2: [-2: ('68 -d 2205) <<1-u>u=8(—,—) 1(7) 11 ._8___ 4 __. $6 .10 avg-x .I-i-u ' __ [—u __ '19 '02 ‘6? '81? '19? .9? '9'"? '3'? '31? '11? ‘01? '68 '83 416 EQUATIONS. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. If' If' 1—x—./1—a.~/1:x=2a, 1 (12+ 3)”-—2(a:”+ 3)=2.r(w+1)2, x=a(3—4a2), or —a. 11 (1-2)Jm-(1+2)JM= “#225 1:3 , If If If If $7525- . M— = If If If If If 1 1 1 1}% x 2+ 4 a N/1+(z2—a~/1+a:2_a ' x_1 or _1_ l—3a” J1+22—.v\/1+a2_ ’ , a “2‘3 . __ _ a (a+x)~/a2+x2=6(a-.r)”, w=(9='=4~/2).',i . a—JQax—mgz w 33:61, or g. ‘ 7 a + , Max—m” a—x M i/m‘+1 nJa2+x2+ (n-l)Jxg-2(n — 1)a2= 2n --1, 2=J(114Z-_1.JT-72)2-52. , (Comp. p. 40.) 2 "nu-Hz. .ra 1+nr =P’ _ L1 1 2)_11\/<1._1>2 (1 Q2 1 n_<2 r>+<2+a 2 2 r + 24-60—22. (1+w+x2)2=Z—:(1+x2+m‘), (App. p. 319.) w=g4 (2)4, a+x+J2ax+xz=Jax—x2+J2a2—aw—xg, (App. p. 323.) 1 — a - x—JxQ—ag ~/x+~/x’-a2 =Jxz— a2.{Jxg+ ax—JwQ-aw}, (App. p. 324.) _ .2{ a .-_ Jib-*7. m#~/a”+2}. EQUATIONS. 417 65. 66. 67. 68. 69. 70. 71. 72. 73. 74:. 75. 76. 77- a2+ -1— If _________2x+’\/2(1+x).=a+l, (Comp. p. 41.) ‘7”: - a] 4- l—x a _. <~/a+ 7;) If (a+b)./a2+62+x2—(a-b)~/a2+b2:x2=a2+b’, (Comp. p. 42.) w: \/i if (a2—62)—-ab. 1 ag+as3 2 _ am 2 aha?)2 If 21.—ix 5 >+12§—.(? =( 6 , (Comp. p. 42.) a:=a, or (2*,~/§).a. 5=1=~/n2—1.c —_—_.. _ .a. ciJn”—1.b a n 1 ' 1 If 2”‘--1=a -—- =\/—*,\/ — . .1; +371", .2: 2 a+LL If' ,f/(a+a:)2+ :/(a—x)2=3:/a”:~x2, (Comp. p. 43.) x: <3*./3)5-25 (3i~/B)5+z5' ' If A/(flbz+cg)(az+m2)=n(ac+bx), (Comp. p. 43.) x: 2 8x13 8 (276“- 8.1:? _ 33/27M8w '5i/5 13 15x3 If , (Comp. p. 4.4.) If abw‘I—Mab) .mP9=(a—b).a:1’, (Comp. p. 44.) x: — > . _=¥‘—— J!) ~/a 2W 1 ‘12—'62 p— q— aib 2i” If ~/$p+q_§.;2:—b2-(~/$+~/$)=O, (007721). p. $=<-a—:F_b)q_p. a’mi1}2 x: _— _ {a2m:F-l x: (1=EJ5)m- 2m (1*./5)"*+z'" ' If (a4m+1)(w%-1)2=2(w+1), If .m/(1+x 2--U(1—.z')2=Z/1—x2, (App. p. 320.) i J? If x“—3:c=2, (App. p. 318.) If x—1=2+ (App. p. 317.) x=1, or 4. x=2, or —1. 27 418 EQUATIONS. 2 2 2 1*,ff5 78. If x--é;=1%, (App. p. 318.) x=-§, or 3 s+4~/f£‘ _HJI? 79. 1'—3=-—x-—, {17— 2 . 80. x2+lé+w+l=41, (12:1, 01' _3iN/5 _ a: a." 2 81. If m”-8(m+1)]5+18x+1=0, w=7i4_\/§, 82. If %.r2(x9+3)=20+5%.x, 3:3, or _2, 83. If'm+7i/E=22, “8, 0,. {_li __lo}._ 84. If (x+3)’—2(x2+3)=2x(x+1)2, x=1, or —3, or —-é-. - 8 7 85. ——=='—__"—-" 32:1, 01' 16- ./ a, We. 86. If x2(x2—23)—10x(x2—24)=6419, x=%{5=*=3~/32_§}. 87. If 2m%(x3+a3)%=2x’(x+2a)+a2(.r—a), x=g, or --a. 88. If x2—7=~/{x2—4:2x+89}, 011:2, or —5. 89. If (w2—5)2=(x—3)2+(x+1)’, 00:3, or -1, or *JG-l. x 63 __220% - _ . 1 -_4 90. If 5+ :5- J5: +49Jx—1196, 00-2401, 01 T6{_7=EJ37}. 1+0? 1+2a*J12a—3 91. m—a, x- 14-4164 2 92. If -———;=a, (App. p. 321), w=pi./p —l, (1+x) where p=2ad=~/2(1+a) . 2(l~a) 1+ac5 93¢ W——a, w_~/4~a+1(,\/§+./&a+1)+[10(6a-1)+2J§(4a+1)%]% ' 4(1—a) ' 94' If Jx2+~/5=x-1+3J5, 1:22. 0r 3—1i158fi— Z. EQUATIONS. 419 .95. 96. 97. 98. 99 l 00. If .r-2 w+2=1+1/m3—3w+2, (App. p. 325.) x=%(3¢JT§), or 9===4~Ji If 2xJ1—w‘=a(1+a:4), (App. p. 327.) 3x2+ — w=*§{}i If'( -1)2 25 3 (n —1)(a‘+ a2x2+ w‘) __ < 1 (n +1)(a‘— a2w2+ x4) _ w_a{\/5n—3+ _2 n—l T2+w<~»-2> 7L , (App. p. 328.) 1 x=3, or —§, or 2(2*./13). (1.1! . (APP- P- 328-) .17 n+1 a 1 \/1 } 72—1}, or . If a%’— 112- 2x(a — cw) (App. p. 329.) 72;- [)2+ 2+ I—a:2 1-—.'1:2 2 g2 (a— can)”: 0, _axb __M c=l=a If (1—x)1\/ a<1+-:-:>—9.=~/;I_1+~/3x—l, (App. p. 330.) _ iJm-l w-qm' IF x2~-xy=10, x=*5, (w—y)”= 4, y=*3- If w’-y2=9, ar=i5, wy—y2= 4, y: i 4. If xy+xy2=12, x+xy3=18, If w2~wy=27,% wy—y’= 3% If' 2y~ 3.r=3x2+2(y—11)2=14, 2. If m’+y’=13, 00:2, or 3, x+y= 5, 9:3, or 2. 4. If 2x2+3xy=26, x==l=2, 3yg+2xy=39, _y==*=3. w=2, or y=2, or *~'aI- S. \c Q N) <54» by “CM .r=13-%-, or y: 4%, or on or 1%; y=10, or x=2, 27—2 420 EQUATIONS. 10. 11. 12. 13. 14-. 15. 16. 17. 18. 19. If 2y+3x= 8, .r=2, or 255, 3y2+2w2=1 1, y=1, 01" If' xg+wy+4yg= 6, x==¥=2, or i%,/F), 3x2+8y2=14~, y==!-_--é-, or ¥_g-N/fi. 13 — If x==E17 01 4yS=4‘]-, iy:i37 or If ar—y=a, } {x=a-b, or O, ys+ay+bx=o, y=—b, 01‘ —a. If W+y§=3, x=8, 01‘ 1, w+_z/=9, y=1, or 8. If ac+y=5, {m=3, 01'2, (m’+y2).(x3+y3)=455,} (comp. p. 45') 31:2, or 3. “ " 61 - If N/‘L -Z=—:+1, = y ---”” M {Zié’ i/F?+./€1/“_=78. . ' If $4 + 3;“ =641,} {x=5, x3y+y3x=290, 9:2. If .r—y=8, } A 331 {#11, w4_y4=14‘560, ( PP° P' ') y: 8. a — a __. x=—* 3+1,0r—3* 3, If m+y=a, } j 2( ~/ ) 6( x4+y4=léx2yg, __ __ ly=g(fl/3+1), or %(3=FJ3). — 1 w=(./5=*=./$>* If N/Zr+~/by=—(w+y)=a+b. (Comp. p. 46. { _ _ ’ 2 ) Puma/w. If xz=y” a:=1, or 9, (x+y)(z-.r-y)=3, (Comp. p. 46.) {9:2, or _6’ (x+y+z)(z-x—y)=7, 2:4. EQUATIONS. 421 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. If x+y+z=216, f/E+y§+Z/§= 12, Z/f+3/?=i/?+f/§5—12. If (w—2)’+(y—3)’+(z—1)’=24,} xy+xz+yz=63, 2x+3y+z=30, w2+wz+z’=28, If w2+wy+y2=37,} y2+yz + 2’ = 1 9, If w3+y3+ xy(ar+y) = 13,} (x2+y2)x”y2 = 4.68, If' w2—6./w2_y=27,} w-2J@= 3, If (a: +y)2 = a24+ x2y2+y4, x4+ 4g“ = 4wy(:2_y”- 402),} If x+Jw2—y2= gQ/w-i-y-lx/ $19 a} (w +y)%-(w-y)§=26, m=125, y: 64, z: 27. w=4, 3:3, 422 EQUATIONS. 30. 31. 32. 33. 34:. 35. 36. 37. 38. 39. 40. If x+./x9—yg=g~(~/M+Q/x_—_y_),1 1/55+i/37—§=y. x=(1+a2)%{1 2a )2} 23/2 +(1+a2 ’ If x2—y2= 17ay +3y2+4xJ2ay—yg,} {x=7a, x’+y”=2ay+8~/§.(x~/5-y~/§), y=a b+2a _— a --- x=T:—’ If w—Jw2-y2=§(~/-v+y+~/w-y).} 3J4“ .- b- (w+y)§—(w—y)%=b. (gm/43. w+y—~/x2+.'/2_2_w 413/1753 If ——————— , w=-_—--.a w+y+h1w2+y2 a 7 w N/a+~'v I l 13* 153 -= , y=—_.a. _y a—y 8 If x+~/@+y+fl+:/Zy+~/y—=210,} {x=144, or 9, ._./.?y+y+J;_./.T3,+./;=126, y=9, or 144- If J;_\/§=Vy+2, ‘12:?) J:+8=8 3+2, =1, If x4+y‘=1+2x_y+3w2y2, {x=2, m3+y3=2y2x+2y2+x+1,} y=1, 3m -_ [w_gy_g(N/5_l): _Z/E - ly—QJ-(Q/5H). 5 If $24-98: w5+y5=2, 1 =i6 *_ 2, ' If w2_.xy=6’ } { g or 3+ 2:61 ~ 1 _ y 3 If ww+y=y4a, = %(4a+1¥J8a+1), =—21-(—1*J8a+1). - [w={-%*Wi 1y={—-;-*%;~/5_7} . } (App. p. 336.) w+y_ a y '“w: If wfmwfi. } (Comp. p. 47.) (J5)%W=(J5)%’ EQUATIONS. 423 41. 42. 43. 44‘. 45. 4.6. 47. 48. 49. 4a 3a .22 = _3_‘ , 01‘ z— , If 5ax+12y(a—x)=0, } x2__y2+a2= O, 6 If 4 2-4=4 bl, my my 4 } (See App. p. 338.) x3”3:wy(x_y)> l I———_ [m=§ {63+ Jgag_b%}s, 13/=';‘{b%—./2a%—b% 8. (Comp. p. 48.) If w’+ .3/x43/2-1- 312+ 3 /m2y4= a9} w+y+33 bag/=6, .9__ If (w’+y”).-m--8%, w=*3, x (Comp. p. 49.) _i2 (mg-y?) ~-=vA-, 3’" ' .9 2 2 __ __. If (a: +y )(w+y)-15wy, } (Comp. p. 49.) {ac-2, or 4, (w4+3/4)(w2+y2)=85x2y8, y=4, or 2. If (x-2)y-JZy_,(y2-1)=2y2—w,} {11:8, or -12,- *EZJ/Z'l-IQ _ 4 4 _ xy—18 ’ y_2’ or "'5 “ __‘ 9$2 " "' 2 If 5—2,\/y+2=~6—;-(~/$*3~/y) ’1 68:41, 1 Z—10\/€=x—16, {yfil ‘ .9 5’ If i/m4' :IJU-—y=:/E, 1 _. { =50 *J‘g)“, ly'sa/ 1*% Jinn. } (Comp. p. 51.) } (Comp. p. 50.) If (b+;¢)w+ay=(w+y)2-2J5§../a_-5.~/fi JI-Jr- =./ b... 2 __ b 2 x=(——b——n {Jn*~/-bf_—_-7-z—(b—n)}, y=6-n. 424 ' EQUATIONS. Com . p. 52, w’(x—2)+6y2w(x_1)+y4=g _x, ( p ) 50. If 424.9%“; } l 2 1 ' __________ (a: i; J4a+2=1=2~/8a2——4a-8b+1, ]y==!=;11- J4a+2i2“./8a2- 442—86 +1. \ I 51- If WWW (a—yXb—y) -=2{b (a-x)(a-y)+aJ(x- b)(b-_y)}, (Comp-@530 wp=4ab, {x=2(a=*=b +Jaziab+ 122), y=2(bia¥,./a2=*=ab +52). 52. If xma"+y“b”=2(ax)§.(by)5,} (Comp. p. 53.) xy=ab, 2n m—n 2 {x=bm{a'2'*}~ If (2 + 4mg - 3.29)“ = 2 —- 4x’yg+ 3.114, 27.112 __ 9.322 + 2.2231 +1 } (App. p. 343.) 32 2 w“ ’ 53/94- x=2, y=1- 426 S EQUATIONS. __ 4 63. If w=y2(1—2y2), w2~1 (2x2-1)(2y2—1)= s, 1 - 1 1 1 __ 1 __ 4:4 5J5, or 4§\/§(14J§;'); y=4-2-J6, or +§\/5+~/33. x4194 ' 64. If 30 ,1, +40 fi=241, } (App. p. 346.) x398 $3+y3 2*}(w421 2 %2_(§)3_9_2 ..5. {1+ . 3xy+216~/a:+xy3 - 6 a: y, (Comp p 7) ar=i—— or =!=—- 3:41 or ='=-§-. 27’ 8’ 8’ 27 65. f I f' (x4+ 25x29 +a2y2)(y4+ 26xy2+ agar”) = 4(a2-bg) (b + c)2(xy)”,} x3+y3= 2mg, (Comp. p. 58.) m=jp2.3/04J02_P2, y ='f/§;2' ‘3 0¥1\/c2-P2: } where pg: a2—2bc—2b2. 612—372: 3xy’ C (J§—./5>,} < W- p- 59.) a -— (Z a w=--2-, or iaJ-2, y=-§, Qri ~/——2' } (Comp. p. 60.) 67, If (1-42)”.(1+ y2)-(1\+w2)"'.(1— y2)=4x2~/1_-1_-?*, 4411=~/3(1-42)(1-.112). J27§—2.:/2J§42 _i [-1 3 @546/5—1) ’y' 357—? :F a): 1. WHAT is the quadratic equation Whose roots are 17 and 22-? ‘ Ans. 3x”—~53a:+34=0. 2. What is the quadratic equation whose roots are 3 and --29 Ans. 5w’-12w-9=0. EQUATIONS. I 427 3. The trinomial ax2+bx+c becomes 42 when x=4, 22 when x==3, and 8 when a2=2; what are the values of a, b, c? Ans. a=3; b=—1,' and c=-2. 4. Solve the equation 02’— 5x—24=0, Without completing the square. Ans. a:=8, or -3. 5. Prove that in simple equations of two unknown quantities there is only one pair of values of the unknown quantities which will satisfy the two equations. 6. If 00,, x2, represent the two values of a: which satisfy the equation x a: 62—2ac ax”+bx+c=0, prove that 4+1: .222 x, ; and verify this formula by the equation 2x2—7x+3=0. 7. If a, ,8, represent the two roots of the equation 1 x2—(1+a)w+-2-(1+a+a2)=0, shew that 012+ [62:41. ' 1 . . 8. Shew that x+ 5 cannot be less than 2, whatever pos1t1ve quantity be substituted for x. 9. If a proposed equation be reduced to the form ax=ax+c, what conclusion is to be drawn as to the value of x? Ans. x=oo . 10. Shew that there can be no more than three distinct values of a: which satisfy the equation aw3+ bx2+cx+d=0, and that their sum = ."_b . a 11. Find the sum of the four roots of the equation a(px’-+ qx+r)2+b(par’+ gm + r)+ c = O. Ans. :g-q- . 12- Is ae-axw-b)++(was-a>}=*+g+ (x— (2)2 an ‘Identity’ or an ‘Equation’? Ans. The former. m+y+z -————-—z . 13. Given x+y+z=}§x=gy, find (Comp. p. 61.) Ans. 2. 14. Given x-y=7z, and x—z=4y, find (Comp. p. 61.) Ans. % . 15. Given m2+y2=1232, and x2—y’=27z, find (Comp. p. 62.) Ans. 60. ' ..__1_ = 16. Gwen that ab 2(cz+b)(p+q)+pq 0,} (Comp. p. 62.) and cd-%(c+d)(p+q) +pq=0; shew that (Pg—q); (a_c)((:;g)f 428 _ PROBLEMS. 17. If the same value of w satisfies both the equations ax2+ bx+c=0, and a'x2+b’a:+ 0’: 0, what is the relation subsisting between a, b, c, a', 6’, 0'? (Comp. p. 63.) Ans. (ac’-a’c)9=(ab’—a’b)(bc’-—b’c). 18. Find the relation subsisting between the coefficients in the equa- tions a1x+ bly=ch a2x+62y=02, a3x+63y=c3, that they may be satisfied by the same values of a: and y. (Comp. p. 63.) Ans. a1(b,c;,- 6,02) +az2 (bag—blag) +a3 (6102— 520,) = 0. 19. Find the relations subsisting between the coefficients, when the equations ax+by+cz=a’x+b’y+c'z=a”x+b’iy+c”z=1, are equivalent to no more than two distinct equations. And shew that in this case the values of .22, y, 2, assume the form 0—:-0. (Comp. p. 64.) a—a’ b—b’ c—c’ Ans' a/_ an = Z)!_ b// = C/___ cl! ' 20. Eliminate b and c from the equations, a+b+c=—p, ab+ac+bc=q, abc=—r. - Ans. a3+pa2+qa+r=0. 21. Find the value of ax2+ byflcz", when ax3=6y3=023, and l + 1+ _1_ = l , (Comp. p. 65.) Ans. (ails? b%+c%)3. d2. 22. Eliminate a, b, c, from the equations em 2’“ imue" é)" . £__.1L__z"_* (a) + + d 7 and am-l-n_bm+n_cm+n- (Comp. p. 65.) _ml 3 mn Ans. x"‘+"+y"‘+"+ z’"+"= d’m . PROBLEMS. 1. I HAVE six times'as many shillings as half-crowns, and together they amount to £8. 10.9. How many have I of each 5’ Ans. 20 half-crowns, 120 shillings. 2. Divide £25. among 3 persons, A, B, 0', so that B shall have twice as much, and C three times as much, as A. Ans. To A, £4. 3.9. 4d. To B, £8. 6.9. 8d. To 0, £12. 10.9. 3. Divide 20 into two such parts that one of them shall be exactly 20 times as reat as the other. g Ans. 19.51;, 4. A boy is exactly one-third the age of his father, and has a bro- ther one-sixth his own age—the ages of all three amount to 50 years. What is the age of each? Ans. 36, 12, 2, years. 5. Of two brothers whose ages differ by 20 years, one is as much above 25 as the other is below it. What is the age of each? Ans. 35, and 15, years. PROBLEMS. 429 6. A is twice as old as B. Twenty-two years ago he was four times as old as B. What is A’s age? Ans. 66 years. 7. Nine years ago A was three times as old as B, but now he is only twice as old. Required the respective ages of A and B. Ans. A is 36, B is 18. 8. Seven years ago a father was four times as old as his son, but in 7 years more he will be only twice as old. What is the age of each? Ans._ 35, and 14, years. 9. The ages of a father and his son together are 80 years,- and if the age of the son be doubled, it will exceed the father’s age by 10 years. Find the age of each. Ans. 50, and 30. 10. A certain party is composed of three times as many men as women,- and when four men have left together with their wives, there remain four times as many men as women. How many were there of each sex at first? Ans. 36 men, 12 women. 11. Divide the number 90 into two such parts, that if half of the greater part be added to the double of the smaller, the result is the original number 90. Ans. 60, and 30. 12. Find two consecutive numbers, such that the half' and fifth parts of the first taken together shall be equal to the third and fourth parts of the second taken together. Ans. 5 and 6. 13. The product of' two numbers is 180, but if the lesser of' the two be increased by 1, the product is increased by 20. What are the numbers? Ans. 9, and 20. 14. Find two numbers in the ratio of 1L to 5, such that if 6 be added to the greater number, and 1 to the smaller, the square roots of the result- ing numbers shall differ by 1. Ans. 24, and 30; or 8, and 10. 15. There is a certain number of which the cube root is one-fifth of the square root: find it. Ans. 15625. 16. A certain fraction becomes 5% if 1 be added to its numerator; but if 1 be added to its denominator, it becomes 1 What is the fraction? In 4 Ans. -- . 15 17. A certain fraction becomes gi- if' 1 be taken from its denominator, and added to the numerator; but, if 1 be taken from its numerator and added to the denominator, it becomes What is the fraction? 7 Ans. — . 11 18. The numerator of a certain fraction being multiplied by 3, and the numerator and denominator added together for a new denominator, the resulting fraction =-;—. Find the original fraction. Ans. 430 PROBLEMS. . _19. A person distributed £5. among 36 persons, old men and widows, glvmg 38. each to the men, and 29. 6d. each to the women. How many .. D were them Of eaCh' Ans. 20 men, 16 women. 20. A person distributed p shillings among 12 persons, giving 9d. to some, and 15d. to the rest. How many were there of each? Ans. 212(512-472) at 9d., and ;§-(4~p—3n) at 15d. 21. Divide the number n into two such parts, that the quotient of the greater divided by the less shall be q with a remainder r. nq+r n—r 1+q ’ 1+q' \ I Q Q 8 Q 22. D1V1de the fraction 5 into two parts, so that the numerators of the two parts taken together shall be equal to their denominators taken together. An 1 11 s. 2, 10. 23. Divide 30 in two such parts that one is the square of the other. Ans. 5, 25. 24:. A certain number of sovereigns, shillings, and sixpences together amount to £8. 6.9. 6d., and the amount of the shillings is a guinea less than that of the sovereigns, and a guinea and a half more than that of the six- pences. Find the numbers of each coin. Ans. 4 sovereigns, 59 shillings, 55 sixpences. 25. What is the number from the 72‘“ part of which if a be taken, a times the remainder is equal to b? ( b) Ans. 12 (1+; . 26. A person sells 0 acres more than the 012"“ part of his estate, and there remain 6 acres less than the 12‘“ part. Of how many acres does the whole estate consist? Ans mn(a—b) ' mn—(m+n)' 27. A labourer is engaged for '12 days, on condition that he receives ence for every day he works, and pays q pence for every day he is idle. At the end of the time he receives a pence. How many days did he work, and how many was he idle? Ans. He worked nq+a , and was idle P+q np—a d . )+q ’ ays 28. A person, being asked what o’clock it was, answered that it was between 5 and 6, and that the hour and minute hands were together. Re- quired the time of day. (Comp. p. 66.) Ans_ 27m, 16fi1—s. past 5. 29. Find the time after 72 o’clock at which the hour and minute hands of a watch are distant d of the minute divisions from each other. (Comp. p. 67.) 12 ADS. PROBLEMS. 431 30. There are two places 154 miles distant from each other, from which two persons A and B set out at the same instant with a design to meet on the road, A travelling at the rate of 3 miles in 2 hours, and B at the rate of 5 miles in 4, hours. How long and how far did each travel before they met? Ans. A travelled 84 miles, and 56 hours, B .......... .. 70 ........................ .. 31. Find a number, such, that whether it is divided into two or three equal parts, the continued product of the parts shall be the same. Ans. 6%. 32. A person bought a certain number of sheep for £94: having lost 7 of them, he sold one-fourth of the remainder at prime cost for £20. How many sheep had he at first? Ans. 417- 33. A farmer buys m sheep for £12, and sells 12 of them at a gain of £5 per cent. : how must he sell the remainder that he may clear 10 per cent. on the whole? Ans 22m—21n £ a h . 2—————Om(m_n).p e c . 34‘. A boy at a fair spends his money in oranges. If he had received 5 more for his money, they would have cost a half-penny each less ; but if 3 less, a half-penny each more. How much did he spend? Ans. 2.9. 6d. 35. A hare is 80 of its own leaps before a greyhound, and takes 3 leaps for every 2 taken by the greyhound, but the latter passes over as much ground in one leap as the former does in two. How many leaps will the bare have taken before it is’ caught? (Comp. p. 67.) Ans. 240. 36. A courier passing through a certain place (P) travels at the rate of 5 miles in 2 hours. Four hours afterwards another passes through the same place travelling the same way at the rate of 7 miles in two hours. How far from the place (P) is the first overtaken by the second? Ans. 35 miles. 37. A person has just a hours at his disposal; how far may he ride in a coach which travels 6 miles an hour, so as to return home in time, walk- ing back at the rate of 0 miles an hour? Ex. a=2, [2:12, 0:4. abc . - (1) Ans. 6+6 miles. (2) Ans. 6 miles. 38. A banker has two kinds of money, silver and gold; and (2 pieces of silver, or 6 pieces of gold, make up the same sum S. A person comes, and wishes to be paid the sum 8 with 0 pieces of ‘money : how many of each must the banker give him ? (Comp. p. 67.) Ans. Of silver, (1.5—3: ; of gold, 6. 39. A certain number being divided into both it and iii equal parts, the product of the 22 parts is 12. times the product of the n+1 parts. Find the number. A S (n+1)”+1 n . . a-c a—b' __- 4.32 PROBLEMS. 40. Two travellers, A and B, set out from two places P and Q at the same time; A from P intending to pass through Q, and B from Q intend- ing to travel the same way. After A had overtaken B, and they had computed their travels, it was found that the distance A had travelled together with the distance B had travelled made up 30 miles ; that A had passed through Q 4 hours before; and that B at his rate of travelling was 9 hours journey distant from P. Required the distance between the two places P and Q. Ans. 6 miles. 41. The rent of a farm is paid in certain fixed numbers of quar- ters of wheat and barley: when wheat is at 55.9. and barley at 33.9. per quarter, the portions of rent by wheat and barley are equal to one an- other: but when wheat is at 65.9. and barley at 41.9. per quarter, the rent is increased by £7. What is the corn-rent? Ans. 6 qrs. of wheat; 10 qrs. of barley. 42. A constable in pursuit of a thief at a uniform pace finds by inquiry that the thief is travelling 1% miles per hour quicker than him- self; he therefore doubles his speed after the first 4 hours, and takes the thief at the end of 6 hours and 20 minutes from the time of his starting. Given that the thief had a start of 1 hour, and never varied his speed, find the rates of travelling of the two parties, and the distance at which the capture took place. Ans. Constable’s speed at first, 8% miles per hour, Thief’s speed throughout 9%- . . .. . . . . . . . Required distance 71% miles. 43. At an election where each elector may give two votes to different candidates, but only one to the same, it is found on casting up the poll, that of the candidates A, B, C, A had 158 votes, B had 132, 0 had 58. Now 26 voted for A only, 30 for B only, and 28 for C only. Hdw many voted for A and B jointly; how many for A and 0; and how many for B and 05‘ Ans. For A and B, 102; A and C, 30; B and C, 0. 44. During a panic there was a run on two bankers, A and B; B stopped payment at the end of three days, in consequence of which the alarm increased, and the daily demand for cash on A being tripled, A failed at the end of two more days. Now if A and B had joined their capitals together, they might both have stood the run as it was at first for 7 days, at the end of which time B would have been indebted to A £4000. What was the daily demand for cash on A at the beginning of the run? Ans. £2000. 45. There is a waggon with a mechanical contrivance by which the difference of the number of revolutions of the wheels on a journey is noted. The circumference of the fore-wheel is a feet, and of the hind- wheel 6 feet; what is the distance gone over, when the fore-wheel has made 12 revolutions more than the hind-wheel? Ans ab n m feet. PROBLEMS. 433 46. A person has two casks containing a certain quantity of wine in each. He wishes to have an equal quantity in each ; and in order to have this, he pours out of the first cask into the second as much as the second contained at first; then he pours from the second into the first as much as was left in the first; and then again from the first into the second as much as was left in the second. At last there are exactly a gallons in each cask. How many gallons were in each cask at first? 11a 5a Ans. —8—~, and —E-,—. 47. A person rows from Cambridge to Ely (a distance of 20 miles) and back again in ten hours, the stream flowing uniformly in the same direction all the time ; and he finds that he can row 2 miles against the stream in the same time that he rows 3 miles with it. Find the velocity of the stream, and the times of going and returning. (1) Ans. —56-ofa mile per hour. Ans. 4 hours. Ans. 6 hours. 48. The Gas Company engage to light a shop, for six days in a week, with 5 large and 3 small burners, but having by them only one large burner, they supply the deficiency with 5 small ones. The shopkeeper, not finding this light sufficient, procures two small burners more, and at the same time agrees for the lights to burn double the usual time on Saturday nights, for which additional gas he was to pay £1. 11.9. How much did he pay a year altogether? Ans. £5. 5.9. 49. A shopkeeper, on account of bad book-keeping, knows neither the weight nor the prime cost of a certain article which he had purchased. He only recollects, that if he had sold the whole at 30.9. per 1b., he would have gained £5. by it, and if he had sold it at 22.9. per 1b., he would have lost £15. by it. What was the weight and prime cost of the article? Ans. Weight 50lbs. Cost 28.9. per lb. 50. A book is so printed, that each page contains a certain number of lines, and each line a certain number of letters. If we wished each page to contain 3 lines more, and each line 4 letters more, then there would be 224 letters more than before in a page; but if we wished to have 2 lines less in each page, and 3 letters less in each line, then the page would contain 145 letters less than at first. How many lines are there in each page, and how many letters in well line? - Ans. 29 lines, and 32 letters. 51. When wax candles are half-a-crown a pound, a composition is invented of such a nature, that a candle made of it will burn two-thirds of the time in which a wax candle of the same thickness and one-fourth as heavy again will continue burning. Supposing the two candles give an equally bright light, what must be charged per lb. for the composition that it may be as cheap as wax? (Comp. p. 68.) Ans. 28. 1d_ 52. A, B, C, D, E play together on the condition that he who loses shall give to all the rest as much as they have already. First A loses, then B, then C, then D, then E. All lose in turn, and yet at the end of the 28 434 PROBLEMS. fifth game they all have the same sum, viz. £32. How much had each before they began to play? ‘ Ans. A£81, B£41, C£21, D£11, E£6. 53. A man at his death leaves property to the amount of £5850 to be divided among three _sons, four daughters, and his widow, in manner following:--viz. the share of two sons is to be equal to that of three daughters, and the mother’s share half that of a son and a daughter taken together. Find each person’s share. Ans. Each son £900, daughter £600, mother £750. 54. A and B engaged to reap equal quantities of Wheat, and A began half an hour before B. They stopped at 12 o’clock, and rested an hour, observing that just half the whole work was done. B’s part was finished at 7 o’clock, and A’s at a quarter before 10. Supposing them to have laboured uniformly, determine the times at which they commenced. (Com!)- P' 69') Ans. A at 91» past 4, B at 5 o’clock. 55. A cistern can be filled by three different pipes; by the 1st in 1,1,- hours, by the 2nd in 3%,- hours, and by the third in 5 hours. In what time will this cistern be filled when all three pipes are opened at once? Ans. 48 minutes. 56. If A and B together can perform a piece of work in a days, A and C together the same in 6 days, and B and C together in 0 days; find the time in which each can perform it separately? (App. p. 350.) c) Ans. A in —————2abc B in —————-———~abc C in Qabc ac+bc—-ab’ ab+bc-ac’ amc’ days" 57. In a tithe-commutation.the rent~charge was apportioned so as to be 3.9. an acre, and in the 1st year the rates payable on the rent-charge wanted £6 of 10 per cent. on the whole receipts. The next year the rates were doubled, and amounted to 15 per cent. on the receipts: what was the number of acres? Ans_ 1600. 58. A and B drink from a cask of beer for 2 hours, after which A falls asleep, and B drinks the remainder in 2 hours and 48 minutes: but if B had fallen asleep, and A had continued to drink, it would have taken him 4 hours and 40 minutes to finish the cask. In what time would each singly be able to drink the “711016? Ans. A in 10 hours, B in 6 hours. 59. There is a number composed of two figures, of which the figure in the unit’s place is triple of that in the tens’, and if 36 be added to the num- ber the sum is expressed by the same digits reversed. What is the number? Ans. 26. 60. The fore-wheel of a coach makes 6 revolutions more than the hind-wheel’in going 120 yards; but, if the circumference of each wheel be increased 1 yard, the fore-wheel will make only 4 revolutions more than the hind-wheel in the same distance. Find the circumference of each wheel. (App. p- 351-) Ans. 4, and 5, yards. 61. The mail-train upon a railway starts a certain time after a luggage-train from the same terminus, and the time is so adjusted that, PROBLEMS. 435 before arriving at the other terminus, the trains will exactly escape col- lision and no more. It happens, however, that from an accident to the engine the speed of the luggage train is suddenly reduced one-half after performing two-thirds of its journey, and a collision takes place a miles from the end of it. The proper speeds of the trains being 722 and 92 miles per hour, (m>n) find the length of the railway, and the difference of times of starting. (Comp. p. 69.) ‘ (1) Ans. 3<2—--:£)a. (2) Ans. 3.7" n( —f-) a. 77272 62. Three persons divide a certain sum of money amongst them in the following mannerz—A takes the nth part of the whole together with gi: ; B takes the 12“ part of the remainder together with. g£; C takes the _ . a . . n‘“ part of what now remains together With 5£3 and then nothing remains. Find the sum. 3122— 312 + 1 W 63. Find two numbers whose product is equal to the difference of their squares, and the sum of their squares equal to the difference of their cubes. (App.'p.353.) 1 _ 1 Ans. -2-,/5, and ;(5+J5). 64. A pack of p cards is distributed into n heaps, so that the num- ber of pips on the lowest cards, together with the number of cards laid upon them, is the same number m for each heap', and the number of cards remaining is found to be 7‘; required the number of plps on all the lowest cards. (Comp. p. 70.) Ans. (m +1)n+,-__p. Ans. . a 65. If a men or 6 boys can dig m acres in 12 days, find the number of boys whose assistance will be required to enable a—p men to dig m +1) acres in {—7) days. (Comp. p. 70.) Ans. p_b(1+ m+n a a 12—}; 'iii ' 66. Supposing the sum of 51 cards in a common pack to be 10n+a, (where a<10), prove the value of the last card to be 10—a, the court- cards reckoning for 10, and the others for as much as is the number of pips upon each. Find also the value of 72. (Comp. p. 71.) Ans_ ,,___33. 67. If a oxen in m weeks eat acres of grass, and c oxen eat d acres in a weeks, how many oxen will eat 6 acres in p weeks, supposing the grass to grow uniformly? (Comp. p. 72.) 972—1) use 12—]: mae m—n'pd m—n' pl) ' Ans. ' 1 68. The distance between two places is a, and on the first day 51th . 1 of the journey from one to the other is performed; on the 2nd day 2th 28—2 436 PROBLEMS. I 1 Q of the remainder ; then 7-n-th and %th of the remainders alternately on succeeding days. Find the distance gone over in 2p days. (Comp. p. 73.) Ans. a{1-(1-%)p.(1-;15)P}. 69. Two labourers A and B, whose rates of working are as 3 to 5, were employed to dig a ditch ; A worked 12 hours and B 10 hours a day: B being called away, A worked one day alone in order to complete the work: when they were paid, B received as many pence more than A as the number of days they worked together. Now had B been called away a day sooner, A would have received 3.9. 11d. more than B at the conclu- sion of the work. What are their respective daily wages on supposition that each is paid in proportion to the work performed? (App. p. 354.) Ans. A’s daily wages 1.9. 6d. B’s 2s. 1d. 70. To complete a certain work A requires 112 times as long a time as B and 0 together; B requires 22 times as long as A and C together; and 0 requires ,0 times as long as A and B together. Compare the times in which each would do it, and prove that l 1 I ~ —— -——— ———=1. Com). . . m+1 n+1 +p-i-l ( 2 p 73) _ 71. 81, S2, S,,...S,,+, are n_+1 stones placed in a straight line a yard from each other, and X is another assumed station in the same line pro- duced ; two persons set out from 8,, the one to carry the stones separately to 8,, and the other to X; find the distance from Sm.l to X, that the latter may travel exactly twice as far as the former. (Comp. p. 74.) n(n + 2) 212 +1 ' 72. A steam-boat sets out from London 3 miles behind a wherry, and having got to the same distance a-head it overtakes a barge floating down the stream, and reaches Gravesend 112 hours afterwards. Having waited to land the passengers %th of the time of coming down, it starts to return, and meets the wherry in % of an hour, the barge being then 5%,— miles a- head of the steam-boat, and arrives at London in the same time that the wherry was in coming down. Find the distance between London and Gravesend, and the rate of each vessel. (App. p. 356.) (1) Ans. 30 miles. (2) Ans. 9, and 3, miles per hour. 73. Two clocks are striking the hour together, and are heard to strike 19 times. There is a difference of two seconds in their time, and one strikes every three, the other every four, seconds. What is the hour they strike? it being observed that, when the clocks strike in the same second, the sounds cannot be distinguished, so as to determine whether one or both strike in that second, and that this is the case with the last stroke of the faster clock. (Comp. p. 75.) Ans. 11 o’clock. 74. ' A and B travelled on the same road and at the same rate to London. At the 5Olb mile-stone from London A overtook a flock of geese, INEQUALITIES. 437 which travelled at the rate of 3 miles in 2 hours; and 2 hours afterwards he met a stage-waggon which travelled at the rate of 9 miles in 4 hours, B overtook the flock of geese at the 45th mile-stone from London, and met the stage-waggon 40 minutes before he came to the 31st mile-stone. Where was B, when A reached London? (App. p. 357 Ans. 25 miles from London. 75. The hold of a vessel partly full of water (which is uniformly increased by a leak) is furnished with two pumps worked by A and B, of whom A takes three strokes to two of BS, but four of B’s throw out as much water as.five of A’s. Now B works for the time in which A alone would have emptied the hold. A then pumps out the remainder, and the hold is cleared in 13 hrs. 20 min. Had they worked together, the hold would have been emptied in 3hrs. 45 min., and A would have pumped out 100 gallons more than he did. Required the quantity of water in the hold at first, and the horary influx at the leak. (Comp. p. 76.) Ans. Quantity in the hold 1200 gallons. Horary influx 120 gallons. IN EQUALITIES. 1. IF 4x-7<2x+3, and 3x+1>13-.r, find Ans. x=4. 2. What is the integral value of .r, when i(x+2)+%x< é-(m—4) +3, 1 and >%(x+1)+§? Ans. x=5. 3. Which is greater w—y or (Jag/5y? Ans. The former, if w>y. b 4. Shew that Zq,+-d—,>-(1;+%, if a+b be positive. (Comp. p. 78.) 1 1 a+-—_- 6+}- 0 ? Ans. The latter. . . 1 5. Which is greater 1 or (id-"'6' 6. Shew that ,/a2—b2+Ja2—(a~b)2>a, if a>b. (Comp. p. 78.) 7. If x2=a’+bz, and _y’=c'“’+d”, which is greater, .xy or ac+bd? (COmP- P- 79-) Ans. my. 8. Which is greater, 1284-1 or n’+n? (Comp. p. 79.) Ans. 1134-1, unless n=1. 9. Which is greater, 3(1+a2+a4), or (1+a+a‘“’)’? (Comp- p, 79-) Ans. The former, unless a=1. 438 RATIOS, PROPORTION, AND VARIATION. 10. Which is greater, 2(1+a2+a“), or 3(a+a3)? Ans. The former, unless a=1. 3 8 ll. Shew that b(%+g+2)> or or <6. (Comp. p. 79.) 12. Shew that (Comp. p. 80.) n2-n +1 2+ lies between 3 and -l- for all real values of n. n n+1 3 13. Shew that abc>(a+b—-c)(a+c—b)(b+c—a), unless (L=b=0. (Comp. p. 80.) 14. Shew that abc>(2a-b)(2b—c)(2c—a), unless a=b=c, each of the factors being positive. (Comp. p. 81.) 15. Shew that ab(a+b)+ac(a+c)+bc(b+c) is between Gabe, and 2(a3+53+03); a, b, a being positive quantities. (Comp. p. 82.) 16. Shew that (a+b+c)3>27abc, and <9(a,’+bs+c"), unless a=b=c. (Comp. p. 82.) 17. If a<.v, shew that (x+a)3—x3<7aa:’. (Comp. p. 83.) 18. Shew that ,3/iz->,4/n+1, for all values of 12 not less than 3. (Comp. p. 83.) 19. Shew that (a2+62+1)(02+d2+1)>(ac+bd+l)2, unless a=c, and b=d. (Comp. p. 83.) ’ 20. Shew that i(a+b+c+d)>:/abcd, unless a=b=c=d. (Comp. p. 84.) 21. If (1,, (1,, a,,...a,,, be positive quantities, shew that iz'gl(al+a2+a3+"'+an)>'\/a1a2+~/ala3+~/aga3+&cs (Comp. p. 84.) 22. Shew that ETIT— a2+ 62 : a+b. RATIOS, PROPORTION, AND VARIATION. 439 4. Find the ratio compounded of a : .r, x : y, and y : b. Ans. a : b, 5. Find the ratio compounded of x+a : a+b, and a(.r+b) ; b(.r+a). us. a : b. 6. What quantity must be added to each of the terms. of the ratio a : b, that it may become the ratio 0 : d? ad_bc Ans. C_d 7. Prove that, if a : b is a greater ratio than c: d, a+cz b+d is a less ratio than a : b, but a greater than c : a. (Camp. p. 85.) 8. What is the proportion deducible from the equation ab=a2—a:2? Ans. a: a+x :: a—x : b. 9. What is the proportion deducible from the equation x2+y2=2ax9 Ans. a: :y z: 3] : 2a—a. 10. Four given numbers are represented by a, b, c, a'; required the quantity which added to each will make them proportionals. Ans. 60- ad M' 11. If four numbers be proportionals, shew that there is no number which, being added to each, will leave the resulting four numbers pro- portionals. 2a+3b _ 2c+3d 4a+5b_ 40+5d' 12. If a : b=c: d, shew that 13. If four quantities of the same kind be proportionals, prove that the greatest and least together are greater than the other two together. 14; If (am-b)2 : (a—b)’ :: (2+0: b-c, shew that a : b :: m : 15. If a : y :: a8 : 63, and a : b :: Um :Jri-Fy‘, shew that cy=dx. 16. If a ; b :: c : d, shew that a(a+b+c+d)==(a+b)(a+c). 17. If a: 6:: c: d, shew that m :JZZ—d :: JE-JE ; Jay?!“ J5+,/Z . Jam? 18. If (a+b+c+d)(a—b-c+d)=(a-b+c—d) (a+b—c—d), shew that a : b :: c : cl. 19. If a : b :: c: d, shew that .1. + 1 3...; __1. a b 6 ma n—b+pc qd_bc'{ 20. If al : b, :: a2 : b, :: a3 : b, :: &c., shew that (Comp. p. 86.) (a,”+ (222+ a32+ &c.)(b,2+ 629+ 532+ &c.) = (a16l + a2b2+a363+ &c.)2, cl §+5+fi+E} . (Comp. p. 86.) and J(a,+a,+a,+&c.)(b,+b,+b,+ 8:0.) =J¢Ta+,/a,_b;+,/Z;5,+&c. 440 RATIOS, PROPORTION, AND VARIATION. 21. If the difference between a and b be small when compared with ' either of them, shew that the ratio Uri—Uzi: ,Z/ii—Z/I; is nearly equal to 712/53 mZ/E. (Comp. p. 86.) 22. If M=M ‘w—‘y, shew that .172 _I/ :: 1*J5:2. Jw+m~/y WWW—y ‘ (Comp. p. 87.) 23. Find the number to which if 1 and 3 be successively added, the resulting numbers are in the proportion 2 : 7. 1 A o__'o HS 5 24. Find two numbers in the ratio 3 : 4, and of which the sum : the sum of their squares :: 7 : 50. Ans_ 67 and 8_ 25. Distribute .9 soldiers among 15 towns in proportion to their respec- tive populations p,, p,, p,,...p,. Ans. 2%.9, Pia, 1113’s, . . . 15.9, where P=p,+p,+p3+...+p,. 26. If m shillings in a row reach as far as n sovereigns, and a pile of shillings be as high as a pile of q sovereigns, compare the values of equal bulks of gold and silver. (Comp. p. 87 Ans. Val. of gold : val. of silver :: 20n2q : map. 27. A person in a railway carriage Observes that another train running on a parallel line in the opposite direction occupies 2 seconds in passing him—but, if the two trains had been proceeding in the same direction, it would have taken 30 seconds to pass him. Compare the speeds of the two trains. Ans_ 7 , 8_ 28. A person, having travelled 56 miles on a railway and the rest of his journey by a coach, Observed that in the train he had performed one- fourth of his whole journey in the time the coach took to go 5 miles, and that at the instant he arrived at home the train would be 35 miles farther than he was from the station at which it left him. Compare the rates of 'the coach and train. (Comp. p. 88.) Ans_ 1 , 3 29. Given that poem, and when x=2, y=10, required the equation between a: and y. Ans. 31:556. 2 30. Given that yiocaz—ag; and when w=Jas~bi, g/=%; find the equation between a: and y. __11 —,:——, Ans. y-aJa as. 31. Given that soct", when f is constant; and socj, whent is con- stant: also 2.9=f, when i=1. Find the equation between f, s, and t. 1 A o ="_ ‘2. ns .9 2f RATIOS, PROPORTION, AND VARIATION. 441 32. Given that y is equal to the sum of two quantities, one of which varies as .r, and the other varies inversely as we; and when a=1, 2, y=6, 5, respectively, Find the equation between a: and y. Ans y_2x+ 4 . —— I'm—'2 0 33: Given that y is equal to the sum of three quantities, the 1st of which is invariable, the 2nd varies as .12, and the third varies as $2. Also when x=1, 2, 3, y==6, 11, 18, respectively. Express y in terms of .r. Ans. y=3+2x+a~2. 34. Given that aoc-l-m, and yes-2;; also when w=a, z=c; find the equation between .1: and 2:. (Comp. p. 89.) Ans. a2m"=c’"“a'. 35. If y=r+.9, whilst room, and seq/.70; and if, when w=4,_y=5, and when .r=9, _1/=10; shew that 6y=5(a:+~/a_). (Comp. p. 89.) ‘ 36. If a+boca--b, prove that az+bgocab; and if aocb, prove that ag—b'iocab. 37 . If §oc.1”+y, and gear—y, shew that avg—y” is invariable. 38. If ax+by=ca+dy, prove that aocy. 39. If aocb, and bocc, shew that (a2+62)3occ3. 40. If A ocB, and BocU, shew that mA+izB+pCocm',/AB+n'~/BC+]9'~/AC: (Comp. p. 90.) 41. If AocB, and Bmm shew that CwyA'Z—B4f/ABE, and that mi/AB— ni/BCocpJA—+ (Comp. p. 90.) 42. If mA+nBocpA-qB, and m’A—n'Cocp'B—i-q’C; shew that (a,,/a_s,,/R+7,,/6)ZM,A+B,B+v,c. (Comp. p. 91.) 43. A sphere of metal is known to have a hollow space about its centre in the form of a concentric sphere, and its weight is; of the weight of a solid sphere of the same substance and radius; compare the inner and outer radii, having given that the weights of spheres of the same substance OC(rad11)3. (Cami). p. Ans. 1 z 2. 44. Two globes of gold whose radii are 1', r’, are melted and formed into a single globe: find its radius, having given that the volume of a globe oc(radius)3. Ans, Ursa. 71% 45. A locomotive engine without a train can go 24 miles an hour, and its speed is diminished by a quantity which varies as the square root of the number of waggons attached. With four waggons its speed is 20 miles an hour. Find the greatest number of waggons which the engine can move. (Comp. p. 92.) Ans_ 14,3_ 442 ARITHMETIOAL PROGRESSION. 46. The value of diamonds 0c the square of their weight, and the square of the value of rubies co the cube of their weight. A diamond of a carats is worth m times the value of a ruby of b carats, and both together are worth c£. Required the values of a diamond and of a ruby, each weighing a carats. (Comp. p. 92.) i on (m+1)bf . 47. If the prices of two trees containing p and (1 cubic feet of timber be a£ and bi: ; required the price of a tree containing r cubic feet, sup- posing the values of the timber and bark to be respectively proportional to the mm and nth powers of the quantity of timber in the tree. (Comp. p. 93.) Am (bP"—aq")r;‘+(¢fnqj—bp’")r“_ P q ~11 q 2 Ans. Value of diamond = (mliilz-Sg, ; value of ruby-:- ARITHMETICAL PROGRESSION. 1. FIND the 64th term of the series 4, 6%, 9, &c. Ans. 161%. 2. Find the 30th term of the series —27, -20, —13, &c. Ans. 176. 3. Find the sum of 50 terms in A.P. of which the first is 3, and the last 199. Ans. 5050. 4. Find the sum of 100 consecutive whole numbers beginning from 1. ' Ans. 5050. 5. Find the 7th term, and the sum of 7 terms, of the series%, é, %, &c. 1 (1) Ans. — 5, (2) Ans. 0. 6. Find the 6‘“ term, and the sum of 6 terms, of the series %, 125- , d 0 4: 1 T5, 8w, (1) Ans. —§, Ans. 1. 7. Find the nth term of 1+3+5+7+ . . . . . . . . . Ans, 2n_1. and of2+2,l,-+2-§-+ . . . . . . . . . Ans. é-(n+5). and of 13+12§—+12-1_.;+ . . . . . . .. Ans. é-(40-m). 8. Sum the following series: (1) 1+2+3+4+... tonterms. Ans. 12.7221 . (2) 1+3+5+7+. . . to 1: terms. Ans. n2. (3) 2+2.};+2§-+.... ton terms. Ans. g(n+ll)- ARITHMETICAL PROGRESSION. 443- (4) 5+ 4ÿî—+4«-%+&c. to 21 terms. . Ans. 52%. (5) 13+12§+ 12%;+. . . to n terms. Ans. gCYQ—n). (6) 2è+2ë+ 3ä—+&c. t0 13 terms. Ans. 58%. 1 2 Il n (7) 2 3 —-Î .... .. to n te1ms. Ans. E(13—7n). 1 3 1 8 —-+ - — ...... .. . 31 ( ) 4 8 +2+ to n terms Ans. 16(n+3). 5 (9) ï+1+1%+ ...... .. to n terms. Ans. %(n+4æ). . l 5 4 n ' (10) -3—+Ë+-ä+ .... .. to n terms. Ans. ñ(3n+1). (11) —-9-7—5— &c. to 20 terms. Ans. 200. 8 2 2 (12) -9-+ 6% +6—3+ &c. to 7 terms. Ans. -2%. 5 2 (13) ÿ+ô+&c. to 101 terms. Ans. —24.~l2%. (14) 2-1.;+3-2-51-'+4;Tl§+ &c. to 24 terms. Ans. 297%. a-b 311—26 72 n+1 1 —__+————— ' . . —— . --——— ( 5) a+b “+6 + tontelms Ans a+6 {na 2 b}, (16) E+fl+n_3+...to n terms. Ans. 73—1 _ n n n 2 l -1 l - (l7) (l—Ï)+(-—Zz———)+(——-n—Ê)+ .... .. to n terms. a a: a a: a a: Ans. 2-2— n .n+1 _ a a: (18) (s+a)+(s+a+d)+(s+a+2d)+ .... .. to n terms, Where n s=(2a+n 1.d)-2-. Ans. (n+1)s. 9. Prove the following forms in Arithmetical Progression: a=gi‘ / —258, _ __l______s_ _ _(l+a)(Z—a)_ b_2'{n-1 n(n—1)}’ b—m’ l-a l+a =—2—b—(l-—a+b); n=——b—+1. 444 ARITHMETICAL PROGRESSION. 10. Given the first term of a series in Arithmetical Progression, the common difference (6), and (s) the sum of the series to n terms; find 12. An n_1 er\/Pé§-r<_1__2)” 8' “E b b 2 b ' Explain the meaning of the two signs in the value of n. 11. How many terms of the series 19, 18, 17, &c. amount to 124? Ans. 8, or 31. 12. The sum of a series in Arithmetical Progression is 72, the first term 17., and the common difference —-2; find the number of~ terms, and explain the double answer. Ans. 6, or 12_ 13. Given s=40, a=7, and 6:2. Find 11. Ans. n=4, or —10. 14. In any Arithc. Prog“. of' which a is the first term, and 2a' the com- mon difference, prove that the number of' terms, which must be taken to . s . s , make a sum 8, 1s a , .9 being assumed such that 5 1s any square number, but no other. 15. Find the series in A.P. in which 7 and 5 are the 5th and 7th terms rGSPeCtiVeIY- (Comp P- 94-) Ans. 11, 10, 9, 8, 7, 6, 5, &c. 16. Insert 40 Arithc. Means between 0 and 20. 20 40 ABS. 5, 4—1, 1%, &C- . . . . “191%. 17. Insert 15 Arith". Means between 3 and 47. Ans. 5533;, 8%, 1121-, &c ..... "4471-. 18. If the Arith°. Mean between a and_b be twice as great as the Geomc. Mean, shew that a : b :: 2+~/ 3 : 2-J3. (Comp. p. 94.) 19. If the Arith". lVIean between a and b be m times as great as the Harmonic Mean, shew that a : b :: ~/m+~/m—1 : JE-Jm—l. (Comp. p. 95.) 20_. Shew that if the same number of Arithc. Means be inserted between every two contiguous terms of an AR the whole will be in A.P. (Comp. p. 95.) h 211. Find the series in A.P. having 29 terms of which the first is 3 and t e 4313 17. Ans. 3, 391—, 4, 4%, &c....17. 22. ’Find the series in A.P. of 12 +7 terms of which the sum of the first 12 terms 1s 40, the sum of the next 4 is 86, and that of' the last 3 is 96. Ans. 2, 5_, 8 11, 14, 17, 20, 23, 26, 29, 32, 35. 23. The first two terms of an A.P. are 189% and 103%; and the sum of all the terms is ~147§~ ; what is the last term, and what the number of terms? (1) Ans. —-238I"l2_. (2) Ans. 6. ARITHMETICAL PRGGRESSION. 445 24. Divide 7-,;(n+4) into 12 parts, such that each part shall exceed the one immediately preceding by a fixed quantity. (Comp. p. 96.) I" Ans. 1, 1%, &c. Q. {i 2 25. The n‘11 term of an Arith". Prog“. is -é.(3n —-1), prove that the sum of 1: terms is +1), and find the series. (Comp. p. 96.) 1 5 4a 11 ABS. '3', 6, '5, '67, 8Z6. 26. The sum of the first 72 terms of a series in Arith". Prog“. is n+1 n , a—b 3a—2b (na— ——2- .b>m, find the serles. Ans. m +721? +&c. 27. There are two series in Arith". Prog“., the sums of which to n terms are as 13—712 : 3n+1; prove that their first terms are as 3 : 2, and their second terms as —4 : 5. (Comp. p. 97.) ————~ . . . . ma-nb . 28. The n+1‘b term of a series in Ar1th°. Prog“. 1s a b , required ' ____.. ma—1 b the sum of the series to 2n+1 terms. Ans. a ,2 .(2n+1). —) 29. Shew that the sum of the m—n‘“ and m+nth terms of any Arith”. Prog“. is equal to twice the 112m term. 30. The sum of a certain number of terms of the series 21, 19, 17, 8:0. is 120. Find the last term, and the number of terms. (1) Ans. 3, or -—1. (2) Ans. 10, or 12. 31. In the two series 127, 120, 113, 1, and 2, 5, 8, 56, shew that the number of terms is the same in both, and find the number. Ans. 19. 32. Determine the relation between a, b, and c, that they may be the 72‘“, q“, and r‘“ terms of an Arithmetical Progression. (Comp. p. 97 Ans. (q—r)a+(r~p)b+(p-q)0=0. 33. In an Arithmetic Progression, if the p+qlth term =m, and the _“ 1 p—qlm term =12, shew that the 29‘“ term =§(m+n), and the (1m term =m-Q(m—n)é%. (Comp. p. 97.) 446 GEOMETRIC‘AL AND HARMONICAL PROGRESSION. 34. If the mth term of an Arithmetic Progression=n, and the nth term-=m, of how many terms will the sum be é(m+n)(m+n-1), and what will be the last of them? (Comp. p. 98.) (1) Ans. m+n, or m+n—1. Ans. O, or 1. 35. The sum of 111 terms of' an Arith“. Prog“. is n, and the sum of n terms is m; shew that the sum of m+n terms is -m+n, and the sum of m_-n terms is (m-n)<1+€)—?). (Comp. p. 99.) 36. In an Arithmetic Progression a is the first term, 6 the common difference, and S, the sum of n terms, prove that (Comp. p. 100.) ' b Sn+Sn+1+Sn+2+ 810- to "n terms=(3n-1)Zg-z + (7n-2)(n -1)% . 37. 8,, 8,, S3, .... “S, are the sums of p Arithmetic Progressions to n terms,- the first terms are 1, 2, 3, &c. and the common differences 1, 3, 5, 7, &c.- Shew that Sl+S,+Sa+...+Sp=(np+l)’—IQB. (Comp. p. 101.) 38. How far does a person travel in gathering up 200 stones placed in a straight line at intervals of 2 feet from each other—supposing that he fetches each stone singly and deposits it in a basket which is in the same line produced 20 yards distant from the nearest stone—and that he starts from the basket? Ans. 195% miles. 39. In the two series, 2, 5, 8, &c. and 3, 7, 11, &c. each continued to 100 terms, find how many terms are identical. (Comp. p. 101.) Ans. 25. GEOMETRIOAL AND HARMONIOAL PROGBESSION. 1. FIND the 12th term of the series 30, 15, 7 91-, &c. Ans. . . 3 2. The first two terms of“ a series in Geomc. PrOg“. are - and - - find the Common Ratio, and the third term. (1) Ans. 25%. (2) Ans. 15—252. 3. Find the Common Ratio and the fifth term of 33-, 2%, 191-, &c. * 2 2 1 An . - . . — . ( ) s 3 Ans 3 4. Sum the following series: (1) 1—2+4—8+... to n terms. i Ans_ ;:{1_(_2)..}. 1 2 4 3 2 " ~__ —___IID t t . A Q _ -‘ __ (2) 5 15+45 o n erms ns 25{1 ( GEOMETRICAL AND HARMONIOAL PROGRESSION. 447 .1 1 4 1 " __ .__.... ~ . A ._{ _(__\ . (3) 1 4“+16 to n teims ns 5 1 4/ [(4) é+é+g+m to n terms. Ans. §{g;—1} . 1" 45 . . (5) 5+—7i+ 21; zvgfi Ans. 8%. 5 . . (6) §+1+§+m zn 2n . Ans. 4-16. 2 1 3 . . 3 (7) §—§+§+... m zrgfi Ans. 21 . — 7 8) 2+ 4 8+ 2+ &c. to 12 terms. Ans. ——-_= . < f ,/ My, 2+1 1 . . — (9) :/[§—_—1+2—_J;+-%+...zn uy‘". Ans. 4+3,‘/2. fig—1 (10) ap+ap+q+a1’+gq+... to n terms. Ans. (1‘2; 1 . a? 93 “=2 2/ " (11) w—y+——-;+... to n terms. Ans. .{1—<—'— a: a: x+y .r (13) g\/§+\/§+\/-2:+m~ Ans 1 -(2x)2_(3a)? a: 2 x 3 ' ' 1 "—"°1 (2.20%- 3 ' 62.(3a) 2 a: ( a 5. Prove the following forms in Geometric Progression :— l(s—l)""-a(s—a)”"‘=0, writ—.2 , n—l n_] “'— JF- a" s 1 l =--—-'._—'_——: n—'—— n- ——=0- s fli/l-“I/a , r _Zr +s_l 6. Of each of the following series find the nth term, and the sum of n terms :— (1) 1+5+13+29+61+... (1) Ans. arr-3. (2) Ans. 4(2"—l)-3n. (2) 2+6+14+30+... (1) Ans. 2"+1-2. (2) Ans. 4(2"——1)-2n. (3) 1+3+7+15+31+... (1) Ans. 2"—~1. Ans.2(2"-—])—n. (4) 3+6+11+20+... (1) Ans. 2"+n. (2)Ans.2"+‘+-;-{n”'+n—-4}. l 3 15 2n—1 1 '1 +5 +g+?+oon ADS. 2n_l - Ans. 2(7Z—1)+Tn_—1 . 14 44 134 1 9 1 (6) 2+4+—é—+-9-+-%+... (1) Ans.5—3n_,. (2)Ans.5n—--2-(1—§1)' 448 GEOMETRICAL AND HARMONICAL PROGRESSION. 7- Sum (8—a)+(8—a+ar)+(s-a+ar+ar2)+...to n terms, and in Where s=a.r_ (Comp. p. 102.) r— nar" (1) Ans. r_1-(1_—cf_——7;),(r"-1). (2) Ans. ar u-o” 8. Find the sum of (13-1)+(23-2)+(33—3)+&c. to n terms. Ans. i-(n—1)n(n+l)(n+2). 9- Given az—1+2(.r—2)+3(.r-3)+&c. to 6 terms=14; find .12. Ans. ar=5. 10. Find the sum of 1+22+3+42+5+62+&c. to .2: terms, when x is an odd number. Ans. 1i2(a:+1)(2x2+ w+3). 11.. The first term of a Geometric Series, continued in iig‘I, is l, and any term 1s equal to the sum of all the succeeding terms; find the series. 1 1 1 ADS. l+§+z+§+ - - - - ~- 12. Insert three Geometric Means between 1 and 9. Ans. 31,-, 1, 3. 13. Insert seven Geometric Means between 2 and 13122. Ans. 6, 18, 54., 162, 486, 1458, 4374. g and 2. Ans. g i 9’ 3' 15. The sum of an infinite Geometric Series is 3, and the sum of its first two terms is 2%; find the series. . , 1 14. Insert two Geometrlc Means between 4 — 0.. . a o u a .- 9+ , 01 3 9 16. There are four numbers in A.P., which being increased respec- tively by 1, 1, 3, and 9, are in G.P.; what are the numbers? Ans. 1, 3, 5, 7. 17. In a Geometric Progression, if the p+qth term =m, and the __ H l .11 p-q‘h term =n, shew that the pi“ term =,/mn, and the qth term=m(:-7-I)gq Ans. 2+§+2 4" Also, if P be the p“ term, and Q the qth term, shew that the nth term "-4 _1_ =<‘g;1;>1’-9. (Comp. p. 102.) 18. Find the relation between a, b, and c, that they may be the 12‘“, q“, and rm terms of a Geometric series. (Comp. p. 103.) Ans. aq"b’“PcP" =1. GEOMETRICAL AND HARMONICAL PROGRESSION. 449 19. Required the sum of the first p terms of the series whose nth term is na+a". aP—l Ans. -é-p(p+1)a+a.a 1 . . 2 5 20. Given 5, —-, the first two terms of an A.P., find the sum of 15 terms; and if the same quantities be the first two terms of a G.P. find the sum of 15 terms. 28 "f5 15 (1) Ans. 15. (2) Ans. -€.{fi —- } 21. If a, b, c, d are quantities in G.P., prove that 412+ 122+ 02>(a—b+ c)2; and that (41+!)+0+d)2=(a+b)2+(c+d)’+2(b+c)2. (Comp. p. 103.) 22. If there be any number of quantities in G.P., r the common ratio, and 8,, the sum of the first 122 terms, prove that the sum of the 7” w products of every two =fi1.8m.8m_1. (Comp. p. 104.) 23. Prove that in any G.P. the sum of the first and last terms is greater than the sum of any other two terms taken equidistant from the beginning and end of the series. 24. If 8,, 8,, 83,...8n be the sums of n terms of n Geom°. Progns., of which all the first terms are 1, and common ratios 1, 2, 3,. . .n, respec- tively, shew that S,+S,+2S,+SS,+. . .+(n-1)S,,=1"+2"+3"+.. .+n". 25. The first two terms of a series in Harmonical Progression are a, b ; continue the series to three more terms. ab ab ab Ans‘ a’ 6’ aa-b’ 312—26, 4a_ab' 26. Given a and b the first two terms of an Harmonic Progression, find the n‘“ term. 615 A Q 0 ns (n—1)a—(n—-2)b 27. Insert two Harmonic Means between 6 and 24. Ans. 8, 12. 6 28. Insert six Harmonic means between 3 and -2_3. U o 3 _1_ __ Z __ _ 29. If a, b, c be three terms in Harmonical Progression, they being' su osed to have the same si n shew that a2+ 02>2b2. Com . . 105. PP g a P P 30. The sum of three consecutive terms in Harmonical Progression is . I . I 1 1 111;, and the first term is —; find the series. Ans. -, -- , -- , &c. 2 2 3 4 31. Shew that the Arith°., Geom°., and Harm". Means between any two quantities are themselves in Geom". Prog“. 29 450 PERMUTATIONS AND COMBINATIONS. 32. From each of three quantities in H.P., what quantity must be taken that the three resulting quantities may be in G.P.? Ans. Half the 2“‘1 term. 33. If a, b, c be in A. P., and a, mb, 0 in G.P., then a, 77225, c are in H.P. 34. If between any two quantities there be inserted 2n—1 Arith°., Geom°., and Harm". Means, the nth Means are in G. P. (Comp. p. 105.) 35. There are four quantities, of which the 1st three are in Arith". and the last three in Harm". Progn., prove that the 1St : 2nd :: 3rd : 4th. 36. If there be five quantities a,, 0,, a3, a4, as, such that a“ a,, as, are in A.P., a,, a3, a,, in G.P., and a3, a4, a, in H.P., shew that a,, a3, (15, are in G.P. (Comp. p. 106.) 37. Find the relation between a, b, and c, that they may be the pt“, q“, and 1““ terms of an Harm". Progn. (Comp. p. 107.) Ans. (p-g)ab+(r—p)ac+(q—r)bc==0. 38. If a, b, c be in G.P., shew that log,n,10g,n,log,n, are in H.P. (Comp. p. 107.) 39. Find the sum of 2]) terms of the series whose n‘h term is n{[(—1)”+1]n+2}. (Comp. p. 107.) 2p Ans. 3 (2p+1)(2p+5). PERMU'I‘ATIONS AND COMBINATIONS. 1. THE number of Permutations of n things taken four together=six times the number taken three together; find 72. Ans. n=9. 2. The number of Permutations of 15 things taken 0* together = ten times the number taken r—l together; find 1'. Ans. r=6. 3. How many days can 5 persons be placed in different positions about a table at dinner? Ans. 12a 4. The number of Permutations of 271:1 things taken n—l together : number of Permutations of 212—1 things 12 together :1 3 t 5 ,' find 11. Ans. 12:4. . 5. How many different sums may be formed with a guinea, a half- gumea, a crown, a half-crown, a shilling, and a sixpence? (Comp. p. 108.) Ans. 63. 6. In the Permutations formed out of a, b, c, d, e, f, g, taken all together, how many begin with ab .9 How many with abc? How many with abcd? (Comp. p. 109.) (1) Ans. 120. (2) Ans. 24. (3) Ans. 6. PERMUTATIONS AND COMBINATIONS. 451 7. Of the Combinations of 10 letters, a, b, c, &c. taken 5 together, in how many will a occur? (Comp. p. 109.) Ans. 126. 8. How many different Permutations can be formed from the letters of the word ‘Algebra’ taken altogether? Ans. 2520. How many from the letters of ‘ Proposition’ .9 Ans. 1663200. 9. At an election, where every voter may vote for any number of candidates not greater than the number to be elected, there are 4 candi— dates, and 3 members to be chosen ; in how many ways may a man vote? (Comp. p. 109.) , Ans. 14. 10. From a company of 50 police-men 4 are taken every night to guard the police-station. On how many different nights can a different selection be made ; and on how many of these will any particular man be engaged? (1) Ans. 230300. Ans. 18424. 11.. How many combinations are there of 52 things taken 13 together; that is, how many different hands may a person hold at the game of whist? .Ans. 635013559600. _ 12. Find the number of different triangles into which a polygon of n sides may be divided by joining the angular points. (Comp. p. 109.) Ans. -16n(n—1) (12-2). 13. Prove that the total number of different combinations of 72 things taken I, 2, 3, . . . n at a time is 2"—1. 14. The total number of combinations of 212 things =65xthe total number of combinations of n things; find 72. Ans. 11:6. 15. Shew that the total number of combinations that can be formed out of n+1 things is more than twice the number that can be formed out of n things. 16. If there be any unknown number of beans in a bag, prove that the chance of bringing out an odd number taken at random is greater than that of bringing out an even number, excluding the case of bringing out none at all. (Comp. p. 110.) 17. In how many ways can 8 persons be seated at a round table, so that all shall not have the same neighbours in any two arrangements? (Comp. p. 110.) Ans. 2520. 18. Out of 17 consonants and 5 vowels, how many words can be made having two consonants and one vowel in each? (Comp. p. 1 11.) Ans. 4080. 19. Find the number of words of 6 letters which can be formed from an alphabet of 19 consonants, and 5 vowels, each word containing two vowels and no more. Ans. 27907200. 20. Find the number of combinations that can be formed out of the letters of the word ‘Notation’ taken 3 together. (Comp. p. 111.) Ans. 22. 29-2 452 BINOMIAL THEOREM, ETC. 21. If there be two dice, one of which has n and the otherm faces, each die being marked in the usual manner from 1 upwards, find the number of different throws which can be made with them. (Comp. p. 111 Ans. é1z(n+ 1 ) +121". 22. Find the number of Permutations with. repetitions (that is, allow- ing quantities which recur to be combined as if they were different) of n things taken r together. Ans. n'. 23. Find' the total number of different combinations of 12 things taken 1, 2, 3, . . . 22 together, of which there are p of one sort, q of another, r of another, &c. (Comp. p. 112.) ' Ans. (p+1)(q+1)(r+l).&c.-1. BINOMIAL THEOREM, ETC. ., 7 1. EXPAND (a+f))8, (a—b)’, (2x—3y)‘, (5—4w)‘, and (1%). (1 ) Ans. a8+8a7b +28a652+ 56a5ba+ 7 0a464+ 56a355+28a266+ 8ab7+ b8. (2) A ns. a7—7asb+21a552—35a4b"+35a86‘- 21 agbfi+7ab"-b7. (3) Ans. 32x5— 240x‘y + 720x3y2—1 080x2y3+ 810xy‘-243y‘. (4) Ans. 625—2000x+2400x9—1280.r3+256x‘. 7 21 35 35 21 7 1 2. Required the coefficient of x5 in the expansion of (.r+a)8. Ans. 56a”. 3. Expand (JEAN/FY. Ans. a2+6ab+b’*4(a+b),,/c_z_b_. _2 *2 ‘ s 2 1 . .2. ‘ 4. Expand <1 2) . Ans. 1+x+4x+2x+16w+ .... .. 5. Find the 5th term of the expansion of (a2—b’)". Ans. Z495al“bf‘. 6. Find the 7‘“ term of the expansion of (a3+3ab)9. . i \ Ans. 61236a1556. 7. Find the 5":1 term of the expansion of (3612—7103)”. Ans. 13613670a8x". 8. Find‘ the 6th term of the expansion of (ax+by)‘°. [ Ans. 252a555w5y5. 9. Find the middle term of the expansion of (a’"+.z’” ‘2. Ans. 924a8mx6'". 10. Find the middle term of (a%_+ 6%)”. Ans, 7Qaibi_ BINOMIAL THEOREM, ETC. 453 11. Find the two middle terms of (a+.r)‘a. Ans. 1716a726, 1716a6a7. 1 2% 2 xi 14 mg 12. Ex and ———. Ans. —-—+—-.——+-.—:1--—.—+ .... .. P ,a/ax-x" (ax)i¥ 3 a% 9 0% 81 a? 1 i 21 i 562' 13. Expand 777-3?- Al'lS. 'i- f +——-—3-+ . . . . .. 14. Find the middle term of the expansion of (1+.x)2". ‘ 1.3.5...(2n-1) A I —————_—-_— O n. ns n " ‘1’ l 15. Find the 1'“ term of the expansion of (3a—2x)‘. 1.4.9...(5r-11) 22y“ L, Ans' 1.2.2. .. (r—l) '(F {3‘05 ' l __ 16. Expand W5 to 5 terms; and find the 5+rlth term. a —.1: 1 1 x5 3 21° 11 2‘5 44 22° Ans. ;+-5--56+2—5.F+-].—2—5-.F+6_25.Zfi+ . . . . .- 1.6.11.16 . . . . .. 1 20+“ JATIS. ( 6+5r).-%-.£2—1T . 1.2.3.4 .... .. (4+1) 5+r a H 17. Find the term which involves a‘i’f)7 in the expansion of (a8+3ab)9. Ans. 78732a‘3b7. 18. Expand (a +2b—-c)3 by means of the Binomial Theorem. Ans. a3+ 6a25+12a52+853—3agc—l2abc—l2bgc+3a02+6509——08. 19. Find the term which involves 2452 in (7a2—3al1+4b’)3. Ans. 7774114112,. 20. Find which is the greatest term of the expansion of (1+ Ans. The 2nd. - ll 21. Find which is the greatest coefficient in (1+.r)~/§ . Ans. The 5th. . i . 22. At what term does the ser1es for (1+ 1% begin to converge? Ans. The 3rd. 23. Find an approximation to the cube root of 31 by the Binomial Theorem- Ans. 3'14138. 24. Find an approximate value of ,Z/ 108. (00111)). p. 112.) Ans. 1'95208. 454 BINOMIAL THEOREM, ETC. 25. Find the sum of 1+g+§+ g+&c. 1, a, ,6’, 'y, &c. being the coefficients taken in order of the expansion of (a+b)". (Comp. p. 113.) ' 2"+I-1 Ans. -—-—-. n+1 26. Prove that the coefficient of the r +1L11 term of (1+ .10)“ is equal to the sum of the coefficients of the r“ and r+1‘h terms of (1+w)". 27. If a, b, c, d be the 6th, 7th, 8““, 9‘11 terms respectively of‘ an 52- 4 expanded binomial, shew that 3,173,0- = 5%. (Comp. p. 113.) u v 0 I o 28. Fmd the coefficient of x" 1n the expanswn of (P x), . (Comp. p. 114.) Ans. -é.(n—6)(n’—1). z 29, If r be the greatest whole number contained in 12, (2+2)? has the first r+2 terms of its expansion positive, and the 1'+mth of the same 2‘ ‘__ O 0 sign as (—1)’"; but (a—x)“ has the first r+1 terms alternately posmve and negative, and all the rest of“ the same sign as (—-1)"“. (Comp. p. 114.) 30. Given that the coefficient of the 5:1“ term of the expansion of (1+.r)" is equal to that of the 1E3“ term, shew that p=g—1. (Comp. p. 115.) 31. What is the meaning of (1+:v)‘/5? Has the Binomial Theorem been proved in any sense to extend to such a quantity? 32. If x>a, prove that the sum of all the terms of the expansion of (a: +a)", after the first two, is less than (2"-n-1)ax"". (Comp. p. 116.) 33. If ]1)(P;1)(7;_2) ' ' ' ' ' '(p;r+l) be represented by 1),, prove that (p +q),=p,+p,_,q,+p,_,q,+ . . . . . . +p,q,_ 1+ (1,. (Comp. p. 116.) 34. If 1202 ‘;1)(’1+2) ' ' ' ' ' '(n+r;1) be represented by nP,, shew that 0 0 O o Q s . . O ,,P,+,,+,P,_,=,,+,P‘. (Comp. p. 117.) 35. Prove that 1.3.5...(2r—1)+ 1.3.5...(2r—-3)._3_+ 1.3.5...(2r— s). 3.5+ .... 0 Ii 11:}. L1. LEE 12 fis equal to 2’(1+r). (Comp. p. 117.) I BINOMIAL THEOREM, me. 455 36. Shew that if t, denote the middle term of (1+.r)”, then will t,+z,+z,+ .... ..=(1-4.x)—i. (Comp. p. 118.) 37. If x be very small compared with 1, prove that Jm+,8/(1—x)’=1 l+.r+ 1+2 -%a: nearly. (Comp. p- 119) 38. If c=a-b, and is very small compared with a and b, shew that azb’ (a’ — a"’x’+ 62.x”)é =a—2c+3c.z:2 nearly. (Comp. p. 119.) mm"- mr" 39. If a: be nearly equal to 1, shew that m_n (Comp. p. 120.) 40. If a> b, prove that a"-b”>nb'*“(a—b) and <\/ Ans. §(~/3~l). 2. 7211276. Ans. N4/1—2—1-1/5'. 3. 752-7271“. Ans. ,‘lTs-Q/é. 4. 375+746. Ans. ~"/2_o+f/5. S" 458 EVOLUTION OF SURDS. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. Does it follow that 22. 2 3a a"b2 a: r:2 O 38 44 b .. at", 12ab _4al1 Ans. (in, a: 02.2: c“ ' 6‘ Ans. Ja—bh/ 402—41”, ab + 402- d’+ 2 J4abcg- abd’. Prove that .,/bc+2b,/bc---f)’-1-,~/(1c:-—2b,\/bc:»—bg is equal to =1=2b. Extract the cube root of 7 —5~/2. Ans. 1-~/2. 4542972. Ans. 3472. 148+46,/ii. Ans. f/2{2+./ 11}. ‘" ' 1 Extract the 4‘“ root of 14+8,,/-3. Ans. Extract the 5th root of 41+29J2. Ans. 1+J2. 22s+1s2,/§. Ans. (1+,/§)i/§. Extract the 6th root of 2889-1292J5. Ans. 2—,,/5. Extract the 7th root of 239+169J2. Ans. 1+,J2'. Extract the square root of 9 + 2 J 3 +2 J5+ 2 J15. Ans. 1+J3+J5. Extract the square root of 6 + 2 J2+ 2 J3 +2 J 6. Ans. 1+(/2'+,,/3. 2-[2' 4. J2 Sh th t ( _) l t . . . ew a 2+J3 1s equa 0 1+ J5 (Comp p 129.) If x=:/r+,,/r’+q3+i/r-,/r2+ q3, shew that .ra+3qx—2r=0. Prove that ~/ 2 +J8 may be expressed by 1+~/:I+~/1-~/:1_. 2+~/8 is “impossible”? Prove that Jada/5 may be reduced to the form UEsf/js', when a . 1—-5- 1s a complete square. 23. 24. Prove that (20+~/3_$-)—2-)%¥+(20--,/392)it is equal to 4. a _____ " .._____ . ,r 2 3 r 2 8 If x=\/——2-+\/%—g§+\/—-2——\/%—§q¢7, and r=0, shew that the values of x are 0 and (Comp. p. 129.) INDETERMINATE COEFFICIENTS. 459 25. If( -'%>.{b-y—(a—x)p}=c’, and p=\/'Z,E%Zx, prove that ~/(b—y).r—,/(a-.'r)_y=c. (Comp. p. 129.) 26. Given 2{a:2+y2—x—y}+l=0; find the real values of x and y. (Comp. p. 130.) ‘ _1_ 2 . 27. Given (x+y(/——1)’=a+b~/-_-_1-; find the real values of .r and y, (Com. .130. 1 —,--, 1 2 2 P p ) Ans. x: éfJa +6 +0), y= '2-(001‘1'6 "'a)' ss+2y,/—_1 15 find d —- = ——, O 0 0 0 5,/-1-2 as+sy,/-1 “3 an y ( amp p ) 1 Ans. .2= 5 , y= 28. Given Ans. x=1, y=3. 29. Find the relations subsisting between a, b, c, d, when the square root of a+(/5+~/E+,,/d can be expressed in the form Joi+~/B+~/;. (Comp. p. 131.) Ans. Each of the quantities fl, fl, \/ 956-1 rational, and the sum of them equal to 2a. INDETERMINATE COEFFICIENTS. BY the method of Indeterminate Coefficients shew that, 3+2.r__3 11 7211 ,_72><11 , 7S><11 4 1. m-5—5§.r+ 5, .2: 5, .z:+ 5, .r- .... .. 2. —-1-ir—3=19+2’.x+3”..1:2+42.w’+52.w‘+ .... .. <14») 3 2+3 __ 4 _ 1 ' (x-1)(w+2)_3(.n-1) 3(a:+2)' 4 2+1 __ 3 ___1_ ' ’- 2(x- 2) 2x ' 5 1 __ 1 _ 1 ' wz—(a+b)x+ab_(a~b)(x—a) (a—b)(.r-b)' 6 x+1 _ 5 4 ' .r".-7x+l2 _x-4 .11—3 ' 7 3x—5 _7_ 1 _l 1 ' .x’—6x+8_2'x—4 2.2-2 1 460 ~INDETERMINATE COEFFICIENTS‘. 8 x9 -- 1 _._%_.+____9____ " (x+1)(x'+2)(a2+3)_2(x+1) ar+2 9. 3x”—7x-;-6= 2 a__ 1 2+ 3 . (at—1) (rs—1) (x—l) w—l 1 1 1 l 10. a‘— x‘ = 4a”(a+ x) + 4a3(a —.x) + 2a”(a”+ w’) ' 11 -x2—x+1_1 2+ 3 ' :r”(a:+1)__:v2 a: a'+l' x- bx3+ dx" _—-—_—___= - 3 2— - 5 a O I I O. 12. 1_ax2+0x4 x+(a b)a: +(a ab c+cl)a;+ 13 1 __1_{_L __1__ w—Q _w+2} ' .r°-1_6 .r—1_a2+1 w’-x+1 w’+w+1 ‘ ~14 m9+px+q ' _ a’+pa +q + bg+pb +q ' (x—aXx—bXx— c) '— (a—b)(a-—c) (ac—a) (b—aXb—cXx— b) ' + cg+pc+q @-oe—w@-o' , 2 8 _ 1 _3_ 2 _5_ ,3 15. J1+x+x+x .... .._1+2x+8a:+16a, 16. Resolve 7x’-—6a:-1 into two factors of the first degree. (Comp. p. 132.) . Ans. (x—1)(7a.'+1). 17. Resolve QxQ—Qlxy—lly2-w+34~y-3 into factors of the first de- gree- (CGmP' P' 132') Ans. (w—lly—i-lXQxfly-S’). 18. Given y=ax+bwg+ cxa+ dx‘+..., find an in a series of powers of y. A _ 1 b 2 Elf—ac , 5b“—5abc+a’d 4 19. Given y“-a.r_y-b"=0, find 3} in a series of powers of x. 0.2: raga:a a‘w‘ , Ans. y=b+-g—b-—§%—5+W-... 20. Given w;1z-}2-n’+—;-n3—i-n4+..., find 12 in a series of powers of x. ' x9 m“ x4 Ans. n=x+i—2-+l 23+1.234,+... . 1 1 —— . . 21. Given x=2—§z"+Ez“-..., and y:z,J1-_y’, ‘find 3 m'a series of 8 x5 :3 " 1.2.3 + 1.2.3.43 " ' powers of x. k Ans. y=x CONTINUED FRACTIONS. " 461 CONTINUED FRACTIONS. . . . 251 3 1. FIND a serles of fractions converging to m; also to I595; . 1 22 23 114 E? Fi’ifi’éli’ 716 1 1 2 Ans‘ '5! Z, a" 55: 5,7) (2) Ans. &c. 84 . . . 2. Express Egg—7 1n a contmued fractlon, and find the convergents. Ans. The quotients are 2, 1, 2, 2, 1, 3, 2. The conver ents are1 g 7 L0. 37— ii g 2’ 3’ s’ 19’ 27’ 100’ 227' 3. Find the convergents to the continued fraction Whose quotients are, 1, 4, 9, 2,1, 1, 4. Ans 1 i 4_6_ g 143 240 1103 '1’ 4" 37’ 78’ 115’ 193’ 887' 4. Find a series of' fractions converging to J5; also to ~/4_5. 1 3 7 17 4:1 6 7 20 47 114: (l)AnS.-1—, é“, '5, 'l—é, é—9,&C- (2)AHS. I, 3, 7, "1—7—, &c. s+J§ 5. Express ' 2 in a continued fraction. Ans. The quotients are 2, I, 4, 1, I, l, 4, &c. 6. Find the convergents to 0'2422638 .... .. 1 7 8 49 47 Ans. E, -2—-9, 5%, T61, T91, &c.] 7. From the last example deduce an explanation of' the Julian and Gregorian corrections of' the Calendar, having given the true length of the year to be 3652422638 .... ..days. (Comp. p. 133.) 8. Prove that filig- difi'ers from 314159 by a quantity less than 000001. 9. The lunar month, calculated on an average of 100 years, is 27'821661 days. Find a series of vulgar fractions approximately nearer and nearer to this decimal fraction. 27 82 7 65 3907 . Ans'T’ 3’ 28’ 143’ 10. The sidereal revolution of Mercury is 87'969255 days; and that of Venus 224'700817 days. 'Represent these quantities approximately by less numbers. \ &c. 87 §§ 2815 Ans. For Mercury -1—, I , -3—2—-, &c. ' 2Q4~ 225 6 4 15 3 ForVenusT, T, g , ——72-, c. 462 INDETERMINATE EQUATIONS AND PROBLEMS. 11. Find the least fraction with only two figures in each term, ap~ 1947 A 11 C ns. -_ O 3359 19 proximating to 12. Given 2“=6, required a: in the form of a continued fraction; and find the convergents. 2 3 5 13 31 Ans. x=2+ Convergents are I, -1-, 5, F, ~15, &c. 1+-—— 1+l+ 2 Q.’ 7 ” 3 13. If 5 =1, find w. (Comp. p. 134.) Ans. 0'53. 14. If 3"=15, find :3. Ans. 2465. L z 3 15. If <§>=3, find at. Ans. 0‘737. 16. Approximate by continued fractions to the roots of the equations, (1) 5x’-3=0, x”-— 5x+3=0. (Comp. p. 135.) Ans .3- 5: 2_7 5!. 2L3 g: c ' 5’ 5’ 35’ 40’ 275’ 315’ ' 4 13 43 142 1 1’ 1 A118- '9'“, I6, Tag, &C. and 682 and less than 2889 - and that 305 1 292 ’ 17. Shew that J5 is greater than it difi'ers from the latter fraction by a quantity less than ————————2X 30 51X1292 - (Comp. p. 136.) IN DETERMINATE EQUATIONS AND PROBLEMS. 1. 14x-5y=7; find the least positive integral values of .1: and y. Ans. 32:3, y=7. \ $2._ 27x+16y=1600; . . . . . . . . . . . . . .. Ans. .z:=48, 9:19. 3. 19x-117y=11; . . . . . . . . . . . . . .. Ans. w=56, y=9. 4 3x+5y=26 ; find all the values of a: and y in positive integers. Ans. ar=7, 2. y=1, 4. 5. 11x+13y=190g Ans. x=9, y=7, 6. 13x+16y=97; Ans. w=5,y=2. INDETERMINATE EQUATIONS AND PROBLEMS. 463 7. 11a:+ 7y=108 ; find all the values of a: and y in positive integers. Ans. x=6, y=6. 8. Shew that there is no solution in whole numbers for the equation 49.23— 35y=1 l. 9. Given 00:4, y=9, one solution of 2x+3y=35, find all the solu- tions in positive integers. A {x= 1, 4, 7, 10, 13, 16. y=11, 9, 7, 5, 3, 1. w=1, 2, 3, 10. Given w+2y+3z=20 2:3, 4, 5. 11. Given 6x+7y+4z=122 and 11x+83-62=145}; find .r, y, z, Ans. a:=9, y=8, 2:3. 12. Find all the positive integral‘solutions of 20x—21y=38, and 3y+42=84. Ans. x=4, y=2, 2=7. 13. Find all the positive integral solutions of my+zv’=2x+3y+29. (Camp-p.137) Ans. x=4, 5. y=21, 7. 14. Find all the positive integral solutions of 7.1:y—5x=3_y+39. 31:1, 3, 5’ 21’ Ans' {y=ll, 3, 2, 1. 15. Find the number of' solutions of 11x+15y==1031 in positive inte- gers. Ans. 7. 16. Find the number of solutions of 3x+7y+17z=100 in positive integers. (Comp. p. 137.) Ans_ 12 17. Find the number of solutions of 20x+15y+6z=171 in positive integers. Ans. 6. 18. Find two fractions having 7 and 9 for their denominators, and . r 4 them sum Ans. §, 19. Find three fractions with denominators 3, 4, and 5, of which the sum is 12'? Ans 2 -3- i 60 ' ' 3’ 4’ 5 ' . . . 4 2 3 4 20. Find the three fractlons whose sum 1s 5%; . Ans 5 , , 5 . 21. Find a number which upon being divided by 39 gives a re- mainder l6, and by 56 a remainder 27. Ans. 1147, or 3331, &c. 464 SCALES OF NOTATION. _ 22. Find a number consisting of two digits which shall be equal to four times the sum of' the digits. Ans. 12’ 01. 24., or 36, or 43_ 23. Find the least number which, upon being divided by 11, 19, and 29, gives the remainders 3, 5, and 10, respectively. Ans, 4,123, 24. Find a number less than 400 which is a multiple of 7, and upon being divided by 2, 3, 4, 5, 6, always leaves 1 for a remainder. Ans. 301. 25. Shew that the solution of ax+by=c in positive integers is always possible, if a be prime to b, and c>ab—(a+b). (Comp. p. 138.) 26. In how many different ways is it possible to pay £20 in half- guineas and half-crowns? Ans_ 7, 27. A certain sum consists of £3. 31 shillings, and its half of fly. a: shillings; find the sum. (Comp. p. 139.) Ans. £13_ 68, 28. Find two numbers such that their sum shall be equal to the sum of their squares. An _1_5 3 s' 13’ 13' 29. What value of a: will make ax2+bm+ c’ a complete square? . L \_(Comp. p. 139.) Ans. x=bn2 2012?: . m —an 30. What integral values of' a: will make 2.v’+x+8 a complete square? (Comp. p. 139.) Ans. 8, —4, -1, and 23. 31. What value of b will make b’—4ac a complete square? (Comp. p. 140.) Ans. b=am+7% . 32. Find three square numbers which are in Arithmetic Progression. (Comp. p. 140.) Ans. (m’—n”-2mn)2, (1122+ n2)”, (m2-n2+2mn)”. SCALES OF NOTATION. 1. 17486 is in the denary scale, find the equivalent number in the “senary scale. Ans. 212542. 2. 215855 is in the denary scale, find the same number in the duodenary scale. 'Ans. t4¢ee_ 3. 3t432 is in the duodenary scale, find the same number in the denary scale. - ' ~ Ans. 80198. 4. Transform 1534 from the senary to the denary scale. Ans. 418. 5. Divide 14332216 by 6541 in the septenary scale. Ans. 1456. SCALES 0F NOTATION. 465 6. Divide 95088918 by U4 in the duodenary scale. Ans. t4lee. 7. Multiply 64ft. 6in. by 8ft. Qiin. Ans. 565ft. 8’. 7". 6’”. 8. The difference between any number (in the denary scale) and that formed by reversing the order of the digits is divisible by 9. Prove it. 9. Extract the square root of 25400544 in the senary scale. ' Ans. 4112. 10. Extract the square root of 32375721 in_tl1e duodenary scale; and then verify the result by squaring it. Ans. 6256' 11. The number 124 in the denary is expressed by 147 in another scale, required the radix of the latter. Ans. 9. 12. ‘In what scale of notation will a number that is double of 145 be expressed by the same digits? Comp. p. 141.) Ans. Radix=l 5_ 13. Find the scale to which 24065 belongs, its equivalent in the denary scale being 6221. Ans_ Radix=7_ 14. The area of a rectangle is 29ft. 4in., and its length is 12ft. Sin. ; .find its breadth. Ans_ gft_ 31m 6'_ 15. The area of a rectangle is 971 ft. 120in., and breadth 24ft. 91m; find Its lensth- Ans. 39ft. 3in. 2’. 3". &c. 16. The area of a square is 17ft. 54111., what is the length of the side? Ans. 4ft. 2in. 0'. 2". 10’”. &c. 17. What is the cube of6ft. 6in.? Ans. 274ft. 10801n. 18. Prove that any number of 4 digits in the denary scale is divisible by 7, if the first and last digits be the same, and the digit in the place of hundreds be double that in the place of tens. 19. Any number is divisible by 4, if the last two digits, taken in order to form a number, be divisible by 4. 20. Any number is divisible by 8, if the number, consisting of the last three digits in order, be divisible by 8. 21. There is a certain number consisting of two digits, which is equal to four times the sum of its digits; and if' to the number 18 be added, the digits will be reversed. What is it? Ans. 24. 22. There is a certain number, a multiple of 10, which exceeds the sum of its digits by 99 ; find the number. Ans_ 100, 23. Prove that the sum of all the numbers which are composed of the same digits is divisible by the sum of the digits, when the digits are all different. 30 466 PROPERTIES OF NUMBERS. 24. Find the greatest and the least numbers of 4 digits in the senary scale, as expressed in the denary scale. Ans. 1295, 216. 25. A certain number consists of two digits such that when the digits are reversed the number is divisible by 3, and is to the former number as 23 : 32. Required the number. (Comp. p. 141.) ADS. 96- 26. Any number consisting of an even number of digits, in a system whose radix is r, is divisible by r+1, if the digits equidistant from each end are the same. 27. If N, N’ be any two numbers in the denary scale composed of the same digits differently arranged, prove that N~N' is divisible by 9. (Comp. p. 141.) 28. The square of any number which has less than 10 digits, (in the denary scale) each of which is 1, will, when reckoned from either end, form the same Arithmetic Progression whose common difference is 1, and greatest term the number of digits in the root. PROPERTIES OF NUMBERS. 1.‘ PROVE that 123 divided by 4 cannot leave 2 for a remainder, n being any of the natural numbers. (Comp. p. 142.) 2. No number can be a square which has any one of the numbers 2, 3, 7, 8 for its last digit. 3. Prove the following properties of a square number :— (l) A square number cannot terminate with an odd number of ciphers. (2) If a square number terminate with 5, it must terminate with 25. (3) If a square number terminate with an odd figure, the last figure but one will be even; and if it terminate with any even figure except 4, the last figure but one will be odd. (4) N0 square number can terminate with two figures the same, except they be two ciphers, or two 4’s. 4. Any number divided by 6 leaves the same remainder as its cube divided by 6. 5. IF m be any odd square number greater than 1, prove that (m+3)x (111 +7) is divisible by 32. ' 6. If each of the quantities a, b, n be a whole number, shew that {2a+(n-l)b}% is always a whole number. (Comp. p, 14.2,) 7. Shew that every perfect cube number is of one of' the forms 41:, or 411*1. PROPERTIES OF NUMBERS. ~ 467 8. Shew that w“—- 5w3+4w is divisible by 120, whatever positive whole number a: may be. (Comp. p. 143.) 9. Shew that %1(2x2+w+3) is a whole number, if .11 be odd. (Comp. p. 143.) 10. The difference of the squares of any two odd numbers is divisible by 8; and the difference of the squares of any two prime numbers, of which the less exceeds 5, is divisible by 24. 11. If n be any whole number, one of the three 12’, 1z2+1 , ng—l, is divisible by 5. 12. If n be an odd number not divisible by 7, either 123+], or 123—1 is divisible by 14. 13. Shew that, when m is any even number greater than 2, m’(m”- 4) is divisible by 192; and, when m is any odd number greater than 3, 112(m2—1Xm2-9) is divisible by 1920. 2- 14. If n be a prime number greater than 3, 72241 (Comp. p. 143.) 15, If' a and b be prime numbers, the number of' numbers prime to ab and less than ab is equal to (a—-1)(b—1)-—1. 16. If there be two binomials each of which is the sum of two squares, their product is the sum of two squares. (Comp. p. 144.) is an integer. 17. Neither the sum nor the difference of two irreducible fractions, whose denominators are different, can be an integer. (Comp. p. 144.) 18. If n be any number, and a the difference between 12 and the next greater square number, and b the difference between n and the next less square number, shew that n—a6 is a square. (Comp. p. 144.) 19. Prove that the product of two different primes cannot be a square. (Comp. p. 145.) 20. Decompose 831600 into its prime factors; and find the multiplier which will make it a perfect cube. (1) Ans. 11X7X52X24X33. (2) Ans. 118580. 21. Find the number of divisors of 1000. Ans. 16. 22. Find the number of divisors of 30030. Ans. 64. 23. If N is a number of the form amb", where a and b are prime _1 o 0 numbers, shew that N. a .7 IS the number of integers not greater than a N and prime to it. (Comp. p. 145.) How many numbers are there not greater than 100 and prime to it? , Ans. 40. How many less than 360 and prime to it? Ans. 96. 30—2 468 CONVERGING AND DIVERGING SERIES. 24. If'one number (A) have exactly as many places of figures as another (B), and also have more than the first half of its figures identical with the corresponding figures in B, shew that the difference between 2/2 and ,2/5 will be less than i, if n be any whole number not less than 2. (Comp. p. 146.) a VANISHING FRACTIONS. f J;—~/Zz_+,/x-a ' Ans ___1_ when x=a. 1. FIND the value 0 N/x2—az ' . flat ‘ s__ a 2. Find the value of 92+;- when a=b. Ans. 52- . a —b 2 3. Find the value of $12—27 when a=b. Ans. gam‘". r+l_ r+l 4. Find the value of L—b— when a=b. Ans. TH . a'b’(a—b) .1. 5. Find the value of 3;(“—n_-_) when x=0. Ans. 5227:, . 6. Find the value of MM when'x=a. Ans. 5a. a—Vhm . 3 2 2 8 7. Find the value of w +5“: Z461 iii—2“ when x=a. Ans. 29 . a: -a 2 8. Find the value of Ja2+ax+x —"/€2:_aw+x when x=0. a+x-,,/a—a: Ans. Jd. 9. Find the value of Jxg+a_.Jifji+‘/a fit when x=a. Jag—ma , 2 Ans. 3a . CONVERGING AND DIFERGING SERIES. ‘ a: x“ .22” . . 1. PROVE that 1+I+fi+m+ . . . . .. W111 begin to converge at some point for any value of w, however great. 2. Shew that the series for (1+.r)“, given by the Binomial Theorem, will always be “convergent” when x< 1; and determine after how many terms the convergency will begin. (Comp. p. 14.-7.) Ans' After terms? 7' being the next whole number which > ' ) LOGARITHMS. 469 3. Shew that both the following series are convergent, (Comp. p. 148.) 5 x7 x2 w‘ :66 w” w 1— L'é-"l- E- E-l- - . . - H J7—E-l-E—‘EZ-l— . s . . .- and if x=264, at which term will the former series begin to converge? x966 Ans. . —— . an 4. In the expression ax”+b.z~"“‘+cx""+ .... .. if a be any fixed quan- tity however small, and b, c, d, &c. any fixed quantities howeVer great, shew that a: may be taken so great that ax" shall contain bx’“'+c.z:""+ .... .. as many times as we please. (Comp. p. 148.) 5. How small must a: be taken, so that the third term of the infinite series l+3x+5xz+7azs+ .... .. may contain the sum of all that follows 500 times at least? (Comp. p. 149.) Ans. w<_l_ _ ' L 702 LOGABITHMS. APPLY the Logarithmic Tables to find, .1. 26“. Ans. 18446750000000000000. 239xS27x543 . 2. A118. 83969 32. a. ,6/235'78. Ans. 2485522... 9 21 4. (g) . Ans. 1186322... 5. , / Ans. 1'295695... " 1.a2><(7-356)9 6. __ Ans. 14425972... #385)“ (1-05)’-1 . 7a m - Ans. 5 79... 5 23 8 8. ~/ 417 . Ans. 017577... 9. £15. 7.9. 3%d.x(1'03)5°. Ans. £67. 78. 1d. 10. Given 20”=100, find as. Ans. x=1'537244. -l b 11. Given c’"=a.b"’”‘, find x. Ans. a2:- IOga 0g mlogc—nlogb g 470 INTEREST AND ANNUITIES. p log é-(1+~/5) 12. Given a‘“+ a’”=a6’, find 50- . Ans' I: 2 log a 13. Given a“+-—l;,=b, find .r. Ans. x: . a loga 4ar.log b C(r log b—log a) ' 14. Given (,I/Zéfr-b‘I-m, find :0. Ans. x= 15. Given xy=y"’, and xP=yq, find a: and y; (Comp. p. 150.) A '3 7 = 0 ns .1: q , J q 16. Given 3".5é”'4=7“‘1.11"”, find 11:. (Comp. p. 1.51.) Ans. .r=1'242073... 17. Given (a4-2agbg+b4)‘"l=ga—_Qf find .21. (Comp. p. 151.) (a+b)” ’ _loga-—b Ans. x- - . log a+b O 18. Given 5‘”+1—5”_3+14%-=4789—51“‘+%—5“"”, find it. (Comp. p. 151.) Ans. 4'25. 19. Given 5x3’-3><2~”=30000, and 3X3m+6><29=20000,} find a: and y. (Comp. p. 152.) Ans. 20. GiVen 3x2-4”+5=12OO, find x. (Comp. p. 152.) ,Ans. .r_=4'33, or -—0'33. 21. Given a‘.a3.a5.a7.&c.=p, find the number of factors a1, a3, as, &c. (Comp. p. 152.) Ans. log p . log a INTEREST A ND AN N UITIES'. 1. WHAT principal put out at simple interest for 3 years, at the rate of 5 per cent. per annum, will amount to £828 P Ans. £72 0_ 2. A person borrows £450 at 5 per cent. simple interest, and re- turns for it £517. 10s.; for what time was it lent? Ans. 3 years 1NTEREST AND ANNUITIES. 471 3. A person returns £610 for the loan of £600 for one month, what is the rate of interest allowed? Ans_ 20 per cent_ 4. What will a capital of £12000 amount to in 10 years, at 6 per cent. per annum compound interest, the interest being paid half-yearly? Ans. £21673. 6s. 8d. 5. Prove that the amount of £1 in 12 years at compound interest is given within less than a farthing by the first four terms of the expansion of (1+r)“.£, the rate of interest being not greater than 4 per cent., and u not greater than 10. (Comp. p. 153.) 6. Find in what time a sum of money will double itself, put out at 5 per cent. per annum, reckoning compound interest. Ans_ 14,-2 years, 7. Find in what time at compound interest, reckoning 5 per cent. per annum, £100 will amount to £1000. Ans. 47 '193 years. 8. What is the amount of one farthing, for 500 years, at 3 per cent. per annum, compound interest? Ans_ £273L 28_ 5gd. 9. Shew that the common rule for determining the equated time of payment of several sums due at different times is in favour of the payer. 10. Required the discount on £160 for a quarter of a year, reckoning interest at the rate of 5 per cent. per annum. Ans_ £1; 19,._ 6d_ 11. What will be the amount of £1212 per annum left unpaid for 76 years, reckoning 4 per cent. per annum compound interest? Ans. £566702. 17s. 4d. 12. What will an annuity of £250 amount to in 7 years, paid half- yearly, allowing 6 per cent. per annum simple interest? Ans_ £2091. 5 8_ 13. What annuity improved at the rate of 8 per cent. per annum, compound interest, will at the end of 10 years amount to £3000? Ans. £207. 1s. 9d. 14. What is the present value of an annuity of £20 to continue for ‘40 years, reckoning interest at the rate of 6 per cent. per annum? Ans. £300. 18.9. 60'. 15. An annuity of £20 for 21 years is sold for £220; required the rate of interest allowed to the purchaser. , Ans_ £6_ 168. 5d. per cent 16. If a lease of 55% years be purchased for £100, what rent ought to be received, that the purchaser may make 5% per cent. per annum for his money? Ans. £5. 168. 47 2 INTEREST AND ANNUITIES. 17. An annuity A is to commence at the end of In years, and to con- tinue q years; find the equivalent annuity to commence immediately and to continue (1 years. (Comp. p. 154.) A ADS. mp o 18. The discount on a promissory note of £100 amounted to £7. 108. and the interest made by the banker was £5'405405...per cent.; find the interval at the end of which the note was payable. Ans. 1% years. . 19. A person puts out Pf. at interest, and adds to his capital at the l . end of every year Eth part of the interest for that year ; find the amount at the end of 11 years. (Comp. p. 154.) mr+m+r}" Ans. P.{ m ‘ 20. The lease of an estate is granted for 7 years at a pepper-corn rent, with the condition that the tenant at the expiration of the lease may renew the same on paying a fine of £100. What is the value of the land- lord’s interest in the estate immediately after any such renewal, allowing compound interest at the rate of 5 per cent. per annum? Ans. £245. 12.}. 10d. 21. A person spends in the first year m times the interest of his pro- perty; in the second 2m times that of the remainder; in the third 3m times that of what remained at the end of the second; and so on. At the end of 2p years he has nothing left. Shew that in the p“ year he spends as much as he has left at the end of that year. (Comp. p. 155.) 22. The reversion of an estate in fee simple producing £60 a year is made over for the discharge of a debt of £577. 48. 5d. How soon ought the creditor to take possession, if he be allowed 5 per cent. per annum interest for his debt? Ans. 15 years. 23. A person puts his Whole fortune P£ out at interest, at the rate of anti per 1.6 per annum, and requires for his annual expences p£ more than the whole interest of P£. In how many years, at this rate, will he have spent the whole? and if, when he has spent half his capital, he dimi- nishes his expenditure one half, how much longer on this than on the former supposition will he continue solvent? (Comp. p. 155.) Pr 1 logPr+p—logp _2‘ +17)“ 031’ (1) Ans. (2) Ans. log 10g 1+1“ log1+r 24. Find the present value of an annuity of £20 a year, to commence in 10 years, and then to continue 11 years, reckoning 4 per cent. per annum compound interest. Ans_ £118_ 78_ 3gd_ 25. What sum ought to be paid for the Reversion of an annuity of £50 for 7 years after the next 14, that the purchaser may make 6 per cent. .per annum of his money? Ans, £123, 98_ 1d, CHANGES, AND LIFE ANNUITIES. ’ 473. 26, If a person purchase the Reversion of an estate after 20 years for £500, what rent ought it to produce that he may make 6 per cent. per annum of his money? Ans, £96, 48_ 3d_ 27. A debt of a£ accumulating at compound interest is discharged in a _ _ log (l—mr) .11 years by annual payments of %£, prove that n-—W . 28. If M 8, M, represent the sums to which an annuity would amount in n years at simple and compound interest respectively, prove that M,__2_2I (n—1)r+2 M,“ 2 ' (l+r)"—1 ' 29. If two joint-proprietors have an equal interest in a freehold 'estate worth 17.56 per annum, but one of them purchase the whole to himself by allowing the other an equivalent annuity of q£ for 11 years, shew that -B__ ___l__) . q-2 (1 0+0“ . (Comp. p. 157.) 30. If P represent the population of any place at a certain time, and (every year the number of deaths is 1 th, and the number of births 1th, of the whole population at the beginning of that year; required the amount of the population at the end of 12. years from that time. (Comp. p. 157.) . Ans. P{1+B;(l}n. P9 31. In the last problem, if p==60, and q=45, shew that the popu- lation will be doubled in 125 years, nearly. (Comp. p. 157 32. What must be the annual increase in the population of any country, that it may double itself in a century? 1 l A . B t -— d -_ ns e ween 14‘3r and 144th. CHANGES, AND LIFE ANN UITIES. 1. What is the chance of drawing the four aces from a pack of cards in four successive trials? (Comp. p. 158.) 1 ” ns' 270725' 2. There are 4 white balls and 3 black placed at random in a line, find the chance of the extreme balls being both black. (Comp. p. 158.) 1 Ans. 5; o 3. Of two bags one contains 9 balls, and the other 6, and in each bag the balls are marked a, b, c, d, &c. If one be drawn from each bag, what is the chance that the two will have the same letter-mark? (Comp. p. 158.) Ans. l . 9 474 CHANGES, AND LIFE ANNUITIES. 4. A plays one game with B, and another with C; the odds that he does not win both games are 4 to 1; the odds that he beats B are 3 to 2. What are the odds in his game with C? (Comp. p. 159.) Ans. Odds against him 2 to 1. 5. From a common pack of cards 12 are dealt to as many persons, one to each, the cards collected, shuflled, and the same repeated. What is the chance that a given person will on both occasions have the same card dealt to him ? What is the chance that the two cards dealt to him will have the sum of their numbers equal to 3? (Comp. p. 159.) 1 2 (1) Ans. 55 . Ans. 6. Two dice are placed together at random so as to form a parallelo- piped: determine the chance that two or more adjacent faces will have the same marks. (Comp. p. 160.) 7 A118. 5; . 7. From a bag containing 2 guineas, 3 sovereigns, and 5 shillings, a person is allpwed to draw 3 of them indiscriminately : what is the value of his expectation? (Comp. p. 160.) Ans_ 39.13,, shillings, 8. A shilling is thrown upon a chess-board, a square of which will just include 4 shillings, find the chance of its falling clear of a division. “Jump. p. 161.) Ans 1 a Z I 9. Three men, A, B, C, in succession throw a die, on condition that he who first throws an ace shall receive £1; what are the values of their several expectations ? (Comp. p. 161.) Ans. A’s=7s. 103—IQd. B’s-=68. 7%d. U’s=58. 5311. 10. There are 5 persons, out of which 4 are going to play at whist. They all cut, and the lowest sits out. What is the chance that two speci- fied individuals will be partners? (Comp. p. 161.) A 1 HS. '5 . 11. There is a lottery containing black and white balls, from each drawing of which it is as likely a black ball shall arise as a white one, what is the chance of drawing 11 balls all white? (Comp. p. 162.) Ans. ——1 . 2048 12. There is a lottery of 10 green, 12 white, and 14 red balls. Let 2 have been drawn, what is the probability that they will be green and Whlte? (OOmP' P' 162) Ans. 17 to 4 against it. 13. A die is thrown time after time: in how many times have we an even chance of throwing an ace? (Comp. p. 162.) Ans. Not quite an even chance in 3, but more than even in 4, times. CHANGES, AND LIFE ANNUITIES. 475 14. Two witnesses, on each of 'whom it is 3 to 1 that he speaks truth, agree in affirming that a certain event did happen, which of itself is equally likely to have happened or not. What is the chance that the event did happen? (Comp. p. 163-) Ans. 9 to 1. 15. A speaks truth 3 times out of 4, B 4 times out of 5, C 6 times out of 7; what is the probability of the truth of what A and B agree in asserting, but which C denies? (Comp. p. 163.) Ans_ 2 to 1_ 16. Thirteen persons are required to take their places at a round ,table by lot; shew that it is 5 to 1 that two particular persons do not occupy contiguous seats. (Comp. p. 163.) 17. P bets Q £10 to £590 that three races will be won by the three horses A, B, C, against which the betting is 4 to 1, 3 to 1, and 2 to 1, respectively. The first race having been won by A, and it being known that the second race was won either by B, or by a horse F against which the betting was 6 to 1, find the value of P’s expectation. (Comp. p. 164.) Ans. 56‘- 51-5jd’. 18.. Supposing the House of Commons composed of m Tories and n Whigs, find the probability that a Committee of p+q selected by ballot will consist of p Tories and q Whigs. (Comp. p. 164.) mcpxnaqlfl m-HLCp-l-Q. 19. There are 3 balls in a bag, of which one is white and one black, and the third white or black ; determine the chance of drawing 2 black ones, if 2 be taken. (Comp. p. 165.) Ans 1 O '6' I 20. In a lottery all the tickets are blanks but one“: each person draws a ticket, and retains it. Shew that each has an equal chance of drawing the prize. (Comp. p. 165.) - Ans 21. A collection is made of ten letters taken at random from an alpha- bet consisting of 20 consonants and 5 vowels; what is the probability that it will contain 3 vowels and no more? (Comp. p. 165.) Ans 60 ~ ' 25a' 22. At the'game of Whist, what is the chance of dealing one ace and no more to a specified person? And what is the chance of dealing one ace to each person? (Comp. p. 166.) Ans. -—9-1-§-9—, or g nearly. (2) Ans. 2197 or 1 n I 20825 73 ear 3" 20825’ 9 23. There are two bags each containing 4 White and 4 black balls. Four are taken at a venture from one of them and transferred to the other. Then 8 being drawn from the latter, 6 of them prove white and two black; what is the chance that, if another be drawn, it will be white? (Comp. p. 167.) _9i Ans. 457 . "‘ .C. signifies the Number of Combinations of n things taken 1' together. 476 CHANGES, AND LIFE 'ANNUITIES. 24. At the game of whist what is the chance of the dealer and his partner holding the four honours? (Comp. p. 167.) Ans. 115 or 2 nearly 76—68" is - 25. In dealing a pack of cards, what is the chance that all the hearts will be found in the first 20 cards dealt, first without regard to order, and secondly in the order of their value? (Comp. p. 168.) salsa 1294912 (I) Ans. W5 (2) Ans. 26. A person puts his hand into a bag containing 12 balls, draws out a certain number at random, and transfers them without examination to a second empty bag. He then puts his hand into this second bag, and draws out a certain number in the same way as before. Shew that the odds in favour of his drawing out an odd number from the second bag are nearly 341 to 340. (Comp. p. 169.) 27. Supposing it an even chance that, on A aged 46 marrying B aged 36, they will both be alive at the end of a: years; find .r, when, according to De Moivre's hypothesis, of 86 persons born together one dies annually until all are extinct. (Comp. p. 169.) Ans. .r=13, nearly, 283* A person 35 years of age wishes to buy an annuity for what may happen to remain of his life after 50 years of age. What is the Present Value of the annuity reckoning interest at 4 per cent. and using Dr. Halley’s Table'l‘? (Comp. p. 170.) Ans. 4% years’ Purchase nearly. 29. An annuity of £10, for the life of a person now 30 years old, - is to commence at the end of 11 years, if another person now 40 should be then dead. Required the Present Value of the annuity, reckoning interest at 4 per cent., and using Dr. Halley’s Table. (Comp. p. 170.) Ans. £17. 16.9. 30. An estate or annuity of £10 for ever will be lost to the heirs 'of a person now 34, if his life should fail in 11 years. What ought he to give for the Assurance of it for this term according to Dr. Halley’s Table; reckoning interest at 4 per cent.? (Comp. p. 171.) Ans, 6643-8, 31. A person now 40 is willing to pay £200 down, besides an annual payment for 10 years, to entitle him to a life-annuity of £44 after he attains the age of 50. What ought the annual payment to be, reckoning ‘interest at 4 per cent. and using Dr. Halley’s Table? (Comp. p. 171.) Ans. £8'55. * This and the three following questions are taken from Price’s Annuities. ‘ -1- See Page 299. ' MISCELLANEOUS EXAMPLES. 477 MISCELLANEOUS EXAMPLES. 1st SERIES. 1. FIND the G.C.M. of (ax + by)”— (a— b)(x + z)(a.r + by) + (a_-b)2.rz, and (ax—byf—(a+b)(.r+z)(ax—by)+(a+b)’.:rz. (Comp. p. 172.) Ans. b(.r+_y). 2. There are four numbers in Arith". Prog“. The sum of the two extremes is 8, and the product of the means is 15. What are the numbers? (APP- P- 353-) Ans. 1, 3, 5, 7. 3. There are three numbers in Geom°. Prog“., whose product is 64, and sum 14. What are/the numbers? (App. p. 354.) Ans. 2, 4, 8, or 8, 4, 2. 4. In what proportion must substances of “specific gravity” a and b be mixed, so that the “specific gravity” of the mixture may be 0? (App. p. 348.) Ans. 2%; cubic feet of the latter to 1 of the former. 5. From a vessel of wine containing a gallons 6 gallons are drawn off, and the vessel is filled up with water. Find the quantity of wine remain- ing in the vessel, when this has been repeated 12 times. (App. p. 349.) Ans (a—b)" ail—1 6. The advance of the hour-hand of a watch before the minute-hand is measured by 15% of the minute divisions; and it is between 9 and 10 o’clock. Find the exact time indicated by the watch. (App. p. 349.) Ans. 28 min. before 10 o’clock. 7. In comparing the rates of a watch and a clock, it was observed on one morning, when it was 12“. by the clock, that the watch was at 11“. 59‘“. 498.; and two mornings after, when it was 9“. by the clock, the watch. was at 8“. 59m. 58". The clock is known to gain 0'18. in 24 hours, find the gaining rate of the watch. (A pp. p. 350.) Ans. 4'9000055‘.... in 24 hours. 8. The product of two numbers is p, and the difference of their cubes is equal to 912 times the cube of their difference. What are the numbers? (APP- P- 352-) Am 1 _~/ (4m—1_)g_+~/§ll, and AW, 2 Jm—l 2 "1‘1 I 9. Shew that .r"—na""w+(n—1)a" is divisible by (:r-a)’, if n be a whole number. (Comp. p. 172.) 10. Shew that .rL-y" is divisible by w+yg, when p is an even num-, ber. (Comp. p. 173.) 478 ~ MISCELLANEOUS EXAMPLES; 11. If N and n be nearly equal to each other, shew that N N 1 N+n N _7-z_=m+-4:.-_, very nearly. . N 1 N+n Also If 17;”, and z. have their first p decimal places the same, this approximation for Mg may be relied on up to 2]) decimal places at least. (Comp. p. 173.) 12. Find the value of a: which, when 2: is indefinitely increased, makes (4m+1)(22+1)2=5(3x+1)(2'+3)2; also find the values of .r and y 222+ (02- (1)2 + 25(w— 20) 322-14 y — b)z + 3a(y-- 3c) (1) Ans. x=1. Ans. x=a+2c, y=b+3c.. which make independent of 2. (Comp. p. 174.) 13. Shew that .x‘+ x3+ x2+rx+s can be resolved into two rational P 7 (Comp. quadratic factors, if —s be a perfect square, and equal to p. 1 7 5.) Hence solve the equation .r‘—6x3+ 5x2+ Sir—4:0. (Comp. p. 175.) Ans. x=2, or --1, or -1§(5i,/.17). 2— 1 2 2 . . 14. Prove that the square root of ( -E>+(y—;-) IS a ratzonal _ . 1 a I) ax-b P function ofa and b, If x=z6. 29. Find the series in 11.12. of which the sum of the first 11 terms is equal to 11*, whatever be the value of n. Ans. 1, 37 5., 7, &c_ SECOND SERIES. 483 30. From the last Ex. deduce the integral solutions of the equation .r’=y."+ a". Ans. .r=5, x=13, x=25, 5:4, y=12, y=24, &c. z=3. z: 5. 2= 7. 31. Shew that “Fifi/Ebert lies between the greatest and least of the quantities Iii/a, I/b, Z/c, (M. 32. The sum of two numbers is 45, and their L.C.M. is 168 ; what are the numbers? Ans, 21, and g4“ 33. The aura. of two numbers is 16, and their L.C.M. is 192 ; what are the numbers? Ans. 64, and 48_ 34. Find a: when 18x—3x’>24. Ans. x=3. 35. Find the values of a: which satisfy the inequality x2<10.r-16. Ans. .r=3, 4, 5, 6, 7._ w2+14x+9 O A O 4. 002+ 2x+3 us 36. Find the greatest value of .r’+ 34x—71 -,_——— can have no value between .22 +2.1: -.7 37. If .r be real, prove that 5 and 9. Prove that 1.2.3. . . 39. Two smiths begin to strike their anvils together. The one (A) gives 12 strokes in 7 minutes, the other (B) 17 strokes in 9 minutes. What strokes of each most nearly coincide in the first half hour? Ans. The 11th of A, and the 10th of B. 40. There are n points in a plane, no three of which are in the same straight line, with the exception of p which are all in the same straight line: find the number of triangles which result from joining them. Ans- than—1)(n—2)—P(z>—1)(P~2)t lfili'lza 1331' Us lp,+1:,+n,+.. .+n, .... .. L731 42. Prove, without the aid of the Binomial Theorem, that, if ,0, re- present the number of combinations of 12 things 7' together, ,0,+,,O,+,,0,+ . . .+,,0,,=2"—1. 41. Prove that is an integer; and deduce that is an integer. 31-2 484 MISCELLANEOUS EXAMPLES. 43. Along a straight line are placed it points. The distance between the first two points is one inch; and in general the distance between the + . . 1 . 1"]? and (r-i-l)th pomts exceeds one inch by 7n-th of the distance between the rm and (r—l)th points. Find the distance between the last two points, and between the first and last points. _ Ex. If m=n=10, shew that the distance between the extreme points is 9'87654321 inches. _ -("-1) (1) Ans. (2) Ans. l—m-(M) {n—1—7n“‘.———————} —-m l 1--m‘1 1— m‘1 44. Two straight rods, each 0 inches long, and divided into m and 12 equal parts respectively, where m and n are prime to one another, are placed in longitudinal contact with their ends coincident. Prove O - Q Q C I that no two d1ViSions are at a less distance than E; inches ; and that two pairs of divisions are at this distance. Ex. If m=250, and n=243, find those divisions which are at the least distance. 45. Two bells toll together for an hour: one tolls 244 times, and the other 251 times, the first and last tolls of each taking place at the be- ginning and end of the hour respectively. Of the strokes (excluding the first and last) find those which are most nearly simultaneous; and determine a person’s station in the straight line joining the bells that those which are most nearly simultaneous of all may appear to him ab- solutely coincident; given that sound travels at the rate of 1080 feet per second, and the bells being a miles asunder. The 105th of I, {also the 140th of I, and 108th of II; and 144th of II. (2) Ans. See Companion. (1) Ans. 46. Three bells commenced tolling simultaneously, and tolled at intervals of 25, 29, 33 seconds respectively. In less than half an hour the first ceased, and the second and third tolled 18 seconds, and 21 seconds, respectively after this, and then ceased. How many times did each bell toll? Ans. 49, 43, and 38. 47, Two particles, (1, 6, start simultaneously from the same point, and move in the same direction, along the same straight line; a moves with the uniform speed of 2 feet per second, whilst the speed of bis such that it moves over 1 foot the first second, and the number of feet described by it in any time varies as the square of the time. Prove that, during the motion, the particles will be three times at any given distance, less than a foot, from each other: and interpret the fourth result obtained in the algebraical solution of the problem. - SECOND SERIES. 485 48. Supposing the receipts on a railway to vary as the increase of ' speed above 20 miles an hour, whilst the cost of working the trains varies as the square of that increase, and that at 40 miles an hour the expenses are just paid; find the speed at which the profits will be the greatefl- Ans. 30 miles. . . . . a2 a8 49._ Determine whether the infinite series + + + . . . _ . - m+ p m+2p m+3p ls convergent 01' dlvergent- Ans. Convergent or divergent as a< or >1. 50. Prove that every quadratic surd, supposing the unit to be a line, may have an exact geometrical representation. 51. A person continually walks at an uniform speed, and always in the same direction, round the boundary of a square field: and another continually walks, at the same uniform speed, from one end to the other of a diagonal of the field. Prove that according to their relative initial posi- tions, they will either (1) never meet at all, or (2) meet once; but they can never meet more than once however long they continue to walk. 52. A sum of money in £. 8. d. is multiplied by a certain number; the pence are now half what they were before, and the shillings and pounds each what the shillings were at first. What is the sum, and the multi- plier? (1) Ans. £9. 19s. 8d. (2) ADS. 2. 53. If E, l—V? are any two consecutive convergents to 9, shew that I)1 D, . b . . . Nl _ a 1 1 the eiror in taking E fOI 5Dm. 54. If p=, /a"+[32 be defined to be the ‘modulus’ of a binomial of the for-m a+16~/-1, where a, )8 are rational; and p“ p,, p3,...pn be the moduli of n such binomials; prove that the modulus of the symbolical product of these n binomials is p,p,p,...p,,. 55. Explain the notation of ‘funo/ions’; and shew that, if F(.r)=a‘, F(x)>)\/1__i1‘_ a 4 .y 2 ' 5. A railway train travels from A to C passing through B where it stops 7 minutes; two minutes after leaving B it meets an express train which started from C when the former was 28 miles on the other side of B: the express travels at double the rate of the other, and performs the journey from C to B in 191- hours; and if on reaching A it returned at once to C it would arrive 3 minutes after the first train. Find the dis- tances between A, B and C, and the speed of each train. Ans. AB=31-21- miles, BC=63 miles: speed of the ordinary train 21 miles, . . . . . . . . . . .. express......42 6. To meet a deficiency of 112 millions in the revenue of a country, an additional tax of a per cent. was laid upon articles exported, and the tax upon imports was diminished 0 per cent.: in consequence of these alter- 488 COLLEGE EXAMINATION PAPERS. ations the value of the imports was increased so as to be n times as great as the exports, and the deficiency was made up. It was afterwards found that if the additional tax upon the exports had been a’ per cent., and the tax upon imports diminished 0' per cent., the values of the articles being altered as before, the deficiency would not have been made up by 912' mil- lions. Find the values of the exports and imports after the alteratiOn .of the tax. 112' (a—a’) +n(c'--c)x 10058; Ans. Val. of exports= m’n Val. of imports= (a _a,)+n(0,_0)x100£. 7. Fifty thousand voters, who have to return a member to an assem- bly, are divided into sections of equal size, and each section chooses an elector, the member being returned by the majority of such electors. There are two candidates, A and B. In those sections which return elec- tors favourable to A, the majority is double the minority, while in those favourable to B, the minority forms only a tenth of the whole. After the primary elections a third candidate C comes forward, and is joined by so many electors of each party, that he is returned by a majority of 3 over A, and 14 over B. If 0 had not come forward, A would have been returned by a majority 19 less than the whole number of votes actually polled by C, and if the elections had been by the 50,000 voters directly between A and B, B would have had a majority of 6000. Find the number of sections. Ans. 100. II. ST JOHN’S COLLEGE. June, 1849. 1. ab{,/.E_+_a-b}+ec{,/Eiz+c}=a; find A 2 / Ans. x=(b=!=c)2+i.(z:_c)§ . 2. Jm+m=J4x—3+J5x—4; find as. Ans, w=], 3. 3x514. 31=13wi+45x2i ; find .22. Ans. x=3‘, or (g): or (Pg). y—w+J2xy—3w2_3 (2.l/—3w)%+w% I g_2x (2.21—3wli" “7% , find a: and y. 1 —- 3 4 — — ‘13—, + 8—1 =2 ,' find :0. m2+ a 1 :— ' 1 1. (m+a) (1+ W)+J 2a.r(1- Ans. m=1*~/2a—a2. 1 I .= 8.112— 6.2: +1 ; find x. + _, 6x’—7x+2 12w“-—l7.r+b 1 '/1*~/§.§ , ’=-- :i: . Ans a 2{1 8 } 3. (x’+g/2+ c’)§+ (ar—y-l- c)§= “Margy, 1_1 1’ findxandy. y; a" w=%(1i~/3).c, Ans. { 1 y=§('1*fi)-c- 4~. 2(x’+.r_y+y’-a’)+J§(x’-y’)=0, 2(m’-—a:z+z*-b”)+J§(x”_g’)=0, find :0 and y. y”— c”+3(yz’- 3)=0; ya 22-(m+.3/803—m8), z=-;-(m--:/803—m3), Ans. { 1 1 x=N—/g_;—i{a+b—-é(~/§~l):/Sea—ma}. where 1)z=(J§+1)(a-b). 5. A lends one half of his money to B at 5 per cent. per annum simple interest, and the remaining half he invests in the three per cents. at 90. B pays the interest regularly during the first five years, but afterwards neglects to do so till other five years’ interest is due, when A calls in all his money, and B becomes a bankrupt paying 108. in the pound. A sells out when the funds are at 81, and then he finds that the whole sum he has received as principal and interest in the ten years ex- ceeds the sum that he originally possessed by £34. 138. 4d. How much did he‘lend B? Ans_ £320. 6. A, B and C are three villages. The road from A to B is level, and C is on a hill above A and B. The distances AB, BC, and 0A are respectively 24, 14"}: and 2&8 miles. P walks up hill i-th slower and down hill érd faster than when the road is level. Q walks up hill #11 COLLEGE EXAWHNAJHON PAPERS. 491 slower and down hill 21;th faster than when the road is level. P travels round in the direction ACB in 30 minutes less than Q requires to go round in the opposite direction. Q travels round in the direction A CB in 1 hour 48 minutes more than P takes to make the circuit in the opposite direc- tion; find the rates of each on a level road. Also supposing them both to start from A in opposite directions, find their points of meeting. (1) Ans. P’s rate 12 miles, Q’s 10 miles, per hour. (2) Ans. At C, or between B and C at a dist. miles from C, according as P or Q takes the direction ABC. 7. Suppose that in- the course of any one year the number of births in Ireland is on an average 32 for 1000 living in the island at the mom mencement of that year, the number of deaths and emigrations to the colonies, 21 in 1000, and of migrations to England 1 in 100. In England suppose the number of' births in the course of any one year to be 3 for every 100 inhabitants living at the beginning of the year, the number of deaths and emigrations to the colonies 289 in 10,000, and of migrations to Ireland 1 in 10,000. If the number of inhabitants in England was twice the number in Ireland at the beginning of 1850, in what year will the population of' the former be three times that of the latter, according to the law above stated? Ans. A.D. 1957. IV. ST JOHN’S COLLEGE. Jane, 1851. .1"+2x+2 + x2+8x+20 __ x2+4w+6 +.r’+6x+12 _ x+1 w+4 '— w+2 x+3 ’ find it ' Ans. .r=0, or ~22}? 2. (5x’+.z: +10)”+ (.z"+ 7x+1)’= (3x2- .1: + 5)"’+ (“2+ 5x+ 8)”,- find an Ans. x=2=l=./§, or 3ifi. 3. (x’+4~.r—2)2+3=4.r(3.r2+4); find .r. Ans. w=%{2+~/52_*~/16—7~/é}. 14—92 2 2_zg=w____ x‘l'y 2 ’ find .2, y, and z. 1 4 = __ 492 COLLEGE EXAMINATION PAPERS. 5. A derives his income from a fixed rental, B from his profession, .C from both. In the first year A pays as much income-tax as B and 0 together, but in the second year B’s and C’s professional incomes being doubled, B pays as much as C, which is gths of what A pays; also the total amount of their incomes in the two years is £5500. Assuming that the income-tax is higher for a fixed rental than for professional in- come in the ratio 3 : 2, find the incomes of A, B, C in the first year. Ans. A’s £1000, B’s £600, C’s £700. 6. A and B start at the same time in a boat-race ; A has 100 yards start and has to row to a post D, B to a post C. At first A’s rate : B’s :: 40 : 39, but when the distance between A and B is %th of the remaining distance A has to row, A’s speed is diminished in the ratio 79: 80, so that two minutes afterwards the distance between them is three yards more than half the remaining distance B has to row. At this point of B’s course, his rate which has hitherto been uniform is increased eight yards a minute, while A’s is still further diminished six yards a minute; and in one .‘minute more B arrives at the post C, A being then three yards from the post D. Find the distance between C and D. Ans. 116 yards. 7. Three men, A, B, C walk in the same direction in the circum- ferences of three concentric circles, starting simultaneously from points where they are at their least distances from each other. A walks his circuit in an even number of hours, (greater than four), B and C their circuits in one hour and two hours less respectively. Whenever A and B are at their greatest distance from each other, they alter their rates in such a manner, that the times they would take to walk their circuits at the rates they are then going are interchanged; and whenever A and C are again at their least distance their times are interchanged in a similar manner. When A and B are at their greatest distance the first time, A has walked a distance equal to twenty-two times C’s circuit; and when they are at their greatest distance the third time, B has walked a distance equal to forty-two times A’s circuit, and C has then walked ten miles less than forty times B’s circuit, and is at his least distance from B. Required the rates of A, B, C at fil‘st- Ans. 8, 4, and 5 miles per hour respectively. V. ST JOHN’S COLLEGE. Jame, 1852. 1. (a+b)’.r+(aafb>s.(4—g)=2ab; find x. a 30b —l.lb Ans. $=W, or W . ‘2. (.r+2~/.;:).i- (.r—2,,/a_3)‘-l=2(.r2—4ix)%; find at. Ans. .r=-0, or 2(l*,,/§). OOLLEGE EXAMINATION PAPERS. 1493 3. (12x-1)(6.r -1)(4x—1)(3x—1)=§1-6; find an Ans. .r=—212(5 =l=1\/ 5 i2,,/5). A w=6, or 2, find a: and y. “5' {9:2, or 6_ 5. A and B set out at the same time from the same place, and walk in the same direction. Whenever the distance between them is an even number of miles, A increases his speed i- mile per hour, and when it is an odd number of miles, B increases his speed @- mile per hour. YVhen A is 4 miles in- advance, B has walked 30% miles, and A has walked 1% miles more than he would have done in the same time, had he walked uniformly at his first rate. Required the rates of A and B at starting. Ans. A 4 miles, B 3 miles, per hour. 6. Three vessels, A, B, C contain liquids in the proportion 3 : 2 : 1, and their prices per gallon are as 1 : 2 : 3. A certain quantity is poured from A into B, and the same quantity from the mixture in B into C; the same quantity is now poured from the mixture in C into B, and from the new mixture in B into A. The value of the liquid in A is thereby in- creased in the ratio of 35 : 27; but had the quantity poured out each time been one gallon more than it was, its Value would have been increased in the ratio of 3 : 2. Find the quantity of liquid in each vessel. Ans. Quantity in A, B, C, 3, 2, 1, gallons respectively. 7. Reckoning meat by the stone, wheat by the quarter, and hay by the load, the price of hay at first was equal to that of wheat together with 4 times that of meat. The rise or fall of meat is %th of the rise or fall of wheat, together with 1—16th that of hay, except during the first month, when from scarcity of fodder, hay rose 8.9., and the efi'ect on meat from this cause alone was a fall of 6d. The variation in hay was 8.9. each month, and wheat rose 2.9. a month from the first, until, after a certain number of months, there was the same relation between the price as at first; wheat being now 50.9. and remaining stationary. After as many months more, hay was 12 times the price of meat. Required the prices of wheat and hay at first, that of meat being 68. Ans, Wheat 4.48, Hay 636-, 494 COLLEGE EXAMINATION PAPERS. VI. ST JOHN’S COLLEGE. May, 1853. 1(x+1)(.r-3) 1(.r+3)(.r—-5) 2 (x+5)(a:-7) _22_ Ans. .r=1='=,‘/l_§. x+6 .r-4 2 .r— x+9)2__ a'2+36 2’ (.r-O'Xm _2'.r’—36' Ans. w=0, 0r g(l*,,/2—6). a. (a: -1)’+ (a-1)2=2(ax+1) + ,Jscv +a)2+ 4am. Ans. .r= (a+2)*~/8a+3. 4. mag-I- y’+b"=J§. {x (a +y)-—b(a— y)}, w’—a’-y’+ b’=~/§ -~{~'v-(a w) + b (a +y)}- Ans. {swag/2+6. y=bfl+a. 5. A rectangular field is divided by two lines parallel to two of its adjacent sides into four parts, of which the least has its longer side in the shorter side of the field. The ratio of the perimeters of the greatest and least parts is 4 : 1, and that of the other two is 3 : 2. _Had however the sides of the least part been double their present lengths, the ratio of the areas of those parts which would then have been greatest and least would have been 4 : 1, and of the other two, one would have contained an acre more than the other. Find the number of acres in the field. Ans. 55?— 0'. 6. Four points A, B, C, D move uniformly with velocities in Geo- metrical Progression in four equidistant and equal parallel lines, whose extremities are situated in two parallel lines, from one of which they start together, and when they reach the 'Other extremities, they return, and so on continually. After a certain interval B, C, D are in a straight line for the first time; after twice that interval A, B, C are in a straight line for the first time; 36 seconds after this, A, C, D are in a straight line for the second time, and the space that has been passed over by B is 14 inches more than that passed over by A when A, B, C, D are again in a straight line. Required the velocities of A, B, C, D. Ans. 1, 2, 4, 8 inches per. second respectively, .OOLLEGE EXAMINATION PAPERS. '495 7. A town is supplied with gas at a stated price per thousand cubic feet whenever the annual consumption and price of coals are cer- tain fixed quantities; but it is agreed, that in any year when they differ from these values, a variation in the former at the rate of (112) per cent. shall cause an opposite variation in the stated price of gas at the rate of (n) per cent. (12 < m) and a variation in the latter at the rate of (p) per cent. shall cause a like variation in the stated price of gas at the rate of (g) per cent. After (1‘) years the average annual consumption and average price -of coals have accorded with their assigned values, and yet the town has paid' £P more than it would have done, had there been no variation from the assigned values of the annual consumption and price . of coals. But if the above-mentioned relative variations in the consump tion and price of gas had been (p) and (q) per cent., and those of the prices of coals and gas (122) and (n) per cent. the town would have paid £Q less. Supposing that in any year (the a“) there is a variation in the consumption and price of coals from their fixed values of (.12) per cent. ,- find the greatest sum the town could pay for the gas it consumed in any given year. Ans. The greatest cost of gas in the tth year 1*. 2 -221" w(1+l—) {1_(1'._2)J_} S'_q_’_£2' 100' mp100’ p8 7212 ‘ where S=12+22+32+ . . . . . .+r’. I VII. ST JOHN’S COLLEGE. June, 1854. x+1 .r—2 .r—3 n+4 , ~ .__AA. .AAA, .r-1+x+2+w+3+.r—4 4' Ans x 2 1 5) 2. (a +x)8+ (a'+ x2)‘=c‘.r‘. —-r Ans. w=g{m-l=l=~/(m+l)(me-3)}, where m=<\/8+éc—a,-3)i. 3. (a:+1)(.r’+])(.r°-1)=p.v‘(.r-1). Ans. x=l, or .§(m*J1n5:4), where m=%(-1=1=~/5=*=4-'JFTI)' 496 .OOLLEGE‘ EXAMINATION PAPERS. a3 63 C3 _ 4. 3,- + 3yz=g + 32.22:;- +3.r_y=.r_y +92 +2.20. .2: y __ z __ 1 m+aa _ m+b3 m+ 03 33,722+ Qmprppz ’ Ans. where m = ¢ 2—3 — W P2 , Pl =’aa+ba+ca, pa=bacs+caas+asbs, p3=a3b803. I 5. A person starts to walk, at an uniform speed, without stopping, from Cambridge to Madingley and back, at the same time that another starts to walk, at an uniform speed, without stopping, from Madingley to Cambridge and back. They meet a mile and a half from Madingley ,- and again, an hour after, a mile from Cambridge. Find their rates of walking and the distance between Cambridge and Madingley. Ans. Their rates are 4 and 3 miles an hour: the distance is 3% miles. 6. Sovereigns in number equal to four hundred times the number representing the ratio of the weight of pure gold to the weight of dross in a Sovereign contain as much pure gold as 133 Napoleons together with Napoleons in number equal to six hundred times the number repre- senting the ratio of the weight of pure gold to the weight of dross in a Napoleon: also there is as much pure gold in 400 Sovereigns as in 503 Napoleons: and 11000 Sovereigns weigh as much as 13581 Napoleons. How much of(1) a Sovereign (2\ a Napoleon is pure gold? Ans. (1) gas. (2) -l%ths. 7. Suppose that each of the University Presses at Cambridge and Oxford has a fixed demand, every new year’s day, for 5000 copies of an edition of the Bible containing 30 sheets, the price of the paper for which is 12s. 6d. per ream of 500 sheets, the expense of setting up the type £140, and the cost of press work 5.9. for every thousand sheets printed. The custom is, at Oxford to keep the type continually stand- ing, at Cambridge to take it down as soon as the printing of what is thought proper at any particular time is completed: in consequence it is .necessary to purchase (suppose) twenty times as much type at Oxford ,as at Cambridge. Assuming that at Oxford the 5000 copies are always printed just before they are wanted, and that at Cambridge a supply is printed at one time for the most advantageous possible number of years (simple interest being reckoned at the rate of 5 per cent. per annum on all money laid out in the material and workmanship of any stock kept on hand one or more years) the capital required for the above purpose at Oxford is £211. 13s. 4d. less than the average capital required for the same purpose at Cambridge. What is the cost of the type for this edition of the Bible at Oxford and at Cambridge? . COLLEGE EXAMINATION PAPERS. 497 VIII. ST JOHN’S COLLEGE. June, 1855. \/.r”—-2zv+3 + \/.r2+2w+4_2 a2+ 2.1: + 4 avg—2.x + 3 _ col»- 4 Ans. .v=2, or 5, or &c. _1 i 2 4 ba—cs>i 2. wa+3axy+ys=ba} A Jim? C 6_3 a—c ’ ns. .x-l-y =0 1 2 4 [73—08 f, Fe 0* '57:; l 3 15+»- xi aa—3a / 4 - 1 , = :l: ‘ —.___---———— h ' =—-———— o Ans a! 2 {I 1 (a3_3a)2}, w em a f/Q-l 4. xiii/2+ 22: 3 8 a: = 5, 2x+3y+52=29 Ans. y=3, or &c. 15x9+10f+6z2§12xz+12yz+297 2:2. 5. A person who copied a manuscript in a regular manner found that the number of lines copied in the first half hour was less by ten than the square root of the whole number of lines in the manuscript: and that the square of the number of lines copied in the first 49 minutes was equal to twice the number of lines then remaining to be copied. How many lines were there in the manuscript P Ans, 4900. 6. A, B, C walk uniformly in three roads, starting at the same moment from their point of intersection. The roads in which A and B walk are at right angles, and the road in which C walks lies some- where between them. A line joining the positions of A and C at any time is parallel to the road in which B walks; and a line joining the positions of B and C at any time is parallel to the road in which A walks. When 0 has walked 8 miles more than A he begins to return, and when he has walked 5% miles more than B, all three are in the same straight line. Find the relative speeds of A, B, and C. Ans. 7. A farmer has between 270 and 280 quarters of wheat, which he agrees to sell to a miller on the following terms. Not less than 20 and not more than 30 quarters (the exact number being at the option 3:4:5. ./ 32 498 ' OOLLEGE EXAMINATION PAPERS. of the farmer) are to be delivered at the beginning of each month of the year, for which the farmer is to be paid according to the market price of wheat at the time of each delivery. The price of wheat at the beginning of June is the same as the price at the beginning of July; and the price increases uniformly a certain number of shillings per quarter each month during the first six months of the year, and decreases uniformly the same number of shillings per quarter each month during the last six months of the year. The farmer, by making the most of his wheat under this con- tract, receives £7. Ss. more for it than he would have done had he deli- vered the same average quantity each month. If, however, the number of quarters to be delivered each month had been not less than 20, and not greater than 25 (the exact number being at the option of the miller), the miller, by arranging the delivery in the way most advantageous to himself, would have paid £11. 12s. less than he actually paid: and the amount paid in April and September together in the latter case would have been £33. 129. less than the amount paid in May and August toge- ther in the former case. How much wheat had the farmer, and what did he receive for it? Ans. 276 quarters, for which he received £766. 8.9. § IX. ST JOHN’S COLLEGE. May, 1856. 5 4 21 5 4 . — ~— "——_"=__'_‘ _‘ me Q ==h 2, =E 3n 1 x—l +.:r+2+.r—3 n+1 ~|_.r—2+.:r:+3 Ans x J or "/ ‘2 ‘._. 2. %+§§=10(g—;). Ans. ar=6 or -2, or 3i~/2l. (a;+a.-1-b)5+(.T+c+d)""___(a+b-~c—d)2 (.r+a+c)"+(.r+b+d)5 (a-b+c-d)”' Ans. wg- %(a+b+c+d), or -~-;-(a+0+c+d)=1=U=I=5(a+b-c—d)2(a-b+c-d)2. 2 2 1- % (ac+1)(.r +1)=(a +1)(.ry+1) , xzi _l, or _l_+_a’ or __f_z ’ .r+1 y+1 l—a 1+a Ans. ‘Z 2 _,___ .- (ac+1)(y +1)=(c +1)(.ry+1) . y=*~/_l, or 1+0 7 or 1 c . y+1 n+1 1 0 1+0 5. Each of three cubical vessels A, B, C, whose capacities are as 1 :8 :: 27 respectively, is partially filled with water, the quantities of water in them being as 1 : 2:: 3 respectively. So much water is now poured from A into B and so much from B into C as to make the depth of water the same in each vessel. After this, 128% cubic feet of water is COLLEGE EXAMINATION PAPERS. '499 poured from C into B, and then so much from B into A as to leave the depth of water in A twice as great as the depth of water in B. The quantity of water in A is now less by 100 cubic feet than it was originally. How much water did each of the vessels originally contain? Ans. 500, 1000, and 1500 cubic feet respectively. 6. A fraudulent tradesman contrives to employ his false balance both in buying and selling a certain article: thereby gaining at the rate of 11 per cent. more on his outlay than he would gain were the balance true. If however the scale-pans, in which the article is weighed when bought and sold respectively, were interchanged, he would neither gain nor lose by the article. Determine the legitimate gain per cent. on the article. Ans. 10 per cent. 7. A metallic lump m, is compounded of the metals a and 6, another lump m, of the metals 0 and c. The ratio of the weight of a in m, to that of c in m, is three times the ratio of the weights of b in m,, m2 respectively; and three times the weight of m, is ten times the weight of m2. In two other lumps In, #2 of the same volumes and compounded of the same metals as 121,, m2 respectively, the ratio of the weights of a and b in p, is equal to the ratio of the weights of b and c in 112,; the ratio of the weights ofb and c in ,u, is equal to the ratio of the weights ofa and b in m,; and five times the weight of ,u, is eighteen times the weight of 101,. Having given that the weights of equal volumes of a, b, c are as 3 : 2 : 1 respec- tively, determine the proportions in which each of the lumps is com- pounded. Ans. ml and 1122 are compounded of equal volumes of the metals: a, is compounded of volumes of a and Z) in ratio 4 : 3, and ,u, of volumes of b and c in ratio 3: 4. In ml, weight of a : weight of b r: 3 :2; in 112,, weight ofb : weight ofc::2 :1; is ,u,, weight ofa : weight ofb :: 2 : 1; and in #2, weight ofb : weight ofc :: 3 z 2. X. ST JOHN’S COLLEGE. Jane, 1857. I. (.r+1)5+(.r—1)5=19{(.r+1)’+(.T-1)3}. Ans. x=0, or *JIB, or *2,/-1. 2. (x+l)(.r+2)(a:+3)(x+4)=(.r+1)’+(.v+2)2+(.r+3)2+(x+4)”. Ans. .r=- 1 =0 1, *2, -a, - 14,/_. . 3. (w’+1)(_y’+1)=10,} Ans lx ’ 2( 15) (xi'Z/X'Tfl-ll" 3' y=—3, *2, 1, 0: glad-15)- 32—2 500 COLLEGE EXAMINATION PAPERS. ' 5 l .r-g, or g , 4. 412+ y9-z”=(x+y-z)2+ 2, $ 1 5 xa+y°—'zs=(.v+_y—z)3+ 9, Ans. 43%;: 0" g ’ .v‘+y“- z‘=(.r +y— 2)‘+ 29. 3 Lz=§ . 5. Alfred, Edward, and Herbert come each with his pail to a well; when a question arises about the quantity of water in the well: but none of them knowing how much his pail will hold they cannot settle the dis- pute. Luckily Mary comes up with a pint measure, by aid of which they discover that Alfred’s pail holds half a gallon more than Edward’s and a gallon more than Herbert’s: but before the precise content of any pail is found out an accident happens and Mary’s measure is broken. They are now however in a position to ascertain the quantity of water in the well: for they find that it fills each pail an exact number of times ; and that the number of times it fills Edward’s is greater by eight than the number of times it fills Alfred’s, and less by forty than the number of times it fills Herbert’s. How much water was there in the well? Ans. 15 gallons. 6. Seven—ninths of the stronger of two glasses of wine and water of equal size is mixed with two-ninths of the weaker, and the remainder of the weaker with the remainder of the stronger. The stronger of these two new glasses is a certain number of times stronger than the weaker; and the stronger of the two original glasses was twice the same number of times stronger than the weaker. Compare the strengths of the two original glasses: the strength of a glass of wine and water being defined to be the ratio of the quantity of wine to the quantity of the whole mixture in the glass. Ans. 4: 1._ 7. Two Tyrolese J tiger agree to shoot at a mark on the following terms: each is to shoot an even number of times fixed for each beforehand, and for every time that either hits he is to receive from the other a number of kreuzers equal to the whole number of times that he misses. They have two matches without varying the conditions. In the first match, the second J ager misses as often as the first hits in the second match, and the first Jager misses twice as often as the second hits in the second match; and the second J ager has to pay to ‘the first a balance of 4 kreuzers. In' the second match, each hits exactly the number of times most favourable to him, and the second has to pay to the first a balance of 36 kreuzers. How many times did each hit and miss in each match? Ans. In the first match, the first J ager hit 4 times and missed 16; and the second Jager hit 6 times and missed 10. In the second match, the first J ager hit and missed 10 times ; and the second J ager hit and missed 8 times. MISCELLANEOUS EXAMPLES. THIRD SERIES. [Solutions will be found in the Companion] SOLVE the following equations I—-- , 3 s 1 I. 0 £4 1 . 1 2. .ri‘-3.r=a8+&§. Ans. m=a+5, or &c. 3. (12+ a)(.n+ 2a)(.r+ 3a)(.r +4a) =c‘. Ans. .r=-§2£*~ / §Za-='=,/c‘+a‘. 4. m3+par2+ (p—l+p—ii>.r+l==0. Ans. .v=l—p, or &c. . 2 3 3 __ ._ =-—1— 5. (p—1).r+p.r+(p l+p_1 .r+1-0. Ans. .r 1_p, or &c. 6- wry-awn ._ Rz-a“ y”(z—.r)==b", Ans' _ (R+ba)(R—~08)'R’ z’(.r-y)= c”. 8_ R2_ be R ‘9 _(R+ca)(R--a3)' ’ R2__06 3=-'—————*- .b. . ’ (Ream—arm 8__ a c where R - a-———-——-,+b,+ C, . ~7. x”(_y+z)=a3, ' (a’+ ca)(aiba-c‘) A . 8: , y”(z-1:x)=b’,} ns x (03+03)2 wys=ci a: (53+ Ca)(asba_ Co) 3’ (a3+c")2 ’ z,=(.r+ Canaan” (a363_ 06>: 502 MISCELLANEOUS EXAMPLES. 8. xyz=a”(y+z)=b’(z+.r)=c’(x+y). I l I n+1“? Ans. w”=2. b c a , 1 1 1 1 1 1 final???) and similar expressions for y and 2'. 9. xyz='a(y2+ z”) = b(zi+.r“") = 0(x2-1-y2). 4' ,_ ‘ . . Ans. .v_ 1 1 1><1 1 1 , and Similarly y and z. (a b c a b a) 10. (.v+y)(.n+2)=a2, (y+z)(.y+‘r)=bga (2 +.v) (z +y) =02, Ans. x=iL—zb—c ——1-+-1— +1) and similarl and z. 2 a2 62 62 , y y 11. x3+y3+z3=3xyz, 3a*w+3=3b_y+x=3c_.z+y.} ADS. xza-b, y=b__c, 2=c__a. 12. Obtain in the simplest form the value of x which satisfies the equation , (x+a—b—c—d)(x—a+b—c—d)(.r-a—b+c—d)(zv—a—b-c+d) =(.r—2a)(.r-2b)(.r-2c)(.r—2d), and investigate three equations independent of a: which include all these relations betwen a, b, c, d, for any one of which the equation becomes an identity. ~ (2) a+b=c+d, Ans. (l) x=~é-(a+b+c+d). a+c=b+(l’ a+d=b+a 13. Two numbers are as 3 : 5, and their G.0.M. is 555: what are the numbers? Ans. 1665 and 2775. 14. Each of two points moves uniformly in the circumference of a circle: one goes round 5 times while the other goes round twice; and at the end of 21590 days they return for the first time to the place from which they started. How long does each take to go round? Ans. 4318 days, and 10795 days? a 15. Four curonometcrs, which gain 6, 15, 27, 35 seconds a day respectively, shew true tIme. After how many days will this happen ' P agam- Ans. 43200 days. 16. Two points revolve in the circumferences of two circles. At starting they are in a common diameter of the circles; and before arriving again simultaneously in their initial positions, they have been in common diameters n times or 122 times, according as they have revolved in THIRD SERIES. 503 the same or in opposite directions. Prove that m+n and m—~n must have a common divisor 4, and no higher common divisor, and that the ratio of the periods of revolution of the points is m-l- n : m- n. 17 . Three particles start at the same point to move uniformly in the circumference of a circle, and the period of the first : that of the second : that of the third :2 a prime number : a second prime number : a third prime number. The first and second arrive together at the starting point 6 min. before the first and third, and 28 min. before the second and third arrive there together; also when the first and second arrive there together, they have been together as many times as the second and third when they arrive there together; and all three arrive there together 1 h. 55 min. 30 sec. after starting. Determine the periods of revo- lution- Ans. l-é-min” 34min” 5221-min., respectively. 18. If the G.C.M. of m+n and nz-n be 4: prove that the G.C.M. of m and n Wlll be either 2 or 4; and that when It is 2, each 0t 2;, 5 1s odd, and when it is 4, one of 77;; , g is odd, and the other even. 19. From the longer of two rods a piece equal to the shorter is cut off: with the shorter and the remainder of the longer a similar operation is performed; and so on, until the rods are of equal length. Prove that if this last length be taken for the unit of measurement, the lengths of the original rods will be represented by numbers prime to each other. Explain that there are many relative lengths of the original rods for which the cut rods will never become equal, however long the operation be continued. 20. Define commensurable magnitudes; and prove that magnitudes which are commensurable with the same magnitude are commensurable with one another. _ 21. The adjacent sides of a rectangular parallelogram are respec- tively equal to the hypothenuses of two right-angled triangles whose sides are commensurable with the unit of linear measurement. Prove that its area will be commensurable or incommensurable with the corresponding unit of square measurement according as its sides are or are not commen- surable with each other. 22. Shew how to find geometrically lines equal to the G.C.M. and the L.C.M. of two given commensurable straight lines. 23. Prove that ' (14(02— 02) + 64(02—a2) + 04(a2— b") a2(b-—c) + 02(c—a) + 02(a— b) 24. Obtain the continued product of a+b+c+d, a+b—~c—-d, a—b—c+cl, a~b +c—d; =(a+b)(b+c)(c+a). also of —a+b+c+(l, a—b+c+d, a+b—c+d, a+b+c—d: and shew that the sum of these products=16abcd. 504 MISCELLANEOUS EXAMPLES. 25. Find the continued product of n such trinomials as x”— ax+a’, .r‘- airs-Pa“, a8—a‘x‘+ a“, x‘6—a8w8+ a“, &c. x2m+ amwm +a2m Ans. ~—-,+———,—-, where m=2”. a: +ax+a 26. Find the sum of n such fractions as 2x—a 4w3- 2a2a: 8x7» 4a“.i:8 F- ax + a2 ’ .v‘—a”.v"+ a4L ’ x8— a4a4+ a? ’ &c. 2mx’”‘-1+ma’".v”‘“‘ 2x+a w2m+ a’”.v”‘+ aim .v’i-I- ax + a2 Ans. , where m=2". 27. Prove that (a2_ be)a+ (ba_ 02)3+ (ce___ a2)s (a— b)3+ (b -c)‘°’+ (c— a)3 =(a +b)(b +c)(c+a). 28. For what unit of time will the durations 115743. and 360360“. be represented by numbers prime to each other? Resolve the same question for 36 hours and 2'76 hours. Ans. (1) 18 seconds. 432 seconds. 29. Divide 1+2a.'2"+‘+.r“"+2 by I+2a:+x’: writing down the (2n—-m+1)th and (2n+nz+1)th terms in the quotient; the (2n—m)th and (2n+m)th remain- ders ; and the complete quotient. Ans. (1) (-1)’".(2n—nz +l)..r2"_"'. . . . (—1)’".(2n —m+1).a:2"+’". . . . (—1)”‘ {(2n— m + 1).r2”-"‘+ (2n -m)a:’"""+‘}+ 2x2"“+ at”. . . . (4) (—1 —m +1 )2"+’"+ (212 -m + 2).r2”+”'+1}+.r”‘+2. . . . (5) 1— 2.2: + 3x”-- . . . + (2n+1).r2"-2n.r’"+1+ . . . +.v“". 30. Eliminate .v, y, z from the equations (w—yXr—zXz—w) =a8. (w+y)(y+z)(z+x)=ba, (w’+r’)(a’+ z’)(z’+ n = (AH/“1 (y‘+ Z"0044+ w‘) =11". Ans. 406* 4JW= {63 * J2_(:“-—as}2. 31. Eliminate a: from the equations 5 3 323=(-’5) +10§+5(5) ,1 a a a a: A... 1=(£+2)i_ (L2)? , 5 a a c a, c ’ 329.1(5) +109+5(-’-’-) C a: J! a THIRD SERIES. .505 32. Eliminate .r, y, a from the equations r+2+i=%1 y s .v .v y 2_ v _ E+.5+;_ ,, ADS. org—1+7. (max-2 +5X5+£ =,, y z z a: a: _1/ J 33. A symmetrical form of the condition that the equations ax+a’==b.r+b’=c.v+c' may be simultaneous is a’(b-c)+b'(c-a)+c’(a—b)=0; and a symmetrical form of the value of .v is aa’(b- c) + 66'(c-a)+ cc’(a—b) (a—b)(b-e)(c—-a) ' I a-a’ b—b’ 0—0' 34. If a,__a,,=b,_b,,= CLO” ; prove that ab’-a’b bc'-— b’c ca'-c'a b/b//__ a//b/_ brc//_ bus!“ C/a//___ cllal and = each of the former; and each of these six __ (a+b+c)-~(a'+b’+c’) “ (a'+ b’+ 0') -(a"+ b”+ c”) ' 35. Prove that the six equations a,(b,e,— 630,) + a,(6,c,- b 103) + a,(b,c,—- 6,0,) 0, 61(02a3— caaz) + 02(03a1-01a3) + 63(01a2— czal) = O, 01(a263- aabg) + 02(a361— (1,63) + 03(a162— 0:26,): 0, d,(b,c_.,- 530,) + d,(b,c,- 6,03) + d,(b,c,— 6,0,) =0, d 1(02a,— esaz) +cl,(e,a,— alas) +d3(c,a2— 02m) = 0, d1(a253-— cab.) +d2(0361—a163) +d3(a,6,- a,b,) = 0, are equivalent to only two independent equations : and find them in the most simple symmetrical shape. a, as 51 52 01 c2 Ans. ——- 506 MISCELLANEOUS EXAMPLES. 36. There are 72 coins 0,, 0,, 0,. . .0,,. If m, of the coins 0,=n2 of 0,; m, of 02=n3 of 0,; m, of 03:12.; 0t 0,, &c. ; m,_, of 0,,_,=np of 01,: how many 0f Cp=ni 0f 6'1.P n,.n,.n3. . . 12,, Ans. . o 0 Establish the “chain rule". 37. There are two amalgams of the same bulk, each composed of mercury and gold, in the ratios of 2 : 9 and 3 : 19 respectively. If they were fused together, what would be the ratio of mercury to gold in the resulting amalgam? ' Ans. 7 : 37. 38. At noon on a certain day a clock and a watch, each of which goes uniformly, are set to true time. It was calculated that the clock would be as much wrong when it should shew any time as the watch would be when it should come to shew the same time, and that it would be midnight by the clock one second before it would be midnight by the watch. Find, in fractions of a second, the daily gaining and losing rates of the clock and the watch. A 86400 8 d 86400 5 ns. 8———6400_1 sec ., an ———86400+1 sec. 39. A shilling’s worth of Bavarian kreuzers is more numerous by 6 than a shilling’s worth of Austrian kreuzers, and 15 Austrian kreuzers are worth a penny more than 15 Bavarian kreuzers. H ow many of them respectively make a shilling? Ans. 30 Austrian, 36 Bavarian. Enunciate the problem indicated by the negative result. 40. From a vessel which will contain a gallons, filled with a fluid a,, a gallon being drawn off, the vessel is filled up with a fluid 01,; a gallon being drawn off from the mixture, the vessel is filled up with a fluid as; and so on, until the vessel contains a portion of each of the fluids a,a,a,. . .a,. How much of each fluid does it contain? 1 n—l 1 "-2 1 n—3 1 2 1 Ans. “(l—5) , <1—5 , <1—Z) ,...(1—Z>, (1—5), 1, gallons respectively. 41. Prove that (a+b)(b+0)(c+a)>8abc, and l, the former or the latter according as w<7-n—: when w<1, the former or the latter according as . The former. (3) The former or the latter according as the greatest and least of a, b, 0, d, . < together 1s > the other two together. THIRD SERIES. 507 43. {If a, b, 0, be in Harmonical Progression, prove that a b c b + 0 ’ 0+ a ’ a + b are also in Harmonical Progression. 44. The relative powers of the instruments a,a,a,...a,, are expressed by saying that generally a,a,a3...a,.._, working together can do as much work in one day as a, can do in p days. If a,, can do a piece of work in one hour, how long will it take a, and a,, where r>1, to do it? Ans. (1) -1-<1+l)n_ hours. -<1+—1->n_r hours. P P P 45. If m and n be any two positive integers of which m is the greater, and p=2”‘, q=2"; prove that .r"’1’+a1’.r1’+a’p is exactly divisible by both miq+aqxq+ a?“ and .v’Y—aYSH- a”. n 46. If 0,: —,—~— , prove that law a $13 29— +(_1)n_1=1+4+1+ +-1- | 2 3 0 a o g a o . n 0 47. If 12,— 125:5. ' prove that _ 2.4.6 . . . 2r P2n+1+PI‘P2n+P2'P2n—l+ ' ‘ ' +P11—I'Pn+2+Pn'Pn-l-l=%' 48. If to, t,, t,, t,, &c., represent the terms of the binomial expansion of (a +x)", prove that (to_ t2+ t4- &Co)2+ ([1_t3+ tb_ &C.)2 = x2)“. 49. If no, a,, a,, a,, &c. be the coefficients in order of the expansion of (1+.r+ .102-1- . . .+.:tP)", prove that. ' (1) their sum =(p+l)"; (2) a,+ 2a,+ 3a,+. . .+np.a,,,,=-%np(p+1)". 50. Apply the general term in the Multinomial Theorem to prove that the coefficient of x’P“ in the expansion of (ao+ a,.v+a,a:”+. . .)2 is 2.(ao.a,,,+,+ a,.a,,,+ a,.a,,,_,+ . . . + a,.a,,+,). 51. In the expansion of (l+.r+.r”+...+.r')2", where n is a positive integer, prove that (1) the coefficients of terms equidistant from the beginning and the end are equal; (2) the coefiicient of the middle term is the greatest; the coefficients continually increase from the first up to the greatest. 52. If a,, a,, a,, a,, &c. be the coefficients in order of the expansion ~ of (l+.r+.v”)", prove that , a”,— a’,+a9,-~ 023+ . . . +(— 1)""1012..-i =%a»i1"("1)na~}' 508 MISCELLANEOUS EXAMPLES. 53. Shew that if m be not less than n, the greatest coefficient in the expanSIon of (a,‘+a,+a,+...+a,,,)" is '2; and that the number of terms . I 9 O m which have this coefiiment Is ————-— . mm 54. Prove by the method of Demonstrative Induction that the num- ber of different throws which can be made with n dice, on the supposi- tion that the sum of the numbers turned up is r, is equal to the coefficient of .v’ in the expansion of (x+.r2+ x3+x4+ x5+x6)". Enunciate the corresponding proposition when each die has p faces, and the numbers marked on the p faces of each die are a,, 0,, 01,. . .ap. Ans. Number of throws =coefficient of .r’ in the expansion of (4011+ .raz-l- . . . +.v%)". 55. Prove that the numerators of any two consecutive convergents to a continued fraction are prime to each other, as also their denominators. 56. The Cambridge Lent Term always ends on a Friday, suppose at midnight. Prove that if it commence on the 2pth day of the week, it will divide either on the (p—l)‘h day of theweek at midnight, or on the (3+p)th day of the week at noon: but if it commence on the (2p+1)“‘ day of the week, it will divide either on the pm day of the week at noon, or on the (3 +p)th day of the week at midnight. 57. Prove that (1) 1+3+5+ .... ..+(2n—1)=n2; (2) 13+23+33+ .... ..+n3=(1+2+3+ .... ..+n)2. Apply (1) to find solutions in positive integers of flags—22, where n is a positive integer; and apply to find solutions in positive integers of x3=92_ 220 58. Prove that the product of any r consecutive integers is divisible by El. [3,. B,.'.....Lg£, if a1+a2+a3+ . . . . ..+a,,=7‘. 59. In how many ways can a line 100800 inches long be divided into equal parts, each some multiple of an inch? Ans. 124. 60. How many different rectangular parallelopipeds are there satisfy- ing the condition that each edge of each parallelopiped shall be equal to some one of n given lines all of different lengths? Ans. n(n+1)(n+2) . ‘ 1 . 2 . 3 61. Prove that if, in any scale of notation, the sum of two num— hers is a multiple of the radix, then (1) the digits in which the squares of the numbers terminate are the same; (2) the sum of this digit and the digit in which the product of the numbers terminates is equal to the radix. ' 62. A certain number when represented in the scale of 2 has each of ' its last three digits (counting. from left to right) zero, and the next digit THIRD SERIES. 509 towards the left different from zero; when represented in either of the scales of 3 or 5, the last digit is zero, and the last but one different from zero; and in every other scale (twelve scales excepted) the last digit is different from,zero. What is the number? Ans, 120_ 63. Prove that every even power of every odd number when divided by 8 leaves 1 for a remainder. 64. If n be a prime number, prove that 1"+2"+3"+ .... ..+(rn)" is a multiple of n. 65. If n be any prime number excepting 2, and N any odd num- ber prime to n, prove that N"“-1 is divisible by 812. 66. A gentleman being asked the size of his paddock replied :— Between one and two roods: also were it smaller by 3 square yards it would be a square number of square yards; and if my brother’s paddock, which is a square number of square yards, were larger by one square yard, it would be exactly half as large as mine. What was the size of his paddock? Ans. 1684 square yards. 67. A walks at a uniform speed, known to be greater than 3 and less than 4 miles an hour, between two places 20 miles apart. An hour having elapsed since A’s departure, B starts after him from the same place, walk- ing at the uniform speed of 4 miles an hour. Shew that the odds are 2 to 1 against B’s overtaking A. 68. When 2n dice are thrown, prove that the sum of the numbers turned up is more likely to be 7n than any other number; and when 2n+1 dice are thrown, prove that the sum of the numbers turned up is more likely to be 7n +3 or 7n+4 than any other number, these being equally probable. ‘ 69. A handful of shot is taken at random out of a bag: what is the chance that the number of shot in the handful is prime to the number of shot in the bag ? Ex. Let the number of Shot in the bag be 105. (1) Ans. <1—2) (1—%)<1—%> .... ..where the number of shot in the bag =a1’6q0' .... ..; Ans. 70. A coin is to be tossed twice: what is the chance that head will turn up at least once? Point out the error in the following solution by D’Alembert :—-Only three different events are possible ; (1) head the first time, which makes it unnecessary to toss again, (2) tail the first time and head the second, (3) tail both times: of these three events two are favourable: therefore thel . . 2 required chance is g . 71. In each of n caskets are p jewels worth 1, 2, 3, .... ..p guineas respectively: a person being allowed to draw a jewel from each casket, 510 MISCELLANEOUS EXAMPLES. find (I) the most probable collective value of the jewels he will draw 5 the value of' his expectation. (3) What would be the value of his expecta- tion, if he were allowed to draw 1* jewels from each casket? If 12:2, and one jewel be drawn from each casket, what is the chance (4) that the value of the jewels drawn is n+1“ guineas? (5) that the collective value of the jewels drawn is that collective value which is most probable? Ans. (1) g(p+1) guineas, if either n be even or p odd; and g(p+1)i;21- guineas, n being odd and p even, each of these being equally probable. (2) ZQJH) guineas. (3) 7;273(p+1) guineas. (4.) l l, 2“'|:.L§—'r (5) m _. (204.6000‘...7Z)z n or 4 6 __1L 4 6 ___1 , according as n is even or odd. 2. . oouunon_ X2. . a a n o “72+ 72. A person borrows £0 on the following terms. It is to be paid off in 12 years: and at the end of each year is to be paid interest at a given rate 1' on the sum remaining unpaid at the beginning of the year, together with such a portion of the principal that the whole sum paid on account of principal and interest together shall be the same for every year. Inves- tigate a formula for the sum to be paid every year. Ans. c.—————r(l+r)n . (1+r)”-1 78. From the gold fields are brought 212 specimens of' gold dust, no two of which are of the same degree of fineness. Each specimen is divided into as many equal portions as is necessary to the following operation, viz. to form as many different mixtures as possible by taking a portion of dust from some specimen and mixing it with a portion from some other speci- men. Each of these mixtures is now divided into as many equal portions as is necessary to the following operation, viz. to form as many new mix- tures as possible by mixing together portions from any n of 'the former mixtures. Prove that 1. 8 . 5. 7 .... ..(212—1) of the latter mixtures will be of the same degree of fineness as a mixture formed by mixing together all the dust of 2n specimens exactly like the original specimens. EASY EXERCISES. EXERCISES. A. IF a stand for 10, b for 3, and a: for 7, what is the value of each of the following quantities? (l) a+b+an (5) Qa—w. (9) Qab-Sx. (2) a+b-a:. (6) 4a+36-2x. (10) 2a+5-36x+100. (3) a-b+a'. (7) 7a+Qb—2x. (11) 7ab—abx. (4) a-b —a:. (8) 5a—4b— 41x. (12) 3a+ bar-xx. (13) What is the coqflicient of a: in Sam? (14) What is the eoefiicz'ent of a: in fiqbx? (15) What is the coqflicient of bx in Gabx? (16) What is the coqflicz'ent of a in each of the quantities 2a, Qab, abx, Saba, ma, axx, pam, abxy? (17) What is the coqficz'ent of 25 in 125? (18) What is the difference between 3+x, and 3.2:, when .2: stands for 7? (19) What is the difference between 3a+w, and 341—m, when a stands for 10, and a: for 6? (20) What is the difference between 3a+w, and 3am, when a stands for 3, and a: for 2? Find the value of each of the following quantities, when a stands for 10, b for 3, and a: for 7:--- (21) Sax-+7. a—x 3a: 461' (22) 3ax+7b. (25) b ‘ (28) 4a+2+10a-16‘ 2a+x 3a aba: 2a+4b a—Qb (23) b . (26) 7+2w—21-7. (29) 3$_a_b- ‘16-!) , 3b+3x 5a+x 5b+a ma 126 pm (24) a i (27) b _2x-3b' (30) 6+.rTa-a: a—b' If a stand for l, b for 9, and c for 8, find the value of each of the following quantities: (31) a2+b’- 0’. (33) 5abc—226’+3c’. (a2) 13a’+3b’-4c’. (34) a’b+b’c. ' 512 EASY EXERCISES. (35> <36) (37) (38) (39) (40) (41 ) (47) (48) (49) (50) (51) What is the difference between J; and 5!; 1 2aba+ 20a’b - 2602 b” a2 c2 2; +7; ‘26; t 80b” Que9 Te?“ _ 25? ‘ ma”+ nbg—pc’. 2JZ-J25. J57: +,[a“—b. a+J5—JZZ+Q./Qbc. (42) ./2c+b-J2b~2a. (43) m\/§+n\/%€—p,/2ac. (4A) 355+,j1li—22/5. (4.5) Jb+c-a-f/ab-2c-aa. (46) ya+Ja.%5_41/m. What is the difference between 3a and a3, when a stands for 2? i What is the difference between 2J5 and 2+J§, when x is 100? What is the difference between 3,]; and y}; when axis 64'? What is the difference between ,Ja-i-b and JE-i-b, when a stands for 1, and b for 8? for 16, and b for 4? , when a stands Add together (1) a+b, (2) a+b, (3) (1-6: and a+b. and a—b. and a-b. (4) a—b+c, and a+b-c. (5) a—b+c, and a+b+c. (6) 1—2m+3n, and 8m—2n+1. (7) 5m+3, and 2112—4. (8) 3xy~2x, and xy-i-Gx. 7x—6 , --x-3_y, -x+y, —2.r+3y, and x+8y. 3-a, —8-a, 7a—1, —a—1, and 9+a. 2a—5b+3c:—-d, and a+5b-c+2d. a”+2ab+b", and 2a2-ab—3b’. 3w2—6x+5, 2x—3-x’, and é—x—Qwg. ac+bd, bd—cd, and ac+cd. 3x2-4y2, ital-y”, and éwa-hé-yg. xa— 3ax2+ 3a’x —a”, and 3a? 2a2x + 4ax2— x3. (17) (18) (19) (20) (21) (Q2) (23) (24> EXERCISES. B. (9) 4p—2q+1, and 7—3p+q. (10) 5ab—Qbc, and ab+bc. (11) 2ax+363h and tax-by. (12) 3a—26+4c, and 2a—3b+c. (13) wy+x—-7, and 3xy—Qa:+3. (14) p+q-pq, and 2pq—3p+2q. (15) P2+2P<1+q22 and P”—2Pq+q’- (16) 7ab—5ac+1, and ab+6ac-2. EASY EXERCISES. 513 (25) (26) (27) (28) (29) (30) (31) (32) (1) (2) (3) (4) (5) (6) (7) (8) (9) (1°) (11) (12) (13) (14) (15) (16) (17) (18) (19) of each? (20) Sum-rm, and 1-n—7mn. 9x-8y-7, and 3z—9x+6y+7. aa—Qabg+agb, 3abg—2agb, and 63+c8. aa—glab", bs—é-aib, and ab2-é—agb. ;11.r2+2wy, gwg— xy+y”, and mx+ny. acl—l-Qbd—3cd, éad— ébd, and %a6+20d—ac. I received m shillings from my father, the same from my mother, and n shillings from each of three friends, express the whole sum. A certain sum is divided between A, B, and C; B receives a pounds more than A, and C receives 6 pounds more than B ,' if A receives .2: pounds, find an expression for the whole sum divided. From From From From From From From From From From From From From From From From From From EXERCISES. G. a take b—w. a+b-c-d take a—b+c-d. Ga—Qb—fa’c take a—26+20. a+x—b-5c take x+Sb—5c. 3x+2y—52 take 2x+3y+4sx 2ax+by—c take ax—by+c. 3bc—ab+a take 260+ab—a. xy+x2+y2 take xy—x’+y’. Qxy + 3x9+ 432 take wy — ZxB-y". 2mn+5m—-3n take mn+m+n. —2.r_y+mw-p_y take —3x_y—2mx-p_y. 5abc-2ab-3ac take 2abc+ab—ac+l. a2—b2+ 02 take a2—262— 202. 4aw—3a2+2x2 take 2ax—a2+4x2. .‘a’a"b+2a”c—5c2 take azb—azc—7c”. 2my+3a—agb+5 take Qa-a2l2+6. gum—la: +g take lax gar —1 s 's y a a +3 9 3' I 1 e1 a+b-c take 561—5 b—éc. The united ages of a father and his son make 60 years, and the father was 30 years old when the son was born, what is the age Divide 1 into two fractional parts, so that one part shall exceed the other by 33 514 EASY EXERCISES. EXERCISES. D. Multiply _ (1) may by b. (3) -2.r_y by 4a. (5) %ab by 20. (2) 311m by -p. (4) —2.ry by --4.~£l. (6) 51m: by mp. (7) m+u—p by 3. ' ‘ (24) a+2x by a-Sx. (8) az+bm' by p. I (25) 7x-1 by 5.2—4. (9) ad+2l1d by 2a. (26) 2a.:r-Sby by 4y—3x. (10) 4ag—2axy by am. (27) 1—2mn by 2m+n. (11) 3x-2xy+6 by -—.r_y. (28) ag—bc by ac—b'. (12) 1—2ax+36x’ by —-3n. (29) 1+2m+3y by .z—y. (13) 2ab—3ac+5bd by -2.r. (30) a+.r—y by b-y. (14) 21y—3 by 7.1:. (31) ac—bc+ad by 2a—b. (15) 2ax+by—cz by 2xyz. (32) a’+a'+a+1 by a—l. (16) 2a’—b.r+d by by. (33) .r”+a.r'+a’a:+ a3 by .rl—a. (17) n+1 by b+y. (34) 4w’-6.r+9 by 2x+8. (18) 6x+4 by .r—l. (35) 4+2x+x’ by 4—2a:+.r’. (19) x-4~ by x+3. (36) as—2x’ by a'—.z’. (20) 2.1—5 by 3.1—2. (37) .r‘+3x'+9a:+27 by 1—3. (21) 1-.r by w+l. (38) 2a‘m’+3b'y by 2a‘.r’—3b‘y. (22) l—x by .1—21’. (39) 2a’—3ab+b’ by 2a’+Sab—b’. (23) ax+by by 2.1—3]. (4-0) a°+a‘—a—1 by l—a+a’-a‘+a‘. EXERCISES. E. Divide (1) 70: by 7. j (12) —§ab.x'y by —§a.ry. (2) 7.2: by .r. m, (13) 3ac—Qabd by a. (3) 70.1: by a. a, (14) 4ac—2abd by 2a. (4) 7a: by 7x. i - (15) Elf—6.1:}; by —2.r. (5) Saba- by ab. (1 —6b’c' by 36c. (6) Sabc by 360. (17) 4-a’.r —8ab.r—-2aa: by -2a.r. (7) -a.ry by m. (18) a”.r’-5abx'+6a.r‘ by ax“. c (8) axg by wt. (19) .r'+3.r+2 by 2+2. (9) 6a’mn B} —2mna. (20) ac—bc+ad-bd by a—b. (101gsua’wy' by 7a’_y. . (21) 6+ 30-26-11!) by 2+a. (11) -7mn'p.r by émnp. (22) 4.0%] 51’— 4-ax by 20+3x. v (23) a’—a.r—6.r’ by a-3x. (24) 2ab+6abc—8abcd by 1+3c-4cd. EASY EXERCISES. 515 Divide / (25) (26) (27) (28) (29) (30) (31) 3w’+16x—35 by w+7. 3x‘+14.r3+ 9x+ 2 by 22+ 5.2: +1 . ab +2a’— 3b’- 4bc—ac- c9 by 2a + 3b + c. 1 5 a‘+ 1 Oaiw+ 4a’x’+ 6ax8— 3.2:4 by 3a’— x’+ 2am. qP3+ 3P’q’— 2M“— 22‘ by P —2- a2x8+ a‘-2aba:8+ b2x8+ aabi— 2a‘b by ax —-bx +a’— ab. 32xs+24~3 by 2.2+ 3. EXERCISES. F. Find the G.G.M. of (1) (2) (3) (4') (5) (6) (7) (8) (9) (10) 128, and 84,, (11) abxy, and 2acxy. 125, and 900. (12) éag and gab 80’ 100’ and 140' (13) abd, acd, and bed. ax, and bl" (14) pay, may”, and apzv. bx”, and bgwa. (l5) xg—yg, and mg+2xy +51”. apx’, and a’px. (16) ax+bx, and ay+by. 56226.2}, and 20abxy. (17) aa—ab’, and ab2+b8. 15a6b’, and 362%“. (18) 2a3—2ab, and 5a3—5ab. Qa‘bgc", and 27a66309. (19) 14a2—7ab, and 10ac—5bc. 1412121219”, and 7mnp. (20) (x+y)3, and (wi—yg)’. (21) (22) (23) (24) (Q5) (26) (Q7) (28) (29) (3°) (31) ara-2.r-1 , and x’+ 2.2: +1. .26— 3x‘+3a:’-2, and x‘—3x2+2. 2x3— 3x’- 9x+ 5, and 2x2—7x-1-3. _ 4.2% 3x—10, and 4wa+7x’- Sui—15. flaw, and 4x“—25x’+20a: + 25. 0.3+ 3, and x2+ 540+ 6. xa-f— 2x, and 223+ 3x’+ x. x”— 302—2, and x‘- x3+ av—IO. 2x8+x’-.1:+3, and 223+ 5x2+.r—3. m’+ xy -1 23’, and .r’— 5.231 + 6y”. 6.1:2—12a: +6, and 3x2— 3. 33-2 1516 EASY EXERCISES. EXERCISES. G. Find the G.C.M. of (1) (2) (3) (4) (5) (6) (7) (8) (9) (19) (1 1) (12) 4» +12x+9502, and 2+13x +1 5.1.". 03+ a’b- ab2— b3, and 0.3— aeb — ab2+ b8. 3x2+ 8.x + 5, and 2x3- .22— 3.2:. ass—1 123+ 39x —45, and 3x’-22x+ 39. 6x2—7w—20, and 4x3-27ar + 5. 3x3—13x’+23.r- 21 , and 6x3+ x’- 44w + 21. 3x2—16x-12, and 24:34 6x”—24x+288. 8x3+ 6x2- 4.1!— 3, and 1223+ 5x2+x+ 3. 3.r-1 -2a:’, and 1-x—2x’. x‘—41a:’+16, and x‘-7m3+28x-1 6. wa—l, avg—2.1: +1 , and :02—1 . 22+ 2ab + b2, a’- b2, and a3+ 2agb + 2abg+ b3. Reduce the following fractions to lowest terms :— 2ax E; . 4abc Eze- . 20:212.: 13? ' 3abx’ 6am ' 7561.1393 15a?3 ' abgx 2ab‘a:2 ' mar-m: mnx 2x’— 3.1: 5.2: 14a8+21a’ 7a’b 4bc+2c 2ac (24) W' (84) L 2 2 -2 (Q5) arm-Z ' (35) (26> “.1122”. (2) (27) (37) (28) (as) w, 29> (39> (so) (4°) ‘32) <2 “5332:2314 EASY EXERCISES. 517 EXERCISES. H. Find the L.C. M. of (1) 21, and 24. (2) 12, 16, and 20. (3) 4, 7, 8, and 14. (7) as, and bx. (4) 4, 7, 14, 21, and 24. (8) ax, and 223/. (5) 1, 2, 3, 4, 5, 6,7, 8,9. (9) 2x, 6x, and 8x. (6) 21, 22, 2a, and 24.. (10) ab, ac, and be. (11) x”, y’, and 2mg. (12) bd, c’d, 0d“, and be. (13) 2a, (1+2, and a—x. (14) 2a, 4b, 1+.v, and 1-x. (15) 3xy, x2+xy, and y’+.vy. (16) a362, agbs, ab—bx, and a’+a.'v. (17) 1+a, l—a, 1+a+a’, and 1—a+a’. (18) l-x, 1+.r, 1—m’ and 1—2x+a:’. (19) x’+3.v+2, 15(a:+1), and 20(w+2). (20) .223—7x—6, m’—2.r—3, and w’- x—6. (21) 8x3-14.w:+6, 4w’+4x—3, and 422+2x—6. 'EXERCISES. I. Add together a: 2a: 3.1: 1 2 3 <1) '5’ '5" and? (1°) 5’ .76, abc' 2ab ab x—5 -4 3“, and "-6- . J7, 2 , and 3 . 2a a 1 a: 7x—6 4x+1 (3) —3-, g, and 5. (12) 6, 3 , and 12 a+w a—a: a 4.2-5 2x 7x+6 --—5"’, and T. , 75-, and 2x+1 4.2—5 3 l 4: (5) 7 , and 7 (14) ;, 3—5, and 5. 2x+1 4.2—5 4 1 3 (6) 7 , and 21 (15) 3y, 7?, and 1 2 3 a: a, a, and —, E, %, and - 1 2 3 avg—ab ivy—be (8) a, b, audit-Z. (17) ab , be ,and2. 2 1 1 a—b b—C 0—11 (9) a? a” and 5v (‘8) 7.7." 77;" n “a? 518 EASY EXERCISES. Subtract 4a: 9.1: (19) ~23- from 16 . (21) 9 from 18 (22) 4 from 8 (23) -——10——- from (24) (25) (33> <24) ‘" . (35) i, y ,and y (36) (20) 7; from .r. 10w+17 5.2—1 3y+ar+13 3x+y 5 15+3a: from 7+ 24 w+1 w+1' 2 4 3 5 —+— from —+-—. a: a: a: a: Add together w and w w+2’ arr-2' and 1. (2+1! a—a: 2+9 w—l an w-I-‘2’ “'29 2+2 w—3' 1 __ {Ed-y (37) (38) (39) (40) 1 (26) w from 35— w+1 2+2' 2x—7 3w+7 T from O w 4! from <28) <29) <30) <31) (32) a from — b . x+y b+cx iii—y from 2 3+2w 1+0: H from w. x+y x-y 2a: 1 (2x-1)3’ (2.1:- a+ar —————2(a_x),, and x2+ g a: 4.2;?” w+y“: and y x—2 1--.r+.r2 ' I d-——.. 3 is" and (41) 2.2:’—4.r+2’ 2w2+4x+2 ’ an 1 1 l—w‘ (42) 20(x2+3.r+2)’ 15(a:+1)’ and 1 20 (a: +2) ' EXERCISES. (1) Multiply g by 2. (2) ....... .. 3’25”- by 2. (2) ....... .. 94‘? by 2. .... ..... gby 6. (5) ..... 9%” by 4... J. (6) Multiply '1715’ by 60. 2x . . . Q . - . .. fi- 84‘, (8) ....... .. sags by 6. ' 12+9x (9) ------- -- 16 by 80. (10) ....... .. 8’7“ by 9, '1? 1+w”+ 2.1: ' + 11x—13 '1'6 25 25 ' EASY EXERCISES. 519 (11) 6x+13 12 2a:-1 ....... .. 7, by 15. 2 3x+4 Multiply by 15. by 11. I O O I O I I l. 3a: . . . . H I 5. 3.x: . . - - u I 6- 21am .... .. T};- by 7a. 2mn .... .. —- b 2n. 2, Y j . '1 1 (31) MultIply 2+5 by 2%}. 1 1 1 _—_—_—_ _- 1+2: I—a: 2' 2a 2a -— +—-—. 1+a l—a . . l 2 DIVIde 2+— by 1-— . a: x 2—2; a: 7 Y1??? b—3a 2a—b 2—25 5'15“ a: 4-x (25) Multiply $4- (16) Multiply if by (17) ....... .. 3E'xby (18) ....... .. i'éfbyé. (19) ....... .. g} y%. (20) ....... .. Zbgby y—%. (26) Divide gig/5’1 by 2.1-. (27) .... .. 3a+46ab by 3a. 5xy‘ 2.2: (28) .... .. -2- by y . 2abc ac . . . - an W —-b—a . aim _5g (30) ' _ 2bc 4a: ' — l —2 Uby_+m O 2 3 3 a 1 b b 1 a o u u o - s - H a—§,-5, ag—ax b2 . . . . . . . .. b a2+ax +22 a—a: . . . - . . . 00 y m 0 . . 1 1 x (43) DIVIde g by 5— g . 1 1 ~ . . . .. m -- m . 62 (45) .... .. ab by a_x . aa-x’ . a—a: . . . - 00 m, a+x . 520 EASY EXERCISES. EXERCISES. K. Square each of the following quantities :— (1) 5aw. -3.ry2 :‘i (2) 52.29. (10) 2.22 ' (19) “2' (3) "7ab- 4‘ (20) mx+n. (A) 2266- <11) 5.2.3 2 a (21) 2mx—n. (5) -7abb. (12) 2+1. i ab 0 abx+c. (5) __ (1.5) ab+1. 3“ (14) 2+3. (23) airy-” (7) 5g. (15) 21y. (24) §ab+c. azb 2172—". 20 ' 2x_33/_ a2—2ab-2b’. “if” P (2'?) 221—221- (9) Fix—39" (18) iii—5' (28) x—%y+1. Cube each of the following quantities :— (29) 3.z:+2. Q_b (35) bx+cx’. (so) 2.2-3. 6 ' (36) I 2 I .1_ _.a — a+—, (31) 2‘” 3f (34) 2.-‘5+s.-6-. “ (32) 2ab+3c. b a (37) a’+a._.1. Find the 4th power of each of the following quantities:— (38) 2"” 2+ 2-17- ‘ (40) ax—b. Extract the square root of each of the following quantities :—- (41) 4a’b’. (45) 4a2b’ (49) 4a2+ b2—4ab. (42) 9.73294. 9222‘ 4 (50) 922+ 6.r+1. 1 7722.27 1 100agb4ce_ Z . W - 1) {02+ 23+ 1 . Qa’ac’3 47) 1+.r2—2x. 2 1 (44) 7117' (48) 422+42+r (52) “r + 234' Complete the squares in each of the following cases :-—- - (53) Jig—12.22. L 25: 2 5 (54) x,_ 1%. (59) a: 7 . . (62) a: — 6.2:. (55) 212112. 1 3 (56) 2222. (60) w’+—2-.r. (52) .2- If, (57) .19-x. 4x 1 7x (58) w’+ —5— . (61) w’— 5 x. (64) w’-- 1—6 . (EASY EXERCISES. 521 Extract the square root of each of the following quantities :— (65) a6— 4a5— 2 a.‘+1 2113+ 9a’. (66) 1 4 a_ _ _ .1: +2.22 x+4 (67) 4.1:“— 1 2x‘fy + 29w4y2— 30x3y3+25x’y‘. (68) 9a9b4—1 2a3b3+ 34a‘b’— 20a‘b + 25a“. (69) 92.624722222221924. 4- _9__Z i9 2 (70) 16 6“+81“° (71) z a: —2.x‘_;r"+ 53;". Extract the cube root of (75) 8—36a+54a’—27a3. (76) (16+ 3a5— 5a3+ 3a —1 . 4x2—4x+1 2 ~——-. (7) 922+6x+1 9—12a'+4.r:’ 25+302+92*' (73) (24+ b4+ 2abs+ 2a‘b (74') 3 + asba (77) 1 25x3— 3 OOx’y + 240.1:3/‘L 64:92a EXERCISES. SK; Simplify the following expressions: (1) (2) (3) ' (4) (5) (6) 2a~/b' +3a_(/b‘+a~/b. 7i/Z+2§/Z—8,f/Zi. 572 +10,J2- 5J2. I Egg/5.272; J21+,,/54—~/9_6. 2JIS-a./§+2JBS. a 5 - I - (7).; \/5 +,/so-;(/20. (8) (9) (10) 5 '26 8,\/;—_6:+10,\/2-—5—2 131;. ERIK/5427132 aZ/ZH+2.3/625-4i/226. N/EQF-m +7551 267275-755'17m5. ~[42% 1425225 -,/ 8 1226. JW+W+JW. Um +i/81—5‘5. W+J§fi W+Jfi JW>~JW- \/4a3-8a2+ 4a + N/ZE 3.2:3 3.1:3 ' .(4/ 32a +,:/l_6_2_d-1/51_2;. (11> (12> (12) (12> (15> <16) <17) <18) (19) (20) 522 EASY EXERCISES. (1) (2) -(3) (4) (5) (6) (7) (8) (9) (10) EXERCISES. L. Simplify the following expressions :_-- (11) 2J2X3J3X5J5- (12) Ja’+abx,~/a-“—ab. aflxbfixcfi- <12) <2./27+3./2>><<./5-./2>- Jlxf/ixflf/g- (14.) (2,/'1'2_.,1_./21)X(2Jf2+~/6). 332mm; - <15) (JETS/627811525). g- .ryx-Za'A/Z-x-g-EW. (16) (@JW*~/m)' fi/Exi/ZEZXf/m (17) J8+~/_35X~/8”~/3__9- afimi/Zwfi/E. (18) JaB-JZS-Tfxdaufliié. sf/Px2JIwf/Z. (19) (a+(/.E)><(b+.,/§). 2 a2bsxJbqb. (20) (W—Jm)xJm. . EXERCISES. M. Simplify each of the following expressions (1) (2) (3) (4) (5) (6) (7) J 2a.:vy -:-/ 6.. / 23b +2.11%}. 1 Zak/1 25303+3aJlIZ 143 92+ 2715. Jail—fl. a +3 Z13. (8) _ (9) (10) mf/LZE/EE. (11) JuaflbfiiflZ‘. (12) a\/mn+mx___m\/ap—ax n . ————-p . 3.... am. WI (72222” Sby8 ' EXERCISES. N.“ Simplify the following expressions :—-- (1) (2) (3) (4) (5) (335)”- (. /2a."’bc2 3. (JEi/Wfi (Z/rbb)‘. (.51)- (7) i/Zfii (8) Jai/SIZF. L (9) 351/531 <6) 6711282 >2- (10) Z/(ax-I- b)". ' w '(11) ’fi/(252-122+9)5. (12) Ua3nb2ncn. (13) (32275)". <12 (.7 1472)“- (15> UTE-17?)”- EASY EXERCISES. 523 EXERCISES. O. Find expressions equivalent to each of the following with rational denominators :- (1) (2) (3) 2ab (4) (5) a m. .22... 4755' 2% 377' a 353' 2 ‘6) (9) (10) 1 11 ~——-—-. < ) 2 a+3,,/._r- (12) . EXERCISES. P. 3—./§ W' 1 :/§+:/Z' 1 EDS-2' (12> (12> <15) <16) (17) I JY-I-I+,,/2::I' 1 Ca) 1 a_~/a2_w2 ' Extract the square root of each of the following quantities :— (1) (Q) 2 (3) (4) (5) (6) (7) (8) (9) (1°) 7+QJIC. 11+6J2. 30~10J5. 27+20J5. 4é—gfi. 28+5JI2. 3\/1. 4. 2' (11) (12) (13) (14) (15) (15) (17) (18) (19) (20) ..;-27.111 4x+2~/-4x7:1_. 2a—2 2ax—a’. q+b-c-2 ab—ac. 4a—y—2Jm, (Hm. aa—2am. 1+JI35. él—(a +b)+./(a+2)('5-' 2" )1 2+2(1-.2),,/I+2.2-.2’. 524 'EASY EXERGISES. EXERCISES. Q. Find the value of a: in each of the following equations :— (1) (2) (3) (4O (5) 6x—10=5x—4. 13x+1=9x+ 5. 3x+30=2x+36. 4x—2m=24—x. 7x—11+5=8x—9. 15—2w+6=3x+1. 3x—6=12—4x—4. 4r +_‘= I a: 2 6 a? 3m+g=4x—6. Q! a 011%: WIS l w P :18 wlii | wlH QOIND ' ll be RBI»— 6x+2(11-w)=8(19-x). 3(x+1)+2(x+2)=32. 3x-2 (5x+4) =2 (4x— 9). 5(2x—2)-3 (2x+1)=27. 6(3-2w)=24—4(4x-5). 45—4 (w-2)=5(x+ 2). (8) (9) (10) (11) (12) (13) (14) (26) (27) 028) (29) (30) (31) (32) (33) (340 (35) (36) (43) (44) (45) -————— (45) 12- 8x=15-3.r—8. 121=14x +1-3JS+10- 500=80x+12+32m— 8. 7x—2x+5=13x—4x—1 5. 12x—6x+4x=3x+84. 2x+-§—=3x— 15x—3%=3~é-+a:. &Elz 19 3 16 15 20 II to 3 l “1 1 ‘ 1 ' =41%. EASY EXERCISES. 525 6 5 5_—+1=;+ . 1 .. 1 1 _ (49) -— + —=13. 50—-—=, () +x41 b 2 2 (51) E+Zr=a +6 . (52) (1) (2) (3) (4) (5) (6) (7) (8) (9) (53) (54) (55) (56) (57) gar—16 _12— 4x x-A- 36 "4-5.r+ 4 7x+16 x+8 a: 21 _4x—11=_3_' x_—Z+ 1 __2a:—~15 x+7 2(a:+7)_ 2x—6 ' a 2 5 Z_Z¥I=I(E¥I—)' 17 10 1 6.v+17_ raw-10" I—Qw' EXERCISES. R. Find the value of a: in each of the following equations :— ~/12+w=2+~/;. —— 2 fl+~/4.~+x=:/—i. 2+J§é=¢5x+4. J;+~/9+$=:/—;l5_:—; . ./8+w-fl=2 1+.x. J2w-45=3JG—J2T1:T J2x—27a=9~/——,\/_2_x. Jx+4la+4Ib=2 b+x—,/§:. 28+J5_9+SJ5 28-fl_9+2J5' .ls (10) (11) (12) (13) (14) (15) (16) ~/Izzz:+l+,~/n_.‘r__,\/;+l Jnx+1—-~/;z; Jfi—l . JQTT+JS-2x_§ ,J_2_x—~/3-2x_ 2 . 13-2J.?l_5 _3_ ~— - 13+2Jw-5 25' JEE-JTI;=2./x-2. Jx2+2x +~/a:2- 2a: =2J5. A/4.'+.r=:/a¢”+20.12:+9. U2+%x=%:/16w”+ 8x+320. EXERCISES. S. Find the values of x and y in the following equations :— (3) 5x+ y =32, (I) x+y=17,} 2x-y=19. (2) 4w—7‘2/=26,} 4x+5y=50. 3x—2y=14~.l (a) 3x- 79:2,} 11y-3x=2. 526 EASY EXERCISES. (21) (22) (23) (24) (25) _ (13) (5) 3x+4y=11, 150:— 2y=11 } (6) 13x—6y=31,} 11w-3y=47. 7x—6y=10,} 6x—7y= 3- 35w+2y=76,} lfly— a:=34. (9)' 5x+2y=16,} 9y+2zr=31. Ila—79:72,} 7x—113/= 0- 36x—45y= 0,} 2x+ 5y=1-%. Qx+ 5y=65,} 7w—2%.y=25. (7) (8) <10) <11) (12) 3(4w- 5y)=2(ar+y) + 3,} 4(3w-2y)=5(x—y)+1 1. 6y—2x_ 4 __ 1.. 3x+3-36,} 8. 3w—2y _3_ Qw-y 2 ' 4 ’ 5x—4y_3_4x—3y 2 ‘T' flee—3+ _ 2 y—7,} (26) (27) (28) (29) (30) (14> <15) <16) (17) (18) 09) (20) me wife H ,.. 1r colt" H. N) + + H 00 ~1|<§ H l“ II II PP to 553 .P‘ W 1537-y=14!3:} 35y+x=255. 11x—13y=16,} 20x-19y=4I3- 45w+ 8y=350,} 21y-13w=132- 101w—24y=63,} 103x—28y=29. 64w+ 90y=237,} 3%.x—4%.y=12,} 7x+ 9y=60- 2%.x+ 3-y= 415:} 4%.x+10y=216- 4;}..r— 3%.y=6,} gw-J— =2. 10|1Q II C» 0 gm \-—-v_l I“: H to 00 $0” +y)=§(2w+4),} %(~'v—y) =%(w ~24“)- l 1 -2-(3.r-5y) + 3 = 3 (2a:+y),} 1 1 1 8 Z (m—2y)=§x+§y- %(3w—73/)=%(2==+y+1), 8-%(~'v—y)=6- } EASY EXERCISES. i 527 x—Q lO-a:__z/-10 (31) T— 3 4 , (36) 2y+4~_2x+y_x+l3 a 8 _ 4 ' (32) 22,i+6=2-”2-“+4%, (37) 3x+y ._ar+3'z/+l3 "T “"76"— 2 Z_ (38) (33) 3+4_4, 7 3 -23-11=_-+y. 1 1 (39) (34.) Em+-7—y=20,} 7y+4x=584t 1 1 (40) (35) §¢*§3/=1, } 6(x+y)~3(x—y)=39. EXERCISES. T. Find the values of x, y, 2', in the following equations: w+z=8, y+z=1a (2) Qx+y=ga } (I) w+y=6, } 2y+m=15, 22+y=27. 4x+22=26, (a) 3x+5y=34,} 5y+z'= 32. (4) 9x+10y=180, 10y+112=178,} 112+12y=196. (5) y+g=4d, l (6) Z Z y+-5-34. x+y+z=15, x+z—y=5, .r+y—z=—3. 528 EASY EXERCISES. (7) (10) %(x+,9)+22=21,1 9-9-7‘”=110' ax_%(y+z)=65. (8) ;-x+y+-;-z=162, x+%(x+y—2)=38.J ix+éy=2fij (11) ggfi=€r+5a 1 5y=4z. %y+%z=3, (9) $+éy+-;Z=IO, ém+éy+éz=l7j %(w+z)+y=9, (12) 3x+2y+-i—=14, 3;(w—z)+7=2y-l 3(w+y)+g=18, 2x=3y.l EXERCISES. U. (1) What number is that which added to its half makes 24? (2) What number is that which increased by two-thirds of itself becomes 20? (3) What number is that of which the half exceeds the third part by 3? (4) What number is that of which the fourth part exceeds the fifth part by S? (5) There is a certain number which, upon being diminished by 6, and the remainder multi lied b 6, roduces the same result as if it were . . P y P . diminished by 4, and the remainder multiplied by 4. What IS the num- ber? (6) Divide 40 into two such parts, that one-tenth of the smaller part taken from one-fifth of the greater will leave 5 for a remainder. (7) Divide 25 into two such parts that one shall be three-fourths. of the other. (8) Find two numbers which produce the same result, 7. whether the one be subtracted from the other, or the latter be divided by the former. (9) Divide £1 among 4 children so that the oldest shall have 1.9. more than the second, the second ls. more than the third, and the third 1.9. more than the youngest. (10) Divide a line '38 feet long into 4 parts, the second of which is 1%- feet greater than the first, the third 2%- feet greater than the second, and the fourth 3% feet greater than the third. EASY EXERCISES. 529 (11) A banker was asked to pay £10 in sovereigns, and half-crowns, and so that the number of the latter should be exactly twice that of the former. How must he do it? (12) Thirteen shillings is the sum of exactly the same number of shil- lings, sixpences, pence, and half-pence. What is the number? (13) I went to the bank with a cheque for 6 guineas, and asked to have for it exactly the same number of sovereigns, half-sovereigns, shillings, and Sixpences. The banker was puzzled: what is the number? (14) I have exactly 5 times as many shillings as half-crowns; and altogether my money amounts to £3. How many have I of each coin? (15) A father is 4.1 times as old as his son; but 3 years ago he was 7 times as old as his son. What is the age of each? (16) The ages of two brothers, who differ only by a single year, when added together amount to the age of their father; and if the father’s age be increased by one-fourth of that of the elder brother, it will amount to fourscore years. What is the age of each? (17) The ages of a man and his wife together amount to 80 years, and 20 years ago the woman was exactly two-thirds the age of the man. What is the age of each? (18) There is a certain fraction whose denominator is greater than its numerator by 1; and if 1 be taken from the numerator and added to the denominator, the fraction becomes equal to What is the fraction? (19) A certain fraction has its numerator less than its denominator by 2, and if 1 be taken from the numerator, and the numerator be added to the denominator to form a new denominator, the resulting fraction is equal to I. What is the fraction? (20) A boy being asked to divide one half of a certain number by 11, and the other half by 6, and to add together the quotients, attempted to obtain the required result at one step by dividing the whole number by 5; but his answer was too small by 2. What was the number? (21) Find the time between 12 and 1 o’clock when the hour and minute hands of a clock point exactly in opposite directions. (22) A person, being asked what o’clock it was, answered that it was between 1 and 2, and that the hour and minute hands were together. Required the time of day. (23) A servant is dispatched on an errand to a town 8 miles off, and Walks at the rate of 4. miles an hour: ten minutes afterwards another is sent to fetch him back, walking 4% miles per hour. How far from the town will the latter overtake the former? (24) A student has just an hour and a half for exercise. He starts off on a coach which travels 10 miles an hour, and after a time he dismounts, and walks home at the rate of 4 miles an hour. What is the greatest distance he can travel by the coach, so as to keep within his time? (25) A cistern which holds 820 gallons is filled in twenty minutes by 8 pipes, one of which conveys 10 gallons more, and another 5 gallons less, per minute, than the third. How much flows through each pipe per minute? 34 530 EASY EXERCISES. (26) A man and a boy engaged to draw a field of turnips for 21s. but when two-fifths of the work was done, the boy ran away, and the man then finished it alone. The consequence was that the work occupied 1%; days more than it should have done. Now the boy could do only half a man’s work, and is paid in proportion. What did each receive per day? (27) The date of the accession of GEO. III. is represented by 1800—21, that of GEO. IV. by 1800+%.2x, that of WILL. IV. by 1800+-12~-3$; and if GEo. III.rd’s reign be increased by 2.21, it will amount to 100 years. What are the actual dates? (28) Her Majesty QUEEN VICT0RI_A_lvas born May 24, AJ). .22, and Prince Albert was born Aug. 26, A.D. a:+1. Now their united ages on the 26th of Aug. 1848 amounted to three times the age of Prince Albert on the birth-day immediately preceding his marriage, which took place Feb. 10, 1840. What is the year of our Lord in which each was born? (29) The interest of the National Debt being reckoned at 30 millions sterling per annum, and 3 per cent. the average rate of interest paid, what reduction in the rate of interest would give the same relief to tax- ation as the paying off 200 millions of debt, and allowing the interest to be paid on the remainder to continue the same? (30) Says Charles to William, if you give me 10 of your marbles, I shall then have just twice as many as you: but says William to Charles, If you give me _10 of yours, I shall then have three times as many as you. How many had each? (31) A man, who has two purses containing money, receives £10 to add to them, and finds that if he puts £5 into each, one will then contain exactly twice as much as the other, but if he puts the whole £10 into that which already contains the most, its contents will be just three times the value of the other. How much was there in each purse to begin with? (32) A party consists of men and women, and there are 6 men to every 5 women; but if there had been 2 men less and 2 women more, the number of each would have been the same. How many are there? (33) A clergyman, who had a dole of £5. 10s. to distribute amongst a certain number of old men and widows, found that, if he gave them 3s. each, he would be Is. out of pocket; but, if he gave each of the men 2s. 2d. and each of the widows 3s. 6d., he would have 6d. to spare. How many were there of each? (34) There is a certain fraction which becomes equal to %, when both numerator and denominator are diminished by 1 ; but, if 2 be taken from the numerator and added to the denominator, it becomes equal to l. What is the fraction ? (35) What is_ the fraction in which twice the sum of the numerator and denominator 1s equal to three times their difference? (36) Find two numbers such that one shall be as much above 10, as the other is below it, and one-tenth of their sum equal to one-fourth of their difference. EASY EXERCISES. 531 (37) Find two numbers such that the half of one added to a third of the other is 12, but a third of the former added to half the other is 13. (38) i A person has two casks with a certain quantity of wine in each. He draws out of the first into the second as much as there was in the second to begin with: then he draws out of the second into the first as much as was left in the first: and then again out of the first into the second as much as was left in the second. There are then exactly 8 gallons in each cask. How much was there in each at first? (39) In the course of last century the change took place, called ‘the change of Style’, which consisted in beginning the year with Jan. 1, instead of March 25, as heretofore, and for that year only, calling the day after Sept. 2, the 1451‘, instead of the 3’“. Now the year of our Lord in which this happened, possesses the following propertiesz—The first digit being 1 for thousands, the second is the sum of the third and fourth, the third is the third part of the sum of all four, and the fourth is the fourth part of the sum of the first two. Determine the year. . (40) Iron, worth £10 in its raw state, is manufactured half into knife- blades, and half into razors, and is then worth £444. But if one-third of it had been made into razors and the rest into knife-blades, the produce would have been worth £30 more than in the former case. How much is the value of the original material increased by these respective manu- factures? EXERCISES. V. Find the values of a: in each of the following equations:— 8x2 $2 x2 x2 2_ = o '—_' _ '1'- 2 (2) (.r+l)2=2.r+17. (s) l3§-g—=2.r2--8-§~. (3) (.r+2)”=4w+5. (9) 3 3 =8 1+a: l—a: ' (A) (2x—5)2=x2~20x+7_3. 10) 1 a _ 5 3x2_ , 2w,_ 5 ( .212 3502+] _ 4(3w”+1) ' m2———-5-——=3-——3—. (11) 14w’+16_ 2x’+8 __2_af 21 awe-11" a ' 25094-10 50+.r8 3 g 1 6 = - . __)=_ ( ) 15 7 25 (12) < 4 4" (13) x"’=3x+10. (19) 7a.’-a:"‘==(i. (14) x2=5x—4. (20) 'a:=ar’-30. (15) x”—9.r=:c—16. g a; (16) xQ-14x=120. (21) w+§=3' (17) l2ar—20=.r2. (18) 4w-ac’=4. (22) at? =27. 34-.2 532 EASY EXERCISES. (23) ea (25) (26) (27) (28) (29) (30) (31) ea (33) (34) (35) (36) (37) (38) (39) x’+9-2"5=6a (4.0) 9x—5w’=2%. 4:1 7w+3xz=6. ( ) $53221. (22) (4'3) trims. 3 (a) llxg-Qm=11;],-. 322—5a:+2=0. (45) 1 1 —ar’——a:—22=O. 2 3 3 (4,6) 1 2 _1_ 3.. §$~3w+7§—8. 3 2 Zme-gamlg. (4J7) 5(22+1)—3(a:-1 )=22. .r’—4=16—-(a:—2)9. (48) 3(x—2)2—3=8(a:+2). 2 2_ _1 __ 29 4,(ac 3)-8(.r ( ) 3(2—w)+2(3—w)=2(4+3x”). (50) 5222+ (x+1)2= +1). (51) w 7 w+60 " sup—5' (52) as _165 w+3 _w+10' 4x+4Jx+2=7. ,/x+s+./x+6=e,/5:. 3i 3§§:~fl_ 5—a: a: _ 396—7 4(x—2l) W __—5_' 15. 7.3% 3.2—5 +135_ 3x+5 3x+5 176"ar-5' 3.2:+2 + 3.11—2 _ ] 5x+11 32-2 3x+2 _ 3.1342 ‘ %(-r—1>(w—Q)=2%(w-2%). 2a:+3__ 2x 6, 10—.r_25—-3a: 2' 2+8 5 3x+14~ .¢.~+12Jr 2+2: 3x+8' EXERCISES. W. Find the values of x in each of the following equations :— (1)* 3.1:”+2.'v= fizz—92+ 2%=0. Tag—112:6. (a—b)w"’— (a+h).r+ 25:0. w+6~/.ii=27. w‘—6a:”= 27. (2) (3) (4) (5) (6) (7) 85. - 2 2J$+fi=5' * This and the three following by the Hindoo method. EASY EXERCISES. 533 (8) w+4+,/.?f4=12. (9) file/FE. (10) $45,112:]. (11) 22+J2T1743=2 (12) 22+6Jm=11+2m (13) 222+'sx-5,/2_wm+s=0. (14.) 92-422+,/W1=5. (15) sx(s-x)=11-4,/5*~—35T5. EXERCISES. X. (1) Find the two consecutive numbers whose product is 156. (2) Find the three consecutive numbers whose sum is equal to the product of the first two. (3) Divide 20 into two such parts, that one is the square of the other. (4) Divide 210 into two such parts, that one is the square of the other. (5) Divide 25 into two such parts, that the sum of their squares shall be 313. (6) Divide 30 into two such parts, that the diference of their squares shall be 300. . (7) The product of two numbers is 144, and if each number be increased by 2, their product will then be 200. What are the numbers? (8) Find the number whose square exceeds the number itself by 156. (9) Find the fraction which is greater than its square by 217. (10) The sum of £4. 10s. is equally divided among, a certain number of persons, and each receives as many half-crowns as there are persons altogether. 'What is the number? (11) A person bought a lot of pigs for £4. 168. which he sold again at 13s. 6d. per head, and gained by the whole as much as one pig cost him. What number did he buy? (12) A gardener, who had no knowledge of Arithmetic, undertook to plant a certain number of trees at equal distances apart, and in the form of a square. In the first attempt, when he had finished his square, he had 11 trees to spare. He then added one of these to each row, as far as they would go, and found that he wanted 24 trees more to complete his square. How many trees were there? (13) A printer, reckoning the cost of printing a book at so much per page, made the whole book come to £16. It turned out however that the book contained 5 pages more than he reckoned, and an abatement also was made of 2 shillings per page. He received £13. 10s. How many pages did the book contain ? Q 534 EASY EXERCISES. (14) There are 4 consecutive numbers, of which if the first two be taken for the digits of a number, that number is the product of the other two. What are the 4 numbers? (15) Two trains start at the same time to perform a journey of 156 miles, but one travels a mile an hour faster than the other and reaches the end of its journey just one hour before the other ; at what rate did each train travel? (16) A student travelled on a coach 6 miles into the country, and walked back at a rate 5 miles less per hour than that of the coach. He found that he was 50 minutes more in returning than going. What was the speed of the coach? (17) A person distributed £5 in equal portions among a certain num- ber of poor men; and another person did the same, but by giving each man a shilling less, relieved 5 more. What was the number of recipients in each case ? (18) A person distributed £36 in equal portions among the poor of a certain place. The next year the same amount was distributed, but the number 'of recipients was diminished by 6, and consequently each received 1s. 8d. more than in the year before. What was the number of recipients in each year? (19) Two travellers A and B start at the same time from two places distant 180 miles to meet each other. A travelled 6 miles per day more than B, and B travelled as many miles per day as was equal to twice the number of days before they met. How many miles did each travel per day ? (20) Twenty persons contribute to send a donation of £2. 8s. to the Society for Promoting Christian Knowledge, one half of the whole being furnished in equal portions by the women, and the other half by the men ; .but each man gave a shilling more than each woman. How many were there of each sex, and what did each person contribute? (21) A person, who can walk forwards four times as fast as he can walk backwards, undertakes to walk a certain distance (one-fourth of it backwards) in a stated time. He finds that, if his speed per hour back- wards were one-fifth of a mile less, he must walk forwards 2 miles an hour faster, to gain his object. What is his speed? EXERCISES. Y. Find the integral values of a: in the following ‘Inequalitz'es’:-— (1) 2x-5>31, and 3x—7<2x+13. (2) x+720. (3) 7a:-15>4m+30, and %..r—%w<3. (4) 2-,lg.x+%.r<8, and 3%..r—%.r>5. EASY EXERCISES. 535 Solve the following ‘ Inequalities’ :— .(5) 1(6) (7) (8) (9) (6:27. @182 (10> (11> (12) as (14) a w. 5 5 What-'6' 8 6 7—5x<9—3x. 3w+2>2x+1 2 3 3 2' a a If E z If a and a: be both positive, shew that a3+23>aa”+a”a, unless Q or “ii? 3/3 U5 Shew that 2.2-3> or or <1. If a>b, and both positive, shew that a3- 53< 3a’ (a— b), and > 362 (a —6). Which is greater or J3+J5P mx+ny any-I- 12a: be any fraction whatever, shew that + g>2. Which is greater Shew that > the least and < the greatest of the fractions EXERCISES. Z. “Find the value, or measure, of each of the following Ratios : (1) 3a : 15a. (7) agpc : 3aca. (2) 2x : 10x2. (8) 3233/2 : 1222f. (3) aa: : bar. (9) ac+hc : 02. (4) abc : be. (10) 2aar+at2 : ma. (5) my : 2.2:. (11) 1-w2: l—a'. (6) 3ahx: 2a2a:. (12) a2-62 : a+h. Simplify each of the following Ratios : (13) 5am : 4.2. (17) 7M1? , 5222 (14) 162.19 : 20.222. 1X2><3 2X3><4 ' (15) gar : 36a. 7, n__1 (16) 2,02” , Ewe (18) —(i22—)a23 : nagwe. (19) Which is the greater 15 : 16, or 16: I7 ? (20) Which is the greater 2am : 363/, or 3a : 26, when a: : 3/ :: 2 : 1? (2.1) What is the ratio compounded of 2a: b, and be: : a? 536 EASY EXERCISES. (22) Compound the ratios a+a ': a, and a:2 : a2— a”. (23) Compound the ratios a : 1, 2a : 1, and 3a: 1. (24) Divide 27 into two parts in the ratio of 7 : 2. (25) Divide 20 into 3 parts such that the ratio of the first two shall be 2 : 5, and that of the last two 5 : 3. (26) Find two numbers in the ratio of 1%- : 2%, and such that, when each number is increased by 15, they shall be in the ratio of 1%- : 2%. (27) The numbers of boys in the three classes of a school were as the numbers 5, 7, 8. At the next inspection the first class was found increased by 4 boys, the 2nd had gained two-sevenths of its former number, the 3rd was doubled—and the whole number of additional scholars was 34. What were the numbers in the classes at the 1st inspection? EXERCISES. Za. (1) Find a 4“1 proportional to a, 2a, and 3a. (2) Find a 4L11 proportional to 1-21, 2%, and 3%. (3) Find a 4‘h proportional to a—a, ag—are, and a-Fa'. (4) If a : b :: c: d, shew that 5a : 66 :: 5c : 6d. (5) If a: b :: c : d, shew that T;‘3,-a 1%11 :: g-c : féicl. (6) If 2a : b :: h : 2e, shew that a : c :: 4a2 : he. (7) Convert the proportion a : a+x :: a-x : 6 into an equation. (8) Convert a: : y :: y : 2a—x into an equation. (9) If a+a : a—a :: 11: 7, find the value of a :a. (10) Find two numbers in the ratio 2 : 3, and the sum of which : their product :: 5 : 12. (11) The 1st, 3rd, and 4th, terms of a proportion are am, 30.22, and 6i?” , what is the second term ? (12) There are two numbers in the ratio 3 : 4, and if each of them be increased by 5, the resulting numbers are in the ratio 4 : 5. What are the numbers ? (13) If a: 6:: b : c, and h : c :2 c : cl, shew that a+b : h+c :: b+c : 0+d. (14) If 6a—a : 403—7) :: 3.r+h : 2a:+a, find a. (15) Shew that a : h is double of the ratio a+c : 6+0, if c be a mean proportional between a and 6. (16) Find the ratio of the value of a gold coin to a silver one, when 13 gold coins together with 12 silver ones are worth 3 times as much as 3 gold and 40 silver. EASY EXERCISES. 537 (17) If a : b :: c : cl, shew that (a+b)2 : ab :: (0-1-0?)2 : ed. (18) Find two numbers, the greater of which: the less :: their sum : 42, and :: their difference : 6. (19) Divide the number a into two parts so that one shall be to the other in the ratio n : 1. (20) There are tWO vessels, A, B, each containing a mixture of water and wine, the wine : water in A :: 2 : 3, and in B t: 3 : 7. What proportion of each must be taken to form a third mixture in which the water: wine:: 5 :11? EXERCISES. Z6. (1) Given that your, and when 2:2, y=20, state the resulting pro- portion. (2) If year, and when 2:2, y=4a, find the equation between a and y. b (3) If ylocinversely as a, and when 2:4, y==8, find the equation etween .2: am y. (4) If l+xoc1~—a:, shew that 1+.reocm. (5) If Qa+3yoc4a+5y, shew that aocy. (6) If x2ocy3, and .r=2, when y=3, find the equation between a: and y. (7) If y=the sum of 3 quantities, of which the 1st ocacz, the 2nd com, and the 3rd is constant; and when .r=1, 2, 3, 31:6, 11, 18, respect- ively, find the equation between as and y. (8) If y=the sum of 3 quantities, of which the 1st is constant, the 2nd mm, and the 3l'docx2. Also when 2:3, 5, 7, y=0, —12, ~32, respect- ively. Find the equation between a and y. (9) Given that the solid content of a globe varies as the cube of its diameter, what is the diameter of a globe formed by melting down two other globes whose diameters are 6 in. and 7 in.? (10) What ratio does the solid content of a globe Whose diameter is 4in. bear to that of a globe whose diameter is 8 in.? (11) Given that the illumination from a source of light varies in— versely as the square of the distance, how much farther from a candle must a book, which, is now 8 inches off, be removed, so as to receive just half as much light? (12) Given that the content of a cylinder varies as its height and the square of its diameter jointly, compare the contents of two cylinders, one of which is twice as high as the other, but with only half its diameter. 538 EASY EXERCISES. EXERCISES. Z0. Find the 15th, and the 20th, terms in each of the following series :— 1, 6, 11, &c- (4) ma, 2mm, 3mm, &c. a , i (6) 9 5‘5 53 &c (a) 4.3.1.810- 2’2’2’ ' Find the sum of 20 terms of each of the following series :— (7) 1, 6, 11, &C. (14) max, 2mm, 3mm, &c. (8) 5, 8, 11, 14, &c. (15) 3.1:, 5.1:, 7.2:, &c. (9) 100, 110, 120, &c. 1 2 3 & — — — c. (10) 100, 97, 94, &c. (16) a’ a’ a’ (11) 15, 11, 7, &c. (17) 25a, 24a, 23a, &c. (18) '13, 3;. 1, &c. (18) 10 ’ 10 ’ 10 ’ &c‘ (19) Find the Arz'th. Mean between i and (20) Find the Arz'th. Mean between 1+2, and l—w. (21) Find the Arith. Mean between ‘5 and (22) Insert 2 Arith. Means between 5 and 14. (23) Insert 3 Arith. Means between 1 and 3. (24) Insert 4 Arith. Means between 100 and 80. 25 There is a series of terms in Arith. Prog. of which the sum of the first two terms is 2%, and the 4th term is 2i. What is the series? (26) The first and last of 40 numbers in Arith. Prog. are 113;, and 1%- ; what are the intervening terms? And what is the sum of the whole series ? \ (27) An insolvent tradesman agreed to pay a certain debt by weekly instalments, beginning with 5s. and increasing by 3s. every week. His last payment was £15. 2s. For how many weeks did he pay, and what was the whole amount of his debt? (28) Find the series in A.P. in which the sum of first 5 terms is one- fourth of the sum of the next 5 terms, the first term being 1. (29) How many terms of the series 1, 3, 5, 7, &c. amount to 1234321? (30) The sum of n terms of a series in A. P. is 3n+5n’, find the 6th term. EASY EXERCISES. 539 ~ . EXERCISES. Z01. Find the ‘ Common Ratio’ in each of the following series in Geom. Pr0g.:—- 1 100 200 400 &c. ' () 9 ) s g, 2%, g, &c. (2) 2%, 5, 10, &c. 9 (6) 01, 0'01, O'OOl, &c. (8) 5. 1. a. &c. (7) 125, 2-5, 5, &c. 3 (8) aw, 2a2a, 4asx, &c. 1 1 1 a: na: nga 5) Z) 53 &c. 2:, F, "TT" 856. . . 1 (10) The first two terms of a series In Geom. Prog. are —3-, and 1, what are the next two terms? (11) The first two terms of a series in Geom. Prog. are 125, and 25, what are the 6th and 7th terms? (12) Find the sum of 5 terms of a series in Geom. Prog. of which the 1st term is%, and the 5th is 9. (13) Find the sum of 4 terms of a series in Geom. Prog. of which the 1st term is g; , and the 4th is 2. (14) Find the Geom. Mean between 30, and 7%. 3 (15) Find the Geom. Mean between 21,, and 2,!- . (16) Insert two Geom. Means between 5, and 320. 1 (17) Insert two Geom. Means between 1, and g. (18) Insert three Geom. Means between 6, and 486. (19) Insert three Geom. Means between 100, and 2%. (20) Which is greater the Arith. Mean, or the Geom. Mean, between 1 and g? and how much greater? (21) Are 1;, w, my in Geom. Frog? and if so, what is the ‘Common Ratio’? (22) A series of terms are in Geom. Prog.; the sum of the first two is 1%, and the sum of the next two is 12. Find the series. 23) A farmer sowed a peek of wheat, and used the whole produce for seed the following year, the produce of this 2nd year again for seed the 3rd year, and the produce of this again for the 4th year. He then sells his stock after harvest, and finds that he has 12656;}r bushels to dis- 540 EASY EXERCISES. pose of. Supposing the increase to have been always in the same propor- tion to the seed sown, what was the annual increase? (24) If a servant agrees with his master to receive for his wages, a farthing for the 1st month, a penny for the 2nd, foul-pence for the 3rd, and so on ; what will twelve months’ wages amount to? EXERCISES. Ze. (1) The first 3 terms of a series in H.P. are 3, 4, 6; what are the next 2 terms? (2) The first 2 terms of a series in H.P. are a, and 6, find the next 2 terms. . . . . . 1 (3) Two terms In a series, which 1s m H.P., are 5, and 2,:- , what are the 2 terms immediately preceding? . . 4 (4) Continue as far as 3 terms more the serles 2, g, 1. (5) If a, I), c be in H.P., shew that a : c :: 2a—b : b. (6) Shew that a, h, e are in A.P., G.P., or H.P., according as a-b a a ora h—c_a’ b’ c' (7) Find the Harm". Mean between 2 and 6. (8) Find the Harm". Mean between 7?- and (9) Insert two Harm". Means between 1 and 2. (10) Insert two Harm". Means between 2% and 6. I (11) Insert three Harm". Means between 1 and 3 3 (12) Insert three Harm". Means between I and T6. 13) The Arith". and Harm". Means between two numbers are 2, and 1%, respectively. What are the numbers? EXERCISES. Zf. (1) In how many different ways may a class of 9 boys stand up to read? (2) With a peal of 6 bells what difference will be made, in the number of changes which can be rung, by the absence of one ringer? (3) If the number of permutations of n things 4 together-:12 times the number taken 2 together, find 12. EASY EXERCISES. 541 (4) If the number of permutations of n things 3 together=6 times the number of combinations of n things 4 together, find n. (5) The number of permutations of n things r together : number taken (r-l) together :: 10: 1; and number of combinations 2' together : number (r—l) together :: 5 : 3 ; find 12 and r. (6) Find the number of combinations of 100 things taken 98 together. (7) In how many ways can 4 men be taken out of 20? Out of a dozen friends how many different parties may be made consmting of no more than 8 ? . (9) The number of combinations of (n+1) things 4 together:9 times the number of combinations of n things 2 together; find 12. (10) Find the number of permutations of the letters in the word ‘ Susannah’ taken all together. EXERCISES. Zg. Expand the following powers of a binomial :— (1) (1+a')5. (4.) (a—b)6. (7) (1+2a")6. (10) (Muir. (2) (1+w)7- (5) (a:+2_y)4. (m+n.r)“. (11) (as—.23)". (3) (2%)". (6) (ii-ear. (9) (a2+b")". (12) (x4+a'2y2)5. (13) Find the 6th term (independently of the rest) in the expansion Of' (a+2a)7. (14) Find the 10m term in (2a +b)“. (15) Find the 8th term in (a-a)9. (16) Find the term containing a" in (a—ba)". (17) Find the term containing a“ in (a—a)"°. (18) Find the 4th term, and the 98““, in (a—b)‘°°. (19) Find the middle term of (a—b)‘“. (20) Find the middle term of (2a2—b%)‘°. Expand the following fractional powers of a binomial: (21) (a+a')%. ' (2a) (1+2s)-%. (25) (aa+by)ii. (22) (a—x)2. (24.) (1—3x)%. (26) (2a2—3b’)%. Expand the following negative powers of a binomial :- (27) (a+a)“. (29) (l+2.r)"". (31) (a"- 2)“? (28) (a—.r)"5. i (30) (1-3x “7. i (32) (a’-~'v5)'%- (33) Find the 5‘h term of (ax—byyd". 542 EASY EXERCISES. (34) Firid the a“ term of (than, (35) Find the 4th term of (ax-wgya, (36) Expand (1+ 2r+3x”)‘. (39) Expand (1+ 2a— 3.222)". (37) Expand (a — b — c)". (40) Expand (I-x- :02)“. (38) Expand (1—.\/§+.r)3. ~ (41) Expand (1- a: —.'v2)‘2. EXERCISES. Zh. Extract the square root of each of the following surds:-- (1) 19+8J5. (5) 3%4~/1_0. (9) 3J6+4,/2'. (2) 12-Jm. (6) 1144,77. (10) 8J3—6J5. (a) 2,/2i+22. (7.) 372-4... (11) 1%.W/2—1-2Jé. (4.) 10%-6,/2'. (s) 572-4.,5. (12) 1173—72—1. Extract the 4th root of each of the following :— (12) 49+20J3. (14) 51-2672. (15) 49+20,/6'. Extract the cube root of each of the following :— (16) as+17,/5. (18) 1o-6,/a“. (20) 26+15,/§. (17) 10,/i+22. (19) a-Iafo'. (21) 16+8J5. EXERCISES. Zi. Expand by the method of Indeterminate coefficients:— (1) (4) (7) (a) 55%.- (a 1532- (8) (3) (6) (9) Resolve the following into their partial fractions:— "2 "2? (14‘) (n) u-Qbiiay (13) ( 7 1+§x+3x2 (1+a)’(1+2.r)’ ' EASY EXERCISES. 543 ( 16) Resolve 82’—-22-3 into two factors of the first degree. (17) Resolve 102’- 62—28 into two factors of the first degree. (18) Resolve 522+ 92+3 into two factors of the first degree. 2 $3 IEJ” 1.2.3 .223 x5 "-'_ + ——-_— -...... 1.2.3. 1.2.3.4.5 (19) Revert the series y=2+ + .... .. (20) Revert the series 31:2- EXERCISES. Zk. Express in the form of continued fractions the following: 27 67 251 (1) T9 . {,5 . (5) ,76—4: . 47 365 907 (2) 22' (4’) my <6) 1T564' Find the convergents to the following fractions : 1051 251 182 743 Express in the form of continued fractions the following: (11) ,/1_6. (12) 726 (13) J56. (14) ,/1_0T. (15) [7. (16) 746. (17) 75. (18) J67. Find fractions the nearest to the value of the following fractions having only 2 digits in their denominators: 251 743 5065 13957 (19) (20) 6-1—1 . (21) 1———3891 . (22) “59476 . EXERCISES. ZZ. Solve in positive integers the following equations : (1) 92+ 7y=57. (4) 112+ 59:254. (7) 82—23y=19. (2) 52+213/=240. (5) 72+15y=225. (8) 392—56y=11. (3) 32+17y=121. (6) 32+ 4y=39. (9) 202—31y=7. (10) Find the positive integral values of 2 for which at the same time 2-1 2—2 2-3 . . . _ -—2- , -3— and 7 , are posmve Integeis. (11) Find the positive integral values of 2 for which at the same time 2-3 2—5 2-8 . . . T , T, and , are posrtive Integers. 544 EASY EXERCISES. (12) Find the least positive integral value of 2, which will make 2—1 2—3 22—5 d %(2-1) 14 ’ 5 ’ 9 a an 21 , positive integers. Solve in positive integers the following equations: (14) 2+_2/+2=100, 72 4y 2 E + -8_ + ,2- ._1 90. (15) Find one solution in positive integers of 42—183+272=100. (16) Divide 142 into two parts so that one shall be exactly divisible by 9, and the other by 14. (17) A boy has a certain quantity of nuts, which he knows to be between 200 and 300; he makes them into parcels of 13 each, and finds that he has 9 over: he then makes them into parcels of 17, and finds he has 14 over: how many nuts had he? (18) A person having a basket of oranges, between 60 and 70, takes them out in parcels of 4, and finds he has 1 over: he then takes them out in parcels of 3, and finds he has none over: how many had he? (19) A farmer bought a lot of cows 'and calves for £135 ; each cow cost £8, and each calf £3. How many were there of each? (13) 32+ 5y +7s=560, 92+25y+492= 2920.} (20) Thirty persons in an excursion spend 30 shillings altogether; each man spends 5s., each woman 1s., and each child 3d. ; how many men, women, and children, were there? (21) An officer of police finds that if he sends his men out 2, 4, 8, or 10, together, he has always 1 left; but if he sends them 6, or 12, together, he has 5 left. How many men had he? EXERCISES. Zm. Find the general solutions of the following equations : (l) 112+5g/=254. (3) 172—2Sy=19. (5) 1311-— 9y =17- (2) 32+7y==39. (4) 72-13y=152. (6) 272—19y=43. (7) 22+3y+ a=15, (8) w+y+z=30, } 102-49+32=10. 72+5%.y+4%.e=180. Find the number of positive integral solutions of (9) 112+5y=254. (11) 72+159=22a l (13) 132-99=17. (10) 32+43/=39. (12) 52+ 8y=42. (14) 22+7y=125. (15) In how many ways may £80 be paid with sovereigns and guineas? (16) In how many ways may £500 be paid in guineas and £5 notes? (17) _ Find the number of ways in which I can mix together 40 gal- lons of Wine, some at 15s., some at 19s., and some at 12s. per gallon, so as EASY EXERCISES. 545 to produce a mixture worth 16s. per gallon, an integral number of gallons of each sort being always taken. (18) How many fractions are there with denominators 3 and 4, whose ' P sum 18 3%. (19) How many fractions are there with denominators 12 and 18, whose sum is l? 36 Solve in positive integers the following equations : (20) 32y— 72=7y+5. (24) y(22+1)=322+1. (21) 52y=22+3y+18. (25) 32y+222=32+2y+5. (22) 2y~(2+y)=34. (26) 22+2y=22+3y+29. (23) 52y—32=24. (27) 2y+22+3y=42. EXERCISES. Zn. Transform the following numbers from the Denary to the Senary scale: (1) 182061, (2) 5002001, (3) 211115600. Transform the following from scale 5 to scale 7: (4.) 4321, (5) 1104.23, (6) 100261. (7 ) Transform 37704 from scale 9 to scale 8. (8) Transform 13256 from scale 7 to scale twelve. (9) Transform 1341120 from the senary to the duodenary scale. (10) Transform 654321 from the septenary to the duodenary scale. (11) Subtract 20404020 from 103050301 in the octenary scale. (12) Extract the square root of the result in the last Ex. (13) Divide 51117344 by 675 in the octenary scale. (14) Find the radix of the scale in which 40501 is equivalent to 5365 in the denary scale. (15) In what scale is the denary number 2704 written 20304 ? (16) Extract the square root of 1010001 in the binary scale, and re- duce the result to the denary scale. I, (17,) Apply the duodenary notation to find the square of 4ft. 2in. 0’. 2 . 10 . (18) Apply the duodenary 'notation'to find the cube of 16ft. 10in. , ‘ EXERCISES. Z0. Transform the following quantities from scale 10 to scale 5: (1) 221242. - (2) 357-234.. (a) 101-265. 35 546 EASY EXERCISES. " (4) Transform 17925 from the denary to the senary scale. (5) Transform 2332 from the denary to the duodenary scale. (6) Transform 2—2- from the denary to the duodenary scale. (7) Transform 21% from the denary to the octenary scale. (8) Transform 7304'513 from scale 8 to scale 4. (9) Transform 3t'97e from the duodenary to the octenary scale. (10) Transform 345'6273 from the octenary to the ternary scale. (11) Transform % from the denary to the duodenary scale. (12) A certain number is 125 in the scale whose radix is 2, 78 in the scale whose radix is y, and 49 in the scale whose radix is 2+9; find the number in the scale whose radix is 10. EXERCISES. Z3). (1) If p and q are any positive whole numbers, and p+q is even, shew that p-q is also even. (2) Shew that the difference between any number and its square is always an even number. (3) Shew that the difference between any number and its cube is always divisible by 6 without remainder. (4) Shew that the product of two odd numbers will always be odd. (5) Shew that the sum of any two consecutive odd numbers will always be divisible by 4. (6) Shew that the product of any two consecutive even numbers is divisible by 8. ('7) Shew that every odd square number, greater than 1, leaves a remainder 1, when divided by 8. (8) Shew that every perfect cube is of one of the forms 7m, or 711241. (9) Shew that upon any number, greater than 12, which is a perfect square, being divided by 12, the remainder, if there be any, is a square. (10) Shew that the difference of the squares of any two odd numbers is exactly divisible by 8. (11) Shew that the square of any number prime to 4 is of the form 4p+1. (12) Shew that the difference of the squares of any two prime num- bers, each of which is greater than 3, is divisible by 24. EXERCISES. ZQ. Find the value of each of the following fractions: 812—1 22+ 52+ 6 1 —~—-— h =1. -_--_-__ =_ ( ) 22+22—3 ’ W en w (2) 22+72+12 ’ When x 3' EASY EXERCISES. 22+22—35 81—a 223—-722+12 2+1— 2 (4“) a,e__7w_,_6 , When wsg' J x_1J , WllEl’l 1?=1 23-22"-2+2 ,/32+1—2 5 --—-——-——- when =2. _______ = , ( ) 23—72+6 ’ x (10) 2—1 ’ When x 1 I .2.~3-l-2(1.v2--a’2—2a3 ,4/52—1—,/2 6 . hen 2=a. __________ }, =1_ ( ) 23—13222+1223 ’ w (11) 2—1 ’ W em” 23— 522+ 32 + 9 O 25- a5 (7) m, when 2=o. (12) m,_a4, when 2=a. . 2+ 2+2 -4 (13) Find the value of Lily—— , when 2:], and y=1. 2+y—2 - - b any ay baa—Fa when 2=a. (14) Find the value of (1) (Q) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) 2y—ay—c2-I-ac ’ EXERCISES. Zr. Given loga find the log. of 3 20 Given log 2:0-30102, and log3=0'47712, find log (5) . 10 Given log 2, and log 3, as in last example, find log . ,/;;. 3 a 47?? ' Given log 2, and log 3, as in Ex. (2), find log 12. Given log n, find the log. of Given log 2, find the log. of Given log 2, find the log. of 64. If amfb"”=c, find 2. If a"5=b, find 2. If 1757-15, find 2. If log 2=%log a—%log b, find 2. If %log2=nloga+mlogb—plogc, find 2. Given log2=0'30103, find the log. of 162°. Given log 2+log,y=g, and log 2—logy=%, find 2 and 3}. Given log2-logy=log n, and ax+by=c, find 2 and y. If logwa=3log,oa-2, find 2. If log 2+logy=loga, and 210g 2—2logy=log 6, find 2 and y. 35-2 548 ANSWERS TO THE EXEEcISES. EXERCISES. Zs. [N.B. log 1'05=0'02119, and log 1'04=0'01703.] (1) What would £200 amount to in 7 years at 4 per cent., compound interest? (2) How much money must be invested at compound interest to amount to £500 in 12 years at 5 per cent.? (3) In how many years will a sum of money double itself placed out at 4 per cent. compound interest? (4) A freehold estate which produces a clear rental of £100 a year is sold for £2500; at what rate is interest reckoned? (5) Find the amount of £100 in 10 years at £100 per cent., com- pound interest. (6) If a person returns 100 guineas for the loan of £100 for 3 months, what is the rate of interest allowed? (7) A person returns £287. 109. for the loan of' £250 at the rate of 5 per cent. per annum, simple interest. For what time was the money lent? (8) Supposing interest paid half-yearly, what will £500 amount to in 8 years at 5 per cent., compound interest? (given log 1'025=0'01072.) (9) How many years purchase should be given for a freehold estate when money is worth 3% per cent.? ANSWERS TO THE EXERCISES. EXERCISES. A. (1) 20. (14) (id). (26) 23. (39) 2. (2) 6. (15) 6a. (27) 14. (40) 6. (s) 14. (16) 2,25,6w,3bx, (28) 1%. (41) 25. (4) 0. m,xx,px,bx_y. 3. (42) 1. (5) 13. (17) 5. (30) m+n—p. (43) 3m+6n—4 6 <3) 8 <4) 5. <5) 7 (6) (22> <23) <24) <25) (26> <27) <28) (7) (8) (9) (10) (11) 262-1.. (29) aarg2j/ <30> 1172i? (31> 57.- 3%? (as) a+l2x' (33) mm—12mp' %- ea g§§§ (25) §;§€% EXERCISES. H. 2cz—1 ZZlZZ‘ .cz:2--awz:+az2 .r+a w2+3ar w—3 ' 2+3x EEIEE' 1—a:+.r2 1+.I:+.a:2 ' x3+x2—-2 2x2+2x+1 ' 44x2+2w-1 3x+1 abar. (12) 602d2. 26mg. (1 3) 2a(a2—a:2) . 24x. (1 4) 4ab(1-w2). abc. (15) 3xy(x +y). 2x29”. (16) a363(a2—:v2). (l7) l-a". (18) 1—w~a:2+.x3. (19) 60(m2+8x+2). (20) x3—7x—6. (21) 8x3-14ar+6. EXERCISES. <7) 5. <8) (9) (10) 602262+3 (11) 19376—23. 34x—23 <12) *1?“ (13) (14) (15) (16) (17) (18) 54422-13 50 ' 62 m . 41 é—Og . 50.1: + acy+ abz abc crwy+cxy abc ' 0. 552 ANSWERS TO THE EXERCISES. .r 5x+35 a: 1 ~34 242+7 (20) (28) 7‘20 . (36) m6. 1 aca: 4x2 '5'); . M Q W. x+5 .229 +3/2 _2_a+x (22) T (30) xy+yz ' (38) 3(a__x)2 5a:-y—3 1 2a: (23) T‘ (31) (39> ..;,- 4?. 2ab. (11) 3 M's—'1' 2417?. (12) aJZiZF. (49) (50) (51) (5Q) (4a+sb)j/§E. 2J2..r. N5. 4.5:?- 22 Z x 3.2: $27.. (13) N/6. (14) 102. (15) a. (16) 2x. (17) 5. ANSWERS TO THE EXERCISES. 555 (1) (2) (3) (4) (1) (Q) (3) (4) (5) (1) (2) (3) (4) (5) (5) (7 ) __ 2a z- o .32 b . 8J§.bgc. 7a —— -2-/18. $3.55. Ea”) 01/53. (2122.222? (725. bofl/ 1 M 2b' 3M2? 8.? Zuni/£5 7 . abi/g. (fl+1)a. QJE- 2J5. JTE—J'G- (18) :08. (19) ab+bJZ+aJ§+J@. (20) a +x—A/a2— :02. (8) (9) (10> <11) (13) EXERCISES. M. <5) i/é?» (6) i/E. (7) 3/52- 12 T (8) \/108a' EXERCISES. N. (6) <7) a. (8) 257553 (9) (10) (ax+b)g. 4x20 EXERCISES. 5- 2J6. 7+3J§ 2 . \/2+J§. QJE-RJE 4~a-9x ' Mme—$5 mar—fly ' ~/§+:/§. <14) <15) <16) <17) (18) 6 '3— (9) (10) 4.17521 (11) JJE+JIL 02> $5553.13- Jean-3. aabzc. (11) (12) (13) (14) F. (15) 1 3 _ 3_ 3 —~ —7-(2J2-J12+J9). 1 3 3 _ m(9~/Z5+6~/5+4). N/mx/m ——-—-—_2 . f/F-Z/ZrJrf/B 01—33 a +Ja2~ x2 x” ' 556 ANSWERS TO THE EXERCISES. a II S31 *1 2e Se *1 a S u u H u u R~ 9° 9° F“ r* Hid-IQ .Q" HHHIeRS-I HUIOO #3 ME :— aéateaa ll H I} II II II II II II II II II II [I ll H-QKO 900 a 9‘ EXERCISES. P. 1+JF-i. ~/2x+1+,/2x—1. Jam. JZ—Jb --c. JQE—JQTQ- N/.. T _$ é+$+ 2 - ,/a.z'-a2—a. \/1+x + I 2 i a . /- 1+- 2+ 1) 2 l-.2: -_-_-- 2 _..1_ ]— w+,,/ 1+ flaw-x“. H 2") 0a 9 (11) <12) (13) (14) (15) (15) (17) (18) (19) (20) EXERCISES. (22) w=5. (23) w=7~ (24) w=7. (25) 40:7. (26) x=14. (27) x=60 (Q8) az=84b (29) #35. (30) 0::5. (31) m=7. (32) arr-2. (33) w==9. (34) sir-=7- (35) w=4. (36) x==8. (37) ar=5. (38) x=5. (39) x=1§~. (40) x=10. (4'1) (4:2) $=4%. Q. (43) (44> <45) <46) (47) (48> <49) (50) <51) (52) (53) (54) (55) (55) (57) x=3~ 03:7. 5% $3 is II II ll 00 H 2*“ E II $1 ll ’3 Iy-a QED-“‘1? :q . 81 II a n H 90% ANSWERS TO THE EXERCISES. 557 EXERCISES. R. (1) .r=4|. (7) x=18a. (12) x=30. (2) (8\ (13) “2%. ‘ ’ 20+b (M) x 21 (3) x=12. (9) w=4. = "Q" (4) x=16. _ 1 ='_7_ <5) .._1 (1°) ‘15) x 12' ‘5' 27 s EXERCISES. S. (l) 00:12,} (8) x=2,) (14) x=5, y=5- y=3l y=3-} (2) x=10,} (9) x=2,} (15) x=6,} y=2. =3. 31:10 (3) #6,} (10) 1.11,} (16) bi} -= . = . y: ' (4) L: y 71 (17) w=3’ y=1.} (11) x 1’} 2/_';-} (5) w=l, =_1__ 18 =6 y=2.} y 5 ( ) (6) #7,} (12) #5,} (19) x=8,} '11:“). 3/=4‘~ y=9_ (7) $.14, (18) w=10, (20) a'=8, y=3-} 9:7. } y=8.} (21) x=2,} (27) w=114~,} (34) x==48,] y=1 y=77. y=56.[ 22 F11, (9.8) x=40, (35) a:= ( ) y=9. } ( ) 9:16.} (36) y=3.} _ 29 ar=12, w=25, (23? y=5.} (24) “=8 (30) x=;3,} (37) w=i,} 7 y: , y: 9%.} (31) .r=7,} (38) #4,} (25) 4 3:10. y=7. 5”: I x==1, (39) x=-é—, 9:21} 'y=2.} (26) x=I44-.} (33) x==6,} (40) x=10,} y=216. y=8- y=16. 558 ANSWERS TO THE EXERCISES. (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11> <12) <18) (14.) (1) (2) (3) (1) (2) (3) (4) x=10, H. l 16. 12. 18. 60. 10. 10, and 30. 10?, and 14%. 1%, and 8%. 68. 661., 5s. 6d., 48. 641., and 8s. 6d. 5, 6%, 9, and 12%, feet. 8, and 16. 8. 4.4. 8, and 40. x=:1:6, l EXERCISES. T. y=8% 2:10. w=1, 9:5, z==9. w=10, y=20>} 2:6. x=64u, y=8m 2:100 (5) w = 1 8,} (6) (7) (8) EXERCISES. U. (15) 24, and 6, yrs. (16) 85, 86, and '71. (17) 44.4, and 86. £1. (18) B- .- 3 (19) 3 - (20) 240. (21) 27-1-3Im. before 1. (22) 5-l—5Im. past 1. (23) 2 miles. (24) 4% miles. (25) 22, 7, 12, gall. (26) 38. Lia-d” ls. 8-24-5—d. (27) 1760, 1820, 1830. (28) 1819, and 1820. EXERCISES. V. (4) w=fi=4. (5) w=*2. (6) x==¥=5. (29) (30) (31) (32) (33) (34) <36) <37) <38) (39) (40> (7) (8) (9) (9) <10) <11) (19) From 3 to 2% per cent. 22, and 26. £15, and £35. 24men,20women. 1 5 men,22 women. .2. 13 ' 1 -5-. 14, and 6. 12, and 18. 11, and 5, gall. A.D. 1752. £1 becomes £53. 8.9. for knife blades, and £35. 8.9. for razors. x=i5. w==¥=3. w: '2". ANSWERS TO THE EXERCISES. (10) <11) (12) (13> <14) (15> <16) (17> (18) (19> (20> (21) (22> <23) <24) (25) (1) (2) (3) (4) (5) (1) (2) (3) (4) (5) (6) x=i2 a,—=l=5) 1 _. I ._ x_14,01 A ar=6, or -10-%-. 8 x=1%, or if). as}; or -—3. m=5, or —5§. 3 a:=1-§—, or T6 . 8 x=2, or -- . 7 26 a:=1, or x=9, or 81. 12, and 18. 3,4,5. 4, and 16. 14, and 196. 12, and 18. 20,and10. a—b' (26> <27) (23) <29) <30) <31) <32) (33> (34> . <35) <36) <37) <38) (39) 00:6, or —1O-21-. a:=6, or -5-§~. w=1-é-, or - 75 2 ar=1 — . , or 3 x=2§, or -—2. x=1%, or --6 - x=2, or —1%-. x=2, or -1;2;:. w=4, or —2. 11 x-7, 01 —§ . x=1%, or -1%. 1 x: 5, or —1-};. 40:2, or —3. x=14, or — 10. EXERCISES. W. (6) (7) (8) (9) (10) w==1=3, or *Jl_3'. 1 33:45 01‘ I. x=5, or 12. a1=i5, or x=4.~, or —1. EXERCISES. X. (7) (8) (9) (10) (11) 8, and 18. 18. 9°9>N>|H (40> <41) lag“ V“) (43) (44> <45) <46) (47) (48) (49) (50) <51) <52) (11) (12) (13) (14) (15) (12> (13) (14> (15) 37:1, _'110 50:41, 01' w=8, or 133%. 00:16, or -1 av: , or 3. x=3, or -4:%. 1 nth—'2, Z. x=%(3=*=,\/5). 800. 4.0. 5, 6, 7, 8, or 1, 2, 3,4. 13, and 12, miles per hour. 560 ANSWERS TO THE EXERCISES. (1) (2) (3) (1) (2) (3) (4) (19) (20) (21> (1) (2) (3) (7) (8) y“ = 2ax—w’. (19) 12 women 26'. each, 8 men 3.9. each. (16) 9 miles per hour. (18) 54, and 4-8. <20) (21) a}: 10 x=6, or 7. a2=16, or 17. 1 3 ' (5) 1 57r- . (6) a 5 - (7) a I - (8) 16 : 17. (22) 8a : 2b. 2.1: : 1. (24) 6a. 55.. (a+x)’. ab = a’—x”. (17) 25, and 20. 18, and 12 miles. 4 miles per hour forwards, and 1 mile backwards. EXERCISES. Y. (4’) “f'g (7) pm}. (5) x>—-— . 14 (10) The latter. (6) x<4.%%. (13) The former. EXERCISES. Z. a a+b 1° 5a: 4. -;,—”- (9) ———c - < a) . 14 4 : 5. . 8b 1 2a+w ( ) y I” 25 ' ( 0) m ' (15) 2a : 86. Egg (1]) (16) 8y : w. i 12 a_b (l7) 28x: 5y. 4y ( ) 1 ' (18) (n—1)x: 2a. .r: a—x. (25) 4, 10, and 6. (27) 15, 21, and 6a8 : 1. (26) 27, and 48. 24. 21 and 6. EXERCISES. Za. (9) 9 I 2- (16) 27 : 1. (10) 4, and 6. (18) 82, and 24. (11) 25y. 2 n 1 ~—-— , —— . (12) 15, and 20. ( 9) n+1 n+1 __ a’—-b” (20) 1 gall. of 1st to (14) x~ Aa-b' 7 of the 2nd. ANSWERS TO THE EXERCISES. 561 (1) <3) (3) (1) (2) (3) (4*) (5) (6) (7) (8) (9) (10) (1) (2) (3) (4) (5) (6) y x: 10 z 1 y=2a32 _. 3. "J— x - 7], and 96. 2, and -8. 5, and 6%. 15mx, and 20mm. 1+29w, and 1+39x. 22a, and 3&2, 970- 670. 8900. .1480. 3- (I) 2. (8) 8. l (9) 2' (10) 2 - 1 3 ° ) _ 1 (l2) 0 1, or T6. (13) EXERCISES. Zb. I (6) 2732493. \ (10) 1 : 3. (7) y=3~2+23+3. (11) 3.8137 inches. (8) y=8+2x—a:2. (12) 1 :2. (9) 824 in. nearly. EXERCISES. Z0. (11) ‘46“ .(21) l(a+b). (12) 57%. 4 (13) 7o, (22) 8, 11. (14') 210mm. (23) 191", 2, 2%- (15) 440.3. (24) 96, 92, 88, 84- 6 210 (25) 1) 1%: 2: <1 ) Ti" (36) 133-. 133.. &c- (17) 810a. and 60. (18) 2a_21, (27) 100,and£767.103. 13 (38) 1, -3, -5, &c. (19) T2“ (29) 1111. (20) 1. (30) 53. EXERCISES. Zd. 2. (14) 15. (20) Thefbrmelf 2a. 1 2 _ b —. E (15) 2 y 9 ’I‘ d 20, 80. Yes. y. 3’ an 9' 1 1 (22) ~1-.l,3.9.&c- 1 1 (17) -, -. 3 —, and—. 2 4 , 25 12.5 (23) 15 tlmes. 133—- (18) 18’ 54" 162'_(21)£5325.33.5;1£d. 2%. (19) 10, 16, 36 562 ANSWERS TO THE EXERCISES. (1) (2) (3) (4') (1) (2) (3) (I) (2) (3) (4) (5) (6) <7) (8) (9) (10) (11) (12) (18) (14) (15) (21) (22) (83) 362880. 600. 12:6. EXERCISES. Z3. (7) 3. (8) Call-1 <11) 5. l 0: “II (12) 1 (18) 3, an Sla (9) (10) (RIO: I-I wit» “00 .p. EXERCISES. (6) (7) (8) Zr- 4950. 4845. 8796. 72 = 1 1 . (10) 5040. 12:7. n=1& r-=6, } EXERCISES. 1 + 5.2: +1 0x9+1 Ox8+ 5x4+ .115. 1+ 73+21x2+ 3533+ 3534+ 21 x5+7x6+x7. a9+ Qasb + 86017!)2 + 8 4a653+ 1 26a554+1 26a455+ 84a3116+ 86a267+ 9a58+ 59. a6— 6asb+ 1 5a‘bz— 20a363+ 1 5a%‘*— 6a 55+ 5“. 324+ 8.2:3y + 24x2y2+ 82wy3+8y4. a5—1 5a‘x + 90a3x2— 270a2x3+ 405ax4- 243.35. 1+ 1 2.2: + 60x2+1 6023+ 24Ox4+192w5+ 6 4x6. m4+ 4m3nx + 6m2n2x2+ 4m 12% 3+ 124.904. a‘°+ 5a"b2 +1 Oa°b4+1 0a4be+ 50268 + 61°. 616+ 66156.1: + 1 5 a462x2+ 20a863x3 +1 5 a264w4+ 6a65m5+ 56x6. (312— 4a9x3+ 6a6x6— 4a3329+ x12. ' 2320+ 55618, 2+ 1 0.221631% 1 OxMyS-l- 5xmys+ xmym. 6 72a2x5. (1 6) 2801266306. (1 9) (17) 1225a2x‘8. ,5 (1 s) -1 61 70039733, and (20) - 30613333. -—1 61700613697. (4) (5) Zg. 1287041858. 22061269. -—86a2a:7. . 5 15 5 a3+—a5.r+-§ai11x2+—1—6 a'éxs-l- . . . . . . . . . 3-1 -I ..l 3 2-1. 3 8_ a 2 a: 861 .2 16a .3 . . . . . . . .. 1 2 1 . 5 1 1+3: 2w+2x §x+ . . . . . . . .. ANSWERS TO THE EXERCISES. 563 4 . (24.) 1-23- 33- .3. f—gfi— . .. ’5. i 4 6 2 62 2 41 63 8 (25) a525+-5-.-,_—y-;-2—5.~§-y~§+i-2-g. Llyg~ . . . . . . . .. a5x5 a5x5 ‘ a5w5 4 6 (26) 2 2.33-i332+_2l.5+~37-22- ....... .. J2 3J2 a 32J2 a (27) 61”"— 4a'5x+10a“622—2Oa_7xa+85a'8x‘*—. . . . . . . . . (28) a“5+ 5a“6.r+15a‘7wz+ 352‘823-1- 7 001—924-1- . . . . . . . . . (29) 1—12x+84x”-448x3+2016x‘- . . . . . . . .. (30) 1+213+25232+226323+ ....... .. (31) a_g+§ ~gx2+-— J23 ‘+lZ—~a_1;x6+ 4 32a a‘ . - - n . - ~ .- (82) a“1+—2a”6.25+ 55a'11210-1— -i12—15—a"160015+ . . . . . . . .. 7 3.0. 14. __E E 33 15 . “14 4. ~--——— 3 8. ——- 8 3. ( ) '7 (av) (by) (34) 2432 a: (35) 81a .2 (86) 1+ 8.2+ 36x2+10423+214x4+812xs+ 824ws+2l6at7+ 81.28. (3 7) 613- 3a% -- 31220 + 8252+ 6250 + 3acg— 53+ 3520 - 8502— 03. (88) 1+62+622+23~8xi7xg~8x5 (39) 1 +1 Ox +25w2— 40433—19024-1- 9225+ 57 0.26— 860.27— 67 5338-18 1 0309—24821". (4.0) 1+2+232+323+ 524+ . . . . . . . .. (41) 1+22+522+1023+2Ox4+ . . . . , . . . .. EXERCISES. Zk. _. 4 4'“ <1) 2+J3- (5) 11.21615. <9) Mia/6;. 4 (2) (10) K/g(K/5_N/3)' . ~52 . 4 __ 1 3 J51 (6) {7 i- (u) “mm—‘7‘”) U +' (I) .522- __ .- .- .-_ .- (12) j36§-\/§. (4.) 14-2J2. (3) JIS-Js. 4 11 02 ./§+./3- I <14) ~/‘12~i/§. I (15) JEWE- (16) 21,/'5'. (18) 1~~/3-_~ (20) 2+J3T __ 2 _. (17) 1+J7. (19) 1-\/-3-- (21) J5+1. 36—2 'SEISIOHHXEI @1111. 0.1. SHEIMSNV 1799 gig (9 (l 190]: 911 QI 8 (L) ¥+I 8 -—-I +3 5 {+8 I +9 'f‘Hl’ *“1'8 I +1. "T11 'H I +3 1 +38 I +1 1 +03 +8 I +1 .______ +I I (9) I (9) I (1’) .______+8 —-—[—-+5 “Ti-5 I I +1 I +9 I +5 + I (8) I (a) I I (I) 'olz ‘SEISIOHHXH Q Q n o .... ..+’51;__85%+86% -/Z=w (61) -I%+wI{ (14) 1 33 3.3 115 251 3’ 67’ 70’ 347’ 764' 7 22 29 51 182 i" 3’ '4’ 7’ 25 ' 1 5 3 11 .11 45 153 122 245 ’4’ 5’ 9 ’ 14’ 37’ 125’ 162’ 611' 1 3+;1 (12) 5+ 1 1 (13) 7+ 1 “5+... 1°+I5+... 14+174+... 1 1 10+ (15) 2+ (16) 6+ 20+-— 1+ 3+ 20+... 1+ 1 1 12+ 1 1 1+4+8zc. 3+I2+ 7+ 1 (18) 8+ 1 1 4+ 1 5+ 1+——-— 2+— ..1 1.1 1+&c. 1+&c. 28 45 85 28 - . ~—. 2 _ . _ EXERCISES. ZZ. 3:4,} (4) a:=19, 14, 9, 4,} (7) 5:11, 34, 57, &c.} 31:3, 3: g, 20, 31, 42. y: 3, 11, 19, &c. w=27, 6,} (5) 00:80, 15,} (8) 2:29, 85, 141, &c.} y: 5, 10. y= 1, 8. y=20, 59, 98, &c. w=12: 29:} 37:9: 5’ 1:} 37:5: 36: 67: &c- y= 5, 2. y=3, 6, 9. y=8, 28, 43, &c.} .23, or 23+80t, for all positive integral values of t. 173, or 173+885t, ........................................ .. 48. (15) 2:16, (19) 12 cows, } x=15, 50, y: 1, 18 calves. y :g’ z: 2' (20) 8 men, 2_ ’ ' (16) 72,70. 11 women, .2: 5, 10,15, 16 h.1d y=42, 24" 6, (17) 269. C 1 ren. 2:53, 66, 79. (13) 57. (21) 161. 566 ANSWERS TO THE EXERCISES. (I) w=24_<-5t,} y=11t--2. (2) x=13-7t, y=3t (3) x=23t—7,} (4) w=31+13t, y: 5+7t. (5) x=2+9t, y=1+13t. (6) w=19t-16,} y=27t-25. (7) x=13t—12, y=4t—1, z=42—38t. (1) 3522515. (2) 255113225. 325430585012. (4) 1465. (5) 14.156. (6) 12230. (1) 184113233... (2) 24112104111... (3) 401-113030“. (4) 45513. EXERCISES. Zm. (8) (9) (10> <11) <12) (18> <14) <15) (16> (17) (18> <19) x=2L y=45—-5t, 2=3t—15. $Q°ZQI§9QEOSFSQQO$ } EXERCISES. Zn. ('7) (8) (9) (10) (11) (12) 61415. 2095. 377:0. 56239. 62444261. 7071. EXERCISES. Z0. (5) (6) (7 ) (8) 1e-3t0e62 . .. 0'84. 25'1 82301022112. (13> <14) <15) (16> (17> x: 3, 5, 9:13, 5.} a7: 1: 3: 7: y=10, 2, 1. ar=2, 6, 8,36. y=36, 8, 6, 2. m=12, y: 1, 3. x=3. 9:2} w=1. y=6. x: 4', 5. 9:21: 7'} w=1,8,5,9,13. y=10,6,4,2,1. 572641. 6. Senary. 1001 , and 9. 17ft. 41%in. (18) 4.76%. 10in. 1124". (9) (10) (11) (12) 566341 22111'21 . .. 0'79215. 85. ANSWERS TO THE EXERCISES. 567 (1) (2) (3) (4*) (5) (6) (7) (8) (1) (2) (3) EXERCISES. Zq. l ’5‘ 3 ’n) --1 . (12) 5.1-. 2 ' k j 5 \ gfi 4 3 —1. --— . l3 4 (6) 5 (10) 5. ( ) (7) _ (14) _c. "é 2 (11) 4 —-— y 5 ' (8) 108 4 EXERCISES. Z7“. m 10g 6 2 _ no Eloga. (9) x: (loga . (15) x-na+b } 3'5218. __]0g a _ 0 1220818. (10) “101%” b y_’m+b' 5 J; as _- , 1 =1— . = __ 12 10g” (1 ) w b (16) m 100 ' 1-07918. “211527” __2__ 0-09691. (12) w— 021’ ‘ (17) Iva/ff): 080618. (13) 24.-0824.. _ g‘j’ y_ I ' ,_ 10g 0 (14) w=31-62,1_ J a—mloga—i-nlogb' y=10. I EXERCISES. Zs. £263. 38. 7%d': (4) 4 per cent. (7) 3 years. £278. 88. 2%d. (5) £024.00. (8) £724. 58. 17 '6 years. (6) 20 per cent. (9) 28%. NOTES. NOTE 1. (p. 49.) In Ex. 7 it will not fail to be observed that the remainder is the same expression as the dividend, with a written in the place of at: this we shall Shew to be true in all such cases. Let it be required to divide the expression Po$"+P1xn-1+P2wn—2+ . . . . . .+Pn_lx +2)" by x- a, and let Q be the quotient and R the remainder: then, writing for the sake of convenience for the dividend, we have, f(a:) = Q . (.x—a)+ R. Now R cannot contain .22, for otherwise the division would not have been completely performed, and therefore R remains unaltered whatever value .2: assumes: also Q being evidently of the form q0a2"'1+ qlxn'g-i- r1200" "3+ . . . +g,,_1, cannot become infinite for any finite value of w. Let then :c==a, f(a)=Q.0+R, i.e. R=f(a), the same expression as the dividend, with a written in the place of x. In order to find the quotient, we have f(w)=Q(w- “HR, z'.e. pox”+p1x"“1+p,x"“°'+. . . +pn_1.z+ pn= (gown-1+ glam—2+ . . .+ q,,_,) (a: —- a) +R, identically ; performing the multiplication, and equating the coefficients of the several powers of a: in each member of this equality, we get Po= qo. l ' grep... P1: Qi_aqm Q1: aqo‘l‘Pu P2: (frag!) l and 3. J ga=aq1+Pm 8m, l K &c., Pie-1: gn—f" “QR-2, gn-1=a q.._.+p.._1, Pn=R*aqn_1;’ L R=aqn_l+p,,. 570 NOTES. From the second of' these sets of equations we discover an easy method of forming the quotient, viz. to obtain any one coefficient, multiply the preceding one by a, and add in the corresponding coefficient in the divi- dend: it also appears that the remainder can be found by this method. EX. 1. Required the remainder when x4—2x8+3m2-w—1 is divided by .r+2. Here a=——2, and the remainder is (- 2)*-2(- 2)3+ a(-2)2- (-2)-1= 4. Ex. 2. Required the quotient and the remainder in the division of 3x5—2x‘-—11:c3-55x+7 by x-3. Here a: 3, and, observing that one term in the dividend is Wanting, and writing the coefficients in order, we obtain 3 -2 —11 0 —55 +7 9 _21_ 30 90 105 7 10.2—0— 35 T15. The quotient, therefore, is 3w4+ 7w3+10x2+ 80x+ 35, and the remainder 112. By applying this method we can easily see that xm-a’” is always divisible by x—a, and also by x+a when m is even; and that wm—l-a’" is only divisible by a:+a, and that when m is odd. In the equality f(x)=Q(x-a)+R, if f(x)=0, when x=a, we have 2R=f(a)=0; if x=a be a root of the equation f(a:)=0, then m—a will divide the expression without remainder. This has already been perceived to be the case in quadratic equations (Art. 205), and ought to be borne in mind. NOTE 2. (pp. 77, '79.) The rules for pointing in the operation of extracting the square and cube roots of arithmetical quantities will perhaps be more clearly per- ceived by considering what part of the square or cube arises from the several terms of the root. NOTES. 571 (i) For square root. Let a be the part already obtained, and b the next digit, so that if there be 12 places of figures after a, the root is a.10"+6.10""1+&0. The part of the square which depends upon this is a2.102”+2a6.102"“1+62.102"“2; whence we see that a is to be determined from the first term, which, as it involves an even power of 10, has an even number of figures following it, and that the introduction of 6 brings in two inferior powers of 10, z'.e. every fresh figure in the root involves two fresh places Of figures in the square; we have therefore to bring down at every trial division two figures. If then we point according to the rule we shall satisfy both the above conditions, and consequently extract the root correctly. (ii) For cube root. If a. 10”+b.10"“l+ &c. be the root, the part of the cube which depends on this is as. 1 03"+ 361.25 . 1 03“_'+ 36102. 1 03"“2-1- 63. 1 Gan-3. The first term has therefore a number of figures following it which is a multiple of 3, and the introduction of a fresh figure in the root brings in 3 inferior powers of 10, and therefore at every trial division we have to bring down 3 figures. These conditions are satisfied by the ordinary rule for pointing, and therefore the operation will be correct. NOTE 3. (p. 87.) The theory of indices is often established in the following manner:— with our original definition of an index there is no meaning to be attached to a fractional or negative one. We are therefore at liberty to assign any meaning we please to them; this may or may not make fractional and negative indices follow the same laws as positive integral ones, and it is evidently most convenient that they should follow the same laws: if then we assume these laws to hold good in all cases we shall be guided to an interpretation of our indices. The fundamental laws that have been proved for positive integral indices are these:—a’".a”=a’"*", and a’"+a"=a”“”, when 12 1s less than m, and =——n}_—m when n is greater than m. Let these hold universally: then we a shall have P P P P 17 . _ _ _ "+-'+u.tOqwrms a”.a".aq .... .-to 9 factors =aq '1 ‘.q =a'1 =ap; g O O O I Q 2'. e. a9 is a quantity such that when multiplied Into Itself 9 times produces a1”, and therefore 2:51;; which gives us the meaning of a positive fractional index. 572 NOTES. Again, for a negative index we have ,1 a'".a ”=a’""‘.=a"-:-a ", or a’".6—z,—,; 1 a. .- a~n= which gives a meaning to a negative index whether integral or fractional. 1 Lastly, we have a".a”"=a°=a".;,=1 ,- .0 (10:1 0 It will be observed that this manner of establishing the theory of indices is the inverse of that pursued in the text. NOTE 4. (p. 224.) The law of the formation of the convergents to a continued fraction may also be proved in the following way :- 4 N4_Q-1N3+N2 N - - ' . . T);_————q4D3+D2, and T); Is In Its lowest telms. therefore, by Art. 258, q,ZV,+Ng and q,D3+D2 are either equal to or are equimultiples of N4 and D,; 5.6. q,N3+N,=lc.N4, q,D,+D,=/c.D,, where k=1, or some other integer. NOW it is observed for the first two or three . . N fractlons that NSDI- N1D2=*1. Let thIs be true as far as 173-, so that 3 We have, as in the text, N3D,—N,D,=-+- . 1 Then N4D3—N8D = Z{(Q4N3+N2)D3_N3N.Nl 2—D‘1—2‘, . . , _ ii'+a therefore, In both cases, a", r" are posmve Integers. And as -"-/-;‘-—- , &c., are all >1, 5', 5', b”, &c. are positive integers, therefore all the symbols in this investigation are positive integers. The equation r'r"=n-a"2, and those similar to it, shew that a”, &c. are less than ~/n, therefore a’, a”, &c. are not >a, which is the greatest whole number Bainés‘s Vaidrgis? - Waterton's Essays onNatural Hist, 24 Goldsmith’s Poems, illustrated - 8 Baker’s Wanderings in Cevlon - 3 Yougtt s Thu Dog - - - .. 24 L; E. L.'s Poetical Works - - 11 BQI'I'OW’S Continental Todr - - 22 The Horse .. .. - 24 limwogd‘s Anthologia Oxoniensis - 12 gains African Travels - - 3 _ yra. ermanica. - — - - er ele ' F _ _ I-Volume _Encyclqpeedias Macaulay's Lag of Ancient Rome 12 Burtou’z séasgzrrifigaf Friuwe- - g and Dictronarre's. Mac Démahi’s Pithin and. Without 13 ‘ C 1;“ l'iegina and Mecca - - 5 . , oems - _ 13 er '8 2'8 k - £12111; iiléufal 8PM.“ ' ' ' 4 * Montgomery's Poetical Works - 15 De Custine’s Rfigsaigd Greecel - 23 Cr 1 a; 1cJIence,Literature? and Art 4 M°ore,sP08tmal works _ _ 15 Eothen _ _ _ _ OP 33‘“ 8. .‘ctmnary 0.5 Medmne " 6 “ Selections (illustrated) - 15 Fer uson’s Swis T 1 - - 22 gregzis glvillfigifineermg - - 8 u Lana ROOM) 1 mean}, Interior: have s - ' "'3 WSIClCte—-- u ~ -"" ‘, '-"" Johnston’s Geographical Dictionary 11 n 1313:1251“ aggdéeiugds : I Forejter s ai‘ggf; affirm? " 22 140119;,“ s Agflcultm'e. 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Lal‘lzg's lgorway - - _ _ 22 s omes 1.6 itur _ _ 7 A an. 3 M 0- _ OtES Of a Traveller - 22 lgefence of Eclzpse of git}: - - 7 r “no sPggilaglggti€ii§ri$¥s - g M‘Clure,s Nth—west Passage ' 17 :i.ls‘l;,1fhne ' - — - - 7 Bourne on the Screw Propeller - 4 MacDoug‘ 'svogéée OftheResomw 13 :Eiafi 0 D8}- ~h§er (The) - - - 19 “ ’s Catechism of Steam-Encrine 4 MFLSOE“ zulus 0- * atal ' " 22 :;c agiselo aith _ - - 7 Boyd’s Naval Cad 813,8 Manual a _ 4 M1168 s Rambles in Iceland - - 22 -~| 18“ man s Greek Concordance 7 Brande’s Dictionary of Science &c 4 OSb-om's Quedah ' _ ' " 16 E _ Heb.&Ch_ald.Concoi-d. 7 “ Lectures on OmanicChemistr - 4 Pfelfierys voyage mum} the world 22 Gxptengnce (The) of Life - - 19 Cresy‘s Civil Envin‘eering - Y 5 “ Second d‘tto ' " " 17 Her 11 e, _- - _ _ _ 19 DelaBeche,sGeo 0gy0momwan &c 7 Scherzer’s Central America - - 18 Harrison 8 Light of the Fora; - 8 De la Rive’s Electricity - , - . 7 seawardg Narrative " " " 19 H°°k 9; Lectures on Passion eek 9 Grove’s Correla. of Physical Forces 8 snow-‘5 Tm” del Fuego _ — 20 0218 a Intrpdnctmn t° scriptures 9 Herschel’s Outlines of Astronom 9 SPOtt‘SWOOde’S EaStem Russia - 20 Huc‘s Clfigzliiilfignt oé'gitto - 10 Holland's Mental Physiology 7 9 Vin Tempsliyvs Mexico and Gua- m ma _ - 10 _ , ema a - - .- - 24 Humphreys.s Parables Illuminated 10 T-lum‘b‘oldt s gages: of 1:1 atufe -_-_ Weld’s Vacations in Ireland - - 24 Ivors - - _ .. _ _ 19 Hunt on Li ght _ _ _ _ 10 “ United States and Canada- 24 JMisonis Surat-Legends _ _ n Lardners Cabinet Cyclopwdia _ 12 Wgme’s African Wanderings - 22 “ Monastm Legends _ _ n Marcet,s (Mud Conversatmns _ _ 15 \Vilberforce's Brazil &Slave-Trade 22 “ lialgignds of the Madonna 11 :Morell’s Elements of Psychology - 16 ployment ures on Female Em~ Moseley-’sEngineei-ingBcArchitecture 16 Works Of Ficfiifln. Jerem T l , - - ~ 11 Our Coal-Fields and our Coal—Pits 22 Kathay_ a};‘ or s Works - - - ll Owen‘s LecturesonComp.Anatomy 17 Cruikshank’s Falstafl" - - .. 6 K..n f“; 811130“ _ - - - 19 Pereira on Polarised Li ht - - 17 l Howrtt’s Tallan em - .. _ 10 L2" 1% s Pictorial Life of Luther - 8 Pescliel’s Elements of P§1ysics - 17 ‘ Macdonald’s Vii a Verocehio - 13 L txc on arsonaae _ - - 19 Philllps’s Fossils of Cornwall, &c. 17 I Melville's Confidence-Men - .. 15 e“ers to m Unknown Friends - 11 “ Mineralogy - - - 17 Moore 5 Epicurean - - .. 15 L G On ' appmese - - - ll " Guide toGeology - - 17 Sir Roger De Cover-lay - - - 19 am err‘nanica -. .- - - 5 Portlock’s Geology ofLondonderry 17 SketcheMThe) Three Tales - 19 Liaguillgegs tBgI:JQRSPlYEtIOB — - IS’oweH’sE?nit§-y Cfiwtgllilds - - 17 Sonthey’s The 60mm &c. - - 20 . _- — - - mee’s cc 0- e - - ’ géaitiand sChurch mGatacombs — l4 Steam-En ine (The mlg’ - - 1: Trel‘l‘ope s gfifggter Time“: - argairet Perelval - - - - 19 'Wilson’s lectric Qelcgraph- - 22 _ ALPHABETICAL CATALOGUE of NEW worms and. 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