Arithmetic: VULGAR, DECIMAL, INSTRUMENTAL, ALGEBRAICAL. In Four Parts: Containing I Vulgar Arithmetic, Both in whole Numbers and Fractions, in a most plain and easy method. TWO Decimal Arithmetic, With the ground and reason thereof, illustrated by divers Examples. III Instrumental Arithmetic, Exactly performing all Questions of what nature soever in a Decimal way, by Scales, with much more ease and facility than can be effected, either by Vulgar or Decimal Arithmetic, the work of Reduction being wholly avoided. Nothing in this kind having been hitherto published by any. IV Algebraical Arithmetic, Containing an Abridgement of the precepts of that Art, and the use thereof, illustrated by Examples and Questions of divers kinds. Whereunto is added the construction and use of several Tables of Interest and Annuities, Weights and Measures, both of our own and other Countries. By WILLIAM LEYBOURN. LONDON, Printed by R. and W. Leybourn, and are to be sold by George Sawbridge at the Bible on Ludgate-hill, 1660. Vera Effigies Gulielmi Leybourn, Philom. portrait of William Leybourn TO THE READER. HEre is presented unto thee a short Treatise of Arithmetic, I confess there are enough, (if not too many) already extant, notwithstanding I have adventured to publish this, more for variety than necessity, not doubting but in the perusal thereof thou shalt find something in this worth thy labour, and what in other Books of this kind is wanting. The whole Treatise is divided into four Parts. The first contains Vulgar Arithmetic in Whole Numbers and Fractions; And in every Rule there are Examples for practice added, and Questions also wrought by those single Rules. And in Division, (which is the most difficult of the four Species,) there are five varieties, so that every man may make use of that which he best understands or fancies: And in the working of the Golden Rule, etc. I have made use sometimes of one kind of Division, and sometimes of another. The Second Part contains Decimal Arithmetic, with the ground and reason thereof, Also Tables of the Moneys, Weights, and Measures used in England, with directions for the making of those Tables, and of any other. And lastly, there are Examples wrought in Decimal Numbers, in all the most usual Rules of Arithmetic, and those Examples are encumbered with as many Fractions as can possibly happen in any Question concerning buying or selling. The third Part is of Instrumental Arithmetic, which performeth any Question Arithmetical in a Decimal way, without the help of Tables, by which the whole work of Reduction is avoided, there being certain Scales of English Money, Weights, and Measures, by me contrived, by which (by inspection only) the Decimal Faction of either Money, Weight, or Measure, may be set down as exactly, and in less time, than they could have been taken out of the Decimal Tables in the Second Part of this Treatise. I have also in this Third Part, (the better to illustrate the use, and commodiousness of these Scales, a figure whereof is inserted between page 246 and 247,) gone through all the most usual Rules of Arithmetic, giving Examples in each Rule, by which the Reader may plainly perceive what labour there is saved by using the Scales, the whole work of Reduction being taken away, and the Fraction immediately set down at first, without Addition, or being already set down, reduced to the known part of the Integer, without Subtraction. Unto this third part there is added an Appendix containing certain Rules of Exchanges, with Tables of the Weights and Measures of foreign Countries compared with the Weights and Measures used in London, with an Example to illustrate the use of each Table; And lastly, there are five Tables calculated at 6 per Cent. Compound Interest, by which the true valuation of any Lease or Annuity, or Money forborn or rebated, may be easily known, with an Example showing the use of each Table. The Fourth Part containeth an Abridgement of the Precepts of Algebra, first written in French by James de Billy, a Translation whereof came to my hands some years since, in the perunsal whereof, finding the Precepts very plain and easy, and considering that we have but very little of this kind of Arithmetic in our English Tongue, I have adventured to insert it here as a Fourth Part, thereby to make this Work the more complete: Unto which Translation there is further added divers Questions of good consequence, which were not in the Original, as by comparing them together may appear. This Treatise thus finished, I present thee with, desiring thy friendly acceptance, and pardon for such faults as may possibly have escaped the Press, or myself, and in so doing thou wilt encourage him, who is A friend to all that are Mathematically affected, William Leybonrn. There is now in the Press, and ready to come forth, Euclides Elements of Geometry, all the 15 books in a most compendious form contracted and demonstrated, together with some other pieces never before printed in English, all published by the care and industry of my loving friend Mr. John Leake, Reader of Mathematics in London. The Contents. The First Part of Vulgar Arithmetic. NUmeration pag. 1 Addition pag. 7 Addition of English money pag. 10 Addition of Troy weight pag. 13 Addition of Avoirdupois little weight pag. 16 Addition of Avoirdupois great weight pag. 17 Tables of Liquid Measures, Dry Measures, Long Measures, and Time pag. 18 The proof of Addition pag. 21 Substraction pag. 22 Questions performed by Addition and Substraction pag. 31 Multiplication pag. 33 Compendiums in Multiplication pag. 41 The proof of Multiplication pag. 42 Questions performed by Multiplication pag. 44 Division pag. 45 A second way of Division pag. 55 A third way of Division pag. 57 A fourth way of Division pag. 60 Questions performed by Division only pag. 65 Reduction pag. 67 Progression pag. 70 Geometrical Progression pag. 75 The Golden Rule pag. 80 The Golden Rule Reverse pag. 88 The Golden Rule compounded of five numbers pag. 92 Of Fractions pag. 96 Numeration pag. 97 Multiplication pag. 100 Division pag. 102 Reduction pag. 104 Addition pag. 110 Substraction pag. 112 The Rule of Fellowship pag. 114 The Rule of Fellowship with time pag. 120 Alligation pag. 124 Position pag. 131 The Rule of Ceres and Virginum pag. 137 Extraction of the Square Root pag. 147 A table of Square Roots from 1 to 1000 pag. 151 Extraction of the Cube Root pag. 162 A table of Cube Roots from 1 to 1000 pag. 165 Some uses of the Square and Cube Roots pag. 177 The Second Part of Decimal Arithmetic. Table's of Reduction of English Coin 185 Of Troy weight Ibid. Of Avoirdupois great weight 187 Of Avoirdupois little weight 188 Of Liquid Measures Ibid. Of Dry Measures 189 Of Long Measures Ibid. Of Time Ibid. Of Dozen 190 The use of the Tables of Reduction 191 Notation of Decimals 208 Addition of Decimals 209 Subtraction of Decimals 213 Multiplication of Decimals 215 Division of Decimals 220 The Rule of Three in Fractions both Vulgar and Decimal 228 The Rule of Three reverse in Decimals 239 The double Rule of Three in Decimals 240 The Third Part of Instrumental Arithmetic. INstrumental Arithmetic what 245 A Figure of the Scales 246, 247 Numeration upon the Scales 247 Addition by the Scales 257 Substraction by the Scales 262 Multiplication by Nepeirs Bones 264 A Figure of the Bones 265 Division by the Bones 27● Examples in the Rule of Three direct 273 Examples in the Rule of Three reverse 275 Examples in the double Rule of Three 277 Examples in Barter 279 Examples in Fellowship 281 Examples in Loss and Gain 285 Examples in Loss and Gain upon time 280 AN Appendix containing divers questions concerning Exchanges of the Coins, Weights, and Measures of one Country, with those of another Country, with divers tables thereunto belonging, also Tables of Interest and Annuities at 6 per Cent. Compound Interest. The fourth Part, being an Abridgement of the precepts of Algebra. A Table of the Cossick Characters 324 The Alegorithm of the Cossick Numbers 325 Addition of simple Cossick numbers 326 Substraction of simple Cossick numbers 326 Multplication of simple Cossick numbers 326 Division of simple Cossick numbers 326 Addition of numbers composed and diminished 327 Subtraction 328 Multiplication 329 Division 330 The Alegorithm of Fractions ibid. The Rule of Algebra ibid. How Equations are found 331 How Equations are reduced 332 When to extract a Root 333 To extract the square root of numbers compounded and diminished 334 How to know whether a question propounded be impossible 336 The Algorithm and use of second roots 337 Addition of second roots ib. Substraction of second roots ib. Multiplication of second roots 338 Division of second roots ib. The extraction and use of second roots ib. The Algorithm and extraction of the root from furred and irrational numbers. 339 Reduction of simple furred roots to one and the same denomination 340 Multiplication and division of simple furred roots 341 How to know whether two furred roots be commensurable, or incommensurable ib. Addition of simple errational roots ib. Substraction of simple irrational roots 342 Addition and Subtraction of numbers, furred, compound, and diminished 343 Multiplication of numbers, furred, compound, and dininished ib. Multiplication of universal roots 346 Division of universal roots 347 Addition and Subtraction of universal roots ib. The Extraction of the root from Binomials and Apotomals 348 The use of Algebra 349 Questions resolved by one simple equatoon ib. Questions resolved by a compound equation 352 Questions resolved by furred numbers 356 Geometrical questions resolved by Algebrae 359 Questions resolved by second roots 362 Questions resolved indefinitely 364 In Appendix. QUestions in Algebra which require the Rule of thee in their operation 369 Examples in Algera concerning Squares 390 Examples relating to Cubes 391 Numeration. NUmeration is accounted the first part of Arithmetic, and it is to know how to read a Sum of figures expressed in writing; or to write down any Sum to be expressed. To the doing of which there are four things necessary. First, to know their number, which is Nine. Secondly, their shapes, which are 1 2 3 4 5 6 7 8 9 Of which the first toward the left hand ever signifieth One, the second Two, etc. Thirdly, to know the value of their places. Lastly, How their proper signification i● altered thereby. The value of their places is thus, When two, three, or more figures stand in one Sum, tha● is, without any Point, Line or Comma betwixt them, as 321, that place next the right hand where the figure 1 standeth, is called the place of Unity, or Unities, and the figure 1 standeth in that place only for one, and the figure 2 when it is found in that first place, stands only for two; and the like of the rest. But in the Sum 321, above expressed, the figure 2 is in the second place, and every place contains the value of that place before towards the right hand ten times; and therefore the figure 2 doth not signify two, but (in this second place) ten times two, that is Twenty. And so the figure 3 if it had been in that place had signified ten times three, that is Thirty; but being here in the third place it signifies ten times thirty, that is Three hundred. And so the whole Sum 321, is to be read, Three hundred twenty and one. Fourthly, It is hereby seen, how their proper significations, which were Three, Two and One, are altered by being thus placed, and the Sum, which otherwise had been but Six, is Three hundred twenty one, as before. In like sort, if their had been more places, as seven, the value is quite through increased ten times by being a place behind towards the left hand; as in the Sum 1̇111̇111̇, The figure 1 in the second place stands for ten times one (that is ten,) in the third for ten times ten (which is one hundred,) in the fourth for ten hundred, (which is called one thousand,) in the fifth for ten thousand, in the sixth for ten times ten thousand (which is one hundred thousand) in the last (here the seventh) place for ten hundred thousand, which is called a Million: and so on, if there were more places, observing the same order. Now to read this readily, make a prick over the place of Unity, another over the third from it, and over every third still towards the left hand, for so those points will be over the places of Unites, Thousands, and Million; and so beginning at the last, that is, at the left hand, read one Million, and because the three following towards the right signify properly one hundred and eleven, but the prick belonging to them is in the place of thousands, call them one hundred and eleven thousand, and three remaining being under the point over Unity, signify only one hundred and eleven; and all three points read together in one sum is, One Million, one hundred and eleven Thousand, one hundred and eleven. In like manner, if this number 73598624 were given to be read (according to former directions) make a prick over every third figure, beginning with the first figure towards the right hand (which is the place of Unity) and then will your number stand thus. 73̇598̇624̇ Then for the ready reading thereof (because the third prick signifieth Millions) call all the figures towards the left hand, standing from that prick, Million, which in this example are 7 and 3, so then this number contains 73 Millions, 598 Thousand, 624 which in words at length we read Seventy three Millions, five hundred ninety eight thousand, six hundred twenty four. Let thus much suffice concerning the planing of large numbers, for the ready reading of them, only take these four Tables following, for illustration of what hath been hitherto delivered in words, the very sight whereof is better than a whole Chapter of information. The first Table is thus to be read] One in the first place signifies One. One in the second place signifies Ten. One in the third place signifies a hundred, etc. as in the Table. The third Table is only certain rows of figures set together and orderly disposed, having the signification or reading of the same numbers in words at length to them annexed, and is only inserted for the better satisfaction of such as shall doubt whether they perfectly understand what hath been before taught. The fourth Table is much like the second, only it consisteth but of one number and extends three places farther than the greatest number in the second Table doth: viz. to twelve places which figures are thus to be read. 736 Millions of Millions, 842 Millions, 708 Thousand, 645. (1 Table) One in the first place signifies 1 one One in the second place signifies 10 ten One in the third place signifies 100 a hundred One in the fourth place signifies 1000 a thousand One in the fifth place signifies 10000 ten thousand One in the sixth place signifies 100000 a hundred thousand One in the seventh place signifies 1000000 a Million One in the vl place signifies 10000000 ten Millions One in the ninth place signifies 100000000 a hundred Million (2 Table) 8 54 762 3.483 97.621 243.794 8.749.807 57.316.248 357.846.903 Millions thousands hundreds (3 Table) number of places 1 8 (eight, number of places 2 54 (fifty four, number of places 3 762 (seven hundred sixty two, number of places 4 3483 (three thousand, four hundred eighty three number of places 5 97621 (ninety seven thousand, six hundred twenty one number of places 6 243794 (two hundred 43 thousand 7 hundred 94 number of places 7 8749807 (eight million, 749 thou. 8 hundred and 7 number of places 8 57316248 (fifty seven mil. 3 hund. and 16 thou. 248 number of places 9 35784698 (three hundred fifty seven million, eight (hundred 46 thou. 9 hund. and 3. (4 Table) 7 hundred million of million 3 ten million of million 6 one million of million 8 hundred million 4 ten million 2 one million 7 hundred thousands 0 ten thousands 8 one thousands 6 hundred hundreds 4 ten hundreds 5 one hundreds Addition. ADdition is the collecting or gathering together of two or more sums, either of one or of divers denominations, into one sum, which is called the (Aggregate) (Total) or Gross sum.) In addition of numbers of one denomination, the order is to set the numbers to be added one directly under the other, that is to say Unites under Unites, Ten under Ten, Hundreds under Hundreds, etc. The Rule. Having placed your numbers to be added in due order, one under another, draw a line under them, and begin at the lowermost figure towards your right hand, and add that to the next figure above, and the sum of them to the next figure above that, proceeding in this order till you have added the whole line together, which when you have done, consider how many ten are contained in that line, and for every ten keep one Unite in your mind, to be added to the next row, but if there be any odd digits, you must set them beneath the stroke, just under the line you added together, having thus finished the addition of one line proceed to the next, and from thence to the third, and so forward, be there never so many. The Examples following will make this plain, Example 1. Thousands Hundreds Ten Unites 7832 5609 376 8547 22364 Let the numbers given to be added together be 7832, 5609, 376, 8547, having thus placed them in order one under another as in the Margin is done, draw a line under them, then begin your addition, at the lowermost figure towards your right hand, saying, 7 and 6 is 13, and 9 is 22, and 2 is 24 now because in 24, there is two ten, and 4 remaining, I place the 4 under the line, and carry the two zens to the next row, saying, 2 which I carried and 4 maketh 6, and 7 maketh 13, and 3 makes 16, in which row there is but one ten contained and 6 remaining, which 6 I set under the line, and carry the ten to the next row of hundreds, saying, 1 that I carried and 5 maketh 6, and 3 maketh 9, and 6 maketh 15, and 8 makes 23, in which 23, ten is contained two times, and 3 remaining, the three I place under the line, and carry the two ten to the next row, saying 2 which I carried and 8 maketh 10, and 5 maketh 15, and 7 makes 22, in which ten is contained two times, and 2 remaining, which 2 I set under the line, and because there is never another row to be added (to which I should carry the two ten) I therefore set it down also under the line towards the left hand, as you see done in the margin, so the total or gross sum of these numbers being added together is 22364. Example, 2. Apple trees 137 Pear trees 76 Cherry trees 107 Plum trees 36 Trees in all 355 A man hath in his Orchard 136 Apple trees, 76 Pear trees, 107 Cherry trees, and 36 Plum trees, and he desires readily to know how many trees he hath in all, place your vumbers one under another as in the margin, and then begin to add them together, at your right hand, saying, 6 and 7, make 13, and 6 make 19, and 6 make 25, place 5 under the line, and carry 2 to the next row, saying 2 and 3 is 5, and 7 is 12, and 3 is 15, place 5 under the line, and carry 1 to the next row, saying 1 and 1 is 2, and 1 is 3 which 3 I set under the line, and because there was not ten contained in that line therefore the total is 355, and so many Trees are in the Orchard. Other Examples for Practice. Addition of numbers of divers Denominations. 1 Addition of English Money The most usual Coins used in England are Pounds, Shillings, Pence, and Farthings, of which Coins 4 Farthings make 1 Penny thus charactered d. 12 Pence make 1 Shilling thus charactered s. 20 Shillings make 1 Pound. thus charactered li. for a farthing we use q. THE RULE. In the addition of numbers of divers denominations this order is to be observed, viz. Place all numbers of the same denomination one directly under another, as Pounds under pounds, shillings under shillings, pence under pence, and farthings under farthings. Then draw a line under them, and begin your Addition with the least denomination first, observing how many times the next greater denomination is contained in the least, and for every time carry one unite to the next denomination, as before you did the ten, setting down the remainder if any be, then adding the next denomination together, take notice how many times the next greater denomination is contained in that lesser, carrying for every time one to the next, thus proceeding till you have gone over all the denominations, be they never so many. Example 1. li. s. d. q. 37 16 9 3. 21 09. 8. 1 13 12 9 2 72 19 3 2 Let the numbers to be added together be 37 li. 16 s. 9 d. 3 q. 21 li. 9 s. 8 d. 1 q. 13 li. 12 s. 9 d. 2 q. Place the numbers as in the margin, draw a line under them, and begin with the least denomination (which in this example is farthings,) first, saying, 2 q. and 1 q. is 3 q. and 3 q. is 6 q. which is one penny and 2 q. remaining, which 2 q. I place under the line, and carry the one penny to the next row which is the place of pence, saying, one penny and 9 d. is 10 d. and 8 d. is 18 d. which is 1 s. and 6 d. (Now against the 8 I make a prick with my pen for my better remembrance to signify that there is one shilling to be carried to the place of shillings,) then go on and say 6 d. and 9 d. is 15 d. which is 1 s. and 3 d. therefore against 9 I make a prick with my pen, and (because that is the last number) I set down the odd 3 d. under the place of pence, and (because I find two pricks in the line of pence, therefore) I carry 2 s. to the place of shillings saying 2 s. which I carried, and 12 s. is 14 s. and 9 s. is 23 s. which is one pound, and 3 s. remaining, I make a prick against 9, and going on, say 3 s. and 16 s. is 19 s. which (being there is no more numbers to be added, and being also less than 20 s.) I set under the line, and finding one prick in the line of shillings, I therefore carry one to the place of pounds, saying, one which I carried and 3 is 4, and 1 is 5, and 7 is 12. set down the 2 under the line (as in addition of numbers of one denomination) and carry one to the next row, saying one that I carried and 1 is 2, and 2 is 4, and 3 is 7, which being the last I set down, and so the total or gross sum is 72 li. 19 s. 3 d. 2 q. Example 2 li. s. d. 29 16. 8. 32 17. 9 81 13 11 144 08 4 Let the numbers to be added be 29 li. 16 s. 8 d. 32 li. 17 s. 9 d. 81 li. 13 s. 11 d. and let it be required to find the total or gross sum. Here in this Example the least denomination is pence, therefore I begin with them, and say, 11 d. and 9 d. is 20 d. which is 1 s. and 8 d. make a prick against the 9 and say 8 d. and 8 d. is 16 d. that is 1 s. and 4 d. make a prick against the 8 and set down the odd 4 d. then (because there are two pricks in the line of pence) you must carry 2 s. to the place of shillings, saying 2 s. which I carry and 13 s. is 15 s. and 17 s. is 32 s. which is 1 l. 12 s. make a prick agianst 17, and say 12 s. and 16 s. is 28 s. make a prick against 16, and (because there is no more numbers to be added) set down the odd 8 s. under shillings, and (being there is two pricks in the line of shillings) carry 2 to the place of pounds, saying 2 and 1 is 3, and 2 is 5, and 9 is 14, set down 4 and carry 1 to the next line, and say 1 and 8 is 9, and 3 is 12, and 2 is 14, which (because it is the last) you must set down, so is the total or gross sum 144 li. 8 s. 4 d. Other Examples for Practice. li. s. d. q. 29 18. 7 3 63 11. 2. 1. 129 4 0 2 3 7 10 1 226 1 8 3 li. s. d. 36 2 8. 29 0 2 31 16. 9 6 2 5 103 2 0 2 Addition of Troy Weight. Troy weight is a Weight used in England, by the which is weighed, Bread, Gold, Silver, Pearl, etc. the most usual denominations of which weight are, Pounds, Ounces Penny weights, and grains, of which 24 Grains make 1 Peny weig. thus charactered pw. 20 Penny weight make 1 Ounce thus charactered ou. 12 Ounces make 1 Pound thus charactered lib for a grain we writ gr. The Addition of Troy weight (and consequently of any other weight or measure whatsoever either Domestic or Foreign) differeth nothing at all from the addition of Coin last taught, if the affinity of one denomination to an other be first known, for whereas in money because 1● d. make 1 s. you therefore observe how many twelves there are in the addition of your pence, and for every 12 you add one shilling to the place of shillings, so in the addition of Troy weight, knowing that 24 gr. make one penny weight, you must therefore in the addition of Grains of Troy weight observe how many times 24 you find in your line of Grains, and for every 24, carry one to the place of penny weights, likewise in the addition of penny weights Troy, you must consider how many times 20 is contained in your line, and for every 20 carry one to the place of ounces, (because 20 penny weights make an ounce.) Also in the addition of Ounces Troy, you must observe how many times 12 you find in your line of ounces, and for every 12 carry one to the place of pounds than lastly add your pounds together, as numbers of one denomination. Example. li. ou. pw. gr. 7 11. 13 19 6 07 16. 19 3 07 09 06 18 02 19 20 Let the numbers to be added together be 7 li. 11 ou. 13 pw. 19 gr. 6 li 7 ou. 16 pw. 19 gr. 3 li. 7 ou. 9 pw. 6 gr. Place your numbers as in addition of money, each under other according to their respective denominations, as in the margin then draw a line under them and begin your Addition with the least denomination first, viz. grains, saying 6 gr. and 19 gr. is 25 gr. which is one penny weight and one grain, make a prick against 19, and carry the odd grain to the number above, saying 1 gr. and 19 gr. is 20 gr. which (because it is less than one penny weight) I set under the line, then finding one prick in the line of grains, I (therefore) carry one to the place of penny weights, saying 1 and 9 is 10, and 16 is 26, which is one ounce, and 6 pw. make a prick against 16, and say 6 and 13 is 19, which (being less than an ounce) set under the line, then for the one prick, carry 1 to the place of ounces, saying 1 and 7 is 8, and 7 is 15, which is one pound and 3 ou. make a prick at 7, and say 3 and 11 is 14, which is one pound and 2 ounces, set down the 2 ounces, and for the two pricks carry 2 pounds to the place of pounds, saying 2 and 3 is 5, and 6 is 11, and 7 is 18, which set under the place of pounds, so is your addition ended, and the sum is 18 li. 2 ou. 19 pw. 20 gr. Other Examples for Practice. i. ou. pw. gr. 32 9 12 16 17 11. 6 9 34 8. 15. 10 8 10 4 7 94 3 18 18 li. ou. pw. gr. 0 10 17. 11 0 6. 0 5 0 0 16 8. 0 5 2 19 1 10 16 19 3 Addition of Avoirdupois little Weight. There is another kind of weight most commonly used in England called Avoirdupois little weight, by which is weighed all sorts of wares or merchandise Garbable, as Sugar, Pepper, Cloves, Mace, etc. This weight is commonly divided into these denominations, Pounds, Ounces, and Drams, of which 16 Drams make 1 Ounce thus charact. cu. 16 Ounces make 1 Pound thus charact. li. for a dram we writ dr. In the addition of Avoirdupois weight, you must observe the very same method and order as in Money and Troy weight, having due respect to the quantity of the denomination, as in the addition of drams to make a prick at every 16, setting down the remainder and for every prick carrying a unite to the next place. The preceding rules being so copious in this particular I shall forbear to make any verbal illustration, but only give you some examples ready wrought, together with the most usual parts into which the Weights and Measures now used in England are divided into: which to the ingenious will be of most validity. Examples of Addition of Avoirdupois little weight. li. ou. dr. 12 11. 09 76 05 12. 32 10. 00 91 07. 13. 32 13 07 246 00 09 li. ou. dr. 06 13. 07. 05 09. 12 06 03 09. 10 00 00 05 07 09 34 02 05 4 Addition of Avoirdupois great weight. There is a weight commonly used in England, by which is weighed all commodities that are sold by the hundred, as Corents, Wool, Flesh, Butter, Cheese, and the like, the which hundred weight containeth 112 pounds, and the hundred weight is divided into quarters, pounds, and ounces, so that 16 Ounces makes 1 Pound thus chaeactred li. 28 Pounds makes 1 quarter of a C. thus chaeactred qr. 4 quarters makes 1 Hundred weig. thus chaeactred C. for an Ounce we writ oz. Examples of Addition of Avoirdupois great Weight. C. qr. li. ou. 37 03. 21 12. 09 01 06 03 33 02 20. 00 10 00 00 00 12 03. 07 03 103 02 27 02 C. qr. li. ou. 05 01. 00 07 03 02 18 06. 00 01 06 08 11 03. 04 00 06 01 10 05 17 01 11 10 I might further proceed to show you Examples of addition of common English measures, viz. of long measures, Liquid measures, and dry measures, as also of Time, Motion, etc. but the preceding Examples being of sufficient extent, I shall forbear to trouble either myself or the Reader with that which I conceive superfluous: Only, before I leave Addition, I will give you a brief view of the most usual measures in England, which take as followeth. And 1 Of Liquid Measures. Liquid measures are those in which all sorts of Liquid substances are measred, of which (according to the Statute of 12 Hen. 7. chap. 5.) a Pint is the least, from which the greater Liquid measures are deduced, according as is expressed in the Table following. 2 Pints make 1 Quart 2 Quarts make 1 Pottle 2 Pottles make 1 Gallon 8 Gallons make 1 Firkin of Ale, Soap, or 9 Gallons make 1 Firkin of Beer (Herring 10 ½ Gallons make 1 Firkin of Salmon or Eels 2 Firkins make 1 Kilderkin 2 Kilderkins make 1 Barrel 42 Gallons make 1 Tierce of Wine 63 Gallons make 1 Hogshead 2 Hogsheads make 1 Pipe or But 2 Pipes or Butts make 1 Tun of Wine. 2 Of Dry Measures. Dry Measures are these in which all kind of dry substances are measured, as Corn, Salt, Cole, Sand, etc. of which a pint is the least. 2 Pints make 1 quart 2 quarts make 1 Pottle 2 Pottles make 1 Gallon 2 Gallons. make 1 Peck 4 Pecks make 1 Bushel Land measure 5 Pecks make 1 Bushel Water measure 8 Bushels make 1 quarter 4 quarters make 1 Cauldron 5 quarters make 1 Wey. 3 Of Long Measures. Long Measure is that by which is measured Cloth, Land, Board, Glass, Pavement, Tapestry, & c● of which measures (according to the Statute of 33 Ed. 1. and 25 El.) a Barley corn is the least. So that 3 Barley Corns make 1 Inch 12 Inches make 1 Foot 3 Foot make 1 Yard 3 Foot 9 inches make 1 Ell 6 Foot make 1 Fathom 5 ½ Yards, or 16 ½ Foot make 1 Pole or Perch 40 Perches make 1 Furlong 8 Furlongs make 1 English mile. 4 Of Time. Time consisteth of Years, Months, Weeks, Days, Hours, and Minutes. So that 60 Minutes make 1 Hour 24 Hours make 1 Day natural 7 Days make 1 Week 4 Weeks make 1 Month of 28 days 13 Months, one day 6 hours make 1 Year. 5 Of Apothecary's Weights. The weights used by Apothecaries are Grains, Scruples, Drams, and Ounces, of which 20 Grains make 1 Scruple thus charact. ℥ 3 Scruples make 1 Dram thus charact. ʒ 8 Drams make 1 Ounce thus charact. ℈ 12 Ounces make 1 Pound thus charact. li. By help of these Tables, and the rules and cautions before expressed, any man may make addition of any of the abovesaid measures one with another, and therefore I shall forbear to illustrate them by Examples, but leave them to every man's own practice, and thus I conclude Addition. The Proof of Addition. Having placed your numbers in order, and added them together, and set the Total under the line, cut off the upper number by drawing a line with your pen betwixt that and the others, then add all the numbers together except the uppermost, and set the Total of them under the Total before found, then add this last Total, and the first number which you cut off with your pen together, and if the sum of those two numbers be equal with your Total sum first found, then is your work right, otherwise not. Example. In the first example of whole numbers, the sums to be added were 7833, 5609, 376, and 8547, these numbers placed in due order and added together, the total or gross sum of them is 22364, now to prove whether this Total be true or not, I cut off the uppermost number, (to wit 7832,) with a dash of the pen, and I add the other three numbers together, namely, 5609, 376, and 8547, and the Total of them is 14732, which number being added to 7832 (the number cut off) the sum of them is 22364 exactly agreeing with the Total first found, clearly evidenceth that the addition was truly performed; but if they had disagreed then the work had been erroneous. The like course must be taken for the proof of those sums which have different denominations as in Money and Weight, as by the examples following will appear. Other Examples proved. 1 Example of Money. li. s. d. q. 37 16 9 3 21 9 8. 1 13 12 9 2 1 Total 72 19 3 2 2 Total 35 2 5 3 Proof 72 19 3 2 li. ou. pw. gr. 32 9 12 16 17 11 6 9 34 8 15 10 8 10 4 7 94 3 18 18 61 6 6 2 94 3 18 18 There are are other ways to prove Addition, by casting away of all the nine in numbers of one denomination, and of all the twelves, twenties, and nine, in pounds shillings and pence, etc. but this as the most certain and easy I embrace, and thus much for Addition and the proof thereof. Subtraction. SUbtrrction is the taking of one or more small sums out of a greater, as 7 s. out of 12 s. or 37 li. out of 100 li. or 137 foot out of 983 foot, and the like. As in Addition the sums to be Added may be either of one or of divers denminations, so likewise they may be in Subtraction, and the manner of placing them is the same, for you must set Unites under Unites. Ten under Ten, Hundreds under Hundreds, etc. Example 1 number given 986 number to be subtracted 234 remainder 752 Of Subtraction of numbers of one denomination, let it be required to subtract 234, out of 986, place the numbers one unde the other as you see done in the Margin, draw a line under them and begin with the first figure towards your right hand, which is 4, saying take 4 from 6, and there remains 2, place 2 under the line, and go to the next figure which is 3, saying, take 3 from 8, and there remains 5, place 5 under the line, and go to the next figure which is 2, saying take 2 out of 9 and there remains 7, place 7 under the line, and your subtraction is ended, and it is evident by the work, that if you take 234 out of 986 there will remain 75●, which you may thus prove, for if you add the 234 to 752, you shall find the sum of that addition to be 986, which is equal to the whole sum from which 234 was subtracted, Example. 2. 96527 2976 9355r Let it be required to subtact 2976 out of 96527, place the numbers one under another as in the margin you fee done, then draw a line under them, and beginning with the first figure towards your left hand, say take 6 out of 7, and there remains 1, place 1 under the line and prodeed to the next figure, saying 7 out of 2 I can not (wherefore you must always add 10 to the number above which in this example is 2, and it makes it 12,) therefore take 7 out of 12, and there remains 5, place 5 under the line, and (because you added 10 to the 2 to make it 12, you must) carry a unite to the next figure, saying, one which I carried and 9 is 10, take ten out of 5, which I cannot, therefore I must add 10 to 5 and it makes 15, and say 10 out of 15, and there remains 5, place 5 under the line, and because you added 10 to 5 to make it 15, you must therefore carry a unite to the next figure saying, one which I carried, and 2 is 3, take 3 out of 6 and there remains 3, place 3 under the line, and because there is no more figures to be subtracted from the number above, you must, say nothing from 9 and there remains 9, set the 9 under the line, and your Subtraction is ended. Other Examples for Practice. li. reams of Paper sheep Lent 5762 bought 9765 from 1000 Paid 378 sold 6529 take 394 Rests to pay 5384 unsold 3236 remains 606 Subtraction of Numbers of divers Denominations, 1 Of English Money. In Subtraction of numbers of divers denominations, you must observe the same order as in Addition, namely, to place every number in due order, with respect to its denomination, as pounds under pounds, shillings under shillings, etc. the greater number always uppermost, and drawing a line under them, begin with the least denomination first, subtracting it from the line above, and setting the remainder under the line as in whole numbers, but if the pence or shillings in the upper row be smaller than those in the nether row, you must add 12 d. or 20 s. to the smaller number, that so subtraction may be made, as by the examples following will appear. Example 1. li. s. d. Lent 269 18 10 Paid 38 12 8 rests 231 6 2 Let it be required to subtract 38 li. 12 s. 8 d. out of 269 l. 18 s. 10 d. Place your numbers as in the margin, then beginning with the least denomination first, (which in this Example is pence) say 8 d. from 10 d., and there remains 2 d. set the 2 d. under the line, and proceed to the next denomination which is shillings, saying take 12 s. out of 18 s. and there remains 6 s. place 6 s, under the line, and go to the pounds, saying 8 out of 9 and there remains 1, place 1 under the line, and say 3 out of 6, and there remains 3, than (because there is no more figures to be subtracted) say nothing out of 2 and there remains 2, which set under the line, so is your subtraction ended, and the remainder is 231 li. 6 s. 2 d. Example 2. li. s. p. Lent 9320 10 07 Paid 2628 16 10 Rests 6691 11 8 Let it be required to subtract 2628 li. 16 s. 10 d. out of 9320 li. 10 s. 7 d. place the numbers in order, and beginning with the pence, say 10 d. out of 7 d. I cannot, (therefore I must add 12 d. (which is one shillings) to 7 d. and it makes 19 d.) but 10 d. out of 19 d. and there remains 9 d. set the 9 d. under the line, and (because I added 12 d. to 7 d.) I must therefore carry one to the place of shillings, saying 1 s. which I carried, and 16 s. is 17. then 17 s. from 10 s. I cannot take, therefore, I must add 20 s. (which is one pound) to 10 s. and it makes 30 s. and 17 s. out of 30 s. and there remains 13, set 13 under the line, and carry one to the place of pounds, saying one which I carried and 8 is 9, take 9 of o I cannot, but 9 out of 10, and there remains 1, set 1 under the line, and carry a unite to the next place, saying 1 which I carried and 2 is 3, take 3 out of 2 I cannot, but 3 out of 12, and there remains 9, place 9 under the line, and carry 1 to the next place, saying 1 which I carry and 6 is 7, take 7 out of 3 I cannot, but 7 out of 13, and there remains 6, place 6 under the line, and carry one to the next row, saying 1 and 2 is 3, take 3 from 9 and there remains 6, place 6 under the line, so is your Subtraction ended, and the remainder is 6691 li. 13 s. 9 d. Example 3. Suppose a man had lent to another man 1000 pound, and that the borrower had paid thereof at one time 127 li. at another time 430 li. 10 s. and at a third payment 50 li. and the creditor would know how much he hath received, and how much is owing of his debtor. Place the numbers as here you see, li. s. d. Money lent 1000 00 00 127 00 00 paid at several times 430 10 00 50 00 00 paid in all 607 10 00 rests to pay 392 10 00 first the sum of money lent, and draw a line under it, than set the sums paid at several times one under another, and draw a line under them: Then add all the sums which have been paid at several times together, which make 607 li. 10 s. which is the sum which the debtor hath paid in all, then subtract this 607 li. 10 s. from 1000 li. and there will remain 392 li. 10 s. and so much is still owing to the Creditor. Other Examples for Practice. li. s. d. Lent 2601 13 6 Paid 98 7 9 Rests 2503 5 9 li. s. d. Owing in all 100 00 00 Paid in all ●6 10 06 Rests to pay 63 10 06 li. s. d. q. Lent 3625 16 08 03 Paid at several times 100 00 00 00 Paid at several times 336 10 06 02 Paid at several times 039 12 09 02 Paid at several times 100 00 00 00 Paid in all 576 03 04 00 Rests to pay 3049 13 04 03 The Proof of Subtraction. The Proof of Subtraction is performed by Addition, for adding the number to be subtracted, to the remainder, the Sum of them must be equal to the number given, if you have truly wrought. As in the first example of numbers of one denomination, the number given is 986 the number to be subtracted is 234 the remainder is 752 Proof 986 Add the number to be subtracted 234, to the remainder 752, the sum of them is 986, equal to the number given. Examples for Practice proved. li. s. d. Lent 62 18 09 Paid 37 19 06 Rests 24 19 03 Proof 62 18 09 li. s. d. Borrowed 100 00 00 Received 36 13 04 Due 63 06 08 Proof 100 00 00 Other Examples in Weight and Measure. 1 Exa. in Troy weight li. ou. pw. gr. Bought of Silver 07 11 13 19 Sold 05 07 03 05 unsold 42 04 10 14 Proof 07 11 13 19 2 Example in Avoirdupois great weight. C. q. li. ou. Bought 37 03 22 11 Sold 13 01 23 06 Rests 24 01 27 05 Proof 37 03 22 11 3 Example in Avoirdupois little weight. li. ou. dr. Bought 84 12 13 Sold 26 08 11 Rests 58 04 02 Proof 84 12 13 4 Example in Time. ho. m. From 364 23 50 Take 76 09 22 Rests 288 14 28 Proof 364 23 50 Questions performed by Addition and Subtraction. Question 1. What number is that which added to 376 shall make 1000? Subtract 376 from 1000, the remainder is 624, the number sought. Question. 2. What number of pounds, shillings and pence must be added to 36 l. 17 s. 3 d. to make that sum up 100 li. subtract 36 li. 17 s. 3 d. from 100 li. the remainder is 63 li. 2 s. 9 d. which added to 36 li. 17 s. 3 d. makes 100 li. Quest. 3. In the year of our Lord 1440, the famous art or mystery of Printing was invented, I would know how long it is since that time to this year of our Lord 1655. From 1655, subtract 1440, the remainder is 215. and so many years are expired since printing was invented. Question 4. An Army consisting of 13721 horse, and 26850 foot, in an engagement there were slain 3760 horse, & 7523 foot, the question is how many were slain in all, & how many horse and how many foot escaped. From the 1372 horse which went out, subtract the 3760 that were slain, there remains 9961, and so many horse escaped, Also from the 26850 foot which went out, subtract the 7523 which were slain, and there remains 19327, the number of foot which escaped, and by adding the 3760 horse which were slain, to the 7523 foot that were slain, their Total is 11283, and so many were slain in all. Multiplication. MUltiplication is that part of Arithmetic which teacheth how to increase one number by another, so that the number produced by their Multiplication shall contain one of the numbers multiplied so many times as there are unites contained in the other. Multiplication may fitly be termed a Compendium of Addition, for that it performeth at one operation the same which to effect by Addition would require many. For instance, if it were required to know how much 7 times 5 is, to perform this by addition, I must set seven five, or five sevens, one under another, and adding them together, I shall find that either of their totals shall contain 35, but this by multiplication is performed with far more brevity, as by examples hereafter shall appear. Before you enter upon the practice of Multiplication, it is necessary to remember the product produced by the multiplication of any one of the nine digits, by any other of the same, as readily to know; that 4 times 5 is 20, 6 times 7 is 42, 2 times 9 is 18, 7 times 9 is 63. 8 times 9 is 72, etc. Which this Table following, will plainly declare, and must be perfectly learned by heart before you attempt to multiply great numbers. Multiplication Table. 2 times 2 makes 4 2 times 3 makes 6 2 times 4 makes 8 2 times 5 makes 10 2 times 6 makes 12 2 times 7 makes 14 2 times 8 makes 16 2 times 9 makes 18 3 times 3 makes 9 3 times 4 makes 12 3 times 5 makes 15 3 times 6 makes 18 3 times 7 makes 21 3 times 8 makes 24 3 times 9 makes 27 4 times 4 makes 16 4 times 5 makes 20 4 times 6 makes 24 4 times 7 makes 28 4 times 8 makes 32 4 times 9 makes 36 5 times 5 makes 25 5 times 6 makes 30 5 times 7 makes 35 5 times 8 makes 40 5 times 9 makes 45 6 times 6 makes 36 6 times 7 makes 42 6 times 8 makes 48 6 times 9 makes 54 7 times 7 makes 49 7 times 8 makes 56 7 times 9 makes 63 8 times 8 makes 64 8 times 9 makes 72 9 times | 9 | makes 81. The use of the Table of Multiplication, and the manner how it is to be read. This Table showeth what the Sum of any two digits multiplied one by another doth amount unto, and is thus to be read, 2 times 2 makes 4, 2 times 3 makes 6, 2 times 4 makes 8: Also 6 times 4 makes 24, 7 times 8 makes 56, 8 times 8 makes (or is) 64, 9 times 9 is 81, etc. In Multiplication there are three terms commonly used, that is to say, The Multiplicand, The Multiplier, and The Product. The Multiplicand is the number to be multiplied. The Multiplier is the number by which the Multiplicand is multiplied, and The Product is the number which is produced by the multiplication of the Multiplicand and the Multiplier together. Thus, if it were required to multiply 8 by 7, here 8 is the Multiplicand, 7 the Multiplier, and 56 is the Product, for 8 times 7, or 7 times 8, is 56. In Multiplication it mattereth not which of the two numbers given is made the Multiplicand, or which the Multiplier, for the Product produced by either will be the same, but the usual way is to make the greater number the Multiplicand; and the lesser number the Multiplier. THE RULE. The numbers to be multiplied must be set one under another. viz. the Multiplicand (or greater number). above, and the Multiplier, (or lesser number) below, the last figure of the Multiplier under the last figure of the Multiplicand, then draw a line under them, and (having learned the preceding Table perfectly by heart) multiply every digit of the Multiplier, into every digit of the Multiplicand, setting the several products under the line, then having finished your Multiplication, draw a line, and add all the products together, and the Sum of those products is the general product of the whole multiplication, as by the following examples will oppear. Example 1. Let it be required to multiply 736 by 7. First, I writ down 736 the Multiplicand, and under it 7, the Multiplyer, and under them I draw a line, than I multiply 7 into every digit of the Multiplicand, saying 7 times 6 is 42, place 2 under the line under 7, and for the four ten keep 4 in mind, then say again, 7 times 3 is 21, and 4 which I kept in mind is 25, place 5 under the line, and keep the two ten in mind, then say again 7 times 7 is 49, and ● which I kept in mind is 51, place 1 under the line, the 5 ten, (because there is no more figures to be multiplied) I set down under the line also, so is the work ended, and the product of this multiplication is 5152. Example 2. Let it be required to multiply 2417 by 5, place the numbers one under another, and draw a line under them as in the margin, then begin your multiplication, saying 5 times 7 is 35, place 5 under the line, and keep the three ten in mind, then say again, 5 times 1 is 5, and 3 which I kept in mind is 8, place 8 under the line, and (because it is less than 10, I keep nothing in mind) then say again, 5 times is 20, place a cipher under the line, and keep the two ten in mind; lastly, say 5 times 3 is 15, and 2 which I kept in mind is 17, which 17 being the last number I place under the line, and so is my Multiplication ended, and the product is 17085. ¶ You may be satisfied of the truth of this work if you will take the pains to set down the Multiplicand 3417, five times one under another, and add them together, as so many several sums, so shall you find the Total of that addition, to be 17085, exactly the same with the product of this Multiplication. Example 3. In the two foregoing examples, the Multiplyer consisted but of one digit, we are now to show how multiplication is performed when the Multiplier consists of more than one figure, therefore in this example, let it be required to multiply 5704 by 37, place your numbers and draw a line under them as you see in the margin, then begin your Multiplication in this manner, saying, 7 times 4 is 28, set 8 under the line, and keep the two ten in mind, then say 7 times nothing is nothing, but the two ten in mind is 2, set 2 under the line, then say 7 times 7 is 49, set 9 under the line, and keep 4 in mind, than lastly, say 7 times 5 is 35, and 4 in mind is 39, which being the last number to be multiplied I set down under the line, so is the mulplication of one of the digits (namely 7) finished. Then begin to multiply the second digit, saying 3 times 4 is 12, place 2 in the second line one place towards the left hand, and keep 1 in mind, then say 3 times nothing is nothing, but 1 in mind is 1, set down 1 by the 2 in the second line, thirdly, say 3 times 7 is 21, place 1 in the second line, and keep the two ten in mind; Lastly, say 3 times 5 is 15, and 2 in mind is 17, which 17, (because there is no more figures to be multiplied) I place in the second line also. Having thus done I draw a line under them, and add these two lines together, as in common Addition of numbers of one denomination, saying 8 is 8, place 8 under the line, then say 2 and 2 is 4, place 4 under the line, then say 1 and 9 is 10, place a cipher under the line, and carry 1 to the next place saying 1 and 1 is 2 and 9 is 11, place 1 under the line and carry 1 to the next row, saying 1 and 7 is 8, and 3 is 11, place 1 under the line, and carry 1 to the next place, saying, 1 which I carry and 1 is 2, place 2 under the line and so is your multiplication ended, and the product is 211048. Example 4. Let it be required to multiply 57325 by 4032, place the multiplicand and multiplierone under another, & draw a line asbefore then proceed to the multiplication as formerly, saying, first, 2 times 5 is 10, set down a cipher and keep 1 in mind: then 2 times 2 is 4, and 1 in mind is 5, place 5 under the line then 2 times 3 is 6, set 6 under the line: then 2 times 7 is 14, set down 4, and keep 1 in mind, than 2 times 5 is 10, and 1 mind is 11, which 11 (being the last) I set down. The multiplication of one of the digits being finished, proceed to the multiplication of the next, saying 3 times 5 is 15, set down 5 in the second line a place more towards the left hand, and keep 1, than 3 times 2 is 6, and 1 kept is 7, set down 7, than 3 times 3 is 9, set down 9, than 3 times 7 is 21, set down 1 and keep 2 in mind, than 3 times 5 is 15, and 2 in mind is 17, which being the last set down also. Two of the figures of the multiplier being finished proceed to the third, which (in this example) being a cipher, you may wholly neglect, and proceed to the multiplication of the fourth figure, only remember to remove the product of the fourth figure one place more to the left hand, as in the example you may see, for the cipher, though it be not written down, yet it must keep its place, and the figures following must be removed a place farther. Then for the Multiplication of the fourth and last digit, say 4 times 5 is 20, set down a cipher (under 9) and keep 2 in mind, than 4 times 2 is 8 and 2 in mind is 10, set down a cipher and keep 1 in mind, than 4 times 3 is 12, and 1 is 13, set down 3 and keep 1, then 4 times 7 is 28, and 1 kept is 29, set down 9 and keep 2, then 4 times 5 is 20, and 2 kept is 22, which 22 (because the multiplicatiou is ended) set down also. Having thus multiplied all the digits severally, draw a line under their products, and add them altogether as in the former example, so shall you find their general product to be 231134400. Other Examples for Practice. Compendiums' in Multiplication. 1 If the Multiplier consist of cyphers in the last place or places, you may omit the multiplcation of them, and place the former figures of the Multiplier under the Multiplicand, thus if it were required to multiply 3257 by 2600. place the numbers as you see in the margin, then multiplying 3257 by 26, the Product will be 84682, to which if you add two cyphers, (because there were two cyphers in the Multiplier) it will be 8468200, which is the true product of the multiplication. 2 If it be required to multiply any number by 10, 100, 1000, 10000, &c, You have no more to do but to add so many cyphers to the multiplicand as there are cyphers in the multiplier, thus if you were to multiply 365 by 10, the product will be 3650, or by 100, it would be 36500, or by 1000, it would be 365000, or by 10000, it would be 3650000, etc. 3 If any number given were to be multiplied by 5, you may abreviate your work thus, add a cipher to the Multiplicand, take half that number, and it shall be the product required, thus if it were required to multiply 8627 by 5, add a cipher to the multiplicand, than it is 86270 the half whereof is 43135, which is the product required. The Proof of Multiplication. The most certain proof of Multiplication is by Division, but because Division is not yet known, I will here show a near way by which Multiplication may be proved. Which is thus, THE RULE. First, take a Cross as in the Margin, then, any sums being multiplied, you may prove the truth of your work in this manner, (1) Cast away all the nine which you can find in the multiplicand, what remaineth set on the right side of the Cross. (2) Cast away also the nine in the Multiplyer, and what remains set on the left side of the Cross. (3) Multiply the figure on the right side of the Cross by that on the left side of the Cross, and out of that product cast away the nine, setting the figure remaining over the Cross, than (4) Cast away all the nine in the product, and if the figure remaining be the same with that which standeth over the Cross, then is your multiplication truly performed, otherwise not. 1 Cast away all the nine in the Multiplicand, saying 4 and 3 is 7, and 2 is 9, which being rejected there remains 4, which I set on the right side of the cross, than 2 Cast away all the nine in the Multiplier, saying 2 and 3 is 5, which (being less than 9) I set on the left side of the cross, than 3 Multiply 4 by 5, saying 4 times 5 is 20 from which cast all the nine, and there remain 2, place 2 over the cross, and 4 Cast away all the nine in the Product, saying, 2 and 5 is 7, and 4 is 11, cast away 9, and there remains 2, which exactly agrees with the figure over the cross, and demonstrates that the multiplication is truly performed. To multiply by any of the nine Digits without charging the memory. To multiply any number by 2, Either double the number in your mind, or add it, by setting it down twice, so 57325 produceth 115750. To multiply any number by 3, To the number given, add the double thereof, the sum is the product, so 57325 produceth 171975. To multiply any number by 4, Double the duplication in your mind, so 57325 produceth 229290. To multiply any number by 5, Conceive a cipher added to the given number, and in your mind half thereof is the product, thus a cipher added to 57325, maketh it 573250, the half whereof is 286625. To multiply any number by 6, Take half adding a cipher, and add to the half, the figure standing next before, thus 573250 produceth 343950. To multiply any number by 7, Take half and add it to the double of the former figure, supposing a cipher added as before, so 57325 thus ordered, produceth 401275. To multiply any number by 8, Double each former figure, and subtract it from the following, so 57325 produceth 458600. To multiply any number by 9, Suppose the number multiplied by 1, then subtract each former figure from the following, beginning with that next before the cipher, the remainder is the product, so 57325 produceth 515925. Questions performed by Multiplication only. Question 1. If a piece of land be 236 perches long, and 182 perches broad, how many square perches are contained therein? multiply 236 the length, by 182 the breadth, the product is 42952, and so many square perches are contained in such a square piece of land. Question 2. In a year there are 365 days natural, and in every day 24 hours, how many hours be there in a year? Multiply 365 the number of days, by 24 the number of hours, the product is 8760, and so many hours be there in a year. Question 3. From London to Coventry it is accounted 76 miles, how many yards therefore is it from London to Coventry? Multiply 1760 (which are the number of yards contained in one mile) by 79, the product is 133760, and so many yards are between London and Coventry. Division. DIvision is the just contrary to Multiplication, for that turns small denominations to greater, as Multiplication turns greater to smaller. Or (in whole Numbers, of which only we yet speak) Division is the ask how many times one Sum is contained in another? and the number which answereth to that question is called the Quotient. And the number containing, which is to be divided, is called the Dividend. And the number contained, or by which the Dividend is to be divided, is called the Divisor. And as often as the Dividend contains the Divisor, so often doth the quotient contain Unity. The ways of performing Division are divers I will begin with that which is most used and taught, which is as followeth. THE RULE. Place the Divisor under the Dividend, so that the figures next to the left hand stand directly one under the other, if the rest of the Divisor be not greater: or if all the Divisor be greater than that above it, than the said Divisor must be devolved one place further toward the right hand; having so placed them, try how many times the lower figures are contained in the upper figures, and write that figure which answereth that question within a crooked line in the margin of the work, which is called the Quotient, and by it multiply the first figure of the Divisor, and take the product out of the figures directly over it, beginning the Subtraction toward the left hand; then cancel that figure of the Divisor, and also, that of the Dividend which hath been already used, with a light dash of a pen, and write the remain (when the product of the first figure multiplied by the quotient is subtracted as before) just over the figure used and canceled; the● proceed to do the like with the second, third, and fourth figure of the Divisor if there be so many; till having canceled it all, and set the remain orderly above the Dividend, you have finished one work. Now if the Dividend have still some figures untouched towards the right hand, then remove the Divisor still toward the right hand, but one place at a time, and then again ask or try how many times the lower may be had in the upper, and write the answer in the quotient whether it be 1 or more, (only it cannot be above 9) or nothing, then put 0 in the quotient, and multiply the Divisor by this new figure, and subtract the product, setting the remainder orderly above, as before, this work must be repeated by removing the Divisor still one place towards the right hand, until Unity in the Divisor stand under Unity in the Dividend, and then the work is done. Example. 1. Let it be required to divide 4096 by 3 Place them thus Ask how many times 3 in 4 the quotient is 1, which is put within a crooked line by itself. Then in your mind multiply the Divisor 3 by the quotient 1. And having said these words Once 3 is 3, presently cancel the 3. And having added these words, Out of 4, cancel the 4. And after these words, and there remains 1, writ 1 just over the 4, as you see here done. Then remove the divisor one place towards the right hand, saying, how many times 3 in 10, the answer is 3, which writ in the quotient; and in your mind multiply the Divisor 3 by the quotient 3, the product is o, wherefore say three times three is nine out of ten, and there remains one, than (having canceled the 10 and the 3, writ over them 1. Again, remove the Divisor 3 one place more, ask how many times 3 in 19? the answer being 6, write 6 in the quotient, and say 6 times 3 is 18 out of 19, and there remains 1, wherefore having canceled the 3, and the 19, writ 1 over 3, and remove the Divisor once more, and ask how many times 3 in 16? answer is 5, which writ in the quotient, then in mind say 5 times 3 is 15 out of 16, and there remains 1, and cancelling the 16 and the 3, write over 3, 1: Now because the Divisor 3 is advanced so far till it is come to stand under 6 in the Dividend, which 6 is the place of Unity there, the said Divisor cannot be removed any more, and therefore the division is ended, and the quotient being 1365, shows that the Divisor 3 is contained in the Dividend 4096, 1365 times; and 1 remaining, which 1 being less than the Divisor 3, doth not contain it once, but one third part of once, which after the Reader hath skill in broken numbers must be joined to the quotient thus 1365⅓. The best proof of this Division is by multiplying the quotient into the Divisor, and to the product add the remain; then if the work be well done, the sum shall be equal to the Dividend. So 1365 multiplied by 3 produceth 4095, to which adding the remain 1, the sum 4096 is equal to the Dividend. Here we divided only by one figure 3, because the first Example being easy and clear, should be a fair Introduction to the second, Note that if the Divisor had been greater than 4, as 5, the work must have begun thus So the quotient would have been 819½, which is one place less. Example 2. Let it be required to divide 1310720 by 4096, Place them thus Now the question to be asked is, how many times 4096 is there in 13107? to find an answer to this question, the Reader which hath but an indifferent faculty of judging; may do it, (for the most part) by considering the first figure in the divisor, as here 4, for presently he knows that 4 times 4 is 16, which cannot be had in 13, therefore the first figure in the quotient must be less than 4. Again, it cannot be much less, because the second place in the Divisor is 0. He may therefore venture on 3, And having put 3 in the quotient, say 3 times 4 is 12, which subtract out of 13 over it thus, say 2 out of 3, and there remains 1, and 1 out of 1 and there remains 0, cancelling the 0 and the 13, and set the remain 1 over the 4, as you see. Then go on saying 3 times 0 is 0, out of 11, and remains still 11, again 3 times 9 is 27, take 7 out of 10 there remains 3, to be set over the 0, which is over 9, and 2 with 1 borrowed (to make the 0 10, from the which the 7 was taken) is 3, and 3 out of 11, remains 8, which writ over the place of 0 in the Divisor, cancelling the 9 in it, and also those figures of the Dividend 110, out of which you have taken any thing; then lastly, say 3 times 6 is 18, take 8 out of 7 I cannot, therefore (borrowing 10,) say 8 out of 17 remains 9, which writ over the 7, then in the next place, the 10 borrowed is 1, and the 10 in the 18 is 1. Say therefore 1 and 1 is 2, out of 3, and the remain is 1, which writ over the 3, having still canceled the figures which you have used orderly as you go. Now the whole Divisor being canceled, it must be removed on one place further, and placed as here. Now ask again how many times 4096 in 8192? it will be found (because of the 0 after the 4) as often as 4 in 8, that is two times, put therefore in the quotient 2, and work as before, saying 2 times 4 is 8, out of 8 remains 0, than 2 times 0 is 0 out of 1 remains 1, than 2 times 9 is 18, take 8 out of 9 remains 1, (which put over the 9,) and 1 out of 1 in the next place remains 0, than lastly, 2 times 6 is 12, that is, 2 out of 2 there remains 0, and 1 out of 1, there also remains 0, cancel and put the remainder over as formerly. Now again, the Divisor being all canceled, should be removed; and ask how many times 4096 in 0000, the answer is 0, which being put in the quotient, the work is all done. And the quotient 320, shows that the Divisor 4096 is contained in the Dividend 1310720, three hundred and twenty times. And whether the Divisor have 2, 3, 5, 7, or more places, the working is still like this, not differing from it at all. Except for brevity in some cases, where the Divisor toward the right hand hath one cipher or more; for those Ciphers may be placed orderly under the Dividend at first, and remain there till the work be done, with the rest of the Divisor, which must needs shorten the Division. As if 2587645 were to be divided by 15000, place them thus And say how many times 15 in 25? answer is 1, which writ in the quotient, and multiply in mind the divisor by the quotient, saying once one is one, out of 2 there remains 1, then cancel the 2, and the 1 under it, and write the remaining 1 over it, as here is done, then say once 5 is 5 out of 5, and there remains 0, therefore cancel the two 5, and over the uppermost write 0. Now remove the Divisor one place, and ask how many times 15 in 108? answer is 7, which writ in the quotient, and multiply, saying 7 times 1 is 7, out of 10 remains 3, which writ over the 0, then say 7 times 5 is 35, take 5 out of 8 remains 3, which writ over 8, and 3 out of 3 remains 0. Now again, remove the Divisor, and ask how many times 15 in 37? answer is 2, which writ in the quotient, and say two times 15 is 30, out of 37 remains 7, and the Division is ended, the remain being 7645. Proof of this. Multiply the quotient 172 by the Divisor 15 860 172 The Product is 2580 Before which put the three Ciphers, And then it is 2580000 To which add the remain 7645 The Total is 2587645 Which is equal to the Dividend, and therefore the work is right. So if one would divide any sum by 10, 100, 1000, 10000, etc. he need but cut off the first, two first, three first, or four first figures towards the right hand, the other figures shall be the quotient, and those cut off, the remain. As if 1587645 be divided by 1000, the quotient is 1587., and the remain 645. If one would know how many pounds Sterling are in 95670 shillings; having placed them thus, that is, 2 under 9, and 0 under 0, then divide by 2, which is very easy. So the quotient 4783, shows that in the number of shillings given, is contained 4783 pounds, and 10 shillings remaining. The reason why the Divisor was 2, is because there are 20 shillings in one pound, and therefore any number of shillings is turned into pounds, by dividing by 20, that is by 2, putting the cipher under unity, only to fill a place at last. And this way of turning any number of shillings into pounds may be easily effected by memory, if you suppose the last figure of your given number to be cut off with a line, or comma, and taking the half of the other figures. Thus, let the given number of shillings be 5739; Imagine the last figure 9 to be separated from the rest by a line thus 573 9, now by memory take the half of 573, by saying in your mind the half of 5 is 2, (and 1 remaining, which makes the 7 following 17,) the half of 17 is 8, (and 1 remaing) the half of 13 is 6, (and 1 remaing, which 1 is 10 s. to be added to the 9, and the whole is 286 li. 19 s. In this way of Division, as in all others, if the remain at last be greater than the Divisor, the quotient is not just, but too little, which may be remedied (without beginning the work again) by dividing the remainder only by the same divisor; for thereof will arise a new quotient, which added to the former quotient, the sum will be the just quotient. So if 7290 be divided by 27, the numbers being placed thus, Because 27 can be had in 72 but twice, put 2 in the quotient, saying 2 times 27 is 54, out of 72, and there remains 18 which writ over 72, cancelling as before is showed; then removing the Divisor, say how many times 27 in 189? answer is 7, but if one should mistake, and write 5 in the quotient, and say 5 times 27 is 135, out of 189, remains 54 and write it over as before; and remove the Divisor, and say, how many times 27 in 540? answer is 20, but should not be above 9, say therefore, 9 times 27 is 243, out of 540, and the remain is 297, which being the last remain, and greater than the Divisor, shows the quotient 259 is too little. Wherefore divide the last remain 297, By the Divisor 27 saying, twenty seven in 29 once, and write 1 in the new quotient; and say once 27 is 27, out of 29, remains 2 which writ over 9, and remove the Divisor, and say, 27 in 27, justly once, so writ 1 in the quotient, so the quotient is 11, which added to the former quotient 259, gives 270, which is the true and whole quotient. A Second way of Division. Although (as I have said) the former way is more used, yet this may seem plainer and more natural to some, I will therefore give one example of it. Example. Let it be required to divide 6477734 by 334, where the first figure in the quotient is easily seen 1, subtract once 334 out of 647, and write the remain under a line, then see how many times 334 in 3137, answer 9, by which multiply the divisor 334 in another paper, the product is 3006, which subtract out of the remain, than the new remain is 131734. Again, how many times 334 in 1317, answer is 4, for the third figure in the quotient; by which multiply the Divisor, the product is 1002, which take out of the later remain, as in the margin, then will remain 31534. Now try how many times 334 in 3153, answer is 9, for the fourth figure in the quotient, by which multiply the divisor (in a by paper) the product is 3006, (as might have been seen above in the second subtraction) which subtracted out of the later remain, there remains now 1474. Lastly, ask how many times 334 in 1474, answer is 4, for the last figure in the quotint, by which multiplying the divisor 334, the product is 1336, which subtracted from 1474, there finally remains 138, which being less than the divisor, shows the division is done. Proof of this. The several subtractions and the final remain added together. The Total equal to the dividend 6477734 If the former (as I have said more usual) way seem difficult to beginners, because the products of the divisor into the several figures in the the quotient are not set down but mentally made; and also because the subtraction of them gins towards the left hand: and lastly, because the remain is still set above: yet this later way which agrees altogether with plain Subtraction before taught, I hope is so plain, that any diligent reader may acquire it, without a Tutor. And yet for the better satisfaction and help of the young Learner, I will add another way or two more of Division. A Third way of Division. There is another kind of division which is very much used, and is in most request with those who have most occasion to divide great numbers, the manner of working is not much unlike the way before taught, one or two examples will make it plain. Example 1. Let it be required to divide 162483 by 1321, set down your numbers as you see them placed in the margin, viz. First, set down 162483 the dividend, then on the left hand thereof set the divisor 1321 with a crooked line between them, then on the right hand thereof make another crooked line which must serve to set the figures of the quotient in, so are your numbers placed in due order, then draw a line under the dividend, and make a prick under the figure 4, because so far the figures of the divisor would extend if they had been placed underneath the dividend, according as in the other examples, this prick serves only to show how far you have proceeded in your work, and must at every division be removed a place further, till at length you come to the last figure of the dividend: your numbers being thus placed with a line under them, you are ready for the work, which must be performed according to the directions of the following rule. THE RULE. Demand how often the divisor may be had in the dividend, and place that number in the quotient, then multiply the divisor by the quotient, and place the product under the line: then subtract this product from the dividend, and set the remainder under the product, then make a prick under the next figure of the dividend, and bring that figure down to the remainder, and then proceed as before. Example, Your numbers being placed as is before directed, you may begin your work in this manner, first, say how many times 1321, can I have in 1624.? say once, place 1 in the quotient, by which 1 multiply the divisor 1321, beginning at the left hand, saying, once one is 1, place 1 under the line, then once 2 is 2, set 2 under the line, then once 3 is 3, place 3 under the line, lastly, once 1 is 1, place 1 under the line, then subtract this 1321, from 1624., and there will remain 303, to this 303 bring down the next figure in the dividend, namely 8, so will that number be 3038, under which draw a line, and repeat the same work again, saying, how many times 1321 can I have in 3038, which may be had two times, place 2 in the quotient, by which 2 multiply the divisor 1321, saying 2 times 1 is 2, place 2 under the line, than 2 times 2 is 4, place 4 under the line, than 2 times 3 is 6, place 6 under the line, lastly, 2 times 1 is 2, place 2 under the line, and subtract this 2642 from 3038, and there will remain 396, to this 396 bring down the next figure of the dividend, which is 3, so is this number made 3963, under which draw a line, and repeat the work once again, saying, how many times 1321 can I have in 3963, which may be had 3 times, by which 3 multiply the divisor 1321, saying 3 times 1 is 3, than 3 times 2 is 6, than 3 times 3 is 9, and lastly, 3 times 1 is 3, which place under the line, and subtract it from the line above, which in this example is the same number, therefore there remains nothing, and the work is ended, but if any remainder had been, that should have been set under the line, as by the examples following will appear. Other Examples for Practice. In this Example where 793058, is divided by 5624, you may perceive that the quotient is 141, and 74 remaining, so that the real quotient is 141 74/5624. The Proof of this Division. This kind of division is proved by Addition, for If you add the several products arising from the multiplication of the several quotients into the divisor, and also add thereunto the remainder (if any be) the total of this addition shall be equal to the dividend, if there be no error in the work. So in the example following, if you add 4325 the first product, and 30275 the second product, and 2796 the remainder, together, in the same order as they now stand in the example, you shall find the Total of this Addition to be 76321, equal to the dividend, which demonstrates the work to be true. A Fourth way of Division. There is a fourth way of division used by some, not inferior to any of the preceding, for that it is no burden to the memory, and it is also proved by Addition. The manner of placing the figures is the same with the third kind of division last taught; And for the performance of the work, this is THE RULE. First writ down the dividend, and on the left hand thereof the divisor, with a crooked line betwixt them, and on the right hand of the dividend make another crooked line wherein to place the figures of the quotient, then draw a line under the dividend, and also make a prick under that figure of the dividend under which the last figure of the divisor would fall, were it to be placed as in the first kind of division. This done, demand how often the divisor may be found in those figures of the dividend, and place that digit in the quotient, then by this digits multiply the divisor, and set the product of this Multiplication directly under the dividend, beginning at the place where you made the prick, then subtract this product from the figures of the dividend, and place the remainder over the dividend, cancelling the figures of the dividend as you proceed, so is the first figure of the quotient finished, then make a prick under the next figure of the dividend, and demand how often the divisor may be found in the last remainder, and the other figure being added thereto, which place in the quotient, and proceed in all respects as before, till you have pointed all the figures of the dividend. Example, Let it berequiren to divide 763258 by 2345, place your numbers as you see in the margin, and because there are 4 figures in the divisor, therefore make a prick under the fourth figure of the dividend, which is 2, and draw a line, then begin your division in this manner: saying, First, how many times 2345 can I have in 7632, (or how many times 2 can I have in 7) say 2 times, place 3 in the quotient, by which 3 multiply the divisor, saying 3 times 5 is 15, place 5 under the prick, than 3 times 4 is ●● and 1 is 13, place 3 under the line, than 3 times 3 is 9, and 1 is 10, place a cipher under the line, than 3 times 2 is 6, and 1 is 7, place 7 under the line, then subtract 7035 from 7632, saying 5 from 12, and there remains 7, place 7 over 2. (cancelling the 5 and the 2,) and bear one in mind, than 1 and 3 is 4, out of 13, there remains 9, place 9 over 3, and cancel 3 and ●, then 1 which I carried, from 6, and there remains 5, place 5 over 6, and cancel 0 and 6, lastly, 7 from 7 there remains 0, which you need not set down, but cancel the two sevens, then will the work stand as above, and the remainder will be 597. Secondly, make a prick under the next figure of the dividend, namely 5, and say how many times 2345 can I have in 5975, answer two times, place 2 in the quotient, by which multiply the divisor, saying 2 times 5 is 10, place 0 under 5, and carry 1, than 2 times 4 is 8 and 1 is 9, place 9 under 7, than 2 times 3 is 6, place 6 under 9, lastly, 2 times 2 is 4, place 4 under 5, so is the product of this multiplication 4690, which you must subtract from 5975, saying 0 from 5 and there remains 5, place 5 over 5, and cancel 0 and 5, then 9 out of 17, there remains 8, place 8 over 7, and cancel 9 and 7, then 1 carried and 6 is 7, from 9 there remains 2, place 2 over 9 and cancel 6 and 9, lastly, 4 from 5 rests 1, place 1 over 5, and cancel 4 and 5, so have you finished your second figure, and your work will stand thus, and your remainder will be 128. Thirdly, make a prick under the next figure of your dividend (namely under 8,) and ask how many times 2345 can I have in 12858, (or how many times 2 can I have in 12,) say 5 times, place 5 in the quotient, by which multiply the divisor, saying 5 times 5 is 25, place 5 under 8 and carry 2, then 5 times 4 is 20, and 2 is 22, place 2 under 0, and carry 2, then 5 times 3 is 15, and 2 is 17, place 7 under 9, and carry 1, then 5 times 2 is 10, and 1 is 11, place 11 under 4 and 6, so is the product of this multiplication 11725 to be subtracted from 12858, saying 5 from 8 rests 3, place 3 over 8, and cancel 5 and 8, then 2 from 5 rests 3, place 3 over 5, and cancel 2 and 5, then 7 from 8 rests 1, place 1 over 8 and cancel 7 and 8, then 1 from 2 rests 1, place 1 over 2, and cancel 1 and 2, lastly, 1 from 1 rests nothing, so is your work ended, which you shall find to stand as in the margin, the remainder being 1133. The Proof of this Division. This kind of division is also proved by Addition, for, if you draw a line under the work, and add all the figures between the two lines together, (in order as they there stand) taking in the remainder (if any be) the Total of this addition will be equal to the Dividend, if the work be true. Other Examples for Practice proved. Questions performed by Division only. Question 1. If a square piece of Land contained 42952 square perches, and one of the sides thereof be 236 perches long, how long must the other side be? Divide 42952 by 236, the quotient will be 182, and so many perches long must the other side be. Question 2. In a yeaar there are 8760 hours, and in every natural day there are 24 hours. I demand how many days be there in a year? Divide 8760 by 24 the Quotient will be 365, and so many days be there in a year. Question 3. The distance from London to Coventry is 133760 yards, and in one mile there is contained 1760 yards, now I would know how many miles it is from London to Coventry; Divide 133760 by 1760, the quotient will be 76, and so many miles is it from London to Coventry. These Questions performed by Division only, are the converse of those that were performed by Multiplication, which I the rather make choice of, that the reader might see how Multiplication and Division prove each other. There are one or two more kinds of Division, something like these last, but I shall forbear exemplifying them; for much variety helps to make a Book rather great then fit. ¶ Here is to be noted that in the following Rules, where there is continual use of Division, I sometimes use one kind of Division, and sometimes another, for variety sake, but the practitioner may use which he is best skilled in, for they all produce the same effect. Reduction. IS twofold, first, that which turns greater denominations into smaller, as pounds into shillings or pence, this is done by Multiplication as followeth. Example 1. Let it be asked how many pence are contained in 729 li. 11 s. 7 d? First, a shilling is contained in a pound 20 times, therefore multiply 729 by 20, or (which is the same, but shorter) by 2, and put o to the product, as in the margin, this shows that in 729 l. there are 14580 shillings. To which add 11 s. it makes 14591 shillings. Again, because one penny is contained in one shilling 12 times, multiply 14591 by 12, it produceth 175092, to which add the 7 pence, so the sum will be 175099, and so many pence are contained in 729 li. 11 s. 7 d. Example 2. Let it be asked how many pints are contained in 4 Tun, 3 Hogsheads, and 27 Gallons? First, 1 Tun is equal to 4 Hogsheads, therefore 4 Tun is equal to 16 Hogsheads, to which add the 3 Hogsheads, so there is 19 entire Hogsheads. Again, because one Hogshead contains 63 Gallons, multiply 19 by 63, it produceth 1197 Gallons, to which add 27, it gives 1224 Gallons. Lastly, because every Gallon contains 8 pints, multiply 1224 by 8, it produceth 9792, and so many pints are contained in 4 Tuns, 3 Hogsheads and 27 Callons. After the same sort might dry Measures be reduced, as quarters to bushels, pecks, or gallons, and likewise all weights and outlandish Coins, of which the proportion of the greater to the lesser is (before) known, or given. Secondly, it is often requisite to turn smaller denominations to greater: this is done by Division, as followeth. Example 1. Let it be asked how many pounds are contained in 80976 shillings? Divide 80976 by 20, the quotient is 4048 li. and 16 s. remaining, which is the true answer. Example 2. Let it be asked how many pounds are in 109754 d? Because a pound contains a shilling 20 times-and a shilling contains a penny 12 times, there fore if 109754 be divided first by 12, the quotient 9146 shillings, and 2 pence over, then if 9146 be divided by 20, the quotient is 457 pounds and six shillings remaining; so that 109754 pence is equal to 457 li. 6 s. 2 d. Or if 109754 had been at first divided by 12 times 20, that is by 240, (which is the number of pence contained in a pound) the quotient had been 457 pounds, and 74 pence remaining, which is all one with the former; for 74 pence is equal to 6 shillings 2 pence. More instances shall not need herein, because the thing of itself is very clear. Progression. IS also of two sorts, the first is of certain numbers in Arithmetical Proportion from 1, that is, such as differ equally, as 1, 2, 3, 4, 5, 6, where the common difference is 1, (as is easily seen,) or 1, 3, 5, 7, 9, 11, where the common difference is 2, or any other, as 1, 8, 15, 22, 29, 36, where the common difference is 7, this is called Arithmetical Progression. 2 Secondly, of certain numbers in Geometrical proportion from 1, that is such as increase by a common Multiplication, as 1, 2, 4, 8, 16, 32, where the common multiplier is 2, that is the first by 2, produceth the second, and the second multiplied by 2, produceth the third, and so on. Or as 1, 3, 9, 27, 81, 243, where the common Multiplier is 3, this is called Geometrical Progression. Both the common difference (in the first) and the common Multiplication (in the later) shall for shortness hereafter be called the common excess. First, now of the first sort, or Arithmetical Progression, the principal use of this is, 1 If the number of places, and common excess be given, to find the last number. 2 When the number of places, and the last number is given, to find the aggregate, or total sum of all the numbers. 3 When the last number, and the total sum is given, to find the number of places. 4 The Number of places, and total sum being given to find the last number. 5 The last number, and number of places given, to find the common excess. 6 The last number, and common excess being given; to find the number of places. I will instance in no more, few of these ever happening to be used. For the first of these, let there be given the number of places 100 The common excess 1 To find the last number also; 100 THE RULE. Multiply the number of places less by 1, by the common excess, and to the product add the first number; the sum is equal to the last number. So here multiply 99 by 1, the product is 99, (for 1 neither multiplies nor divides) to this add the first number 1, it gives 100 for the last number. Or let the numbers be 1, 7, 13, 19, 25, 31, where the common excess is 6. and the number of places also 6. Now (if the number of places less by 1, that is) 5 be multiplied by the common excess, which is 6, the product is 30, to which adding the first number which is 1, the last number 31, is thereby composed. This is so easy that it needs no proof. 2 For the second, which is, The last number, and the number of places given, to find the total sum of all the numbers. THE RULE. Add the first and last numbers together, and multiply the sum by half the number of places, the product is equal to the aggregate or sum of all the numbers added together. So if to the first number 1 be added the last number 100, it gives 101, which multiplied by 50 (which is half the number of places) produceth 5050, which is equal to all the hundred numbers added together. And hereby may that vulgar question be answered, which is, If a man take up 100 stones placed a yard one from another all in a right line, by one at once, and bring them back one by one to his first standing, how many yards doth he go backwards and forwards? It is showed before that he goes forward 5050 yards, and he must needs come back just as much; that is, in all 10100 yards. which is 5 miles and 3 quarters; wanting 20 yards. Or secondly, suppose the numbers were 1, 9, 17, 25, 33, 41. Whereof the common excess is 8, the first and last added gives 42, which multiplied by 3, (half the number of places) the product is 126, which is the sum of them all. 3 For the third thing, that is, by the last number and the total, to find the number of places. THE RULE. Add the first and last numbers, and by the sum of them divide the total, the quotient will be equal to half the number of places. This is so plain it needs no clearing. 4 For the fourth, if the total, and number of places be given, to find the last number. THE RULE. Divide the total by half the number of places, the quotient is a number, from which if 1 be taken, the rest is the last number. As let the number be 1, 3, 5, 7, 9, 11, 13, 15, or any other (in arithmetical proportion) whatsoever. The sum of these is to be 64. And the number of places is 8, the half of it 4. Now if 64 be divided by 4, the quotient is 16, from which if 1 be taken, there remains 15 for the last number. 5 Now for the fist variety, If the last number and number of places be given, to find the common excess. THE RULE. From the last number take 1, and the remain shall be the Dividend; then from the number of places also take 1, and make this later remain the Divisor▪ then the quotient of this Division shall be the common excess. Example, Let the numbers be 1, 4, 7, 10, 13, 16, from 16 take 1. remains 15, for the dividend, then from 6, (which is the number of places) take also 1, remains 5 for the Divisor. Now when 15 is divided by 5, the quotient is 3. And 3 is also the common excess, or difference between 1 and 4, or 4 and 7, etc. 6 Lastly, let the last number, and the common excess be given, to find the number of places. THE RULE. From the last number take 1, and divide the remain by the common excess; then to the quotient add 1, the sum is the number of places. As let the numbers be 1, 5, 9, 13, 17, 21, 25, 29, from 29 take 1, remains 28, which divide by 4, (which is the excess) the quotient is 7, to which add 1, the sum is 8, which is the number of places, as the reader may easily count. Geometrical Progression. I shall not be so large in this as in the former, because these things are of little use to the Arithmetician, except where a number is to be many times doubled, tripled, or the like, which cannot be so easily abridged here, as in the other, because there the last number arising of many additions of the excess to 1, was easily found by one multiplication: but here the last number being made by many Multiplications of the excess, is therefore many times harder than the other The varieties here shall be but two. 1 The common excess and number of places being given, to find the last number. 2 The excess, and last number being given, to find the total sum. The first of these may thus be found. Let the numbers be 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, the exceesse is 2, the places 10, find out the fifth number (which is easily done, for any one may reckon so fare by heart,) that is here 16, and multiply 16 by 16, it produceth 256, which is the ninth number, lastly, multiply 256 by the excess 2, thence ariseth 512, the number desired. So if the places had been more, as 72, having found the 9 number 256, multiply it by 256, thence comes 65536 for the 17th. number, which multiplied by the excess 2, giveth 131072 for the 18th. place, which multiplied by 131072, gives 17179869184, for the 35 place; and that multiplied again by the excess 2, giveth 34359738368, for the 36 place, that multiplied by 34359738368, than the product will be 1180591620717411327424, for the 71 place, which lastly, multiplied by the excess, gives 2361183241434822654848, for the 72 place, which is the last number of the progression required to be found. Perhaps this may seem somewhat tedious but where things cannot be performed without labour, the Reader must content himself with such Rules as make it less; for it is certain that this way is much shorter than to have multiplied still by the excess 71 times, which else he must have done. All this notwithstanding, he is not bound to use the same numbers, much less in other questions where the number of places is not the same, but whereas I began from the place 9, he may begin at 8, 10, or 12, or where he pleases; so as he remembers still where he is; for this is general, if the number belonging to any place whatsoever, be multiplied by itself, the product shall be the number belonging to twice so many places want one place. Now for the second thing, which is to find the sum of all the numbers. THE RULE. From the last number take the first, and divide the remain by the excess want 1, then multiply the quotient by the excess; and to the product add the first number, the sum of them is equal to the sum of all the numbers. So if from the last number, or 72 place be taken 1 remain is 2361183241434 22654847 which should be divided by 1, (that is the excess want 1 for the excess is but 2) but because 1 neither multiplies nor divides that labour is saved; Now multiply this remain by the excess, the product is 4722366482869645309696, to which adding the first number 1, by making the figure 6 next the right hand to be 7, you have the total sum of all the 72 numbers. A Question resolved by Geometrical Progression. A Londoner sojourning in a Country Market Town, in Winter, made himself a new freeze, Suit and Coat, on which were set 6 dozen of buttons of Silk and Silver; A Baker being in his company liked it so well he would buy it of him; the Citizen consented to let him have it, paying for the first button a single barley corn, for the second 2, for the third 4, and so on doubling to the last. The bargain was liked on both parts for the present, but shortly after revoked, for it could not be performed, and no man can be holden to an impossibility. But why this could not be performed: may be judged; first, by enquiring the worth of so much barley in Money: And secondly, the weight of it; and how it should be removed. 1 For the first, allowing 10000 corns to a pint (which is more then enough) then 5120000 corns make a quarter; and yet (for shortening the division) we will allow 10000000 corns to a quarter; by which dividing the whole number of corns (which is done by cutting off the first 7 figures toward the right hand,) the quotient will be 472239648286964, and so many whole quarters there are omitting the remain, as in this case not considerable. Now allowing barley were to be sold at 15 d. the bushel (which is cheap,) it is so many Angels: and therefore dividing by 2 it is 23611832414143482 pounds sterling: which is in words, Two hundred thirty six millions of millions, one hundred and eighteen thousand, three hundred twenty four millions: one hundred forty three thousand; four hundred eighty two pounds, whick I take to be too much for any Tradesman to get or keep. And reckoning land for ever at twenty year's purchase, if this sum of pounds be divided by 20, the quotient is the yearly rent of 11805916207174 pounds. And this divided again by 365 (the number of days in a year) the quotient 32344975918, that is above thirty two thousands of millions a day for ever. So great a vanity may be concluded on for want of a little premeditation. 2 Now, secondly, for the weight of it, if we put 8 Bushels to weigh 2 hundred pound weight (for sure it doth weigh more) then the whole number of quarters multiplied by 2, giveth the weight of all to be 944473296573928 hundred weight, and if this be divided by 20, (which is but cutting off one figure toward the right hand, and dividing the rest by 2) or which is all one, cut off one figure from the number of quarters, the quotient 47223664828696, is so many tuns. And therefore it will require 47223664828 ships of 1000 a piece to carry it: And consequently if every Nation in the World had above 10000 such ships, yet there must be above four millions of such Nations: which I suppose are not to be found in this World. And here I will leave this, having used this long example, (which though it require more labour as all great examples do, yet the same skill will do it, as if the places had been fewer) that the Reader being throughly exercised thereby, may the easier leap over others which are shorter. THE GOLDEN RULE, Or Rule of Three. THis is the most useful and most easy Rule in Arithmetic, and deserves a Golden name. It is when there are three numbers given or known to find a fourth in proportion with them. But 4 numbers are in proportion, and called Proportional, when as the first is to the third, so is the second to the fourth. As if there were given, 3, 4, and 6, to find a fourth which may be to 4, as 6 to 3, that is double, and that fourth number is 8; And this is called Proportion direct: and the Rule whereby it is done, The Direct Rule. There is also another proportion called Reciprocal; which is when as the first is to the third, so is the fourth to the second: As 3, 4, 6, and 2, this is called The Reverse Rule. In direct proportion, the product of the two middle numbers multiplied together, is ever equal to the product of the first and last multiplied together; which serves not only for a Proof, but a ground of the Rule, which Rule shall here follow: the Reverse Rule being deferred till we have done with this. The Rule Direct. Multiply the second term (or number) by the third and divide the product by the first; the quotient shall be the fourth number desired. Example. Let the three numbers given be 2, 6 3, multiply 6 by 3, the product is 18, then divide 18 by 2, the quotient is 9, which is th' fourth number in proportion with 2, 6, and 3. For as 2 to 3, so 3 times 2, which is 6, is to 3 times ●▪ which is 9 And so the product 18 divided by 2, and the quotient 9, causeth that the product of 2 into 9 shall be also 18, and consequently if 2 be the first of the 4 proportional numbers, and 6 and 3 the two middlemost, then 9 is the last. Otherwise. Divide the second by the first, and multiply the third by the quotient, the product shall be the fourth. So if one divide 6 by 2, the quotient is 3, by which multiply 3, the product is 9, for the fourth number, as before, otherwise still this Rule might be expressed; but where the first way is so short and clear, there many other ways would rather trouble then help the person that should use them. In the first way (which here we mean to use and no other) if the first number be 1, than the product of the second and third gives the fourth, without any Division. Or if the second, or third number be 1, then there needs no multiplication, but dividing the greater of them by the first, the quotient (in whole numbers for yet we speak of them) is the fourth number, which was sought. Note 1. To know when to use the direct, or the Reverse Rule, consider, if more, require more; or if less, require still less: then use the direct Rule, but if more require less, or less more, then use the Reverse Rule, this will be easily understood when we come to examples. Note 2. To know how to place the three numbers when they are confusedly given, remember that two of them are always of one denomination, as both pounds, or both sheep, or both yards, or acres; and the other number hath another denomination, now know that this single number is ever the second number in order. And one of the other two, namely, that which hath some relation to this second, is the first; and the other is the third number, whose relation is sought for in the fourth, whence its plain that the second and fourth are also of the same denomination. And having premised these things, let us now exemplify the Rule in some questions. Question 1. If 3 yards of cloth cost 4 li. what shall 21 yard's cost? Set the numbers in order as in the example, if 3 yards cost 4 li. what 21 yards? Here you see that the first number and the third number are both of one denomination, viz. both yards, and the second number is of another denomination, namely pounds, wherefore the fourth number which is sought for, must be also pounds, therefore multiplying (according to the Rule before given) the second number by the third, and dividing the product by the first, the quotient shall answer the question. First, 21 multiplied by 4, (which is the third ●●mber multiplied by the second) produceth 84, which divided by 3 the first number, the quotient is 28 li. and so much shall 21 yard's cost: for 28 is to 4, as 21 to 3, seeing each contains either 7 times. And the work will stand thus, Question 2. If 4 men eat 2 pecks of corn in a week, how many pecks shall serve 100 men? Place your numbers as here you see, then multiply 100 by 2, (that is the third number by the second, and the product is 200, which divided by 4, the quotient is 50, for the number of pecks required. Question 3. If 20 sheep cost 13 pound 13 shillings 4 pence, what is that for every sheep? Turn the shillings and pounds into pence, thus, Multiply 12 s. by 12 the product is 156 And 13 li. by 240 (because 240 pence make one pound) the product is 3120 To which add the 4 d. 4 It makes in all 3280 Then the question will be, If 20 sheep cost 3280 pence, what shall one sheep cost? By the rule before delivered, I should multiply the second number by the third, but in this example, the third number being 1, it doth not multiply; I therefore divide 3280 the second number, by 20 the first number, and the motient 164, is the price of one sheep in pence, which divided by 12, the quotient is 13 s. and 8 d. remaining, the price of every sheep therefore is 13 s. 8 d. Question 4. How many 10 inch tiles will pave a floor that contains 16 square yards? First, remember there are 36 inches in one yard in length; which multiplied into 36, gives 1296, for the square inches in one square yard; multiply 1296 therefore by 16, thence comes 20736, the sum of all the 16 yards in inches. Secondly, seeing every tile is 10 inches in length, and 10 in breadth, multiply 10 by 10, it produceth 100 for the square inches in one tile; See the manner of work. Then by the golden Rule. If 100 inches require 1 tile; what shall 207; 6 inches require? Here because 1 doth neither multiply nor divide (as hath been several times intimated) I therefore divide the third 20736, by the first 100, the quotient is 207, and 36 remaining. So it appears that 207 is too little, and 208 too much to do the work: the just number being 207 36/100, we shall not trouble the reader with this till he know something of Fractions. Question 5. If 100 li. give 6 li. interest for a year, how much shall 750 li. give? Multiply 750 by 6, the product is 4500, which divided by 100, the quotient is 45 li. for the thing required. Question 6. If 750 li. give 45 li. interest for a year, what shall 100 li. give? Multiply 45 by 100, the product is 4500, which divided by 750, the quotient is 6 li. for the interest of 100 li. for a year. Many other questions might be added, but the rule is so plain that it needs them not; and so general, that he which can resolve one, may aswel resolve any other: And for that reason; and because in all the Rules which follow, this rule will be constantly made use of, I will say no more of it here. The Golden Rule Reverse. Question 1. IF 12 workmen do any piece of work in 8 months: how many workmen shall do the same in 2 months? THE RULE. Multiply the first term by the second; and divide the Product by the third, the quotient is the number desired. Here 12 is not the first number, though it be first named; but the three numbers placed in order, stand thus, 8, 12, 2, for the middle term must always be of the same denomination with that which is required. Now multiply 12 by 8, the product is 96, which divided by 2, the quotient is 48, which answers the question, As in the example, For if 8 months require 12 men; then (a fourth part of 8) 2 months shall require (four times 12, that is) 48 men. For here less requires more, that is less time, more hands; and therefore is it wrought by the Reverse Rule. Question 2. How many els of Tapestry will serve to hang a room 3 yards high, 6 yards long; and 5 yards broad; not regarding Doors, Windows, or Chimney, but as if there were no such? First, multiply 6 by 3, the product is 18, which doubled (because there are 2 sides called lengths) is 36 yards for all the length. Secondly (for the same reason) multiply 3 by twice 5, that is by 10, the product is 30 yards, for all the breadth; which added to 36, gives 66 yards equal to all the length and breadth in yards. But now because els, that is, Flemish els (for such measure are hang sold by) is equal to 3 quarters of a yard, that is, their Ell is to our Yard as 3 to 4. Say therefore if 4 give 66, what 3? multiply 66 by 4, it produceth 264; then divide 264 by 3, the quotient is 88 Again, multiply 88 by 4, and divide the product (which is 352) by 3, the quotient is 117, and 1 remaining, to which the divisor 3 being applied; the number justly answering the question is 117 els, and one third part of an Ell. Note 1. Because here we had to deal with things which had equal length and breadth, that is square yards, and square els; therefore one multiplication and division was not sufficient to proportion this; but if instead of working by 4 and 3, we had done it by their squares which are 16 and 9, it might have been performed at once, thus, multiply 66 by 16, product is 1056, which divided by 9, the quotient is 117⅓, as before, but I began not with this way, for I supposed my reader ignorant of squares. Note 2. It might also have been done, by reducing all the terms into quarters of a yard at the first, and after the number is found, reducing them again to els; but because it is more proper to work thus, till fractions have been taught, I leave that, and proceed to another question. Question 3. If 1 close would graze 21 horses for 6 weeks; then (supposing no waste to be made) how many horses would it feed for 7 weeks? Multiply 21 by 6 it produceth 126, which divided by 7, the quotient is 18. At that rate therefore it would keep 18 horses for 7 weeks. Question 4. If 1 close will feed 18 horses for 7 weeks, how long shall it feed 63 horses. Multiply (according to the rule) 18 by 7, the product is 126, which divided by 63, the quotient is 2, therefore 2 weeks it shall keep them. The like way serves for hay, oats, or any other provision for Man or Beast; which may be of use in Garrisons, and such like cases where scarcity may be feared, to proportion either the mouths to the meat, or meat to mouths I will say no more of this Rule; Neither will I treat of the double Rule of Three, as a rule by itself; but come to the Rule of five numbers, which is held an abridgement of the other. The Golden Rule Compound of five Numbers. Question 1. IF a hundred pound weight (that is 112 pound weight) carried 120 miles cost 14 s. how much shall three quarters of a hundred, (that is 84 pound) cost, being carried 40 miles? THE RULE. Multiply the three last numbers one into another, (that is) the third by the fourth, and that product by the fifth; the last product shall be the Dividend. Again, Multiply the two first numbers together; the product shall be the divisor. This Division being made, the quotient will be the number of shillings desired. Example of the former Question. First, place your numbers according to the tenor of the question thus, Your numbers being placed in order, reduce the 13 s. into pence, and it is 168 d. then multiply 168 by 84, the product is 14112, which multiplied by 40 the later product, it produceth 564480, for the dividend. Then multiply 112 by 120 it produceth 13440, for the Divisor. Divide 5644800, by 13440, the quotient will be 42 pence; which is 3 s. 6 d. and answers the question. In this Rule the first number and fourth, also the second and fifth; and also the third and sixth, are of like denomination and nature. Question. 2. If 100 li. for 6 months yield 3 li. interest, what shall 625 li. yeield for 36 months? Place them, 100, 6, 3, 625, 36. Multiply the three last, as before is showed, the later product is 67500 for the dividend; And the 2 first multiplied make 600 the Divisor, then divide 67500 by 600 (or 675 by 6, which is all one) the quotient will be 112 whole pounds; and 300 (or 3) remaining, which because it is half the divisor, signifies the half of a pound; that is 10 shillings. So the answer to the question is 112 li. 10 s. Which might have been given in one denomination, namely 2250 shillings, if before the work the pounds had been turned into shillings, by multiplying them by 20, as hath been showed before. But since most questions, except such as are studied for the purpose, are apt to end in some fraction, I shall next treat of fractions. Only first, having spoken of the double Rule of Three, this may let you know, that all questions which are wrought at once by the compound rule of five; may be done at twice by the single rule of three: and the doing of them so by two operations is is called the double rule. As in our last question there are two things considerable, the difference of money; and the difference of time. First, for the money, Say, if 100 li. give 3 li. what 625 li.? answer 18 75/100 li. Secondly, for the time; Say if 6 li. give 18 75/100 li. what 36 more. answer 112 50/100 li. But this will be better understood anon; and then the Reader may use that which he likes best. Of Fractions. THe word Fraction signifies a breaking or breach of any entire thing into parts; and when a number is broken so, the parts (which must needs be every one less than the whole; and the whole is accounted but One, or Unity) being less than Unity, are called fractions (that is fragments, or pieces) of Unity. Now the Unite, or entire number which is to be broken, may be any thing, as one pound, in respect of which, shillings and pence, and farthings are fractions; or one shilling, in respect of which, pence, and farthings are fractions; or one penny, in respect of which, farthings are fractions; and the like of weights and measures, or any other thing to be broken into parts. In Fractions, we shall treat first of Numeration, then of Multiplication and Division, then of Reduction; then lastly, of Addition and Subtraction. The reason of this Order will soon be seen; for Multiplication and Division are here much easier than Addition, etc. and therefore aught to be learned before them. Numeration. NUmeration is nothing else but the way of writing Fractions; and that this may be done, we must consider that any Unite, or Number representing an Unite, may be broken into two parts equal; and then each of the parts is called one second, or half; or it may be parted into three equal parts, and then each part is called one third; and two of them are called two thirds; and the like may be understood if it were parted into 4, 5, 6, 7, 8, 9, 20, 50, or 100, or how many soever. Now to write these; do thus, Writ one half Thus ½ Writ one third Thus ⅓ Writ one fourth Thus ¼ Writ one fifth Thus ⅕ Writ one sixth Thus ⅙ Writ one seventh Thus 1/7 Writ one eighth Thus ⅛ Writ one nine Thus 1/9 Writ one tenth Thus 1/10 In every one of these 10 fractions, the Number below the line is called the Denominator, and it shows into how many parts the Unite is broken. The Number above the line shows how many of those parts are taken, or contained in the fraction, and is therefore called the Numerator: So in the fraction ⅗: the Denominator ● shows the Unite to be broken into 5 parts: and the Numerator 3 signifies 3 of such parts to be contained in the fraction; which fraction therefore is called three fifths. And here it is plain, that as the Numerator is in proportion to the Denominator: so is the fraction to 1, for 3/3 or 5/5: or any the like, is equal to 1. And therefore all fractions are quotients of lesser numbers divided by greater, as 4/7 signifies 4 to be divided by 7, and as the dividend 4, is to the divisor 7: so is the quotient 4/7 to Unity. And therefore this line of separation which is drawn between the dividend and divisor, doth properly signify division. Hitherto we have spoken only of such fractions as are less than 1, and those are called Proper Fractions: but there are also 2½, 3¾, 5 1/7, 6⅗, and the like mixed Numbers; which so written signify two and an half, 3 and 3 quarters, five and a seventh, 6 and 3 fifths. These by multiplying the whole Numbers, by the Denominator, and to the product adding the Numerators respectively are turned to 5/2, 15/4, 36/7, 33/5, which are called Improper fractions, because every one of them contains more than Unity. These, nevertheless may be multiplied, divided, added, or subtracted in the same way as are proper fractions. And this shall serve for Numeration of Fractions. Multiplication. THE RULE. MVltiply all the Numerators together, the last product shall be the Numerator of the product required: Likewise multiply all the Denominators together, the last product shall be the Denominator of the product sought. Example 1. If ⅗ be multiplied by 4/9, multiply the Numerator 3 by the Denominator 4, the product is 12, for the Numerator of the new product. Also multiplying the Denominator 5, by the Denominator 9, they produce 45, for the Denominator of the desired product, so that product which was required, is 12/45. Example 2. If ½, ¾, ⅘, 5/9, and 3/11 were to be multiplied all together, begin with the Numerators, saying, once 3 is 3, and 3 times 4 is 12, and 12 times 5 is 60, and 60 times 5 is 180, for the Numerator: Then multiply the Denominators: saying 2 times 4 is 8, and 8 times 5 is 40, and 40 times 9 is 360, and 360 times 11 is 3960, for the new Denominator. So that the product of all these is 180/3960, that is equal to 1/22, as shall be seen hereafter in Reduction. And thus it appears that proper fractions being less than one, are still made less by multiplying; as here the product 1/22 is much less than 3/11, which is the least Multiplyer, and the reasons hereof is plain, for seeing Multiplication is but the taking of a Number, a certain number of times, if that number of times be more than 1, than the Number to be taken is increased by being taken more than once; but if the Number of times be 1, it is not increased, nor diminished, but is still the same; Lastly, If that number of times be less than 1, as ½, the number not being taken once, but half of once, produceth a number less by half; that is the half of the number to be taken; and the like reason is of all others. Example 3. Multiply the mixed Numbers 3½, 4⅓, and 5¾: First, (as hath been shown already) turn them to improper fractions: thus, first, say 2 times 3 is 6, and 1 is 7. So the first is 7/2. Secondly, 3 times 4 is 12, and 1 is 13: so the second is 13/3. Lastly, 4 times 5 is 20, and 3 is 23: so the last is 23/4 Now the fractions to be multiplied are 7/2, 13/3, and 23/4; First, for a new Numerator, say, 7 times 13 is 91, and 91 times 23 is 2093, for a new Numerator. Then say, 2 times 3 is 6, and 6 times 4 is 24. So the new Denominator is 23. And the product of all these fractions is 2093/24, that is, if real division be made, 87 5/24. Division. DIvision of one fraction by another, is but cross multiplication of them; that is, the Numerator of the one, by the Denominator of the other, and hereby the proportion of one fraction to another is seen. Example 1. Divide ¾ by 6/8, to do it, set them thus: and multiply as the cross leads, saying, 3 times 8 is 24, for a new Numerator, and 6 times 4 is also 24, for a new Denominator; so the quotient is 24/24, that is 1, which shows the fractions to be equal one to another. Example 2. In division it is to be remembered that the Numerator of the quotient ever ariseth of the Numerator of the Dividend; And the Denominator of the quotient comes of the Denominator of the Dividend, each being cross multiplied, as before. If a Fraction be to be divided by a whole number, multiply the Denominator by that number, the product gives the new Denominator, and the Numerator remains the same. So if ¼ be divided by 9, say 9 times 4 is 36. So the quotient is 1/36. Or if ¼ were to be multiplied by 9, the product (by multiplying the Numerator by 9), will be 9/4: that is, 2¼. Example 3. For 320/8 is equal to 40, and 45/9 equal to 5, but 40 contains 5 eight times. And so in the second example, it may be proved, that as 27 to 20, so is ⅗ to 4/9. For first multiply the two middlemost, then 20 times ⅗ is 60/5, that is 12. Secondly, multiply the first and last, and then 27 times 4/9 is 108/9: that is also 12. Wherefore by that which hath been said in the Golden Rule, the four Numbers 27, 20, 3/5, 4/2, are proportional. Reduction. REductioon of Fractions is threefold. 1 To reduce one fraction (which is not already in the least) to a lesser Denomination. 2 To reduce many fractions of divers denominations, to one denomination. 3 To reduce any fraction from one denomination (as near as may be) to any other denomination desired. For the first of these, divide both the Numerator, and the Denominator, by the greatest common divisor that you can think of; the two quotients being placed respectively in a fraction, that fraction shall be equal to the former fraction, and in lesser terms. So (in the 3 Example of Division) to reduce 2880/360 to 8/1, divide 2880 by 360, quotient is 8, then divide 360 by 360, the quotient is 1, and the new fraction 8/1 is equal to the former fraction, 2880/360, and in less terms; as you may see. But to find the greatest common divisor. The Rule is, Divide the greater term by the lesser (I mean by terms, the Numerator and Denominator) and by the remainder (if any be) divide the divisor, and if any thing still remains, by that divide the last divisor, continuing this course till nothing remain greater than Unity; that divisor which is last of all is the greatest common measure of both terms, by which both being divided, and the quotients placed like a fraction, that fraction shall be equal to the former fraction, and in the least terms. Example. Reduce 148/16 to the least terms; first divide 148 by 16, the quotient is 9, and 4 remains: again, divide 16 by 4, the quotient is 4, and nothing remains; wherefore taking 4, (the last divisor) for the greatest common divisor, by it divide 148, the quotient is 37, and by it divide 16, the quotient is 4. These two last quotients placed orderly in a fraction, make 37/4, which is equal to 148/16, and in the least terms, for no number greater than 1, will divide evenly both 37 and 4. Other ways there are of lessening fractions, as dividing the terms (if they be even Numbers) by 2, and the quotients (if even) again by 2, or else by 3, o● any other Number that will divide them both evenly, that is, leave nothing remaining, but the former Rule being general and easy, shall serve for all. Now secondly, to reduce many Denominations to one common denominator; let the fractions be ½, ¾, ⅘, ⅞, 9/10, to be reduced all to the denomination 3200. THE RULE. Multiply all the Denominators together: saying 2 times 4 is 8, and 8 times 5 is 40, and 40 times 8 is 320, and 320 times 10 is 3200, this last product 3200, shall be the common Denominator. Then to get Numerators for every one of them. Multiply every particular Numerator into all the Denominators except his own: As first, for the first, say 1 time 4 is 4, and 4 times 5 is 20, and 20 times 8 is 160, and 160 times 10 is 1600. For the first Numerator; so the first fraction reduced is 1600/3200. Then for the second Numerator: say, 3 times 2 is 6, and 6 times 5 is 30, and 30 times 8 is 240, and 240 times 10 is 2400. So the second fraction reduced, is 2400/3200. After the same manner may the other three be reduced to 2560/3200 for the third: 2800/3200 for the fourth: and 2880/3200 for the last; these are severally equal to the other, the first to the first, etc. as may be proved; thus, Let the Unity be a pound Sterling, then s. The ½ of it is 10 and ¾ is 15 and ⅘ is 16 d. and ⅞ is 17 6 and 9/10 is 18 s. d. In all 76 6 That is 3 whole Unites, and 16 s. 6 d. over, Turn 16 s. 6 d. all to six pences, it is 33, and because 6 d. is the fortieth part of a pounds, therefore all the fractions are equal to 3 33/40. Now add the new Fractions (which being all of one denomination) may be added like whole Numbers: thus, Which divided by the Denominator 3200, the quotient is 3 2640/3200. Now 2640/3200, reduced to the least terms, as hath been showed how it may, will be 33/40, so the sum of these also is 3 33/40. which is equal to the sum of the fractions given to be reduced, and therefore they are equal in sum, and might be thus proved equal severally, that is, the first of them propounded, to the first reduced. Divide the Numerator 1600 by the Numerator 1, quotient is 1600. Also divide the Denominator 3200, by the Denominator 2, the quotient is also 1600, and so may any of the rest be proved equal by the equality of quotients. But I leave it as plain enough already. Thirdly, Any fraction being given, to change the denomination to any other more requisite, retaining still (as near as may be) the same value. THE RULE. Multiply the Numerator given, by the Denominator required, and divide the product by the Denominator given; the quotient shall be the Numerator required. Example. Let the Fraction given be 7/13 of a pound Sterling, what is that in twentieth parts or shillings? Multiply 7 by 20, the product is 140, which divided by 13, the quotient 10 10/13, that is, 10 s. and 10/13 of a shilling: which may be brought to pence thus, multiply 10 by 12, product is 120, which divided by 13, quotient 9 3/13 d. And again, multiply 3 by 4, the product is 12, which divided by 13, quotient is 12/13 of a farthing, so seven thirteenths of a pound is 10 s. 9 d. and almost a farthing. But he which is resolved to have it in the smallest coin, may do it at first work; for seeing a farthing is the 960 part of a pound, multiply 7 by 960, they produce 6720, which divided by 13, the quotient is 516 farthings, and 12/13 of a farthing: these farthings may be turned to shillings, dividing by 48, or to pence by 4. This Rule though it be brief and plain; is of great use in Arithmetic; either for turning Natural and furred Fractions into Decimals; or any other desired Denomination, with such facility and speed as may be wished. Fractions of Fractions. In reduction of Fractions, some make another, or more parts, as fractions of fractions for one: that is, when there is a part of a fraction, or a part of a part of a fraction, etc. to be valued in one fraction. THE RULE. Multiply all the Numerators together, the last product shall be the Numerator desired: then multiply all the Denominators together, and this last product shall be the Denominator sought. Example. Let the Fractions of Fractions propounded, be ⅘ of ¾ of ½, for so they are usually written; and let the Numerator be multiplied: saying, 4 times 3 is 12, and 12 times 1 is 12, the Numerator therefore required is 12: then for the Denominator, say, 5 times 4 is 20, and 20 times 2 is 40, for the Denominator required; and 12/40 is equal to ⅘ of ¾ of ½. Proof. Let the Unite be 40 s. one fifth of 40 is 8, and therefore ⅘ is 32, of which one fourth is 8, and ¾ is 24, of which one half is 12, and therefore 12/40 is the just sum of all the Fractions, This needs no further exemplifying. Addition. TO add many fractions into one sum, consider whether they be of one Denomination or divers; if of one, then add all the Numerators together into one sum, that sum is the new Numerator: and the Denominator, in this case is not altered. Example. Let the fractions to be added be 2/4, 4/4, 5/4, ¼, Add the Numerators: saying 2 and 4 is 6, and 6 and 5 is 11, and 11 and 1 is 12. So the sum of them all is 12/4, that is 3 Unites. As, let the Unite be 20 s, one fourth is 5 s. and 2/4 is 10 s. and 4/4 is 20 s. which added to 10 s. is 30 s. then 5/4 is 25 s. which added to thirty shilling gives 55 s. And lastly, ¼ is 5 s. which added to 55 s. makes 60 s. that is 3 times 20 s. that is 3 l. or 3 Unites. But if the fractions to be added, be of divers denominations, as let them be 3/2, ¾, ⅘, ⅞, than (by the reduction aforegoing) they must be turned all into one denomination, and then they will be 320/480, 360/480, 384/480, and 420/480, and may be added like those before: thus, So the sum of all is 1484/480, or 371/120, that is, 3 11/120, which if it be money, and the Unite 1 l. it is then 3 l. 1 s. and 10 d. as may be tried thus: First, ⅔ of a pound, is 13 s. and 4 d. and ¾ is 15 s. and ⅘ is 16 s. Lastly, ⅞ is 17 s. 6 d. These all added together, the sum if 3 l. 1 s. 10 d. Subtraction. IN Subtraction of one fraction from another, if they be both of one denomination; it is done by taking the Numerator of one from the Numerator of the other, the remain is the new Numerator, and the Denominator the same as before. So if 2/9 be subtracted from 8/9. the remain is, 6/9 the like of all others. But if they be not of one Denomination, they must first be reduced to be so; then that which is said before is sufficient. The golden Rule in fractions is the same as in whole Numbers, I will give you but one instance. If ¾ of a yard of tape cost ½ of a penny, what shall one inch, that is, 1/36 of a yard cost? Multiply the second by the third, the product is 1/72, which divided by ¾, the quotient is 4/216 of a penny, for the price of 1/36 of a yard. Otherwise, Seeing ¾ of a yard may be turned to 27 inches; Say, if 27 cost ½, what 1? divide ½ by 27, it makes 1/54 for the answer: which is equal to 4/216, and in the least terms. And wheresoever this may be done, to have the first and third Numbers fractions of one denomination, the best way is to work with their Numerators, not regarding their denominator at all: as if ⅔ cost ¾, what 7/3? instead thereof write, if 2 cost ¾, what 7? multiply ¾ by 7, it produceth 21/4, which divided by 2, the quotient is 21/8, and that is the answer in the least terms. And all this while it should have been noted that the fractions are ever written in a smaller figure than the whole Numbers. The Rule of Fellowship. THis Rule is useful for Merchants, and all such as trade in Companies, with a joint stock; and must share a proportional part of the gains, or loss; every one according to his stock which he laid in. The Rule is twofold, with equal time; or with unequal time. That which is with equal time, is commonly called, The Rule of Fellowship without time. Of this we will first speak. THE RULE. As the whole joint stock, is to all the gain or loss; So is each man's particular stock, to his part of the gain, or loss. Example 1. Two Merchants A. and B. buy 700 l. a year Land for ever, (when money is at 8, per cent.) for 14000 l. of which A. paid 8000 l. and B. 6000 l. after 5 years (money being fallen to 6 per Cent.) they sell it for 18700 l. so there is gained 4700 l. how much of this must A. have? First for A. Say, if 14000 gain 4700, what 8000? answer, 2685 10000/14000. Then for B. If 14000, gain 4700, what 6000? answer 2014 4000/14000. As by the following operation doth appear. (I) For A. (TWO) For B. Here note, that this work might have been much abreviated, if from each of the three numbers you had cut off two Ciphers towards the right hand, as hath been formerly showed in the Compendium of Multiplication and Division. Now for the proof hereof, If you add 2685 10000/14000. which is the sum that A. gained; To— 2014 4000/14000. The sum of them is 4700. Which is equal to the total gain. And according to the proportion of these two Numbers: that is, as 8 to 6, or 4 to 3. So they ought to have parted the yearly rent also, all the time they received it: that is, A. aught to have 400 l. yearly; and B. 300 l. Example 2. A. B. and C. join their moneys to make a stock of 25000 l. of which A. laid in 10000 l. B. 8000 l. and C. put in 7000 l. with this (after a certain time in trading) they gained 7500 l. how must this be parted? First for A. Say, if 25000 gain 7500, what 10000? Or shorter, if 25 get ½, what 10? Multiply 7½ by 10, it produceth 75, which divided by 25, the quotient is ●, that is, (restoring the 3 Ciphers) 3000 l. for A. Then for B. Say, if 25000 gain 7500, what 8000? Or shorter, if 250 get 75, what 80. Multiply and divide as the Golden Rule requires, and to the quotient, restore the 2 Ciphers, than it will be 2400 l. for B. Lastly, for C. Say, if 250 give 75, what 70? answer 21, to which put the 2 Ciphers, it makes 2100 for C. And these three 3000, 2400, and 2100, being added together, make 7500. And have that proportion as the particular stocks had: and therefore the work is right. (I) for A. (TWO) for B. (III) for C. And if instead of gaining 7500 l. whereby every one is supposed to have his stock, and a part of the gains; they had lost 7500 l. then their particular stocks had not been due to them, but so much as would be left after their proportional parts of the loss were abated. Example 3. A. B and C. with a joint stock of 25000 l. gain 7500: of which A. gets 3000. B. 2400. C. 2100, what was their stock? This is but the Converse of the former, therefore say, if 5700 require 25000, what doth 3000 require? 10000 for A. and so work for the other two. Many examples are of little use, (except to load the Readers memory) where the Rule is so short and plain, I will therefore add no more to this part of the Rule, but immediately come to the Rule of Fellowship with time. The Rule of Fellowship with time. THis Rule is to be used when the times of the continuance of the particular stocks are unequal, and differ; so that here the difference of time, and also the difference of stock being both to be considered; it can be done no better way than by taking the Power of them both to be the particular stock; and all those Powers added, to be the whole stock, that which I call the Power, is the product of the money of every one multiplied by his time; And then THE RULE. As the sum of those Products, is to the whole gain; so is each particular product, to its part of the gain. Question 1. Three Merchants A. B. C. make a stock of 10000 l. of which A lays in 4000 for 3 months, B. 3000 l. for 6 months; and C. 3000 l. for 8 months, with this they gain 2000 l. what is each man's share? First, for A. multiply 4000 by 3, it makes 12000, let that be accounted his particular stock. Secondly, for B. multiply 30000 by 6, it makes 18000. Lastly for C. multiply 3000 by 8, it produceth 24000, for his stock, add these, they make 54000 l. for the general stock; then say, For A. If 54000 give 2000, what 12000? answer, 666 24000/54000. Then for B. If 54000 give 2000, what 18000? answer 444 36000/54000. Lastly, for C. If 54000 give 2000, what 24000? answer, 888 48000/54000. The three Fractions may be reduced (by dividing each Numerator, and Denominator by 6000) and then the three shares will be 444 4/9, 66 6/9, and 888 8/9, which altogether make 2000, as they ought. Question 2. Three Farmers, A. B. and C. lay out 1000 l. to stock their grounds with , of which A. put in 200 l. for 6 years; B. had 300 l. going for 4 years; and C. 500 l. for 2 years; at the end (by unseasonable times) there was lost 200 l. which made the remain of their stock but 800 l. what had each man left? Multiply 200 by 6, it gives 1200: Likewise, 300 by 4, it gives 1200. Lastly, 500 by 2, the product is 1000: all these are 3400 for the joint stock. Then first for A. Say, if 3400 lose 200, what 1200? answer, 70 2000/3400 for A, to which B is equal, because the power of his stock is so. Therefore for C. Say, if 3400 lose 200, what 1000? answer, 58 2800/3400. So the 3 shares are 70 20/34, 70 20/34, and 28 2●/34, equal to 200. Now because A put in 200 l. and lost 70 20/34 subtract the loss from the stock, remains 129 14/34. And so doing for B, his remain will be 289 14/34. And for C, his remain is 441 6/34. Now these three remains, 129 14/34, 229 14/34, and 441 6/34, make up 800 l. which was the whole remain. Quest. 3. A. rends a close for a year, to pay 80 l. he puts into it 200 sheep: 2 months after B. puts 40 sheep in; and 5 months after that C. puts in 100 sheep; how much must every one pay of the rent? Multiply 200 by 12, it produceth 2400 And 40 by 10, produceth 400 Lastly, 100 by 5, (which is C time) produceth 500 In all 3300 Then for A. If 3300 pay 80, what 2400? answer, 58 600/3300. Then for B. If 3300 pay 80, what 400? answer, 9 2300/3300. And for C. If 3300 pay 80, what 500? answer, 9 400/3300. The whole numbers make 79, and the broken numbers make 1. In all 80. Note Whereas, hitherto we have considered only difference of time and money; it may be noted, that there may be difference of other kinds, as persons or place; but whatsoever they are the power of all is found like these by multiplication; and are to be worked like these, with so many Uses of the Golden Rule, as the question requires. I will therefore add but one question more, which is this. Quest. 4. One leaves a Legacy of 900 l. among four Kinsfolks. A. B. C. D; so as B may have twice as much as A. and C thrice as much as B. and D as much and half as much as C, what is every one to have. Say, if A be 1, B is 2, C 6, and D 9, add these Numbers: 1, 2, 6, 9, together, they give 18, then say, If 18 require 900, what 1? answer is 50. So A is to have 50 l. B 100 l. C 3000 l. and D 450 l. which are their just parts; and altogether are equal to 900 l. and the work right. The Rule of Alligation. THis hath its name from binding, tying, or uniting many particulars in one Mass or Sum, the nature of it will be understood in working some Questions or Examples. Question 1. A Corn Master would mix 4 sorts of grain together, viz. wheat at 4 s. the bushel, wheat at 4 s. 0 d. the bushel; Rye, at 3 s. the bushel, and Barley at 2 s. 8 d. the bushel; so as to make 15 quarters in all, to be sold at 3 s. 6 d. the bushel; How much must he take of each? THE RULE. Multiply the whole Mass to be made, by any particular difference; and divide the product by the sum of all the differences, the quotient shall be the just quantity of that particular kind, whose price standeth against the difference you worked with. Example. First turn the quarters into bushels, by saying, 8 times 15 is 120, then for the quantity of the first sort at 4½: multiply 120 by ⅚, the product is 100, which divided by 2⅚, the quotient is 35 5/7 bushels of that sort at 4 s. 6 d. and working so for every of the other; they will be found to be thus: At 4 s. 6 d. 35 5/7 bushels. At 4 s. 21 3/17 bushels. At 3 s. 6 d. 21 3/17 bushels. At 2 s. 8 d. 42 6/17 bushels. In all 120 Now to prove this right; first multiply the whole Mass 120 bushels, by the desired price 3½ s. omitting the denominations, the sum is 420 s. Then secondly, multiply 38 5/17 by 4½, it is, 158 14/17 And 21 3/17 by 4, it produces 84 12/17 And 21 3/17 by 3, it makes 63 2/17 And 42 6/17 by 2⅔ is 112 16/17 In all 420 But so much all should be worth in shillings. And therefore the question in rightly solved. Quest. 2. One hath 6 sorts of Fruits at several prices, Dates at 2 s. Almonds at 1 s. 4 d. Currants at 10 d. Raisins at 5 d. Prunes at 4 d. and Figs at 3 d. the pound; and would take of every sort some to make a mixed quantity of 30 l. weight to sell one with another for 9 d. the pound, how much must he take of each? So 30, multiplied by 6, produceth 180, which Divided by 38, the quotient is 4 22/38 of a pound weight: and so much must be taken of Dates, at 24 d. Secondly, 5 times 30 is 150, which divided by 38, the quotient is 3 36/38 for Almonds. And working after the same manner with 4, 1, 7, 15, their respective quantities will be found to be these; pounds' 38 parts. Dates 4, 28 Almonds 3, 36 Currants 3, 6 Raisins 0, 30 Prunes 5, 20 Figs 11, 32 In all 26, 152 That is 26 252/32, and the reduction of the fraction will make it 30, as it ought to be, and by comparing the prices of these particulars added, with the price of 30 li. weight, at 9 d. per l. weight, which makes 470 d. this may be proved like the former. But that the Reader may be perfect in it, I will do it here also, as followeth. Say first, 24 times 4 is 96, and 24 times 28 is 672; for the first, set them thus: 96, 672. And 16 times 3 is 48, and 16 times 36 is 576 48, 576 And 10 times 3 is 30, and 10 times 6 is 60 30, 60 And 5 times 30 is 150 00, 150 And 4 times 5 is 20, and 4 times 20 is 80 20, 80 Lastly. 3 times 11 is 33, and 3 times 32 is 96 32, 96 In all 227, 1634 Now this 1634 being the sum of the Numerators of Fractions, whose common Denominator is 38, must be divided by 38, and the quotient will be 43, which added to the whole number 227: the sum is 270. And so much is 30, multiplied by 9, which shows the work to be right. The Combination, or linking of Numbers may be varied at pleasure, as whereas above I linked 24, and 3, also 16 with 4, and 10 with 5, it might have been 24 with 5, and 16 with 4, and 10 with 3. Or 14 with 4, and 16 with 3, and 10 with 4, of which diversity of linking would follow diversity of solutions, but all true, as the Reader may easily prove by himself. Likewise, if the Numbers to be linked were 3, 5, 7, or any odd Number, one of them may be linked to two severally, to make the work even. Example 3. If the Numbers were 12, 10, 8, 6, and 4, and the mean or common price required were 9, you might first link them as you see here, takeing 12 twice, or else you might take any other twice as you shall like; and so the work will be every way right, though not the same; if the differences be rightly set off and orderly used, as is taught before in the first question. Question 3. A Goldsmith would mix 3 sorts of silver, A. B. C. A is 10 d. weight better; B 7 d. weight better; and C, 4 d. weight better, to make an Ingot of 50 li. weight, which should be in finesse 8 d. weight better: How much must be taken of each? 1. Then first 50 multiplied by 4 200 and divided by 9, quotient is 9 2. 50 multiplied by 2, and 50 divided by 9, quotient is 9 3. 50 multiplied by 2, and 100 divided by 9, quotient is 9 4. And again, the same 100 9 In all 450 9 Which is equal to 50, the quantity required, Now the first fraction multiplied by 10, (omitting the Denominator) is 2000 The second also by 10 gives 500 The third by 7 makes 700 The last by 4 makes 400 In all 3600 That is, 1600/9, which is equal to 400, and if the whole Ingot 50, be multiplied by the betterness required, namely, by 8, they shall produce 400 also: So this is proved. In every Alligation, or linking of two numbers, this is evident, that if the sum of the numbers linked be greater than the mean number required taken so many times as there are numbers to be linked, the question would be absurd; and the resolution-thereof impossible. And this shall serve for the rule of Alligation. The Rule of False Position. THis Rule serves to resolve questions, which are not presently fit for the Golden Rule; and therefore in stead of the true Number which is sought: Suppose any number great or small, and make trial of it, whether it resolve the question without any error; if so, it is the true Number: if not note what error is at the end of the work, and whether it be too much or too little: if too much, mark it thus + but if too little, thus − Than suppose again another Number, (it imports not whether it be nearer or further off) and try as before, and mark that error also; And, than THE RULE. Multiply the first Position by the second Error, and the second Position by the first error, and (if the errors be both + or both −) Subtract the lesser product from the greater, and keep the remain for a Dividend, and the difference of the errors for the Divisor: the quotient of that division is the true Number required. But if the errors had been one +, the other −, then the sum of the products added together had been the Dividend: and the sum of the errors, the Divisor; the rest of the work the same as before. Question 1. A man is to drive 48 young Turkeys 40 miles: and for every Turkey which comes alive to the end of the Journey, he is to receive 3 d. but for every one which dies by the way, he is to pay 6 d. at the end, he received 72 d. How many died by the way? Let the first supposition be That by the way there died 20 For them he was to pay 120 d. and for 28 which lived, he was to receive 84 d. so he paid more than he received 36 d. and should have got clear 72 d. Add 108 Wherefore the first error is − 108 Let the second supposition be 10 For these he paid 60 d. and for the rest he received 114 d. the difference is 54, and should be 72: so the second error is − 18 Now 20 multiplied by 18, produceth 360 and 10 by 108 produceth 1080 and the difference is 720 for the Dividend, likewise the difference of the errors is 90 for the Divisor, and the quotient is 8, which is the true Number of those he lost by the way. As may be proved by trial. Question 3. If it was required to make up a pound Sterling of shillings and groats only: and so as the number of groats, may be to the Number of shillings, as 7 to 1, how many shillings must there be? First, suppose the shillings 4 then the groats must be equal to 16 s. that is 48 groats; but the shillings taken 7 times are 28, to which 48 should be equal, but is +20 Secondly, suppose the shillings 2 then the groats (making 18 s.) are 25, which should be equal to 7 times 2, but is +40 Multiply 4 by 40 product is 160, than Multiply 2 by 20, the product is 40, which taken from 160, rests for the Dividend 120 And the difference of errors is 20 Lastly, 120 divided by 29, quotient is 6 The number of shillings therefore is 6 And the number of groats is 42 But as 7 to 1, so is 6 times 7 which is 42 to 6 times 1, which is 6: so the work is done. Question 3. If there be 4 several weights, A. B. C. D. of which D is 24 ounces, and C is double to B, and triple to A, and D with twice A is double to C. and quadruple to B: How much doth every one of these weights weigh? First, suppose A to be then D with twice A is 24, and 16, that is 40, of which ● being the half is ●0, and B 10. 8 Now thrice A is 24, to which C should be equal, but is − 4 Secondly, let A be supposed 4 then D more, twice A is 32, and C 16 and B is 8, but thrice A is 12, to which 16 should be equal, but is +4 Then 8 multiplied by 4, giveth 32, and 4 by 4, produceth 16: both these products gives 48 for the Dividend: and the sum of the errors (because the first is −, the other +) giveth 8, for the Divisor; and the quotient will be 6, to which A is equal, and twice A more D is 36, of which C, (being half is 18, and B is 9, and thrice A is equal to C, namely, 18, and all right. Whereas the first error is equal here to the second, it follows that the Positions were equally false: and therefore their difference which is 4, being parted into two equal parts, 2 and 2, if 2 be taken from 8, the remain is the true number 6, or if ● be added to 4, (which was the second position) the sum will be also 6. And further, whensoever the erronrs be one +, the other −, though they be not equal; yet than if the difference between the positions be parted into two parts, which are in proportion one to another, as the two errors are one to another respectively: then if the first part be taken from the first position (if that be the greater) or add to it (if it be the less, the same number required is thereby had. As, let the last question be resumed, And let the first position for A be 15 Then the first error will be − 18 Then let the second position be 3 And so the second error will be +6 And the difference of positions is 12 which divided into two parts 9 and 3, which have that proportion one to another as have the errors 18 and 6, then if the first part 9, be taken from the first position 15, there remains the true Number 6. Or else if the second part 3, be added to the second position 3: thereby also is made the true Number 6. The way of parting 12, (or any other) into two parts proportional with the errors is easily done by the Golden Rule, thus: As the sum of the Errors 24, is to the difference of positions 12; So is the greater error 18 to the greater part required, namely, 9 Many other questions are in other books exemplifyed and wrought by this Rule; but seeing I intent not to write a great book; and also because some of those questions may be easily resolved without this Rule, I will add no more: ondy mention one of those questions. I there be a Cistern with 4 Cocks, which holls 8 barrels of water, and the first cock will runit all out in 6 hours, the second in 4, the third in 3, and the last in 2 hours: in what time shall all of them run it out? If the first in 6 hours run 8 the second in the same time would ●un 12 the third 16 the last 24 In all 60 Then say, If 60 require 6● what 8? The answer 48/60, that is 4/5 o● an hour; in which time all the 4 Cocks together would run out all the 8 Barrels of water. The Rule of Ceres and Virginum. THis is the most uncertain, and unnecessary Rule in Arithmetic; being seldom used except in sporting questions to puzzle young beginners, with easy problems; such as follow. Question 1. A Caterer bought 8 birds of two sorts, as Geese and Hens for 20 s. the Geese cost 4 s. a piece, the Hens 2 s. a piece; How many did he buy of each sort? This may be done by the Rule of False; and also thus: multiply the whole number 8, into the least price 2, it produceth 16, which taken from the whole price 20, rests 4 for a Dividend; which divided by 2, which is the difference of the particular prices, the quotient is 2, for the number of Geese; and 6 must be for the Hens: the proof is easy. Question 2. If 21 Persons, Men, Women, and Children spend 26 shillings; so that every man pays 2 s. every woman 1 s. every child 6 d. How many is there of each sort? THE RULE. Multiply the number of Persons by the least expense, and take the product of it from the whole expense, the rest shall be the Dividend; which divided by the difference betwixt the greatest and least particular expenses: the quotient is a Number which the Number of men (or they which spend most) comes near to: but cannot exceed: or of the said Dividend be divided by the sum of the greatest and least expenses, the quotient is a Number then which the Number of men or those which spend most) cannot be much less. So here 21 multiplied by 6 d. that is, by ½, the product is 10½, which taken from 26, rests 15½, for the Dividend: and then taking ½ from 2, rests 1½ for the Divisor, and the quotient is 3⅓, which is something more than 10, the Number of men therefore must be but 9 Then turn the Dividend, and the Divisor both into whole Numbers, by multiplying them by the common Denominator 2, so they reduced will be 31 and 3, as before is to be seen in the quotient. Multiply the Divisor 3, by 9, (which is the Number of Men) the product is 27, which taken from 31, (which is the reduced Dividend) the remain is 4, for the Number of Women; and the Children must be 8. Example 1. 9 men at 2 s. each 18 s. 4 women at 1 s. each 4 8 children at 6 d. each 4 In all 21 In all 26 But the number of men may be also 8, which multiplied by the reduced Divisor 3, product is 24, which taken from 31, the remain is 7 for the women; and then the children must be 6. Example 2. 8 men at 2 s. each 16 7 women at 1 s. each 7 6 children at 6 d. each 3 In all 21 In all 26 Or the number of men may be 7, which multiplied by 3, produceth 21, which taken from 31, remains 10 for the women; and 4 children. Example 3. 7 men at 2 s. each 14 s. 10 women at 1 s. each 10 4 children at 6 d. each 2 In all 21 In all 26 So here is already seen 3 various solutions of this question, which makes this Rule the less to be regarded. But further, the number of men may be 10, and not more, for if you put them 11, that multiplied by 3, produceth 33, which is greater than 31, from which it should be taken, but I say it may be 10, and then there is only one woman, and 5 children: this confirms the former part of the Rule. Now for the later part, if the Dividend 31 be divided by the sum of the two extreme expenses (reduced by doubling as the Dividend is) 4, the quotient will be 7¾. And the men may be 7, as hath been showed; but they may be also but 6, and fewer they cannot be: as 6 men, 13 women, and 2 children; for if you put them 5, that multiplied by 3, produceth 15, which taken from 31, there remains 16 for the women, and so there should be no children: which is contrary to the supposition. And further, because the quotient was 7¾, the number of men might be so, if pure Arithmetical division be only regarded: and then the women also are in number 7¾, and the children 5½, as may easily be tried; I need not exemplify it. Question 3. If there be an Exhibition of 900 l. per annum to 30 persons: some Clerks, some Messengers, and some Doorkeepers, at 60 l. each Clerk, 40 l. each messenger, and 20 l. each doorkeeper, how many must there be of each sort? Multiply (according to the Rule) 30 by 20, the product is 600, which taken from 900, remains 300 for the Dividend; and 60 want 20, that is 40, for the Divisor: and the quotient is 7½, and more the Clerks cannot be; Also divide by 60 more 20, that is 80, quotient is 3¾, and much fewer the Clerks cannot be. Not to stand upon the Fractions (in this case of dividing men) the Clerks may be 7, 6, 5, 4, or 3: and the messengers 1, 3, 5, 7, or 9, and the Doorkeepers 22, 21, 20, 19, or 18, that the Clerks cannot (in whole Numbers) be more than 7, or less than 3, may thus be proved; First, let them be 8, than 8 times 40 is 320, which is more than 300, out of which it should be taken: Secondly, let them be 2, than 2 times 40 is 80, out of 300 remains 220, which divided by 20, gives the quotient 11, for the messengers, so the Clerks and messengers being 13, the remain thereof to 30, namely, 17, must be Doorkeepers. but, 2 Clerks at 60 l. each 120 l. 11 Messengers at 40 l. each 480 17 Doorkeepers at 20 l. each 340 In all 30 In all 940 Which is 40 l. too much, therefore the Clerks cannot be two. Note. It may be asked, why the remain 220 should be divided by 20: whereas the like remain in the former Example, namely 16, was taken (without any division) absolutely for the number of Women, or middle number? I answer, although the greatest or first Number being found, (as here to be 2) the residue of 2 to 30, might be rightly parted into two fit parts in the same manner as the first question of this Rule was resolved, or else by the Rule of False: yet to give further satisfaction, the cause of this is, the difference betwixt the two lesser expenses, was there ½, which (before the division) was reduced to 1, which neither multiplies nor divides any Number, but leaves it the same: whereas, in this last, the middle expense (or exhibition) being 40, and the least 20, the difference of them was 20, by which dividing the Remain of the last subtraction: the quotient is ever the Number of the middle persons. Which may serve as an addition to the Rule, where the sorts of things are but three. Question 4. If there be 10 persons of four several Countries. English, French, Dutch, and Spanish, to pay a Debt of 1000 l. So that every English man pays 50 l. every French man 70 l. every Dutch man 130 l. and every Spaniard 150 l. How many is thereof each? The Dividend, (according to the former Rule) is 500 Now to make the Divisor, take his sum that pays least (namely 50,) out of each of the other three 150, 130, and 70, and the Remaines will be 100, 80, and 20. Add the first and last for the Divisor, it is 120. And the quotient will be 4 2/12, and the Spaniards cannot be more. Secondly, add the first and second together for the Divisor, it is 180, and the quotient is 2 14/18, and the Spaniards cannot be less. I mean, they cannot be much more than 4, or less than 2: and therefore, seeing any one solution will serve, let them be 3, and by that multiply 100, and take the product out of 500, there remains 200 for a second Dividend, which divided by (the second remain) 80, the quotient is 2½: therefore the Dutchmen are 2, which multiplied by 80 make 160, take that out of 200, there remains 40 for a third Dividend: which divided by (the third remain) 20, the quotient is 2 for the Frenchmen also; and consequently the English must be 3, because all of them are 10: But the Spaniards may be also 4 or 2. Example. 4 Spaniards, at 150 l. each 600 1 Dutchman at 130 l. 130 1 Frenchman, at 70 l. 70 4 English, at 50 l. each 200 10 In all 1000 2 Spaniards at 150 l. each 300 3 Dutch, at 130 l. each 390 3 French, at 70 l. each 210 2 English, at 50 l. each 100 10 In all 1000 The reason why the Spaniards and English, as also the Dutch and French are equal in number, is because their payments differ equally from 100, which is the mean sum with which 10 men should pay 1000 l. and making it so, this question, and many other of this nature may be answered by the Rule of Alligation: thus, And also, because the other two differences 30 and 30 are equal: the Number of the French is equal to the Number of the Dutch. But both those numbers together are 4, because 3 Spaniards, and 3 English, taken out of 10, the remain must be 4. Wherefore the Number of the French is 2, and the Dutch also are 2. Or thus, If 160 require 10, what 30? Answer is 1 160/160, (that is, in this case 2) for the Spaniards; and consequently 2 English, and therefore the French and Dutch each 3. But where any one of the particular sums, is equal to the mean sum, there this cannot so well be done by Alligation. Example. If one should buy 12 loaves of bread for 12 pence, so that some might be two penny; some penny, some halfpenny; and some farthing loaves: and it be required to know how many he must buy of each; then because of 12 loaves for 12 pence, the mean price is 1, but one of the particulars being also 1, there should be no penny loaves, because there is no difference betwixt the mean price, and a penny. But it may be found by the Rule of Ceres and Virginum, to be either. 4 two penny loaves 8 pence 2 penny loaves 2 pence 2 halfpenny loaves 1 penny 4 farthing loaves 1 penny In all 12 loaves. In all 12 pence. Or else, 3 two penny loaves 6 d. 4 penny loaves 4 3 halfpenny loaves 1 ½ 2 farthing loaves 0 ½ In all 12 In all 12 Extraction of Roots. ANd first for the square Root, that is, a square Number being given, to find the root or side of it in a Number: which root or side, being multiplied into itself, must therefore produce that square Number. The doing of this is showed in (almost) all books of Arithmetic; and the reason of it (in some) which is taken from the fourth proposition of the second book of Euclid; which saith, If a right line be divided by chance; the squares made of the parts, together with the Rectangle made of the parts twice, is equal to the square of the whole. Example. Now let AB, be supposed 20, A 15, and BE 5. Then the square of AB, (which is made by multiplying the root 20 into itself) is equal to 400. And the square A, that is, 15 times 15 is equal to 225 And the square of BE is 5 times 5 25 And the rectangle AO, 5 times 15 75 And the rectangle DO, 5 times 15 75 In all 400 which is equal to the square of AB, as before. If therefore a square Number, as 334084 be given to have the root extracted: first, make a point over 4, (the place of Unity) and another over o, the second figure from it; and so another over o, the second figure from o, towards the left hand, observing the like order still, if there were more secondary figures, point them all as in the Margin; Then look at the figures under the first point towards the left hand which are 33, and take the greatest square root of them, which is 5, (for 6 times 6 is 36) and say 5 times 5 is 25, which take out of of 33, there remains Then multiply the root 5 by 20, it gives 100 for a Divisor; and the Dividend is 840, namely, the figures which reach to the next point, and so the quotient might be 8, but must be but 7, because the square of the quotient, being now 49, must (together with 7 times the Divisor) be taken out of 840. And the remainder will be Add this last quotient 7, being put to the former quotient 5, after the manner used in plain division, it will be 57, which multiply still by 20, the product is 1140 for a new Divisor, and the Dividend (because there remains but one point) is all the figures 9184. And the quotient can be but 8, and must be so much; for 8 times 1140 is 9120 and 8 times 8 64 In all 9184 Which taken out of the remain 9184, there now remains nothing: which shows the square 334084 is justly resolved, and putting the last quotient 8 to the two former, 5 and 7, the whole will be 578, which is the true root required. And the parts of it, if they be 3, are 500, 70, and 8. Or if but two 570 and 8. And the truth may be either way proved by adding the squares and twice the rectangles of the parts; for the sum of them shall be equal to the whole square 334084, as hath been showed before. It is ever certain that there shall be as many figures, or places of figures in the root, as there are points in the square, ordered as before. And by reading this little seriously, may any one be able to find the root of any other square whatsoever: the first operation in all being to take the greatest square contained under that point next the left hand, and put the root thereof for the first figure in a quotient. The Secondary operation, must be repeated so often as there are remaining points; as hath been plainly showed in the foregoing Example. I will therefore add no more, but give you two other squares, and their roots, leaving the Reader to extract them himself. So, if there were given the square 88̇30̇ The root of it according to the former practice may be found to be 94. Or let there be given the square 1̇52̇27̇56̇ The root of that will be 1234. Here followeth a Table of Roots and their Squares from 1, to 1000 1 Cent. R. Square 1 1 2 4 3 9 4 16 5 25 6 36 7 49 8 64 9 81 10 100 11 121 12 144 13 169 14 196 15 225 16 256 17 289 18 324 19 361 20 400 21 441 22 484 23 529 24 576 25 625 26 676 27 729 28 784 29 841 30 900 31 961 32 1024 33 1089 34 1156 35 1225 36 1296 37 1369 38 1444 39 1521 40 1600 41 1681 42 1764 43 1849 44 1936 45 2025 46 2116 47 2209 48 2304 49 2401 50 2500 51 2601 52 2704 53 2809 54 2916 55 3025 56 3136 57 3249 58 3●64 59 3481 60 3600 61 3721 62 3844 63 3969 64 4096 65 4225 66 4356 67 4489 68 4624 69 4761 70 4900 71 5041 72 5184 73 5329 74 5476 75 5625 76 5776 77 5929 78 6084 79 6241 80 6400 81 6561 82 6724 83 6889 84 7056 85 7225 86 7396 87 7569 88 7744 89 7921 90 8100 91 8281 92 8464 93 8649 94 8836 95 9025 96 9216 97 9409 98 9604 99 9801 100 10000 101 10201 102 10404 103 10609 104 10816 105 11025 106 11236 107 11449 108 11664 109 11881 110 12100 111 12321 112 12544 113 12769 114 12996 115 13225 116 13456 117 13689 118 13924 119 14161 120 14400 121 14641 122 14884 123 15129 124 15376 125 15625 126 15876 127 16129 128 16384 129 16641 130 16900 131 17161 132 174●4 133 17689 134 17956 135 18225 136 18496 137 18769 138 19044 139 19321 140 19600 141 19881 142 20164 143 20449 144 20736 145 21025 146 21316 147 21609 148 21904 149 22201 150 22500 151 22801 152 23104 153 23409 154 23716 155 24025 156 24336 157 24649 158 24964 159 25281 160 25600 161 25921 162 26●44 163 26569 164 26896 165 27225 166 27556 167 27889 168 28224 169 28561 170 28900 171 29241 172 29584 173 29929 174 30276 175 30625 176 30976 177 31329 178 31684 179 32041 180 32400 2 Cent. R. Square 181 32761 182 33124 183 33489 184 33856 185 34225 186 34596 187 34969 188 35344 189 35721 190 36100 191 36481 192 36864 193 37249 194 37636 195 38025 196 38416 197 38809 198 39204 199 39601 200 40000 201 40401 202 40804 203 41209 204 41616 205 42025 206 42436 207 42849 208 43264 209 43681 210 44100 211 44521 212 44944 213 45369 214 45796 215 46225 216 46656 217 47089 218 47524 219 47961 220 48400 221 48841 222 49284 223 49729 224 50176 225 50625 226 51076 227 51529 228 51984 229 52441 230 52900 231 53361 232 53824 233 54289 234 54756 235 55225 236 55696 237 56169 238 56644 239 57121 240 57600 241 58081 242 58564 243 59049 244 59536 245 60025 246 60516 247 61009 248 61504 249 62001 250 62500 251 63001 252 63504 253 64009 254 64516 255 6●025 256 65536 257 66049 258 66564 259 67081 260 67600 261 68121 262 68644 263 69169 264 69696 265 70225 266 70756 267 71289 268 71824 269 72361 270 72900 271 73441 272 73984 273 74529 274 75076 275 75625 276 76176 277 76729 278 77284 279 77841 280 78400 281 78961 282 79524 283 80089 284 80656 285 81225 286 81796 287 82369 288 82944 289 83521 290 84100 291 84681 292 85264 293 85849 294 86436 295 87025 296 87616 297 88209 298 88804 299 89401 300 90000 3 Cent. R. Square 301 90601 302 91204 303 91809 304 92416 305 93025 306 93636 307 94249 308 94864 309 95481 310 96100 311 96721 312 97344 313 97969 314 98596 315 99225 316 99856 317 100489 318 101124 319 101761 320 102400 321 103041 322 103684 323 104329 324 104976 325 105625 326 106276 327 106929 328 107584 329 108241 330 108900 331 109561 332 110224 333 110889 334 111556 335 112225 336 112896 337 113569 338 114244 339 114921 340 115600 341 116281 342 116964 343 117649 344 118336 345 119025 346 119716 347 120409 348 121104 349 121801 350 122500 351 123201 352 123904 353 124609 354 125316 355 126025 356 126736 357 127449 358 128164 359 128881 360 129600 361 130321 362 131044 363 131769 364 132496 365 133225 366 133956 367 134689 368 135424 369 136161 370 136900 371 137641 372 138384 373 139129 374 139876 375 140625 376 141376 377 142129 378 142884 379 143641 380 144400 381 145161 382 145924 383 146689 384 147456 385 148225 386 148996 387 149769 388 150544 389 151321 390 152100 4 Cent. R. Square 391 152881 392 153664 393 154449 394 155236 395 156025 396 156816 397 157609 398 158404 399 159201 400 160000 401 160801 402 161604 403 162409 404 163216 405 164025 406 164836 407 165649 408 166464 409 167281 410 168100 411 168921 412 169744 413 170569 414 171396 415 172225 416 173056 417 173889 418 174724 419 175561 420 176400 421 177241 422 178084 423 178929 424 179776 425 180625 426 181476 427 182329 428 183184 429 184041 430 184900 431 185761 432 186624 433 187489 434 188356 435 189225 436 190096 437 190969 438 191844 439 192721 440 193600 441 194481 442 195364 443 196249 444 197136 445 198025 446 198916 447 199809 448 200704 449 201601 450 202500 451 203401 452 204304 453 205209 454 206116 455 207025 456 207936 457 208849 458 209764 459 210681 460 211600 461 212521 462 213444 463 214369 464 215296 465 216225 466 217156 467 218089 468 219024 469 219961 470 220900 471 221841 472 222784 473 223729 474 224676 475 225625 476 226576 477 227529 478 228484 479 229441 480 230400 5 Cent. R. Square 481 231361 482 232324 483 233289 484 234256 485 235225 486 236196 487 237169 488 238144 489 239121 490 240100 491 241081 492 242064 493 243049 494 244036 495 245025 496 246016 497 247009 498 248004 499 249001 500 250000 501 251001 502 252004 503 253009 504 254016 505 255025 506 256036 507 257049 508 258064 509 259081 510 260100 6 Cent. R. Square 511 261121 512 262144 513 263169 514 264196 515 265225 516 266256 517 267289 518 268324 519 266361 520 270400 521 271441 522 272484 523 273529 524 274576 525 275625 526 276676 527 277729 528 278784 529 279841 530 280900 531 281961 532 283024 533 284089 534 285156 535 286225 536 287296 537 288369 538 289444 539 290521 540 291600 5 Cent. R. Square 541 292681 542 293764 543 294849 544 295936 545 297025 546 298116 547 299209 548 300304 549 301401 550 302500 551 303601 552 304704 553 305809 554 306916 555 308025 556 309136 557 310249 558 311364 559 312481 560 313600 561 314721 562 315844 563 316969 564 318096 565 319225 566 320356 567 321489 568 322624 569 323761 570 324900 571 326041 572 327184 573 328329 574 329476 575 330625 576 331776 577 332929 578 334084 579 335241 580 336400 581 337561 582 338724 583 339889 584 341056 585 342225 586 343396 587 344569 588 345744 589 346921 590 348100 591 349281 592 350464 593 351649 594 352836 595 354025 596 355216 597 356409 598 357604 599 358801 600 360000 6 Cent. R. Square 601 361201 602 362404 603 363609 604 364816 605 366025 606 367236 607 368449 608 369964 609 370881 610 372100 611 373321 612 374544 613 375769 614 376996 615 378225 616 379456 617 380689 618 381924 619 383161 620 384400 621 385641 622 386884 623 388129 624 389376 625 390625 626 391876 627 393129 628 394384 629 395641 630 396900 7 Cent. R. Square 631 398161 632 399424 633 400689 634 401956 635 403225 636 404496 637 405769 638 407044 639 408321 640 409600 641 410881 642 412164 643 413449 644 414736 645 416025 646 417316 647 418609 648 419904 649 421201 650 422500 651 423801 652 425104 653 426409 654 427716 655 429025 656 430336 657 431649 658 432964 659 434281 660 435600 661 436921 662 438244 663 439569 664 440896 665 442225 666 443556 667 444889 668 446224 669 447561 670 448900 671 450241 672 451584 673 452929 674 454276 675 455625 676 456976 677 458329 678 459684 679 461041 680 462400 681 463761 682 465124 683 466489 684 467856 685 469225 686 470596 687 471969 688 473344 689 474721 690 476100 8 Cent. R. Square 691 477481 692 478864 693 480249 694 881636 695 483025 696 484416 697 485809 698 487204 699 488601 700 490000 701 491401 702 492804 703 494209 704 495616 705 497025 706 498436 707 499849 708 501264 709 502681 710 504100 711 505521 712 506944 713 508369 714 509796 715 511225 716 512656 717 514089 718 515524 719 516961 720 518400 7 Cent. R. Square 721 519841 722 521284 723 522729 724 524176 725 525625 726 527076 727 528529 728 529984 729 531441 730 532900 731 534361 732 535824 733 537289 734 538756 735 540225 736 541696 737 543169 738 544644 739 546121 740 547600 741 549081 742 550564 743 552049 744 553536 745 555025 746 556516 747 558009 748 559504 749 561001 750 562500 751 564001 752 565504 753 567009 754 568516 755 570025 756 571536 757 573049 758 574564 759 576081 760 577600 761 579121 762 580644 763 582169 764 583696 765 585225 766 586756 767 588289 768 589824 769 591361 770 592900 771 594441 772 595984 773 597529 774 599076 775 600625 776 602176 777 603729 778 605284 779 606841 780 608400 8 Cent. R. Square 781 609961 782 611524 783 613089 784 614656 785 619225 786 617796 787 619369 788 620944 789 622521 790 624100 791 625681 792 627264 793 628849 794 630436 795 632025 796 633616 797 635209 798 636804 799 638401 800 640000 801 641601 802 643204 803 644809 804 646416 805 648025 806 649636 807 651249 808 652864 809 654481 810 656100 811 657721 812 659344 813 660969 814 662596 815 664225 816 665856 817 667489 818 669124 819 670761 820 672400 821 674041 822 675684 823 677329 824 678976 825 680625 826 682276 827 683929 828 685584 829 687241 830 688900 831 690561 832 692224 833 693889 834 695556 835 697225 836 698896 837 700569 838 702244 839 703921 840 705600 841 707281 842 708964 843 710649 844 712336 845 714025 846 715716 847 717409 848 719104 849 720801 850 722500 851 724201 852 725904 853 727609 854 729316 855 731025 856 732736 857 734449 858 736164 859 737881 860 739600 861 741321 862 743044 863 744769 864 746496 865 748225 866 749956 867 751689 868 753424 869 755161 870 756900 871 758641 872 760384 873 762129 874 763876 875 765625 876 767376 877 769129 878 770884 879 772641 880 774400 881 776161 882 777924 883 779689 884 781456 885 783225 886 784996 887 786769 888 788544 889 790321 890 792100 891 793881 892 795664 893 797449 894 799236 895 801025 896 802816 897 804609 898 806404 899 808201 900 810000 9 Cent. R. Square 901 811801 902 813604 903 815409 904 817216 905 819025 906 820836 907 822649 908 824464 909 826281 910 828100 911 829921 912 831744 913 833569 914 835396 915 837225 916 839056 917 840889 918 842724 919 844561 920 846400 921 848241 922 850084 923 851929 924 853776 925 855625 926 857476 927 859329 928 861184 929 863041 930 864900 931 866761 932 868624 933 870489 934 872356 935 874225 936 876096 937 877969 938 879844 939 881721 940 883600 941 885481 942 887364 943 889249 944 891136 945 893025 946 894916 947 896809 948 898704 949 900601 950 902500 951 904401 952 906304 953 908209 954 910116 955 912025 956 913936 957 915849 958 917764 959 919681 960 921600 961 923521 962 925444 963 927369 964 929296 965 931225 966 933156 967 935089 968 937024 969 938961 970 940900 971 942841 972 944784 973 946729 974 948676 975 950625 976 952576 977 954529 978 956484 979 958441 980 960400 981 962361 982 964324 983 966289 984 968256 985 970225 986 972196 987 974169 988 976144 989 978121 990 980100 991 982081 992 984064 993 986049 994 988036 995 990025 996 992016 997 994009 998 996004 999 998001 1000 1000000 Extraction of the Cube Root. ACube is a solid, or Body contained within six equal squares; and may be fitly represented by a Die. When a Cube is given, as is 435̇519̇512̇; point the Number as you see in the place of Unity, and every third figure after; then see what is the root of the greatest Cube contained under the first point toward the left hand; that is, in 435, it will be found 7, (for 8 times 8, taken 8 times is 5 12, which is too much) put this 7 for the first figure in the quotient, having taken the Cube thereof, which is 343, out of 435: thus, And so the first work is done. For the second, take the square of the quotient, that is 49, which multiply by 300, the product is 14700 to which add 30 times 7, that is ∷ ∷ 210 which makes the Divisor 14910 And the Dividend is 92519 So the quotient might be 6, but must be but 5, because the Cube of the new quotient, and 210 times the square of the said quotient, must be allowed in this work, as followeth. The remain is 92519̇512̇ multiply the first product 14700 by 78875 The second quotient 5, and they produce 73500 then multiply 210 by the square of 5, it makes : 5250 to which add the cube of 5. ..:: 125 It makes in all 78875 Which taken from 92519 there remains 13644512 and the quotient is 75, whose square 5625, multiplied by 300 the product is 1687500 secondly, 30 times 75 is :: 2250 And the new Divisor is 1710000 And the third quotient figure is 8, by which multiply 1687500, it is 13500000 Likewise, 2250, multiplied by the square of 8, (64) is 144000 And the Cube of 8 is ::: 512 Altogether are 13644512 Which taken from the second remain, there remains now thirdly nothing. And therefore the last quotient 8, put to the former two, it is 75.8 for the whole Root, as may be tried by multiplying 758 into itself; and the product again by 758, than the last product shall be equal to the whole Cube, which was given at first to be resolved. R. Cube 1 1 2 8 3 27 4 64 5 125 6 216 7 343 8 512 9 729 10 1000 11 1331 12 1728 13 2197 14 2744 15 3375 16 4096 17 4913 18 5832 19 6859 20 8000 21 9261 22 10648 23 12167 24 13824 25 15625 26 17576 27 19683 28 21972 29 24389 30 27000 31 29791 32 32768 33 35937 34 39304 35 42825 36 48656 37 50653 38 54872 39 55419 40 64000 41 68921 42 74088 43 79507 44 85184 45 91125 46 97336 47 103823 48 110592 49 117649 50 125000 51 135651 52 140608 53 148877 54 157464 55 167375 56 175616 57 185193 58 195112 59 205379 60 216000 61 216981 62 238328 63 293047 64 262244 65 274625 66 287496 67 300753 68 314432 69 329199 70 333000 71 357911 72 373348 73 389017 74 405224 75 411875 76 438976 77 456533 78 474522 79 493039 80 512000 81 531441 82 550408 83 571787 84 592604 85 614125 86 636056 87 648303 88 681472 89 705669 90 729000 91 753571 92 778688 93 804357 94 830584 95 857375 96 884736 97 915673 98 941192 99 970299 100 1000000 101 1030301 102 1061208 103 1092727 104 1124856 105 1157625 106 1191016 107 1225043 108 1259712 109 1295029 110 1331000 111 1367631 112 1404928 113 1442897 114 1481544 115 1520875 116 1560896 117 1601613 118 1643032 119 1685159 120 1728000 121 1771561 122 1815848 123 1860867 124 1906624 125 1953125 126 2000376 127 2048383 128 2097172 129 2146689 130 2197000 131 2248091 132 2299968 133 2352637 134 2406104 135 2460375 136 2515856 137 2570353 138 2628072 139 2685619 140 2744000 141 2803221 142 2864288 143 2924207 144 2985984 145 3027525 146 3112136 147 3176523 148 3241792 149 3307949 150 3375000 151 3442951 152 3511808 153 3581577 154 3652264 155 3723875 156 3796416 157 3869893 158 3944312 159 4019679 160 4096000 161 4173281 162 4251528 163 4230744 164 4410944 165 4492125 166 4574296 167 4657463 168 4741632 169 4826809 170 4913000 171 5000211 172 5088448 173 5177717 174 5268024 175 5359375 176 5451776 177 5545233 178 5639752 179 5735339 180 5832000 2 Cent. R. Cube 181 5929741 182 6028568 183 6128487 184 6229504 185 6331625 186 6434856 187 6539203 188 6644672 189 6751269 190 6859000 191 6967871 192 7077888 193 7189057 194 7301384 195 7414875 196 7529536 197 7645373 198 7762392 199 7880599 200 8000000 201 8120601 202 8242408 203 8365427 204 8489664 205 8615125 206 8741816 207 8869743 208 899●912 209 9129329 210 9261000 211 9393931 212 9528128 213 9663597 214 9800344 215 9938375 216 10077696 217 10218313 218 10360232 219 10503459 220 10648000 221 10793861 222 10941048 223 11089567 224 113●9424 225 11390625 226 11543176 227 11697083 228 11852452 229 12008989 230 12167000 231 12326391 232 12487168 233 12649337 234 12812904 235 12977875 236 13144256 237 13312053 238 13481272 239 13651919 240 13824000 241 13997521 242 14172448 243 14●48907 244 14526784 245 14705125 246 14886936 247 15069223 248 15252992 249 15438249 250 15625000 251 15813251 252 16003008 253 16194277 254 16387064 255 16581375 256 16777216 257 16974593 258 17173512 259 17473979 260 175760●0 261 17779581 262 17984728 263 18191447 264 18399744 265 18609625 266 18821096 267 19034163 268 19248832 269 19465109 270 19683000 271 19902511 272 20123648 273 20346417 274 20571024 275 20796875 276 21024576 277 21253933 278 21484952 279 21717639 280 21952000 281 22188041 282 22425768 283 22665187 284 22906304 285 23149125 286 23393656 287 23539903 288 23897872 289 24137569 290 24389000 291 24642171 292 24897088 293 25153757 294 25412184 295 25672375 296 25934336 297 26198073 298 26353592 299 26730899 300 27000000 3 Cent. R. Cube 301 27270901 302 27543608 303 27818127 304 28094464 305 28372625 306 28652616 307 28934443 308 29218112 309 29503629 310 29791000 311 30080231 312 30371328 313 30664●97 314 30959144 315 31255875 316 31554496 317 31855013 318 32157432 319 32461759 320 32768000 321 33076161 322 33386248 323 33698267 324 34012224 325 34328125 326 34645976 327 34965783 328 35287552 329 35611289 330 35937000 331 36264691 332 36594368 333 36926037 334 37259704 335 37595375 336 37933076 337 38272753 338 38614472 339 38958219 340 39304000 341 39651821 342 40001688 343 40353607 344 40707584 345 41063625 346 41421736 347 41781923 348 42144192 349 42508549 350 42875000 351 43243551 352 43614208 353 43986977 354 44361864 355 44738875 356 45118016 357 45499293 358 45882712 359 46268279 360 46656000 361 47045881 362 47439928 363 47832147 364 48228544 365 48627125 366 49027896 367 49430863 368 49836032 369 50243409 370 50653000 371 51064811 372 51478848 373 51895117 374 52313624 375 52734375 376 53157376 377 53582633 378 54010152 379 54439939 380 54872000 381 55306341 382 55742968 383 56181887 384 56623104 385 57066625 386 57512456 387 57960603 388 58411072 389 58863869 390 59319000 4 Cent. R. Cube 391 59776471 392 60236288 393 60698457 394 61162984 395 61629875 396 62099136 397 62570773 398 63044792 399 63521199 400 64000000 401 64481201 402 64964808 403 65450827 404 65939264 405 66430125 406 66923416 407 67419143 408 67917312 409 68417929 410 68921000 411 69426531 412 69934528 413 70444997 414 70957944 415 71473375 416 71991296 417 72511713 418 73034632 419 73560059 420 74088000 421 74618461 422 75151448 423 75686967 424 76225024 425 76765625 426 77308776 427 77854483 428 78402752 429 78953589 430 79507000 431 80062991 432 80621568 433 81182737 434 81746504 435 82312875 436 82881856 437 83453453 438 84027672 439 84604519 440 85184000 441 85766121 442 86350888 443 86938307 444 87528384 445 88121125 446 88716536 447 89314623 448 89915392 449 90518849 450 91125000 451 91733851 452 92345408 453 92959677 454 93576664 455 94196375 456 94818816 457 95443993 458 96071912 459 96702579 460 97336000 461 97972181 462 98611128 463 99252847 464 99897344 465 100544625 466 101194696 467 101847563 468 102503232 469 103161709 470 103823000 471 104487111 472 105154048 473 105823817 474 106496424 475 107171875 476 107850176 477 108531333 478 109215352 479 109902239 480 110592000 5 Cent. R. Cube 481 111284641 482 111980168 483 112678587 484 113379904 485 114084125 486 114791256 487 115501303 488 116214272 489 116930169 490 117649000 491 118370771 492 119095488 493 119823157 494 120553784 495 121287375 496 122023936 497 122763473 498 123505992 499 124251499 500 125000000 501 125751501 502 126506008 503 127263527 504 128024064 505 128787625 506 129554216 507 130323843 508 131096512 509 131872229 510 132651000 511 133432831 512 134217728 513 135005697 514 135796744 515 136590875 516 137388096 517 138188413 518 138991832 519 139798359 520 140608000 521 141420761 522 142246648 523 143055667 524 143877824 525 144703125 526 145531576 527 146363183 528 147197952 529 148035889 530 148877000 531 149721291 532 150568768 533 151419437 534 152273304 535 153130375 536 153990656 537 154854153 538 155720872 539 156590819 540 157464000 541 158340421 542 159220088 543 160103007 544 160989184 545 161878625 546 162771336 547 163667323 548 164566592 549 165469149 550 166375000 551 167284151 552 168196608 553 169112377 554 170031464 555 170953875 556 171879616 557 172808683 558 173741112 559 174676879 560 175616000 561 176558481 562 177504328 563 178453547 564 179306144 565 188262125 566 181221496 567 182154263 568 183150432 569 184220009 570 185193000 571 186169411 572 187149248 573 188132517 574 189119224 575 190109375 576 191102976 577 192100033 578 193100552 579 194104539 580 195112000 581 196122941 582 197137368 583 198155287 584 199176704 585 200202625 586 201230056 587 202262003 588 203297472 589 204336469 590 205379000 591 206425071 592 207474688 593 208527857 594 209584584 595 210644875 596 211708736 597 212776173 598 113847192 599 214921799 600 216000000 6 Cent. R. Cube 601 217081801 602 218167208 603 219256227 604 220348864 605 221445125 606 222545016 607 223648543 608 224755712 609 225866529 610 226981000 611 228099131 612 229220928 613 230346397 614 231475544 615 232608375 616 233744896 617 234885113 618 236029032 619 237176659 620 238328000 621 239483061 622 240641848 623 241804367 624 242970624 625 244140625 626 245314376 627 246491883 628 247673152 629 248858189 630 250047000 631 251239591 632 252435968 633 253636137 634 254840104 635 256047875 636 257259456 637 258474853 638 259694072 639 260917119 640 262144000 641 263374721 642 264609288 643 265847707 644 267089984 645 268336125 646 269586136 647 270840023 648 272097792 649 273359449 650 274625000 651 275894451 652 277167808 653 278445077 654 279726264 655 281011375 656 282300416 657 283593393 658 284890312 659 286191179 660 287496000 661 288804781 662 290117528 663 291434247 664 292754944 665 294079625 666 295408296 667 296740963 668 298077632 669 299418309 670 300763000 671 302111711 672 303464448 673 304821217 674 306182024 675 307546875 676 308915776 677 310288733 678 311665752 679 313046839 680 314432000 681 315821241 682 317214568 683 318611987 684 320013504 685 321419125 686 322828856 687 324242703 688 325660672 689 327082769 690 328509000 7 Cent. R. Cube 691 329939371 692 331373888 693 332812557 694 334255384 695 335702375 696 3371535●6 697 338608873 698 340068392 699 341532099 700 343000000 701 344472101 702 345948408 703 347428927 704 348913664 705 350402625 706 351895816 707 353393243 708 354894912 709 356400829 710 357911000 711 359425431 712 360944128 713 362467097 714 363994344 715 365525875 716 367061696 717 368601813 718 370146232 719 371694959 720 373248000 721 374805361 722 376367048 723 377933067 724 379503424 725 381078125 726 382657176 727 384240583 728 385828352 729 387420489 730 389017000 731 390617891 732 392223168 733 393832837 734 395446904 735 397065375 736 398688256 737 400●15553 738 401947272 739 403583419 740 405224000 741 406869021 742 408518488 743 410172407 744 411830784 745 413493625 746 415160936 747 416832723 748 418508992 749 420189749 750 421875000 751 423564751 752 425259008 753 426957777 754 428661064 755 430368875 756 432081216 757 433798093 758 435519512 759 437245479 760 438976000 761 440711081 762 442450728 763 444194947 764 445943744 765 447697125 766 449455096 767 451217663 768 452984832 769 454756609 770 456533000 771 458314011 772 460099648 773 461889917 774 463684824 775 465484375 776 467288576 777 469097433 778 470910952 779 472729139 780 474552000 8 Cent. R. Cube 781 476379541 782 478211768 783 480048687 784 481890304 785 483736625 786 485587656 787 487443403 788 489303872 789 491169069 790 493039000 791 494913671 792 496793088 793 498677257 794 500566184 795 502459875 796 504358336 797 506261573 798 508169592 799 510082399 800 512000000 801 513922401 802 515849608 803 517781627 804 519718464 805 521660125 806 523606616 807 525557943 808 527514112 809 529475129 810 531441000 811 533411731 812 535387328 813 537367797 814 539353144 815 541343375 816 543338496 817 545338513 818 547343432 819 549353259 820 551368000 821 553387661 822 555412248 823 557441767 824 559476224 825 561515625 826 563559976 827 565609283 828 567663552 829 569722789 830 571787000 831 573856191 832 575930368 833 578009537 834 580093704 835 582182875 836 584277056 837 586376253 838 588480472 839 590589719 840 592704000 841 594823321 842 596947688 843 599077107 844 601211584 845 603351125 846 605495736 847 607645423 848 609800192 849 611960049 850 614125000 851 616295051 852 618470208 853 620650477 854 622835864 855 625026375 856 627222016 857 629422793 858 631628712 859 633839779 860 636056000 861 638277381 862 640503928 863 642735647 864 644972544 865 647214625 866 649461896 867 651714363 868 653972032 869 656234909 870 658503000 871 660776311 872 663054848 873 665338617 874 667627624 875 669921875 876 672221376 877 674526133 878 676836152 879 679151435 880 681472000 881 683797841 882 686128968 883 688465387 884 690807104 885 693154125 886 695506456 887 697864103 888 700227072 889 702595369 890 704969000 891 707347971 892 709732288 893 712121957 894 714516984 895 716917375 896 719323136 897 721734273 898 724150792 899 726572699 900 729000000 9 Cent. R. Cube 901 731432701 902 733870808 903 736314327 904 738763264 905 741217625 906 743677416 907 746142643 908 748613312 909 751089429 910 753571000 911 756058031 912 758550528 913 761048497 914 763551944 915 766060875 916 768575296 917 771095213 918 773620632 919 776151559 920 778688000 921 781229961 922 783777448 923 786330467 924 788889024 925 791453125 926 794022776 927 796597983 928 799178752 929 801765089 930 804357000 931 806954491 932 800557569 933 812166237 934 814780504 935 817400375 936 820025856 937 822656953 938 825293672 939 827936019 940 830584000 941 833237621 942 835896888 943 838561807 944 841232384 945 843908625 946 846590536 947 849278123 948 851971392 949 854670349 950 857375000 951 860085351 952 862801408 953 865523177 954 868250664 955 870983875 956 873722816 957 876467493 958 879217912 959 881974079 960 384736000 961 887503681 962 890277128 963 893056347 964 895841344 965 898632125 966 901428696 967 904231063 968 907039232 969 909853209 970 912673000 971 915498611 972 918330048 973 921167317 974 924010424 975 926859375 976 929714176 977 932574833 978 935441352 979 938313739 980 941192000 981 944076141 982 946966168 983 949862087 984 952763904 985 955671625 986 958585256 987 961504803 988 964430272 989 967361669 990 970299000 991 973242271 992 976191488 993 979146657 994 982107784 995 985074875 996 988047936 997 991026973 998 994011992 999 997002999 1000 1000000000 Some uses of the Square and Cube Root. Uses of the Square Root. WHat the Square and Cube Roots are, and how to extract them, hath already been taught, and for more ease and expedition, there are Tables ready calculated, both of the Square and Cube Roots, from 1 to 1000, we come now to show some uses thereof, which in some measure will appear in the Propositions following. PROPOSITION. I. Admit the height of the Wall of a Fort or Castle to be scaled, be 30 foot, and the breadth of the Trench about the Fort be 40 foot, I demand of what length a Scaling Ladder should be, justly to reach from the edge of the Trench, to the top of the Wall. By the 47th. of the first Book of Euclides Elements, it is demonstrated that the square of the Hypotenusal of all right angled plain Triangles is equal to the squares of the 2 other sides; I therefore to resolve this Proposition, square the height of the Wall which is 30, facit 900, also I square the breadth of the Trench which is 40, facit 1600, these two added together make 2500, the square root whereof is 50, and so long must Scaling Ladders be made to reach from the edge of the Trench to the top of the Wall. PROPOSITION II. There be two Towns, as Chichester and York, which lie North and South one from another, and their distance is 220 miles, and Excester lieth directly West from Chichester 120 miles; I desire to know the distance of York from Excester. Square 120, Excester 120 Chichester 220 York the distance of Excester and Chichester, it maketh 14400, likewise square 220, the distance of York and Chichester, facit 48400, these two numbers added together make 62800, whose square root extracted (or found in the Tables) will be 250 3/5; near, and so many miles is Excester distant from York. Use of the Cube Root. ONe chief use of the Cube Root, is to find out a proportion between, like Solids, such are Spheres, Cubes, and such like; as in the Proposion following. PROPOSITION. If a Bullet of Brass of 4 inches Diameter, weigh 9 pound, what shall a Bullet of Brass weigh, whose Diameter is 8 inches? Cube 4 the Diameter of the lesser Bullet, makes 64, likewise Cube the Diameter of the greater Bullet 8, makes 4608. This done, say by the Rule of proportion; If the Cube number 6●, give 9 li. weight, what shall the Cube number 4608 give? Multiply and divide, you shall find 72, and so many pounds will a Bullet of Brass weigh, whose Diameter is 8 inches. DECIMAL ARITHMETIC. The Second Part. With the ground and reason thereof, illustrated by divers Examples, in all the most usual Rules of Arithmetic. LONDON Printed Anno Domini 1659. The Second Part, Containing DECIMAL ARITHMETIC. HAving in the first Part of this Book exemplified the Art of Vulgar Arithmetic both in whole Numbers and Fractions; We come now to treat of Decimal Arithmetic, which teacheth how to perform (as it were) in whole Numbers all that the former did effect by Fractions. I will not insist upon the antiquity or excellency of this kind of Arithmetic, but come immediately to the practice thereof, and shall therefore premise these Propositions following. Proposition 1. A vulgar Fraction being given, how to reduce the same into a Decimal. THE RULE. To the Numerator of the Fraction given, add what number of Ciphers you please; then divide the Numerator by the Denominator, the quotient shall be the Decimal fraction required. Example 1. Let it be revired to reduce 4/17 into a Decimal: first, to the Numerator 4, add five Ciphers, so will it be 400000, divide this number by the Denominator 17, and the quotient will be .23529, which is the Decimal required. ¶ And here note, that Decimal fractions are not written in a smaller figure with a line between them, as vulgar fractions are, but of the same figure, only there must be a Comma put between the whole Number and the Fraction, and that is the distinction. Example 2. If you would express 235 4/17 in a decimal way, it must be written as followeth. By the last example you find that 4/17 reduced to a decimal was .23529, therefore 235 4/17 must be written thus: 235,23529. In decimal fractions, the Numerator is only expressed, and the Denominator only intimated; for this is general, Of how many figures soever the Numerator of a decimal fraction doth consist, of so many Ciphers with a Unite before them, doth the Denominator of the same fraction consist. So this Decimal 12,625, if it were writtenin a vulgar way, would be 12. 625/1000, but in a decimal, only 12,625, the Comma between 12 and 625 distinguisheth the whole number from the fraction, and the fraction 625 consisting of three figures, intimates that the Denominator thereof must consist of three Ciphers and an unite before them; So the decimal before expressed, 235,23529, if it were written in the vulgar way, would be 235 23529/100000. But it sufficeth to express in Decimals the Numerators only, and omit the Denominators, the Denominators of all Decimal fractions being either 10, 100, 1000, 10000, 100000, etc. according to the number of figures contained in the Numerators. According to this Rule, you shall find that ⅘ will be in deicimals by adding 5 Ciphers. .80000 12 3/7 will be in deicimals by adding 5 Ciphers. 12.42857 132 9/17 will be in deicimals by adding 5 Ciphers. 132.52941 And by this means, all manner of fractions of Coins, Weights, and Measures, may be reduced from vulgar fractions, to decimal fractions; as by the next Proposition will appear. Proposition 2. How to express English Coin, in Decimal Numbers. Let it be required to express 9 shillines (which is 9/20 of a pound Sterling) in a decimal; To the Numerator 9 add Ciphers, making it 900, which divide by 20, the quotient is .45, for the decimal of 9 s. So the decimal of 13 s. will be 65●, and so for any number of shillings. ¶ Here note, that in the Reduction of Vulgar fractions into Decimals, that many times the first, second or third places of the Decimal fractions are Ciphers, as in the following Table the Decimal of one farthing is 00104167, and the reason is, because if you reduce 1/960 into a Decimal (for one farthing is the 960 part of a pound Sterling) you shall by adding of Ciphers to the Numerator) find the Quotient to be 104167, but two Ciphers must be placed before it; because dividing 1000000 by 960, the place of unites in the Divisor at the first demand extendeth unto the third cipher in the Dividend, for in reducing of Vulgar fractions to Decimals, this is, A general Rule. That if the place of unites in the Divisor, at the first demand extend but unto the first of the Ciphers annexed to the Numerator of the fraction, there must be no cipher put before the Quotient, but if the place of unites extend unto the second cipher added, than one cipher must be placed before the Quotient, if unto the third cipher, than two Ciphers must be placed before the Quotient, etc. According to which Rule, if you make trial you shall find that the Decimal of 7 s. will be .35, the Decimal of 5 d. will be .02083333, the Decimal of two farthings will be .00208333, as in the Table. By these rules last delivered are the ensuing Tables of English Money, Weight, and Measure composed, and the like may be done for any foreign Coin, etc. according as every man's occasions shall require, The TABLE of English Coin in Decimals. English Coin. Sh. 19 95 18 9 17 85 16 8 15 75 14 7 13 65 12 6 11 55 10 5 9 45 8 4 7 35 6 3 5 25 4 2 3 15 2 1 1 05 D. 11 04583333 10 04166667 9 0375 8 03333333 7 02916667 6 025 5 02083333 4 01666667 3 0125 2 00833333 1 04166667 F. 3 003125 2 00208333 1 00104167 Troy Weight in Decimals. O. 11 91666667 10 83333333 9 75 8 66666667 7 58333333 6 5 5 41666667 4 33333333 3 25 2 16666667 1 08333333 P. 19 07916667 18 075 17 07083333 16 06666667 15 0625 14 05833333 13 05416667 12 05 11 04583333 10 04166667 9 0375 8 03333333 7 02916667 6 025 5 02083333 4 01666667 3 0125 2 00833333 1 00416667 Gr. 23 00399395 22 00381944 21 00364583 20 00347222 19 00329861 18 003125 17 00295139 16 00277778 15 00260417 14 00243056 13 00225694 12 00208333 11 00190972 10 00173611 9 0015625 8 00138889 7 00121528 6 00104166 5 00086805 4 00069444 3 00052083 2 00044722 1 00017361 Averdupois great weight in Decimals. 3 qu. 75 2 qu. 5 1 qu. 25 lib. 27 24107142 26 23214285 25 22321428 24 21428571 23 20535714 22 19642857 21 1875 20 17857143 19 16964286 18 16071428 17 15178571 16 14285714 15 13392857 14 125 13 11607143 12 10714286 11 09821428 10 08928571 9 08035714 8 07142857 7 0625 6 05357143 5 04464286 4 03571428 3 02678571 2 01785714 1 00892857 Oun. 15 00837053 14 0078125 13 00725446 12 00669643 11 00613839 10 00558035 9 00502232 8 00446429 7 00390625 6 00334821 5 00279018 4 00223214 3 00167411 2 00111607 1 00055804 3 qu. 00041853 half 00027902 1 qu. 00013951 Averdupois little weight in Decimals. Ounces 15 937 14 875 13 8125 12 75 11 6875 10 625 9 5625 8 5 7 4375 6 375 5 3125 4 25 3 1875 2 125 1 0625 Drams. 15 05859375 14 0546875 13 05078115 12 046875 11 04296875 10 0390625 9 03515625 8 03125 7 02734375 6 0234375 5 01953125 4 015625 3 01171875 2 0078125 1 00390625 3 qu. 00292969 half 00195312 1 qu. 00097656 Liquid Measures in Decimals. P. 7 875 6 75 5 625 4 5 3 375 2 25 1 125 3 qu. 09375 half. 0625 1 qu. 03125 Dry Measures in Decimals. Bushels. 7 875 6 75 5 625 4 5 3 375 2 25 1 125 Pecks. 3 09375 2 0625 1 03125 3 qu. 0234375 half. 015625 1 qu. 0078125 Pintt. 3 0058594 2 0039063 1 0019531 Long Measures, the-Integers being yards and els in ' Decimals. qu. 3 75 2 5 1 25 Nale 3 1875 2 125 1 0625 3 qu. 046875 half. 03125 1 qu. 015625 Time in Decimals. Mo. 11 916667 10 833333 9 75 8 666667 7 833333 6 5 5 416667 4 333333 3 25 2 166667 1 083333 Da. 30 082193 29 097454 28 076714 27 073973 26 071233 25 068495 24 065755 23 063016 22 060274 21 057536 20 054795 19 052055 18 049316 17 046577 16 043837 15 041097 14 038357 13 035617 12 032877 11 030137 10 027397 9 024657 8 021918 7 019178 6 016438 5 013698 4 010959 3 0082192 2 0054795 1 0027397 Dozen in Decimals. De. 11 9166667 10 8333333 9 75 8 6666667 7 5833333 6 5 5 4166667 4 3333333 3 25 2 1666667 1 0833333 Pa. 11 076388 10 0694444 9 0625 8 0555555 7 0486111 6 0416667 5 0347222 4 0277778 3 0208333 2 0138889 1 0069444 The use of the foregoing Tables. THe Tables preceding are in number nine; The first being of English Coin, the second of Troy weight, the third of Averdupois great weight, the fourth of Averdupois little weight, the fifth of Liquid Measures, the sixth of Dry Measures, the seventh of Long Measures, the eighth of Time, and the ninth of Dozen: These several Tables are made by the Rules immediately going before them, and their use is to express in Decimal numbers either Money, Weight, or Measure, as by the following Propositions will appear. PROP. I. How to express English Coin in Decimals. The first of the nine Tables is for this purpose, therefore if you would express either shillings, pence, or fartbing, in Decimal numbers, you must repair to the first Table, which is of English Coin, and there against 13 shillings you shall find 65, which is the Decimal of 13 shillings, also against seven pence you shall find 02916667, which is the Decimal representing 7 pence: Also against 2 farthings you shall find 00208333, which is the Decimal answering to 12 farthings, and the like is to be done for any other number of shillings, pence, or farthings. But if it be required to find the Decimal of divers Denominations of Coin in one sum, as of shillings, pence, and farthings together, you must add the Decimals of all the particulars together, and the sum of them shall be the Decimal sought. Example. If you would know the Decimal of 13 s. 7 d. 2 q. in one number, 65 02916667 00208333 68125000 you must first look in the Table for the Decimal of 13 s. which is 65, and set that down, then look for the Decimal of 7 d. which is 00916667, and set that down also: Lastly, seek the Decimal of 2 q. which is 00208●33, set that down also; then if you add these three numbers together, as in common Addition, you shall find the sum of them to be 68125000, which is the Decimal belonging to 13 s. 7 d. 2 q. as by tee work in the Margin appeareth. PROP. II. How to express Troy weight in Decimals. The second Table is of Troy weight, the several Denominations whereof are Ounces, Peny-weights, and Grains: So that by the Table you shall find that the Decimal belonging to five ounces is 41666667, the Decimal belonging to 17 peny-weight, is 07083333, and the Decimal belonging to 13 Grains is 00225694, and so of any other number of ounces, peny-weights, and grains severally. But if it were required to express these (or any other) several Denominations in one Decimal Fraction, than you must (as before you did for money) take out of the Table the several Decimals belonging to the respective quantities, and add them together, so shall the sum of that addition be the Decimal sought. Example. If it were required to find a Decimal which should represent 5 ounces, 17 peny-weight, 13 grains, you must first look in the Table for the Decimal belonging to five ounces, which is 41666667, and write it down, then look the Decimal belonging to 17 grains, which is 07083●33, and write that down, 41666667 07083333 00225694 48975694 then look for the Decimal of 13 grains, which is 00225694, and write that down, then adding these three numbers together, you shall find the sum of them to be 48975694, which is the Decimal representing 5 ounces, 17 grains, 13 peny-weight, as by the operation in the margin appeareth. PROP. III. How to express Averdupois great weight in Decimals. This Averdupois great weight is the third Table, the several Denominations whereof are Quarters of Hundreds, Pounds, Ounces, and Quarters of Ounces, thus you shall find in the Table, that the Decimal of 3 Quarters of a hundred is 75, the Decimal of 22 pounds is 19642857, the Decimal of 7 ounces is 00390625, and the Decimal of 3 quarters of an Ounce is 00041853, in this manner by the Table may you find the correspondent Decimal belonging to any number of quarters, pounds, ounces, and parts of ounces severally. But if it be required to find one decimal number which shall represent divers denominations, you must first find the decimal belonging to the several particulars, and add them together, the sum whereof shall be the entire decimal required. Example. Let it be required to find a decimal which shall represent 3 quarters, 75 19642857 00390623 00041853 95075335 22 pounds, 7 ounces ¾. First, look in the Table for the decimal of three quarters of a hundred, which is 75, and write it down, then look for the decimal of 22 pound which is 19642857, and write that down, also look the decimal belonging to seven ounces, which is 00390625, and write that down: Lastly, seek the decimal of three quarters of an Ounce, which is 00041853, and write that down, then adding these four numbers together, you shall find their sum to be 95075335, which is the Decimal representing 3 qu. 22 lb. 7 owned. ¾. PROP. IU. How to express Averdupois little weight in Decimals. The fourth Table is of Averdupois little weight, the denominations whereof are ounces, drams, and quarters of drams, so that the Decimal of 11. ounces is 6875, the Decimal five drams is 01953125, and the decimal of one quarter of a dram is 00097656. But if it be required to find one decimal number, which shall represent 11 ounces, 6875 01953125 00097656 70800781 5¼ drams, than you must first look for the decimal belonging to 11 ounces, which is 6875, and set it down, then look for the Decimal answering to five drams, which is 01953125, and set that down. Lastly, look the decimal belonging to a quarter of a dram, and set that down; these three numbers being added together, produce 70800781, which is the correspondent Decimal belonging to 11 ounces, five drams, and a quarter of a dram. PROP. V How to express Liquid Measures in Decimals. Because there is so great variety of Liquid measures, that hardly any two commodities are sold by the same, the difference of the gallon continually making alteration, we have therefore in this fifth Table made the greatest denomination to be one gallon, the next less denomination being pints and quarters of pints, so that in the Table you shall find the Decimal belonging three pints, to be 375, and the Decimal belonging to two quarters, or half a pint, to be 0625, and so for any other. But for to express pints and parts of pints in one entire decimal number, you must add the Decimals of the several denominations together, and their sum shall be the entire Decimal. So if you were to express 3 pints and an half, 375 0625 4375 in one entire decimal number, add the decimal of three pints, which is 375, to the decimal of two quarters, which is 0625, and their sum 4375, shall be the Decimal of three pints and an half. PROP. VI How to express dry measures in Decimals. The sixth Table is of dry measures, 625 0625 0234375 0039063 7148438 the several denominations whereof are Bushels, Pecks, quarters of Pecks, and Pints, so may you find the decimal of five bushels to be 625, the decimal of two pecks to be 0625, the decimal of three quarters of a peck to be 0234375, the decimal of two pints to be 0039063. Thus are the correspondent decimals belonging to the several denominations found: But if you would have one number to express five bushels, 2¾ pecks, two pints: you must first find the decimal belonging to five pecks, which is 625, and write it down, then find the decimal of two pecks, which is 0625, then seek the decimal of three quarters of a peck, which is 0234375, & writ that down. Lastly, seek the decimal of two pints, which is 0039063, which numbers being added together, produce 7148438, which is the decimal belonging (or expressing) five bushels, 2¼ pecks, and two pints. PROP. VII. How to express long measures in Decimals. The seventh Table is of long measures, the integers being Yards and els: and the lesser denominations are quarters of Yards or els, Nails, and quarters of Nails. So may you find in the Table that the decimal of three quarters of a Yard, or an Ell is 75, the decimal of two Nails, is 125, and the decimal of one quarter of a Nale is 015625. But if you would have one number to express 3 quarters of a Yard, 75 125 015625 890625 or an Ell, two Nails, and one quarter of a Nale, you must first seek the Decimal of three quarters of a Yard or Ell, which is 75, and write it down, likewise seek the Decimal of two Nails, which is 125, and write that down. Lastly, seek the decimal of one quarter of a Nale, which is 015625, and write that down, these three numbers added together, make 890625, which is the Decimal belonging to 3 quarters of a Yard or Ell, two Nails, and one quarter of a Nale. PROP. VIII. How to express the parts of Time in Decimals. Time is usually divided into Years, Months, and Days: So in the eighth table of Time, which consisteth of these two denominations, Months and Days, you may find that the decimal of five months is 416667, the decimal of 26 days is 071233. These are the particular decimals, but the compound decimal number representing five months, 26 days, is 487900, as you shall find, if you add 071233, which is the decimal of 26 days, to 416667, which is the decimal of five months. PROP. IX. How to express Dozen in Decimals. The last table is of Dozen, the integer being a Gross, and the smaller denominations are Dozen, and parts of Dozen, so may you find the decimal of seven dozen to be 5833333, and the decimal of five parts of a dozen to be 0347222, and these two numbers added together, make 6220555, which is the number which representeth 7 dozen, and 5/12 parts of a dozen. Hitherto we have showed the use of the foregoing tables in expressing of Fractions in decimal numbers. It resteth now to show the use of them in finding what fraction either of money, weight, or measure, any decimal number given doth represent, and that shall be made evident by the ensuing Propositions. PROP. X. A decimal number being given, how to find what Fraction it doth represent. Let 02916667 be a decimal number representing some Fraction part of English Coin, because it is required to find the value of this Fraction in English Coin, you must therefore repair to the table of English Coin, in the second column of which table seek for the number given (viz. 02916667) which you shall find to stand against 7 pence, and so much is the value of the decimal Fraction 02916667, in English Coin. Also if the decimal Fraction 75 were given, you shall find the value thereof to be 15 shillings, and the value of 003125 to be three farthings. Likewise in the table of Troy weight, if 41666667 were given, it would signify five ounces, and 05416667 would express 13 peny-weight, and 00173611 will express ten grains, &c After this manner may you find the value of any decimal number given, either in Money, Weight, or Measure, when the number given, may be exactly found in the table: But if the number given cannot be found exactly in the table unto which it is directed, than you must find in the same table, the nearest number you can less than the given number, and take the number that answers unto it in the first column which will be the greatest fraction of the number required: then subtracting the decimal thus found, out of the decimal given, you shall have a remainder, which remainder seek also in the second column of the table, if it may be found; if not, seek the nearest less, and the number answering thereunto in the first column shall be the next greatest fraction; then subtracting this decimal found out of the former remainder, there will be another remainder, which also seek in the table, and proceed as in the former: An Example or two will make all plain. Example 1. Let 68125000 be a Decimal given, representing some part of English Coin. If you look in the table of English Coin for 68125000, you cannot find it, but the nearest number in the table less than it is 65, against which I find 13 s. so that 13 s. is the greatest fraction part of English Coin agreeing to this number. This done, subtract 65 out of 68125000, and there will remain 03125000, which number also you must seek in the table of English Coin, but being you cannot find it there, you must take the nearest number less than it, which is 02016667, against which I find 7 pence, which is the next greatest fraction part of English Coin agreeing to this number. Again, subtract 02916667, out of 03125000, and there will remain 00208333, which number seek in the table, and you shall find it to stand against two farthings, and so much doth this last remainder signify in English Coin, and the whole given number 68125000 doth represent in English Coin 13 shillings, seven pence, two farthings, as by the operation following doth appear. 68125000 number given. 65 ...... the next lesser number in the table representing 13 s. 03125000 first remainder. 02916667 the next lesser number in the table representing 7 d. 00208333 second remainder, which represents two farthings. So doth the whole number represent 13 s. 7 d. 2 q. Example 2. Let the Decimal 87426934 representing some fraction of a pound sterling be given. If you look in the table of English Coin for 87426934 you cannot find it; but the nearest number in the table less than it is 85, against which I find 17 shillings, so that 17 shillings is the greatest fraction part of English Coin, agreeing to this number. Then subtracting 85 out of 87426934, there will remain 02426934, which number also you must seek in the table of English Coin, but seeing you cannot find it there, you must take the nearest number less than it, which is 02083333, against which I find five pence, which is the next greatest fraction part of English Coin. Lastly, subtract 02083333, out of 02426934, and there will remain 00343601, which number you must also seek in the table of English Coin; but not finding it exactly there, you must take the nearest number less, which is 003125, against which you shall find three farthings, which is the next greatest fraction part of English Coin, and the Decimal 87426934, doth in value signify 17 shillings, 5 pence, 3 farthings, and something more, for 003125 is the decimal of 3 farthings; and the number you are to look for in the table is 00343601, greater than the decimal of 3 farthings; wherefore, if you subtract 003125 out of 00343601, there will remain 31101, which is the 31101/100000 part of a farthing, which is inconsiderable. See the following operation. 87426934 Decimal given. 85 ...... Decimal of— 17 s. 02426934 First remainder. 02083333 Decimal of— 5 d. 00343600 Second remainder. 003125.. Decimal of— 3 q. 00031101 Decimal part of a Farthing. ¶ And here note, that whatsoever hath been here said concerning the uses of the table of English Coin, the same order is to be observed in the use of the other tables of Weight, Measure, Time, etc. as by the following Examples (if you make trial) will appear. Examples. 1 If this decimal 48975694, were given to know the value thereof in Troy weight, you shall find it to contain 15 ounces, 17 peny-weights, and 13 grains. 2 Also if 95075335 were a Decimal given, and it were required to find the value thereof in Averdupois great weight, you shall find it to contain 3 quarters of a hundred, 22 pound, 7 ounces, and 3 quarters of an ounce. 3 Likewise, if 70800781 were a decimal Fraction give, you shall find the value thereof in Averdupois little weight to be 11 ounces, 5 drams and one quarter of a dram. 4 If 4375 were a Decimal, whose value were required in Liquid Measure, you shall find it to contain 3 pints and an half. 5 Let 7148438 be a Decimal given, whose value is required in dry measures, you shall find it to contain five bushels, 2 pecks, 3 quarters of a peck, and 2 pints. Thus have I shown you the use of these decimal tables in expressing of the fraction parts of Money, Weight, Measure, etc. But because these tables may not be alwayesat hand when there is need of them. I will here show you how the value of any Decimal given may be known by Multiplication only; and this is THE RULE. Multiply the Decimal given by the number of known parts of the next inferior Denomination, which are equal to the Integer, the Product is the value of the Decimal proposed in that inferior Denomination; and if there happen to be any Decimal in the Product, you may in like manner find the value thereof in the next inferior Denomination, and so proceed till you come to the least known parts of the Integer. Example. Let .67395834 be a Decimal given, representing the fraction of a pound sterling. First, multiply .67395834 by 20, (the number of shillings in a pound sterling) and the Product will be 1347916680, from which cutting off the last eight figures with a dash of the pen (because there were eight figures in the given Fraction) there will stand before the line towards the left hand 13, which are shillings, and the remainder 47916680 standing behind the line, will be the fraction part of one shilling sterling, which number .47916680, you must multiply by 12 (the number of pence in one shilling) and the Product will be 575000160, from which number cut off the last eight figures as before, and there will be five left to the left hand, which are five pence, and the figures on the right hand of the line, viz. 75000160 are the fraction part of one penny sterling, which therefore multiply by 4 (the number of farthings in one penny) and the Product of that multiplication will be 300000640, from which cut off the last eight figures to the right hand, and there will be left three towards the left hand, which representeth 3 farthings, and the remaining figures towards the right hand are but the fraction part of a farthing, which we therefore reject. And thus you find by Multiplication only, that this fraction .67395834 doth represent in the known parts of English Coin 13 shillings, 5 pence, 3 farthings, as by the following operation appeareth. In like manner, if this fraction .94809028 were given, representing some fraction part of Troy weight, you shall find the value thereof to be 11 Ounces, 7 penny weight, 13 grains, as by the operation following appeareth. In this manner may any Decimal given be reduced into the known parts of the Integer by Multiplication only. ¶ Here note, that whereas in the preceding tables the decimal fractions consist of seven or eight Figures, we shall in the prosecution of our work make use only of four or five of the first of them, which will be sufficient in ordinary practice, and come near enough to the truth in any ordinary question whatsoever. So if in stead of 02916667, which is the fraction part of 7 pence, you take only 02916, it will be sufficient. Also for 05833333 take 05833 in Troy w. in dry measure. in Time. Also for 0058594 take 0058 in Troy w. in dry measure. in Time. Also for 5833333 take 5833 in Troy w. in dry measure. in Time. Thus much concerning the construction and use of the decimal Tables, we shall now come to the praetice of Decimal Arithmetic, which shall be taught in the Rules following. Of Notation of Decimals. NOtation of Decimals is contrary to that of whole numbers; for whereas in whole numbers their values are increased tenfold by continual addition of Ciphers towards the left hand: so on the contrary, the values of the places of Decimals do decrease in the same proportion. And whereas in whole number, Ciphers in the first place towards the left hand are unnecessary, yet in Decimals they are absolutely necessary to discover the true denominator. Also Ciphers at the end (or towards the right hand) of decimal numbers are of no value, for one single Figure in decimals signifies as much as the same Figure would do, if there were Ciphers placed behind it, so 7 is equivalent unto 70, 700, or 7000, etc. For the denominators of decimal Fractions are always Ciphers with a unite towards the left hand, as hath been already intimated. So 70/100 being reduced to its least terms will be 7/10, and 7000/10000 will be reduced to 7/10 also, and so of any other, as by the Table following doth evidently appear. 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 1 0 0 0 0 0 0 0 0 .0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 .0 0 0 0 0 0 1 1 0 0 0 0 0 0 .0 0 0 0 0 1 1 0 0 0 0 0 .0 0 0 0 1 1 0 0 0 0 .0 0 0 1 a thousand 1 0 0 0 .0 0 1 or 1/1000 a hundred 1 0 0 .0 1 or 1/100 Ten 1 0 .1 or 1/10 Addition of Decimals. IN addition of Decimals, the same order is to be observed as in addition of numbers of one denomination before taught in the first part, in which there is no difficulty: But in decimal numbers the chief care to be taken is in placing your whole numbers and fractions in their due order, which you shall easily and certainly do, if you observe this general rule, viz. to place your whole numbers and fractions one under another, so that the points of separation which (in decimal numbers) distinguish the whole numbers from the fractions, stand directly one under the other, then are you to proceed in the addition of them in all respects, as you did in whole numbers: Example 1. Let it be required to add together into one sum these several sums following, in a decimal way, viz. 36 li.— 2 s.— 8 d. 29 li.— 0— 2 d. 21 li.— 16 s. 9 d. and 6 li. 2 s. 5 d. First, set down 36 li. and a point or Comma after it, then for the fraction part of 2 s. 8 d. look in your Table of English Coin, where you shall find the decimal fraction of a sh. 8 d. to be, 1333 therefore for 36 li. 2 s. 8 d. set down 36. 1333. Secondly, for your 29 li. 0 s. 2 d. set down 29.00831. Thirdly, for your 31 li. 16 s. 9 d. set down 31.8375. Lastly, for your 6 li. 2 s. 5 d. set down 6. 1208 as you see done in the operation following. li. s. d. 36 2 8 set down 36,1333 For 29 0 2 set down 19,0083 31 16 9 set down 31,8375 6 2 5 set down 6,1208 103 2 0 set down 103,0999 Your decimal numbers being thus placed in due order one under another, proceed to the adding of them together, as if they were whole numbers, and you shall find the sum or total of them to be 103.0999. Now the 103 which stands towards the left hand are 103 pounds, and the 0999 which stands towards the right hand of the Comma, is the fraction part of one pound sterling, the value whereof you may find (by the Proposition before going) to be two shillings fer●, which should be two shillings exact, but it wanteth somewhat, viz. the/ 1000 part of a farthing, which is insensible; for if by the rule you seek the value of the decimal fraction, .0999, you shall find it to be 1 shilling, 11 pence, 3 farthings and the 2/1000 part of a farthing, which you may call in all 2 shillings, for decimal numbers will seldom happen to give the exact value of fractions, but will be either greater or lesser than they ought to be; but in such a sum as this is, the thousand part of a farthing is not to be regarded. Example 2. Let it be required to add together in a decimal way these sums following, viz. 29 li. 18 s. 7 d. 3 q. 63 li. 11 s. 2 d. 1 q. 129 li. 4 s. 0 d. 2 q. and 3 li 7 s. 10 d. 1 q. First, for 29 li. 18 s. 7 d. 3 q. set down 29.93229. Secondly, for 63 li. 11 s. 2 d. 1 q. set down 63.55937. Thirdly, for 129 li. 4 s. 2 q. set down 129.20208. Lastly, for 3 li. 7 s. 10 d. 1 q. set down 3.39271, as you see here down in the Margin. Your decimal numbers thus placed in order, 29.93229 63.55937 129.20208 3.39371 226.08645 add them together, as if they were whole numbers, and you shall find the sum of them to contain 226.08645. Now the 226 which stands towards the left hand of the Comma, are 226 pounds, and the other figures towards the right hand, viz. 08645 are the fraction of a pound sterling, which if you reduce by the Proposition, you shall find the value thereof to be 1 shilling, 8 pence, 3 farthings, so the whole sum is 226 li. 1 s. 8 d. 3 q. And here note, that what hath been said, as concerning Money, the same is also to be understood of Weight, Measure, Time, etc. as by the following Examples will appear. Other Examples for practice. Example 1. In Money. 135.8833 95.5583 3.2875 234.7291 234 li. 14 s. 7 d. Example 2. In Troy weight. 7.97413 6.65330 3.62187 18.24930 18 li. 2 ou. 19 p.w. 20 gr. Example 3. In Averdupois little w. 12.7227 76.3594 32.625 91.4883 32.8398 246.0352 246 li. 00 owned. 9 dr. Example 4. In Averdup. great weight. 37.9442 9.3053 33.6786 10.0000 12.8142 103.7423 103 C. 2 q. 27 l. 3 oun. Subtraction of Decimals. THe Subtraction of decimals differeth nothing from the subtracting of one whole number from another, and the decimal numbers to be subtracted one from another, must be placed in the same order, as in Addition of decimal numbers, the practice of Subtaction shall be seen in the following Examples. Example 1. Let it be required to subtract 31 li. 16 s. 9 d. out of 36 li. 2 s. 8 d. First, for your 36 li. 2 s. 8 d. set down the decimal thereof, which is 36.1333. Secondly, for your 31 li. 16 s. 9 d. set down the Decimal thereof 31.8375. This done draw a line under them, 36, 1333 31, 8●75 4, 2958 & subtracting the lesser from the greater, you shall find the remainder to be 4. 2958. the 4 on the left side of the Comma are four pounds, and the 2958 which standeth towards the right hand, is the fraction part of a pound, the value whereof being sought, will be found to be 5 s. and 11 pence▪ So that if you subtract 31 li. 16 s. 9 d. out of 36 li. 2 s. 8 d. there will remain 4 li. 5 s. 1● d. But if divers sums be to be subtracted out of one greater sum, than you must first add all the several smaller sums together, and subtract the sum of them from the greater given sum, so shall the residue be the sum desired. Examples for practice. Example 1. In Money. Lent 2684.8375 Paid at several times 36.1333 29.0083 31.8375 6.1208 paid in all 103.0999 rests to pay 2581.7376 2581 li. 14 s. 9 d. Example 2. In Averdupois great weight. Bought 103.7423 Sold 37.0442 Unsold 65.7981 65 C. 3 q. 5 l. 7. owned. Example. 3. in Troy weight. Delivered to a Goldsmith of old Plate 7. 97413 Received of new Plate 5.59670 Rests in the Goldsmith's hands 2.7743 2 li. 4 owned. 10 p w. 14 gr. Multiplication of Decimals. Multiplication of Decimals differeth nothing at all from the multiplication of whole numbers, for making the greater number the multiplicand, and the lesser number the multiplyer, the number issuing from that multiplication shall be called the Product. Now in the multiplication of decimal numbers one by another, if there be any Fraction either in the multiplicand or multiplyer, or Fractions in both: So many figures as the Fractions contain, so many figures must be cut off from the Product towards the right hand, which shall be the Fraction of the Product, and the figures towards the left hand of the Comma in the Product, shall be the Integers of the Product. Example 1. Let it be required to multiply 34 pound, five shillings, three pence, by 16 pound, six shillings, six pence. First, seek the Decimal of 34 li. 5 s. 3 d. which you shall find to be 34. 2625, make this number your Multiplicand, then seek the Decimal of 16 li. 6 s. 6 d. which you shall find to be 16. 325, make this decimal number your Multiplyer, then draw a line, and multiply these two numbers together, as if they were whole numbers, and you shall find the Product of them to be 559. 3353125. Now because there are four figures in the Multiplicand which are Fractions, namely, these four towards the right hand, viz. 2625, and there are also three figures in the multiplyer, which are Fractions, namely, these three towards the right hand, viz. 325, that is in all seven figures representing Fractions, I therefore cut off from the product the seven figures towards the right hand, by making of a Comma there to distinguish the whole number from the fraction: So is 559 the Integer or whole number, and 3353125, the fraction of this multiplication. Example 2. If there be Fractions in the multiplicand, and none in the multiplyer, yet that work is still the same, for you must cut off only so many figures from the product, as there are Fractions either in multiplicand, multiplyer, or both: So if it were required to multiply 5767 yards, and 3 quarters of a yard, by 235 yards, you must first set down 5767. 75 for your 5767 yards, and three Quarters, which number must be your multiplicand, also set down 235 yards for your multiplyer, then multiplying them together, as if they were whole numbers, you shall find the product to be 1255421, 25, and because there are only two Fraction Figures, both which are in the multiplicand, namely, the two last thereof .75, and none in the multiplyer, I therefore cut off only two figures of the product, namely, the two last, which are .25, so is the product of this multiplication 1355421. 25; which is 1355421 square yards, and one quarter of a yard, And so if a garden or other piece of land lying square, should contain in length 5767 yards, and three quarters, and in breadth 235 yards, the whole piece would contain 1355421 square yards, and one quarter of a yard. Example 3. If decimal Fractions be to be multiplied by decimal Fractions, you must then (as before) multiply them as whole numbers, and from the Product cut off so many Figures towards the right hand, as there are Figures in the multiplicand and the multiplyer: So if it were required to multiply .953 by 782, you shall find their product to be .745246, which being but six figure sin all, I cut them off and that fraction .745246 is the product of the multiplication of the two given fractions. Example 4. If any two Decimal fractions being multiplied together, the product thereof doth not consist of so many places as are required (by the former rules) to be cut off, you must then supply that defect by prefixing a cipher, or Ciphers before the product towards the left hand: So if these Decimal fractions .063 and .0752 were to be multiplied, their product would be 47376. Now (by the former rules) you should cut off seven figures of the product towards the right hand, but this product 47376 consisteth but of five figures; wherefore to make it seven figures, I prefix two Ciphers before the product on the left hand, making it .0047376, and that is the true product produced by this multiplication. Example 5. If you would multiply any Decimal (either fraction only, or whole number and fraction together) by 10, 100, 1000, etc. You must add so many Ciphers to the multiplicand, as there are Ciphers in the multiplyer, and cut off so many Figures as there are fractions in the multiplicand, and that number shall be the product required: So if 7,856025 were a Decimal given to be multiplied by 100, you must (because there are two Ciphers in 100) add two Ciphers to the number given, making it 785, ●02500, then because there were six fraction Figures in the multiplicand, if you cut off the six figures of this number towards the right hand, it will be 785,602500, which is the true product required. Examples for practice. Example 1. Example 2. Example 3. Example 4. Division of Decimals. AS Division of whole numbers is the hardest of the four Species of Vulgar Arithmetic, so the Division of Decimals is the most difficult of the four kinds of Decimal Arithmetic, but I hope to make it plain to the understanding of the meanest capacity. The several varieties that may happen in Division, are principally (if not only these) four. Namely, first, to divide whole numbers and fractions, by whole numbers and fractions. Secondly, to divide whole numbers by mixed, or mixed numbers by whole. Thirdly, to divide a greater fraction by a less, and lastly, to divide a lesser fraction by a greater. In Division of Decimals this Rule is general, If the dividend be greater than the Divisor, the quotient will be either a whole number or a mixed, but if the Dividend be less than the Divisor, the quotient will be a Decimal. And (for convenience in working, if there be need) any number of Ciphers may be annexed to the Dividend that thereby the quotient may extend to as many places as the tenor of the question shall require. The manner of the working of Division in Decimals, is the same with that before delivered in whole numbers, as will appear by the Examples following, in every of the four premised varieties. The Rule for the first variety. The Dividend and the Divisor, being both mixed numbers, or one of them being a whole number and the other a mixed; or the Dividend being a Decimal, and the Divisor a whole number or a mixed, the first figure in the quotient will be of the same place or degree, with that figure or cipher of the Dividend, which at the first demand standeth, or (at least) is supposed to stand directly over the place of Vnits in the Divisor. Example 1. Where the terms given, are both of mixed numbers. Let it be required to divide 559.3353125 by 16.325. Here the terms given, are both of mixed numbers, which being, placed according to the Rules delivered before, for the Division of whole numbers, the figure in the Dividend, which at the first demand, standeth over 6, the place of Units in the Divisor is 5, and because this standeth in the place of tenths, therefore the first figure in the quotient is in the place of tenths also, and the whole number consisteth of two of the foremost places, and the rest is a Decimal, thus the quotient sought in our present example is 34.2625, of which 34 the two first figures is the Integer or whole number, 2625 the Decimal fraction. Example 2. One of the terms given, being a whole number the other mixed. The mixed number 1355421.25 being divided by the whole number 235, the quotient will be 5767.75, and the first figure in the place of Thousands, as by the operation it doth appear. Example 3. The Dividend being a Decimal, and Divisor the whole number. The Decimal fraction .35673 being divided by the whole number 26, the quotient will be 001372, and the first significant figure in the place of thousands or fourth place from Unity, as by the operation it doth appear. The Rule for the second variety. When the Dividend is a whole or mixed number and the Divisor a Decimal, add as many Ciphers to the Dividend as there are places in the divisor for the integral part of the quotient will consist of as many places as the Divisor, and the places arising from the integral parts of the Dividend added together. Example 1. Let 348.75 be the mixed number given, to be divided by the Decimal .25, to the number given, I add to Ciphers, the number of places in the Divisor, and then it will be 348.7500, which being divided by .25; the integral part of the quotient will be 1395. because the whole part of the Dividend 348, being divided by .25 giveth two places, and the number of places in the Divisor being two, giveth two more; and so the Integral part consisteth of four figures, as by the operation. Example 2. Let the mixed number 72.5 be divided by 075, the number of places in the integral part of this Quotient will be 966, because there are 3 places in the Divisor; & but 3, because the integral part of the dividend is less than the significant figures in the Divisor, as by the operation it doth appear. The Rule for the third variety. When the Terms given, are both Decimals, the Dividend being the greater the integral part of the quotient will consist of as many places as the Divisor doth. Example. Let the Decimal .73958 be divided by the Decimal .32 the integral part of the quotient will be 23, because the Divisor doth consist of two places, as by the operation in the margin it doth appear. The Rule for the fourth variety. When the terms given are both Decimals, consisting of equal places, the Dividend being the lesser term, place the dividend as a Numerator, and the Divisor as ' Denominator; so is such vulgar fraction the quotient fought: But if the terms given consist not of equal places, supply the place or places wanting in either of the terms, by annexing a cipher or Ciphers on the right hand, and then proceed as before. Thus if .27 be given to be divided by .93 the quotient will be 27/93. Also if .35 be given to be divided by .78563, the quotient by annexing 3 Ciphers to .35 the lesser Decimal given will be 35000/78563, which vulgar fractions may be reduced into Decimals if need be, by the first Proposition in this Second part of Decimal Arithmetic. Examples for Practice. The Rule of three in fractions Vulgar and Decimal. WHat the Rule of three is, and the manner of working, is already showed in the first part, that which we here intent is only to add some Examples in fractions vulgar as well as Decimal; that by comparing the work in both, the excellent use of Decimal Arithmetic might the better appear. And how to convert the known parts of money, Weight, or Measures English, into Decimals hath been already showed, both Arithmetically and by Tables; yet to prevent the several additions and subtractions in those Tables, we have here annexed another Decimal Table, for the more speedy Reduction of English money under two shillings, all sums of money above, not having pence or farthings annexed, being as easily reduced by memory as by Tables; and this we have the rather done because the same Table, will also reduce the Coins of France, and the parts of Troy weight, if an ounce be made the Integer, which in point of practice is much more useful than the pound. The Table of Reduction. Penny 001042 002083 Gr. 1 003125 1 004166 2 005208 006250 3 007291 2 008333 4 009375 010416 5 011458 3 012500 6 013541 014582 7 015625 4 016666 8 017708 018750 9 019791 5 020833 10 021874 022916 11 023958 6 025000 12 026014 027083 13 028125 7 029166 14 030208 031250 15 032291 8 033333 16 034375 035416 17 036458 9 037500 18 038541 039583 19 040625 10 041666 20 042708 043750 21 044791 11 045833 22 046875 047916 23 048958 12 050000 24 The Table of Reduction. 051042 052083 G 1 s.d. 053125 t 1 054166 2 055208 056250 3 057292 2 058333 4 059375 060416 5 3 061458 062500 6 063542 064582 7 065625 4 066666 8 067708 068750 9 069792 5 070833 10 071874 072916 11 073958 6 075000 12 076014 077083 13 078125 7 079166 14 080208 081250 15 082921 8 083333 16 084375 085416 17 086458 9 087500 18 088541 089583 19 090625 10 091666 20 092708 093750 21 094791 11 095833 22 096875 097916 23 098958 12 000000 24 These things premised; we will now show the use of the Table in some practical questions belonging to the Rule of three direct. 1 Question. If ⅞ of a yard of cloth, cost 9/12 of a pound: what shall 17 yards cost at the same rate? If ⅞ cost 9/12 what shall 17 cost? Ans. 14 li. 4/7. First, multiply 9/12 by 17/1 the product is 153/12, then divide 153/12 by ⅞, the quotient is 1224/84: again, if you divide 1224 by 84, the quotient is 14 ●/●4 or in the least terms 14 pound 4/7 of a pound. And the value of this fraction 4/7 of a pound, will be found by the third Rule of Reduction page 107, to be 11 shillings, 5 pence, and 4/7 of a penny, which is somewhat above two farthings: for it is 2 farthings, and 2/7 of a farthing. The same Question in Decimals. If ⅞ of a yard of cloth, cost 9/12 of a pound, what shall 17 yards cost at the same rate? To answer this question ⅞ of a yard, and 9/12 of a pound must first be reduced into Decimals, either by Division, or by the Tables of Reduction: by both which ways of Reduction the Decimal of ⅞ will be 875, and the Decimals of 9/12 will be .75 and then the terms of the question will stand thus; If .875 parts of a yard cost .75 parts of a pound, what shall 17 yards cost at the same rate? If 0.875— 0.75— 17. Here if you multiply the second term 0.75 by 17 the third term given, the product will be 12.75. and this product divided by .875 gives in the quotient 14.57142, that is 14 pound .57142 parts of a pound, or 145 Decades, that is 14 pound 10 shillings, and .7142 parts of a Decade (or two shillings) which by the preceding Tables is 1 s. 5 d. 2 farthings, and .0059 parts of a farthing. 2 Question. If a piece of Gold plate weighing 19 ounces 3 penny weight and 5 grains, be worth 62 pounds, 10 shillings, 6 pence, what is one ounce of the same gold worth. This question in vulgar fractions must be expressed thus. If 1 l. 3437/5260 Troy weight, cost 62 l. 126/240, what shall 1/12 of a pound Troy cost? at the same rate. To answer this question the fractions 1 l 3437/5760, and 6● l. 126/240, must be first reduced into improper fractions, and the fraction 1/12 into the least known parts of a pound Troy, and then the question will stand thus. If 9197/5760 give 15006/240 what shall 480/5760 give? Now because it is necessary the terms given be reduced into their least Denomination, before the question be resolved, therefore the answer may be found, by using the terms given thus reduced, as whole numbers, not having any regard to the Denominators of these fractions; Saying thus, If 9197 grains, cost 15006 pence, what shall 480 grains. And here if you multiply 15006 by 480 the product will be 7202880 which being divided by 9197 the quotient will be 783 pence 1629/9197 parts of a penny, and dividing .783 by 12 it will be 65 shillings 3 pence 1629/9197, or 3 l. 5 s. 3 d. 1629/9197. And although this question is thus more easily answered than it would have been, if the terms had been wrought as vulgar fractions, yet the same terms being reduced to Decimals, the answer of the question will yet be found with more ease, as shall appear by the operation following. The same question in Decimals. If a piece of gold plate weighing 19 ounces 3 penny weight and 5 grains, be worth 62 l. 10 s. 6 d. what is an ounce of the same gold. The Decimal of 19 ounces 3 penny weight and 5 grains, making an ounce the Integer is by this Table 19.16041, for that 19 ounces are 19 Integers, 2 penny weight is one tenth of an ounce, and the Decimal of one penny weight 5 grains is by this Table 06041; and the Decimal of 62 l. 10 s. 6 d. by the same Table is 62.525, and because an Unite or Integer is the third term given, there needs no multiplication, if therefore you divide 62.525 the second term, by 19.16041 the first term propounded, the quotient will be 3.2632 that is 3 pounds, 5 shillings, 2 pence, and somewhat more, as by the operation in the margin it doth appear. If 5 Else and a quarter of linen cloth, cost 2 l. 16 s. 8 d. 3 q. what shall 278 else and a half cost at the same rate? If you would work this Question by whole numbers your easiest way is first to reduce all the terms into their least Denominations, that is to say, the else into quarters, and the pounds, shillings, pence, and farthings all into farthings, so shall your 5 else and a quarter be 21 quarters, and your 278 else and an half will be 1114 quarters, and your 2 l. 16 s. 8 d. 3 q. will be 2723 farthings, and then will your question stand thus in whole numbers. If 21 quarters— cost— 2723 farthings— what will— 1114 quarters cost Then multiplying the second number by the third, that is, 2723, by 1114, the product will be 3033422, which divided by 21, the quotient will be 144448 farthings, which being again reduced into pounds, shillings, and pence, giveth 150 l. 9 s. and 4 pence, as by the operation following doth appear. The Operation. But if you would work the same question by Decimal numbers you may save the labour of reducing the terms to their least Denominations, for 5 Else and a quarter is in Decimal numbers 5.25, and 278 Else and an half is 278.50, and 2 pound 16 s. 8 d. 3 q. is in Decimals 2.8364, and then your question in Decimals will stand thus. If 5.25 Else cost 2.836 pounds what 278.50 Else If you multiply (according to the Rule) the second term by the third, that is 2.8364 by 278.50 the product of that multiplication will be 789.937400, which divided by the first term 5.25, the quotient will be 150.4642, which Decimal (by the Example in page 177) representeth 150 l. 9 s. 4 pence, and so much in money will 278 Else and a half cost. The Operation. I have been the larger in this Rule and especially in this Example, which is encumbered with fractions sufficient, because I would have the Reader the better discern the difference between the Vulgar and the Decimal way, and also to see how expeditious the one is over the other. Now this Example being thus largely explained, I shall with the more brevity pass over the Rules following giving one Example, or two at the most in each Rule; And thus much shall suffice for the Golden Rule, or Rule of three direct in Fractions. An Example in the Rule of Three Reverse. A Lends B 233 l. 6 s. 8 d. for a year without interest, upon condition that B should do the like courtesy for A when required. A hath occasion for money 7 months, how much money ought B to lend A, to requite his courtesy, and save himself harmless? I will not in this place tell you what the Rule of Three reverse is, nor the manner of working thereof, that being already sufficiently declared in the first part, but give you the Example, and the working thereof, which take as followeth: So will the Question be thus stated. Here you see that 12 months and 7 months are whole numbers, and so we let them alone without any reduction, but the Decimal of 233 l. 6 s. 8 d. will be found by the Tables and Rules to be 233.33, which is the middle term in the question, and of the same quality with that, must the fourth term sought be, therefore if (according to the Rule delivered in the first part) you multiply 233.33 by 12, the product will be 2799.96, which divided by 7 giveth in the quotient 39999, which is the Decimal of 400 l. and so much money ought B to lend A for 7 months. Examples in the Rule of Proportion, consisting of five numbers. Question 1. If 100 l. in 12 months, yields 6 l. interest, what interest shall 264 l. 16 s. 5 d. yield in 15 months at the same rate? Set down your numbers in Decimals, as in the Example following appeareth, so shall you find the Decimal of 264 li. 16 s. 5 d. to be 264.8208, all the rest being whole numbers having no fractions joined with them we neglect, and work with them as they are, so will the several numbers of your question (if rightly disposed) stand as followeth. Your numbers being thus orderly disposed, you must according to the Rule before delivered in the first Part, Page 92) multiply the first and second terms together, which in this Example are 100 and 12, whose product is 1200, which is your Divisor; Then multiply the three last terms one into another, as 264,8208 (which is the Decimal of 264 li. 16 s. 5 d.) by 6, and the product thereof will be 1588.9248, which number again multiplied by 15, (which is the last term,) the product will be 23833.8720 which is your Dividend, and this number being divided by your former product, giveth in the quotient 19.8615, which is the Decimal of 19 l. 17 s. 2 d. 3 q. ferè, and so much doth the simple interest of 264 l. 16 s. and 5 d. amount unto in 15 months, after the rate of six per centum for a year. Quest. 2. If the carriage of 23 hundred and 3 quarters of any thing 127 miles, cost 4 li. 13 sh. 6 d. what shall the carriage of 47 hundred and an half of such like commodity cost, being carried 381 miles. Place your numbers in order as in the following Example doth appear, then multiply the first and second terms together for your Divisor, and the three last one into another for your Dividend, and so will the quotient of this division answer the question demanded, and the work will stand as followeth. Here you see that the first and second terms multiplied together produced 3016.25, which must be your Divisor, and the three last terms being multiplied one into another, produce 84605.81250, which number divided by 3016.25, giveth in the quotient 28.050, which Decimal representeth 28 l. one shilling, and so much will the carriage of 47 hundred and a half cost, being carried 381 miles? Thus have I shown the use of decimal Arithmetic in such questions as concern the Golden Rule, or Rule of Three; both Direct, Reverse, and Compounded, by an Example or two in each rule, & those compounded of fractions sufficient, I should now proceed to questions in Fellowship, with and without Time, as also Barter, Alligation; the Extraction of the Square and Cube Roots, etc. but forasmuch as these last mentioned Rules depend only upon the Rule of Three, as by Examples in the first part doth plainly appear, I shall therefore save that labour, and spare my Reader the pains of practising questions which wholly depend upon that which I (by this time,) suppose him perfect in; Yet if the Reader be desirous to make trial of any such question for his own satisfaction, he may either make trial of those questions in the former part of this Book in those several Rules, reducing the numbers there given into Decimals, or if he please, he may frame questions according to his own fancy. And thus I shall conclude this second Part. The end of the second Part. THE THIRD PART Containing INSTRUMENTAL ARITHMETIC. THe Arithmetic, of which we now come to treat, and which I call Instrumental Arithmetic, is not any new kind of Arithmetic, but is indeed the same with Decimal Arithmetic before taught; only, whereas in Decimal Arithmetic, there were certain Tables made of Money, Weight, and Measure, by help of which the Decimal of any fraction of Money, Weight or Measure might be set down (as it were) in whole numbers, here in this Instrumental part, we have contrived certain Scales of Money, Weight, and Measure, equally divided into the several Denominations into which the several Weights and Measures, can be equally divided, unto every of which Scales there is joined a Scale of 100, 1000, or 10000, equal parts, according to the length of the Scale, so that by inspection only you may readily and exactly without addition (as in using the fore mentioned Tables you must necessarily do) set down the Decimal fraction of any part of Money, Weight or Measure, with great celerity and exactness, if the Scale be any thing well divided, though the Scale be but of a reasonable length. Now the Scale, which I have chief made choice of in this Work, as being of most use with English men; (though other Scales may be made for the Coins, Weights or Measures of any other Country as well, and upon the same ground) are chief these, viz. 1 Of Money. 2 Of Troy Weight. 3 Of Averdupois great Weight. 4 Of Averdupois little Weight. 5 Of Liquid measures. 6 Of dry measures. 7 Of Long measures. 8 Of Dozen. Unto every of these Scales, is joined another Scale of 100 or 1000 equal parts, these Scales are made to face one another, so that if you look upon any one Division, in the one, you shall also discern plainly what Division or part of a Division answereth thereunto in the other. English Coin, Two shillings being the Integer Troy weight, Two penny weight being the Integer Averdupois great weight, 28 lib: or one quarter of an hundred being the Integer Averdupois little weight, 16 ounces, or one pound being the Integer Place this between pag 246 & 247 Deliniavit Antonius Thompson Dry measure, S Bushels being the Integer Liquid measure, 36 Gallons, or one Barrel being the Integer Long measure, one Ell or one Yard being the Integer These Scales being thus disposed, as they may easily be upon any Ruler of Silver, Brass, or Wood; but best of all upon a square Ruler, made in form of a Parallelepipedon, will by inspection only give you any Decimal fraction required without Addition, or (on the contrary) reduce the fraction into the know parts of the Integer, by inspection also, without subtraction. Let thus much suffice for a general description of what I mean by Scales, the particular description of them, will more plainly appear, when we treat of Numeration upon the Scales, unto which we shall now proceed. But first take a view of the Scales as they are here disposed, and as they may be set upon such a Ruler as I have here mentioned. Numeration upon the Scales. THe Scales here to be described are in number eight, as hath been already showed, and as by the figure of them appear. Now Numeration upon a Scale is to find upon what part of the Scale, any number upon the same Scale will fall. We will begin with the first, and so proceed till we have given an Example in every one. 1 The first Scale is of English money, and is divided into 24 equal parts, which represent 24 pence or 2 shillings, these parts are numbered with Arithmetical figures, from the beginning thereof, by 1, 2, 3, 4, 5, etc. to 24, each division representing one penny, and the whole 24 divisions representeth 24 pence or 2 shillings, So that where the figure 1 standeth, that part of the Scale representeth one penny, where the figure 2 standeth, it representeth two pence, where the figure 18 standeth, it representeth 18 pence, or one shilling six pence, and so of any other figure of the same Seal. Then because there are four farthings contained in a penny, each of these pence (or divisions) is subdivided into 4 other equal parts by short lines, every one of these representing one farthing, so is the whole Scale divided in all in 96 equal parts, which are the number of farthings contained in two shillings. This if you look into the Scale of Money for 8 pence 3 farthings, you shall find it at the letter a, which letter is here put only for example sake, Also if you would find in the Scale the place of 18 d. half penny, you shall find it at c, and thus may you find the place of any number of pence and farthings under two shillings upon the Scale. Unto this Scale of money (as to all the rest of the Scales,) there is joined another Scale which I shall always hereafter call the Scale of 1000, the use of which Scale is this. When you have found any Number of pence or farthings upon the Scale of Money, you shall find upon the Scale of 1000, what parts of a thousand is the Decimal of those pence and farthings: Thus when in the Scale of money you find at the letter a 8 pence 3 farthings, if you cast your eye directly cross to the Scale of 1000, you shall find 364 to stand directly against 8 d. 3 q. which 364 is the Decimal of 8 d. 3 q. Also, if you find upon the Scale of money 18 d. half penny, which is at the letter c, you shall find against it in the Scale of 1000, this number 771, which is the Decimal of 18 d. half penny. And in this manner may the Decimal of any number of pence or farthings under two shillings be most easily and exactly obtained. Now on the contrary, suppose a Decimal fraction were given, representing some part of English Coin if you look in the Scale of 1000 for your Number given, right against it in the Scale of Money you shall find what Number of pence and farthings is representeth thereby. As for Example. Suppose .364 were a Decimal given, and it were required to find what part of Coin it doth represent. Look in the Scale of 1000 for the Number 364, and right against it you shall find 8 pence 3 farthings. Also if .771 were a Decimal given, if you look in the Scale of 1000 for .771, you shall find against that number 771, 18 pence 2 farthings. And thus of any other. By what hath been already said, it may be easily discerned of what exceeding expedition these Scales thus disposed are of, for I dare affirm, that I will set down 2 (if not 3) numbers by the Scale, as soon as one by the Tables, and if the Scale be but of any reasonable length, altogether with as much exactness, but if I should vary an unite in the last place, in my estimation in the Scale of 1000, it is not any thing material. I have been very tedious in showing the use of these Scales to find the fraction parts of money, but the reason is because I intent to be the briefer in the rest, for weight and measure, the manner of working (when the division of the Scale is known,) being the same in all respects without the least alteration. 2 The second Scale is of Troy weight, two Penny weight being the Integer, which Scale is divided into 48 equal parts or divisions, each of which divisions contains one grain, and are numbered by Arithmetical figures at every three grains by 3, 6, 9, 12, etc. to 24, and at the place where 24 should stand, there standeth PW, which signifieth one penny weight, or 24 grains, this PW standeth in the middle of the line. Then is the same Scale continued farther by Arithmetical figures 3, 6, 9, 12, etc. as before to 24, and there is written PW again, representing two penny weight, or 48 grains. The Scale being thus divided, it is easy to find the place where any number of Grains under 48 shall be upon the Scale; As for example, if it be required to find where 8 penny weight shall fall, look upon the Scale of Troy weight, from the beginning thereof, and count the figures 3 and 6, then count also two of the smaller Divisions, and that makes 8 grains, which you shall find to stand at the letter d, which is the place of 8 grains; Also if upon the Scale you would find the place of one penny weight 10 grains, you shall find it at the letter e, and so of any other number of grains under 48, or two penny weight. But if you had a Decimal given, and would know what number of grains it representeth, if you seek your Decimal given in the Scale of 1000, right against it in the Scale of Troy weight, you shall find the number of grains represented thereby. Example. Let .167 be a Decimal fraction given, If you look in the Scale of 1000, for 167, right against it in the scale of Troy weight, you shall find 8 penny weight. Also if .708 were a Decimal given, if you seek 708 in the scale of 1000, right against it you shall find 1 penny weight 10 grains. 3 The third scale is of Avoirdupois great weight, 28 pounds, or one quarter of an hundred, being the Integer, this scale is numbered by 1, 2, 3, 4, etc. to ●8, which 28 representeth 28 l. or a quarter of a hundred, and each of those is subdivided into four small parts, each representing one quarter of a pound. Now if you would know what is the Decimal of any number of pounds or quarters under 28, if you seek the number of pounds in the scale of Avoirdupois great weight, right against it in the scale of 1000, you shall find the decimal thereof. Thus if it were required to find the Decimal of 8 pound and an half, if you look upon the Scale for 8 pound and an half you shall find it at the letter g, and right against it in the Scale of 1000 you shall find 304, which is the Decimal of 8 pounds and an half. On the contrary, suppose 304 were a Decimal given, and it were required to find what part of Avoirdupois great weight were represented thereby, if you look in the Scale of 1000 for 304, right against it in the Scale of Avoirdupois great weight, you shall find 8 pound and an half. 4 The fourth Scale is of Avoirdupois little Weight, 16 ounces or one pound being the Integer, This Scale is first divided unto 16 equal parts, and numbered by 1, 2, 3, 4, etc. to 16, each Division representing one ounce. Then again, each of these ounces is subdivided into 8 other smaller parts or divisions, each of which divisions representeth two Drams; but if your Scale be large enough, you may have each ounce divided into 16 equal parts, or divisions, each division representing one Dram. Now to find the Decimal belonging to any number of Ounces and Drams, repair to the Scale of Avoirdupois little weight, and on it find the quantity of ounces and drams required, and right against it in the Scale of 1000, you shall have the Decimal thereof. Thus if it were required to find the Decimal of 6 ounces, and 6 drams, if you look this in the Scale of Avoirdupois little weight, you shall find it at the letter h, and right against it in the Scale of 1000, you shall find 398, which is the Decimal of 6 ounces and 6 drams. But if 398 were a Decimal given, and it were required to find the value thereof in Avoirdupois little weight, if you look for 398 in the Scale of 1000, right against it in the other scale you shall find 6 ounces 6 drams. 5 The fifth Scale is of Dry Measures, one Quarter or 8 Bushels being the Integer, This Scale is first divided into 8 equal parts, and numbered by 1, 2, 3, etc. to 8, each of which divisions representeth a Bushel, & each of those parts is again subdivided, first into 4 equal parts or divisions each representing one peck, and then those again subdivided into 4 other smaller parts, representing quarters, halves, and three quarters of a peck. Now if you would know the Decimal belonging to any number of Bushels (under 8 Bushels or one quarter) Pecks and parts of a Peck, if you seek the number of Bushels, Pecks, and parts of a Peck in the Scale of Dry Measures, right against it in the Scale of 1000, you shall have the Decimal required. As for Example, if it were required to find the Decimal belonging to 5 Bushels 2 Pecks, and half a Peck, if you look into the Scale of dry Measures you shall find 5 bushels, 2 pecks, and an half, to stand at the letter k, and right against it in the Scale of 1000, you shall find 702, which is the Decimal answering to 5 bushels, 2 pecks and an half. But if 702 were a decimal given, and it were required to find what number of bushels, pecks and parts of a peck it representeth, if you look in the scale of 1000 for 702, you shall find against it in the scale of Dry measures, 5 bushels 2 pecks and an half. 6 The sixth Scale is of Liquid Measures, the Integer being 36 Gallons or one Barrel, this Scale is divided first into 36 equal parts or divisions, and numbered by 5, 10, 15, etc. to 36, than every of these divisions is again subdivided into 4 other small divisions, each representing a quart, but (if the Scale be large enough) you may sub-divide each Gallon into 8 parts, so will every part represent one pint. Now to find the Decimal belonging to any number of Gallons (under 36 Gallons or one barrel) quarts or pints, repair to the Scale of Liquid measures, and seek there upon the Scale, the number of gallons, quarts, or pints, and against it in the Scale of 1000, you shall find the Decimal thereunto belonging. So if it were required to find a Decimal representing 10 gallons and two quarts, or 4 pints, which is all one, if you seek in the scale of Liquid measures for 10 gallons 2 quarts, you shall find it at the letter m, against which in the scale of 1000, you shall find 292, which is the Decimal of 10 gallons 4 pints. Likewise if 292 were a Decimal given, and it were required to find what number of gallons, quarts or pints were represented thereby, if you look in the scale of 1000 for 292, right against it in the scale of Liquid measures, you shall find 10 gallons, 2 quarts, or 4 pints. 7 The seventh scale is of Long Measures, the Integer being Yards or els, this scale is divided into 4 equal parts, and numbered by 1, 2, 3, 4, representing 1 quarter, 2 quarters, 3 quarters or 4 quarters of a Yard or Ell, these are again subdivided first into 4 other equal parts, representing nails, and those may be again subdivided at pleasure if need be. Now if you would know what decimal belongeth to any number of quarters or nails of a yard if you seek the number of quarters and nails in the scale of Long measure, the scale of 1000 will give you the Decimal thereof. Thus if it be required to find the Decimal belonging to 1 quarter and 3 nails, if you seek this in the scale of long measure, you shall find it to stand at the letter o, against which in the scale of 1000 you shall find 437, which is the decimal answering to 1 quarter & 3 nails of a Yard or Ell. Also if 437 were a decimal given, & it were required to find what quantity of yards or els were represented thereby, if you look in the scale of 1000, for 437, you shall in the scale of long measure find against it one quarter and 3 nails. 8 The eighth and last scale is of representative inches, the whole scale being divided into 12 equal parts and numbered by 1, 2, 3, etc. to 12, and those parts are again subdivided into halves, quarters, and half quarters, as Carpenters-Rules are usually divided. Unto this Scale, (as unto all the other) there is joined a Scale of 1000, this Scale will readily discover what is the Decimal belonging to any number of Inches, halves or quarters, and the use is the same with the Scales before mentioned Thus have I given you a brief description of these Scales and the uses of them, and do now suppose my Reader to be perfectly acquainted with the way of numbering or counting upon them, wherefore I intent only to give you a question or two in the most usual Rules of Arithmetic, and so conclude; for Decimal Arithmetic being already sufficiently explained, I shall not need to repeat the Rules (or the manner of working them) again, but give you one Example, by which the exactness and expedition of these Scales may the more evidently appear, for when we work by Scales, it is supposed that we do not use Vulgar, but Decimal Arithmetic, and Addition, Subtraction, Multiplication, Division, the Rule of Three, and indeed, all the other Rules of Arithmetic, are to be performed, as is before taught, the Scale serving only to avoid Reduction. Addition. WHat Addition is, and the manner of working of it hath been already taught, both in the first and second Parts, we will now come to an Example, which let be in Addition of English Coin, and let the sums to be added be 36 l. 8 s. 8 d. 29 l. 0 s. 2 d. 31 l. 16 s. 9 d. and 6 l. 2 s. 5 d. First, 36 29 31 6 set down 36 l. 29 l. 31 l. and 6 l. one under another, in such order as you see here in the margin, drawing a line by the side of them as you see here done, and also a line under them. This done, seeing that your first number to be set down to 36 l. is 8 s. 8 d. you must for the 8 s. (because two shillings, which we called a Decade, or the tenth part of a pound, is made the Integer, in the Scale of Money) set down 4, which is done by memory, and after it make a comma. Then your next number to be set by 29 l. being 0 s. 2 d. for the 0 s. 36 4, 29 0, 31 8, 6 set down a cipher, thirdly, for your number to be set by 31 l. being 16 s. 9 d. for the 16 s. set down 8 decades, with a comma after it, and lastly, the number to be set by 6 l. being 2 s. 5 d. for the 2 s. I set down 1 decade with a comma after it, and then will your work stand, as here you see. Then take your scale in hand, and seeing your first number of pence are 8 d. look in your scale of money for 8 d. and against it in the scale of 1000, you shall find 333, which set to 36 l. 8 s. behind the comma, than your next number of pence being 2 d. look in your scale for 2 d. and against it in the scale of 1000, you shall find 082, which set to 29 l. 0 s. behind the comma. 36 4, 333 29 0, 083 31 8, 376 6 1, 208 Then your third number of pence being 9 d. look in your scale for 9 d. and against it in the scale of 1000, you shall find 376, which set to 31 l. 16 s. and lastly, your last number of pence being 3 d. look in your scale for 5 d. and against it you shall find 208, which set to 6 l. 2 s. and then will your whole work stand, as here you see. Your sums being thus set down, which is done with more facility than you can imagine, till you make trial and be something perfect therein, you must then add all the numbers together, as in Addition of Decimals, and you shall find the sum of them to be 1034,000, Now to know what this is in money, is as easy as it was to set the several sums down, for the figures 103, which stand behind the down right line, are 103 l. and the figure 4 which stands between the down right line and the comma, l. 36 4, 333 29 0, 083 31 8, 376 6 1, 208 103 4, 000 are 4 decades or 8 s. and being the rest to the right hand are all Ciphers, they signify neither pence nor farthings, so is the total of this addition 103 l. 8 s. 0 d. That the manner of working may appear more plain. I will give you another short Example as difficult as I can invent, which I performed by a Scale of Wood but of 8 inches long. Let the sums to be added together, be these following. li. t. d. q 332 17 4 1 159 6 8 1 217 5 3 3 709 9 4 1 First, set down your several sums of pounds one under another as before, and draw a line by the side of them, 332 159 217 and another under them. So will they stand as here you see. 1 Your sums of pounds being thus orderly placed and lines drawn, repair to your Scale and seeing your first number of shillings, pence and farthings is 17 s. 4 d. 1 q. for your 17 s. set down 8 Decades, which is 16 s. with a comma after it, then will there rest to be set down 1 s. 4 d. 1 q. or 16 d. 1 q. which if you seek in your scale of money, you shall find to stand against it in the scale of 1000 this number 677, which is the Decimal of 1 s. 4 d. 1 q. 2 Your second number of shillings pence and farthings is 6 s. 8 d. 1 q. for your 6 s. set down 3 decades, which is 6 s. and then there will remain 8 d. 1 q. which if you seek in your scale of money, you shall find to stand against it in the scale of 1000 this number 344, which is the decimal of 8 d. 1 q. 3 Your third number of shillings, pence, and farthings, is 5 s. 3 d. 3 q. for your 5 s. set down 2 decades, which is 4 s. with a comma after it, then will there rest to be set down 1 s. 3 d. 3 q. or 15 d. 3 q. which if you seek in your scale of money, you shall find to stand against it in the scale of 1000, l. 332 8,677 159 3,344 217 2,656 709 4,677 this number 656, which is the Decimal of 15 d. 3 q. or 1 s. 3 d. 3 q. and the three sums to be added together will stand as here you see. These sums being added together, according to the rule for Addition of Decimals, you shall find the sum of them to be 7094,677, now to know what this is in money, take notice that the 709 which stands to the left hand of the downright line are 709 pounds, and the figure 4, which stands between the downright line and the comma, are 4 decades or 8 s. but (because the first figure next after the comma is above 5, viz. 6) you must add 1 s. to the 4 decades, making them 9 s.; then will there remain 177, wherefore if you look in the Scale of 1000 for 177, you shall find against it in the Scale of Money 4 d. 1 q. So is the whole sum of this Addition 709 l. 9 s. 4 d. 1 q. as by the preceding work doth appear. ¶ Here note, that when you had set down your 709 l. 4 decades or 8 s. there remained beyond the comma 677, which if you had sought in your Scale of 1000, you should have found against it in the Scale of money 15 d. 1 q. or 1 s. 4 d. 1 q. (which is all one) as before, for it appeareth plainly by the Scale that 500 in the line of 1000 is equal to one shilling. I might proceed farther in giving you Examples in Weight and Measure answerable to the Scales, but that would only make the Reader spend his time to little purpose, for being before acquainted with Decimal Arithmetic, and (as by this time I suppose he is) with Numeration upon the Scales, he cannot be deficient in the applying of the other Scales of Weight and Measure to the same purpose for which they were contrived, I having so largely exemplified the use of the Scale of money. Subtraction. SUbtraction (as hath been before said) is the taking one or more smaller sums out of one greater, I shall only give you an Example or two, as I have taken the numbers from a Scale. Example. Delivered to a Goldsmith of old plate 297 ounces 13 penny weight, 19 grains. Received of the same Goldsmith first 165 ounces, 11 penny weight and 7 grains, and after that received of the same Goldsmith 32 onnces 19 penny weight, and 23 grains, what plate remains in the Goldsmith's bands? Take your numbers out the scale of Troy weight, and set them down, as here you see. ounces Delivered 165 5,646 Received more 32 9,979 Received in all 198 5,625 Rese in the Goldsmith's hands 99 1,271 or 99 ounces— 2 penny w.— 13 gr. Then add the several weights of plate received, together, and they make 1985, 625, or 198 ounces, 11 penny weight, 6 grains, which if you subtract from 2976,896, or 257 ounces, 13 penny weight 19 grains, which was the quantity of plate delivered, there will remain 991271, or 99 ounces 2 penny weight, 13 grains, and so much plate is still in the Goldsmith's hand. And let thus much suffice for subtraction. Now we should proceed to Multiplication and Division, but being when the numbers are taken from the scale and set down, the manner of working doth not at all differ from Multiplication and Division of Decimals before taught. But that we are now a treating of Instrumental Arithmetic (because I will not, altar the method before used, in the first and second parts of this Book;) I will therefore here show you how Multiplication and Division, may be Instrumentally performed by Nepers' Bones. Multiplication by Nepers' Bones. MUltiplication and Division are accounted the hardest parts of Vulgar Arithmetic; but by the invention of the Right Honourable John Lord Nepeir Baron of Merciston, they are both of them made very easy, and more certain than in the ordinary way, the memory not being charged at all, but as in Addition and Subtraction, what this Artifice is, is showed by himself in his Treatise of Rabdologia, as also in that of Mr. Seth Partridge and others, and therefore it will not be necessary to make any large discourse of it now, it will be sufficient here, only to give you a description of them, and how to use them. These Rods or Bones are nothing else, but the ordinary Multiplication Table, commonly called Pythagor as his Table, cut into pieces, with Diagonal lines between the figures of every product in single multiplication; as by the figure following it doth appear. A Figure of the Bones. The Bones are usually made of small pieces of Box, containing in length about 1 inch and 8/10 of an inch, and in breadth about 2/10 of an inch, and in thickness near 1/10 of an inch. One set of these Bones consisteth of five pieces of Box, and are graduated with Diagonal lines and figures, as you see here in the figure of them. On one of the five pieces there is 0 and 9, the 0 being on one side of the Wood, and the 9 on the other side, another hath 1 and 8. the third 2 and 7, the fourth 3 and 6, the fifth 4 and 5. Now these five pieces of Box contain one Table of Multiplication cut in pieces, but so few are not sufficient for use, for you cannot well have less than six sets of them, that is 30 pieces, and so many are usually made. For the placing of these Bones, when you use them, there belongeth a piece of Wood about four inches in length, and two inches in breadth, having 2 ledges of the same thickness with the Bones, on the thin legde which is at the end of this board, there is figured the nine Digits, 1, 2, 3, 4, 5, 6, 7, 8, 9 This board is called a Tabulat, because we usually say when the bones are laid thereon for any the Bones are Tabulated. See the figure thereof. Now these Bones thus ordered and Tabulated doth not only give you the product of every single Digit, multiplied by itself, or any one of the Digits, but the several products of all the 9 Digits together by every of the 9 Digits, as if it were demanded how much 123456789, will amount unto, being multiplied by 6, in this Table it will be found to be 740740734, for the product of 9 multiplied by 6 is 54, the first figure whereof towards the right hand, viz. 4, is placed under the Diagonal line, and 5 being in the place of ten is placed above it, again 8 multiplied by 6 is 48, the first figure whereof towards the right hand, viz. 8 is placed under the Diagonal line as in the other product, and these 2 figures 5 in the upper part of the former product, and 8 in the lower part of this must be added together, and being so they make 13, therefore next unto 4 in the first product, I set down 3, and carry 1 to 4 the figure placed above the Diagonal in this second product, which together with the lower figure in the next product, viz. 2, do make 7, for 4 and 1 that I carried is 5 and 2 the figure under the Diagonal in the next product are 7, and so the 3 first figures of this product towards the right hand are 734, in like manner must the figures remaining between the Diagonals be added together, and their sum if it amount not to ten must be set down in his proper place, but if they make or exceed ten, set down the overplus, and carry the ten in mind to the next figures between the Diagonals as in the last product; So likewise in that which follows 4 and 6 the figures between the Diagonals being 10, I set down a cipher, and carry one to the next figures, which are 3 and 0, now 4 and 1 that I carried is 4, and 4 and 0 is 4 still, and therefore next to the cipher I set down 4 in the paper; the figures in the next Diagonal are 3 and 3. that is 7, wherefore I set down 7 in the paper, the next are 2 and 8 that is 10, therefore I set down 0 in the paper and carry 1, the next are 1 and 2 which with 1 that I carried make 4, therefore I set down 4 in the paper, the last are 1 and 6 which make 7 and so at last the whole product of 123456789 multiplied by 6 is 740740734. Now as the multiplication Table thus ordered doth give the product of all the 9 Digits placed orderly by every of the 9 digits, so being cut in pieces or slips, and having many of them made in Pasteboard of Wood, any other sum may be placed on the Tabulat, and the several products thereof by every of the 9 digits will be found in the like manner; thus if it were required to multiply this number 5767 by every of the 9 digits, placing the slips with these figureson the top orderly as, they are in the following Figure, product of 5767 by 3 will be found to be 17301, the product thereof by 8 is 46136, and so for any other. Also if it were required to multiply 5767 by 749, the product of 5767 by 9 is 51903, and the product thereof by 4 is 23068, and the product by 7 is 40369, and these products being set down and added together, as in ordinary Multiplication, the product of 5767 multiplied by 749 will be 4319483, as by the operation in the margin it doth appear. Of Division by the Rods. LEt it be required to divide 4319483 by 5767 placing this Divisor on the Tabulat, as was directed for Multiplication, the products of your Divisor by every of the 9 Digits is given you, and may be deducted from the Dividend, as hath been already showed Pag. 55, and according to the Rules there delivered, I ask how often is 5767 in 43194, and the answer by my Table is 7 times, and therefore I subtract 40369 the product of 5767 by 7 from 43194, and there rests 2825, which with the next figure in the Dividend, viz. 8, is 28258, in which I find 5767 to be contained 4 times, and therefore I subtract 23068 the product of 5767 by 4 from 28258, and there rests 5190, which with the next figure in the Dividend, viz. 3, is 51903, in which I find 5767 to be contained nine times, and therefore I subtract 51903 the product of 5767 by 9, from 51903 the remaining figures of the Dividend, and there resteth nothing, and therefore the quotient is 749, as by the operation it doth appear. And thus may you do with any other sum (having a competent number of Bones or Rods ready for the work) without any charge to the memory, which in other ways cannot be avoided, and by reason of which many mistakes may happen, which are in this prevented. Also by the Bones there is this commodity, that when it is doubtful what figure to set in the Quotient, the Bones will certainly direct you, so that you cannot possibly mistake. These Rods, or Bones (as they are commonly called) are of excellent use in extracting the Square and Cube Roots, but that being already taught in the first Part, I shall not here insert it again, for he that can extract the Square and Cube Roots any way, cannot be ignorant to effect the same by the Bones, if he can multiply and divide by them. Having done with the four Species of Arithmetic in this instrumental way, I should now proceed to the Golden Rule of three, and consequently to the other Rules of Arithmetic, but having in the former parts of this Book performed that already, both in whole numbers, Fractions, and Decimals, I shall only give you an Example in the most necessary Rules, and so conclude this third part. Example in the Rule of Three Direct. IF 37 els and a half of linen cloth, cost 24 l. 7 s. 9 d. what shall 283 els and an half cost? First, set down your 37 els, then if you look in your Scale for your half Ell, you shall find it to stand against 500 in the Scale of 1000, which 500 may be called 5 only, for the two Ciphers may be omitted. Then set down your 24 l. and for your 7 set down 3 decades, which is 6 s. and look in your Scale for the Decimal of 1 s. 9 d. which you shall find to be 875; Lastly, set down your 183 els, and for your half Ell set down 5 as before, so will your numbers stand thus, Your numbers thus taken out of your Scale, and placed as here you see, if you multiply the second and third together, you shall find the product of that multiplication to be 4475.10625, which divided by the first number 37.5, giveth in the quotient 119.3361, which is 119 pounds, 3 decades, or 6 shillings, and 361, which reduced by the Scale giveth 8 pence 2 farthings, and something more. Example in the Rule of Three Reverse. IF when the price of a Quarter of Wheat is 1 li. 5 s. 6 d. the penny white loaf shall weigh 12 ounces 16 penny weight, I demand what the penny white loaf shall weigh, when the price of the Quarter of Wheat is 7 s. 6 d? If you place your numbers according to the tenor of the Question, they will stand as followeth. 1 li.— 5 s.— 6 d.— 12 où.— 16 pw.— 7 s.— 6 d. But being taken out of the Scales of Money an Troy Weight, they will stand thus, Here if you multiply 1.275 which is the Decimal of 1 l. 5 s. 6 d. by 12.8. which is the Decimal of 12 ounces 16 penny weight, you shall find the product of that multiplication to be 16.3200, which being divided by .375, which is the Decimal of 7 s. 6 d. the quotient will be 43.5, which is the Decimal of 43 ounces, 10 penny weight 3 grains, and so much aught the penny white loaf to weigh, when the quarter of Wheat is sold for 7 s. 6 d. Example in the Double Rule of Three. IF 24 yards of stuff of three quarters broad cost 4 l. 14 s. what shall 328 yards of the same stuff cost being 5 quarters broad. If you place your numbers according to the directions of this Rule, they will stand thus, If 24 yards of 3 quarters cost 4 li. 14 s. what shall 328 yards cost of 5 q. But if you take your fraction numbers out of their proper Scales, they will stand thus, First, multiply the two first numbers, as 24 and 3 together, they make 72 for Divisor, then multiply 4.7, which is the Decimal of 4 l. 14 s. by 328, and the Product is 15416, which again multiplied by 5 the last number, giveth 77080, unto this Product, (that there may be a competent number of figures in the quotient,) I add two Ciphers, making it 7708000, which I divide by 72, and the quotient is 107.055, which is 107 l. 1 s. 1 d. and so much is 328 yards of stuff worth, being 5 quarters broad. Example in Barter. TWo Merchants having two several commodities, are willing to barter, or exchange the one with the other. The one hath Indigo, which he will sell at 4 s. the pound, for ready money, but in Barter he will have 4 s. 9 d. the pound; the other Merchant hath Kerseys, which for ready money he will sell for 3 s. 6 d. the yard. Now the question is, at what price he must rate his Kerseys in Barter, to equalise the 9 d. advance upon the pound of Indigo. The tenor of the Question is this. If 4 s. in Barter require 9 d. what shall 3 s. 6 d. require? Your numbers placed will stand thus, 4 s.— 9 d. 3 s. 6 d. But being taken out of your Scale they will stand thus, Say then by the Rule of three Direct, If 2 decades or 4 s. in Barter require. 357, which is the Decimal of 9 d. what shall 1.57 require? which is the Decimal of 3 s. 6 d. First, multiply .375 by 1.57, the product is .65625, (but being it is a fraction I cut off the two last figures because we require only three figures in the quotient, which divided by 2 giveth in the quotient .328, which is the Decimal of 7 d. 3 q. this 7 d. 3 q. added to his 3 s. 6 d. maketh 4 shillings 1 penny 3 farthings, and so much ought he to rate his Kerseys at by the yard in Barter to save himself harmless. Example in Fellowship. THree Persons A, B, and C bought 4000 Sheep, which cost 48● li. 6 sh. 8 d. of which money A paid 203 li. B paid 165 li. 15 sh. 8 d. and C paid 114 l. 11 s. First, say by the Rule of three Direct. 1 If 483 l. 6 s. 8 d. buy 4000 sheep, how many sheep shall 203 l. (which is A share) buy? Answer 1680. 2 Say, if 483 l. 6 s. 8 d. buy 4000 sheep, how many sheep shall 165 l. 15 s. 8 d. (which is B share) buy? Answer 1372. 3 Say again, if 483 l. 6 s. 8 d. buy 4000 sheep how many sheep shall 114 li. 11 s. (which is C share) buy? Answer 948. Your numbers being taken out of your Scale, proceed as followeth. First for A. Secondly for B. Thirdly for C. The manner of Work. 1 For A multiply 203 l. (which is A share) by 4000, (which is the number of sheep bought) and the product is 812000, which number should be divided by 483.3333, but being it is greater than 812000, I therefore add four Ciphers thereto, that I may have four figures in the quotient, and it makes 8120000000, which divided by 483.3333 giveth in the quotient 1680, and so many sheep belong to A. 2 For B, multiply 165.7833 (which is the Decimal of B share) by 4000, (the number of sheep bought) and it produceth 6631332000, which divided by 483.3333, giveth in the quotient 1372, and so many sheep belong to B. 3 For C, multiply 114.55, (which is the Deeimal of C share,) by 4000, (the number of sheep bought,) it produceth 45820000, which number should be divided by 483.3333, but being it is not large enough to give figures enough in the quotient, I therefore add two Ciphers making it 4582000000, which divided by 483.3333, giveth in the quotient 948, and so many sheep ought Cto have. Now for proof, A 1680 B 1372 C 948 4000 if you add the number of sheep that A, B and C should severally have, you shall find them in all to make 4000 which demonstrates the work to be true. Examples in Loss and Grain. IF one yard of Stuffe cost 6 s. 8 d. and I sell the same again for 8 s. 6 d. what shall I gain in laying out 100 l. upon such commodity? Take the difference between the price that your commodity cost, and the price for which you sell it, that is, in this Example, the difference between 6 s. 8 d. and 8 s. 6 d. which is 1 s. 10 d, then say by the Rule of Three Direct. If 6 s. 8 d. gain 1 s. 10 d. what will 100 li. gain? If you place your numbers according to the Rule of Three Direct, as they are here given, they will stand thus, 6 s. 8 d.— 1 s. 10 d. 100 li. But being taken out of your Scale and placed, they will stand as followeth. Your numbers being placed, multiply 425 which is the Decimal of 1 s. 10 d. by 100 li. and the Product is 42500, to which I add two Ciphers (that I may have a competent number of figures in the quotient) and it makes 42500●0, which divided by .3333 the decimal of 6 s. 8 d. giveth in quotient 127.5, which is 127 li. 5 livers or 10 s. so there is 27 li. 10 s. gained in laying out of 100 l. I will here prove this question by the converse. If one yard of Stuffe which is sold for 8 s. 6 d. there was gained 27 li. 10 s. in laying out of 100 li. I demand what the said stuff cost a yard at the first hand? Add 100 li. and 27 li. 10 s. together, and they make 127 li. 10 s. Then say by the Rule of Three Direct, If 1 27 li. 10 s. give 100 l. what shall 8 s. 6 d. give? Take your numbers out of your Scale and place them as here you see. Here if you multiply 425, which is the Decimal of 8 s. 6 d. by 100, you shall have 42500, to which if you add a cipher, you make it 425000, this number being divided by 127.5, which is the Decimal of 127 l. 10 s, giveth in the quotient 3333, and if you had added more Ciphers to the Dividend, you should have had more three in the quotient, and no other figures, but these four three are enough, and are a Decimal fraction representing 6 s. 8 d. and so much did the yard of Stuffe cost at the first hand. Examples in Loss and Gain upon Time wrought by the Double Rule of Three. IF one Ell of Lockeram, cost me 2 s. 8 d. ready money, and I sell the same again for 2 s. 10 d. the Ell to be paid at the expiration of three months, I demand what I shall gain in 12 months, laying out 100 l. upon that commodity. This and such like questions, although they may be wrought by the Rule of Three Direct, at two operations, yet they are best performed by the Double Rule of Three compounded of five numbers, wherefore the question may be thus stated. If 2 s. 8 d. in three months, gain 2 d. what shall 100 l. gain in 11 months? If you take your numbers out of your scale▪ and place them according as was directed in the first part of this Book, you shall find them to stand thus, Your numbers being placed according to the tenor of the question, if you multiply 1.333, which is the Decimal of 2 s. 8 d. by 3 months, the product will be 3999, which must be your Divisor, then multiply .83 which is the Decimal of 2 d. by 100 l. and it makes 8300, that again multiplied by 12 months, giveth for the product 99600 for your Dividend, wherefore if you divide 99600 by 3999, it will give you in the quotient 25 almost, which is 25 li. for the Decimal fraction remaining is so small, that it wanteth not near a farthing of 25 li. and therefore we call it 25 li, and so it is exactly, as you may try, if you reduce all the numbers to their least denominations, and work as is before taught in Vulgar Arithmetic. I will prove this question by the converse. If one Ell of Lockerum cost me 2 s. 8 d. ready money, for what price shall I sell the same again to be paid at the end of three months. So that I may gain 25 li. in 100 li. for 12 months? Say by the Rule of Three Direct. If 100 li. in 12 month's gain 25 li. what shall 2 s. 8 d. gain in 3 months? If you take your numbers out of your Scale, and place them according to the Double Rule of Three, they will stand as followeth. Your numbers being thus placed, if you multiply 100 li. by 12 months, you shall find the product to be 1200, which is your Divisor. Then multiply 25 li. by 1.333, which is the Decimal of 2 s. 8 d. and the product thereof will be 33325, which multiply again by 3, and the product will be 99975 for your Dividend, this 99975 divided by 1200, giveth in the quotient. 83, which is the Decimal of 2 d, which 2 d. added to 2 s. 8 d. the price which the Ell of Lockeram cost, giveth 2 s. 10 d. and at that price must you sell the same at 3 month's time so that you may gain 25 l. in the 100 li. in 12 moneth●. I might further proceed to show you Examples in divers other Rules; As in Alligation, Position, etc. but those Rules being already handled in the First Part of this Book, it will be easy to apply them to the scales, And (as I intimated at the beginning of this Third Part,) that although I have only made choice of the eight scales there mentioned and described, yet it was easy to contrive scales for the Coins, Weights, and Measures of other Countries, and not only so, but when the value of the Money, Weight, or Measure of one Country, and the Money, Weight, or Measure of another Country is known, it is easy to contrive two scales, which facing one another, shall immediately tell you how many pounds in one place shall be equal to so many Crowns, or other Coin in another place, but this I do only intimate, that such as are desirous, may fit themselves with scales answerable to their most necessary occasions. And thus shall I conclude this third part, referring some things necessary to such a work, to the Appendix following. An Appendix. Section 1. Of Exchange of the Coins, Weights, and Measures of one Country, with the Goins, Weights and Measures of another Country. TO perform this work, there is nothing required more than the Golden Rule, if first the Rate or Proportion between the Coins, Weights, and Measures of any two Countries be known which is best obtained by experience, rather than taken upon trust, all that in this place I shall do, is only to instruct the ingenious in the manner of Work, and make use of such Rates or Proportions as I find set down by Mr. Lewis Robert's Merchant, in his Map of Commerce. Question 1. How many Riders (each Rider containing 1 l. 1 s. 2 d. 2 q. Sterling) shall I receive for 251 l. 6 s. 4 d. 2 q. Sterling? Facit 237 Riders. If 1 l. 1 s. 2½ d. give 1 Rider how many Riders shall 251 li. 6 s. 4½ d. give? Here if you reduce your numbers to their least Denominations, or set them down in Decimals, and multiply and divide according to the Golden Rule, you shall find in your quotient 237, and so many Riders ought to be received for 251 l. 6 s. 4 d. 2 q. Sterling. Question 2. How many French Crowns (each French Crown being valued at 6 s. Sterling) shall I receive for 492 l. 18 s. Sterling? Facit 1643 French Crowns. If 6 s. give 1 F.C. what shall 492 li. 18 s. give. Multiply and divide according to the Golden Rule, and you shall have in your quotient 1643, and so many French Crowns are to be received for 492 l. 18 s. Sterling. Question 3. A Merchant delivered at Paris 1643 Crowns of 6 s. Sterling the piece, how many pounds Sterling ought to be received at London? Answer 492 l. 18 s. Sterling. If 1 Crown give 6 s. what shall 1643 Crowns give? Multiply and divide, and you shall have in your quotient 492 l. 18 s. and so much Sterling Money ought to be delivered at London for 1643 French Crowns of 6 s. the Crown Sterling. Question 4. If 3 yards at London, be 4 els at Antwerp, how many yards at London make 84 els at Antwerp? els Antwerp yards London els Ant. 4 3 84 Facit. 63. And so many yards at London, are equal to 84 els at Antwerp. Question 5. How many yards of London make 27 els of Antwerp, when 100 els of Antwerp make 60 els of Lions, and 20 els of Lions make 25 yards of London? The first work. 75 facit 75 That is 75 yards of London is equal to 100 els of Antwerp. The second work. Question 6. If 100 li. Sterling be 104 l. 6 s. 4 d. Flemish what is one pound Sterling worth? Facit 322 ponce 36/100 of a penny. Question 7. How many els of Frankford make 42¼ els of Vienna in Austria, when 35 els of Vienna make 24 at Lions; 3 els of Lions, 5 els of Antwerp, and 100 els of Antwerp, 125 els at Frankford. First work. els Ant. els Frank. els Ant. 100 1.25 5 5 625 Facit 6.25 or 6 els and a quarter of Frankford equal to 3 els of Lions. Second Work. Facit 50 els of Frankford, equal to 35 els of Vienna. Facit 60.35 or 6 5/14 els of Frankford, equal to 35 els of Vienna. Thus have I given you a few Examples of Exchanges, I will now insert some few Tables derived from Mr. Lewis Roberts his Map of Commerce aforesaid, of the truth of which, I am not a competent Judge, but shall leave that to the scrutiny of such as have occasion to trade into Foreign Countries. A Table showing what one pound of Avoirdupois weight at London, maketh in divers eminent Cities, and other remarkable places. One pound of Avoirdupois weight at London, makes at ANtwerpe .9615 lb. Amsterdam .9 lb. Abbeville .91 lb. Ancona 1.282 lb. Avignon 1.12 lb. Bordeaux .91 lb. Burgoyne .91 lb. Bologna 1.25 lb. Bridges .98 lb. Calabria 1.3698 lb. Callais 1.07 lb. Constantinople .8474 lb. Constantinople Loder; lb. Deep .91 lb. Danzig 1.16 lb. Ferrara 1.3333 lb. Florence 1.282 lb. Flanders in general 1.06 lb. Geneva .9345 lb. Genoa 1.4084 subtle lb. Genoa 1.4285 gross lb. Hamburg .92 lb. Holland .95 lb. Lixborne .881 lb. Lions 1.07 common weight lb. Lions .98 silk weight lb. Lions .9 customers weight lb. Legorn 1.3333 lb. Milan 1.4285 lb. Mirandola 1.3333 lb. Norimberg .88 lb. Naples 1.4084 lb. Paris .89 lb. Prague .83 lb. Placentia 1.3888 lb. Rochel 1.12 lb. Rome 1.27 lb. Rovan .875 by Vicont lb. Rovan .9017 common weight lb. Seville 1.08 lb. Tholousa 1.12 lb. Turin 1.2195 lb. Venetia 1.5625 subtle lb. Venetia .9433 gross lb. Vienna .813 lb. The use of the preceding Table. How much weight at Bologna, will 655 l. Avoirdupois make? Look in the Table for Bologna, and right against it you shall find 1.25, which showeth that one pound Avoirdupois at London is equal to 1.25 l. at Bologna, Therefore say by the Rule of Three: If 1 l. Avoirdupois give 1.25 l. at Bologna, what shall 655 li. Avoirdupois give? Answer 818.75. As by the operation following doth appear. A Table showing what one pound Weight in divers foreign Cities, and remarkable places, maketh at London of Avoirdupois Weight. One pound weight in Ant's werpe makes at London of Avoirdupois weight. 1.04 lb. One pound weight in Amsterdam makes at London of Avoirdupois weight. 1.1111 lb. One pound weight in Abbeville makes at London of Avoirdupois weight. 1.0989 lb. One pound weight in Ancona makes at London of Avoirdupois weight. .78 lb. One pound weight in Avignon makes at London of Avoirdupois weight. 8928 lb. One pound weight in Bordeaux makes at London of Avoirdupois weight. 1.0989 lb. One pound weight in Burgoyne makes at London of Avoirdupois weight. 1.0980 lb. One pound weight in Bologna makes at London of Avoirdupois weight. .8 lb. One pound weight in Bridges makes at London of Avoirdupois weight. 1.0204 lb. One pound weight in Calabria makes at London of Avoirdupois weight. .73 lb. One pound weight in Calais makes at London of Avoirdupois weight. .9345 lb. One pound weight in Deep makes at London of Avoirdupois weight. 1.0989 lb. One pound weight in Danzig makes at London of Avoirdupois weight. .862 lb. One pound weight in Ferrara makes at London of Avoirdupois weight. .75 lb. One pound weight in Florence makes at London of Avoirdupois weight. .78 lb. One pound weight in Flanders in general makes at London of Avoirdupois weight. .9433 lb. One pound weight in Geneva makes at London of Avoirdupois weight. 1.07 lb. One pound weight in Genoa subtle makes at London of Avoirdupois weight. .71 lb. One pound weight in Genoa gross makes at London of Avoirdupois weight. .7 lb. One pound weight in Hamburg makes at London of Avoirdupois weight. 1.0865 lb. One pound weight in Holland makes at London of Avoirdupois weight. 1.0526 lb. One pound weight in Lixborne makes at London of Avoirdupois weight. 1.135 lb. One pound weight in Lions common weight makes at London of Avoirdupois weight. .9345 lb. One pound weight in Lion's silk weight makes at London of Avoirdupois weight. 1.0204 lb. One pound weight in Lion's custom weight makes at London of Avoirdupois weight. 1.1111 lb. One pound weight in Legorne makes at London of Avoirdupois weight. .75 lb. One pound weight in Milan makes at London of Avoirdupois weight. .7 lb. One pound weight in Mirandola makes at London of Avoirdupois weight. .75 lb. One pound weight in Norimberg makes at London of Avoirdupois weight. 1.136 lb. One pound weight in Naples makes at London of Avoirdupois weight. .71 lb. One pound weight in Paris makes at London of Avoirdupois weight. 1.1235 lb. One pound weight in Prague makes at London of Avoirdupois weight. 1.2048 lb. One pound weight in Placentia makes at London of Avoirdupois weight. .72 lb. One pound weight in Rotchel makes at London of Avoirdupois weight. .8928 lb. One pound weight in Rome makes at London of Avoirdupois weight. .7874 lb. One pound weight in Rovan by Vicont, makes at London of Avoirdupois weight. 1.1428 lb. One pound weight in Rovan common weight makes at London of Avoirdupois weight. 1.1089 lb. One pound weight in Sivil makes at London of Avoirdupois weight. .9259 lb. One pound weight in Tholousa makes at London of Avoirdupois weight. .8928 lb. One pound weight in Turin makes at London of Avoirdupois weight. .82 lb. One pound weight in Venetia subtle, makes at London of Avoirdupois weight. .64 lb. One pound weight in Venetia gross: makes at London of Avoirdupois weight. 1.06 lb. One pound weight in Vienna makes at London of Avoirdupois weight. 1.23 lb. The use of the foregoing Table. In 7652 li. weight at Mirandola, how many pound Weight at Avoirdupois. Look in the Table for Mirandola, and right against it you shall find .75, which showeth that one pound Avoirdupois is equal to the 75 or ¾ of a pound at Mirandola, wherefore say by the Rule of Three. If 1 l. at Mirandola, gives .75 or ¾ of a pound Avoirdupois, what shall 7652 l. of Mirandola give? Answer 5739, as by the operation doth appear. A Table reducing English els to the Measures of divers foreign Cities and remarkable places. One Ell at London makes at AMsterdam 1.6949 els One Ell at London makes at Antwerp 1.6666 els One Ell at London makes at Bridges 1.64 els One Ell at London makes at Arras 1.65 els One Ell at London makes at Norimberg 1.74 els One Ell at London makes at Colen 2.08 els One Ell at London makes at Lisle 1.66 els One Ell at London makes at Mastrich 1.57 els One Ell at London makes at Frankford 2.0866 els One Ell at London makes at Danzig 1.3833 els One Ell at London makes at Vienna 1.45 Aulns One Ell at London makes at Paris .95 Aulns One Ell at London makes at Rovan 1.03 Aulns One Ell at London makes at Lions 1.0166 Aulns One Ell at London makes at Callais 1.57 Aulns One Ell at London makes at Venice linen 1.8 Braces One Ell at London makes at Venice silk 1.96 Braces One Ell at London makes at Lucques 2. Braces One Ell at London makes at Florence 2.04 Braces One Ell at London makes at Milan 2.3 Braces One Ell at London makes at Legorn 2. Braces One Ell at London makes at Madera 1.0328 Braces One Ell at London makes at Isles 1.0328 Braces One Ell at London makes at Seville 1.35 Vares One Ell at London makes at Lisbon 1. Vares One Ell at London makes at Castillia 1.3875 Vares One Ell at London makes at Andoluzia 1.4625 Vares One Ell at London makes at Granado 1.3625 Vares One Ell at London makes at Genoa 4.8083 Palms One Ell at London makes at Saragosa .55 Canes One Ell at London makes at Rome .56 Canes One Ell at London makes at Barselona .7125 Canes One Ell at London makes at Valentia 1.2125 Canes The use of this Table. In 632 els at London, how many Braces at Florence? Look in the Table for Florence, and right against it you shall find 2.04, which showeth that one Ell at London, maketh at Florence 2.204 Braces, wherefore say by the Rule of Three. If one Ell at London, give 2.04 Braces at Florence, how many Braces shall 632 els give? Answer 1289.28, as by the operation following doth appear. A Table reducing the Measures of divers foreign Cities and remarkable places, to English Ells. One Ell at AMsterdam makes at London .59 els One Ell at Antwerp makes at London .6 els One Ell at Bridges makes at London .6097 els One Ell at Arras makes at London .606 els One Ell at Norimberg makes at London .5474 els One Ell at Colen makes at London .4807 els One Ell at Lisle makes at London .6024 els One Ell at Mastright makes at London .6369 els One Ell at Frankford makes at London .4792 els One Ell at Danzig makes at London .7228 els One Ell at Vienna makes at London .9896 els One Auln at Paris makes at London 1.0526 els One Auln at Rovan makes at London .9708 els One Auln at Lions makes at London .9836 els One Auln at Callais makes at London .6369 els One Brace at Venice linen makes at London .5555 els One Brace at Venice silk makes at London .5102 els One Brace at Lucques makes at London .5 els One Brace at Florence makes at London .4901 els One Brace at Milan makes at London .4347 els One Brace at Leghorn makes at London .5 els One Brace at Madera Isles makes at London .9681 els One Vare at Sivil makes at London .7409 One Vare at Lisbon makes at London 1. One Vare at Castilia makes at London .7207 One Vare at Andolusia makes at London .7339 One Vare at Granado makes at London .7339 One Palm at Genoa makes at London .2079 One Cane at Saragosa makes at London 1.8181 One Cane at Rome makes at London 1.7857 One Cane at Barselona makes at London 1.4035 One Cane at Valentia makes at London .8247 The use of this Table. In 5727 Braces at Ligorn, how many els English. Look in the Table for Ligorn, and right against it you shall find .5, which showeth that one Brace at Ligorn maketh at London. or half an Ell, wherefore say by the Rule of Three. If one Brace at Ligorn give .5 els at London, what shall 5727 Braces give? Answer 2863.5, as by the work appeareth. Section 2. Concerning Interest and Annuities. The first Table showing what one pound being forborn any number of years under 31, will amount unto, accounting interest upon interest, after the rate of 6 per cent. Y. 6 per cent. 1 1,06 2 1,1236 3 1,19101 4 1,26247 5 1,33822 6 1,41851 7 1,50363 8 1,59384 9 1,68947 10 1,79084 11 1,89829 12 2,01219 13 2,13292 14 2,26090 15 2,39655 16 2,54035 17 2,69277 18 2,85433 19 3,02559 20 3,20713 21 3,29056 22 3,60353 23 4,81975 24 4,04893 25 4,29187 26 4,54938 27 4,82234 28 5,11168 29 5,41838 30 5,74349 The first Column of this Table having Y at the top thereof, beginning at 1, and so proceeding to 30, signify years, and the number in the next Column answering thereunto do show what one pound is worth, being forborn any number of years under 31, which Table is made according to this proportion. As 100 to 106, so is 1 to 1.06 and again, As 100— 106— 106— 1.1236 and thirdly. As 100— 106— 1.1236— 1.19101 Et sic ad infinitum. The use of this Table. What 136 l. 15 s. 6 d. amount unto, being forborn 20 years, after the rate of 6 per centum, interest upon interest. Look in the Table for 20 years, and right against in the broader Column, you shall find 2.20713, which shows that one pound being forborn 20 years will be augmented to 3.20713. Then if you reduce your 136 li. 15 s. 6 d. into a Decimal, either by the Tables in the Second Part, or by the Scales in the Third Part of this Book, you shall find it to be 136.775. Wherefore say by the Rule of Three Direct. If one pound being forborn 20 years will be augmented to 3.20713, to how much will 136.775 li. be augmented to in the same time. Answer, to 438 li. 13 s. 1 d. 1 q. as by the operation following doth appear. Or 438 li. 13 s. 1 d. 1 q. The Second Table, showeth what one pound Annuity will amount unto, being forborn any number of years under 31, at 6 per cent. interest upon interest, the Annuity being to be paid yearly. Y. 6 per cent. 1 1,00000 2 2,06000 3 3,18360 4 4,37461 5 5,63709 6 6,97531 7 8,39383 8 9,89746 9 11,49131 10 13,18079 11 14,97164 12 16,86994 13 18,88213 14 21,01506 15 23,27596 16 25,67252 17 28,21287 18 30,90565 19 33,75999 20 36,78559 21 39,99272 22 43,39228 23 46,99582 24 50,81557 25 54,86451 26 59,15638 27 63,70576 28 68,52810 29 73,63979 30 79,05818 The use of this Table. What will an Annuity of 20 li. payable yearly, be augmented unto in 12 years, being all that time forborn accounting interest upon interest at 6 per cent. per annum. Look in the first column of the Table for 12 years, and right against it in the next column you shall find 16,86994, which shows that 1 li. Annuity payable yearly, being forborn 12 years will amount unto 16.86994, wherefore say by the Rule of Three Direct. If 1 pound Annuity forborn 12 years give 16.86994 what shall an Annuity of 20 pound a year give? being forborn the same term of 12 years? Answer 337 li. 7 s. 11 d. 3 q. fere, as in the operation doth appear. or 337 l. 7 s. 11 d. 3 q. ferè. The third Table showeth what one pound being forborn any number of years under 31 is worth in ready money, rebating yearly, after the rate of 6 per cent. interest upon interest. Y 6 per cent. 1 ,943396 2 ,809996 3 ,839619 4 ,792093 5 ,747258 6 ,704960 7 ,665057 8 ,627412 9 ,591898 10 ,558394 11 ,526787 12 ,496989 13 ,468839 14 ,442300 15 ,417263 16 ,393646 17 ,370364 18 ,351343 19 ,330512 20 ,311804 21 ,294155 22 ,277505 23 ,261797 24 ,246978 25 ,232998 26 ,219810 27 ,207367 28 ,195630 29 ,184536 30 ,174110 The making of the Table. As— 106— 100— .943396— 889996 and again, As 106— 100— .889996— 839619 Et sic ad infinitum. If 356 l. be payable at the end of 7 years, what is it worth in ready money, discounting or rebating after the rate of 6 per cent. interest upon interest. Look in the Table for 7 years, and against it you shall find .665057, being the ready money which 1 l. is worth payable at 7 years' end, wherefore say by the Rule of Three. If 1 l. in 7 years rebate or decrease to .665057, to what will 356 l. rebate or decrease in the same time? Answer to 170 l. 5 s. 1 d. as by the operation doth appear. The fourth Table showeth the present worth of one pound Annuity; to continue any number of years under 31, and payable yearly after the rate of 6 per cent. interest upon interst. Y. 6 per cent. 1 0,94339 2 1,83339 3 2,67301 4 3,46510 5 4,21236 6 4,91732 7 5,58238 8 6,20979 9 6,80169 10 736008 21 7,88687 22 8,38384 23 8,85268 24 9,29498 25 9,71224 26 10,10589 27 10,47725 28 10,82760 29 11,15811 30 11,469●2 21 11,76305 22 12,04158 23 12,30337 24 12,55035 25 12,78335 26 13,00316 27 13,01053 28 13,40616 29 13,59071 30 13,76482 The use of this Table. What is the present Rent or Annuity of 25 pound per annum worth payable yearly, for 21 years, accounting Interest upon Interest at 6 per centum. Look in the Table for 21 years, and right against it you shall find 11.76407, which is the present worth of one pound Annuity, for 21 years, wherefore say by the Rule of Three. If an Annuity of 1 l. per annum for 21 years be worth 11.76407 ready money, what is an Annuity of 25 li. per annum worth in ready money for the same time? Answer 294 l. 2 s. 0 d. 1 q. as by the operation following doth appear. or 294 l. 2 s. 0 d. 1 q. The fifth Table showeth what Annuity payable yearly, one pound will purchase for any number of years under 31, after the rate of 6 per cent. compound interest. Y 6 per cent. 1 1.06000 2 .54363 3 .37411 4 .28859 5 .23739 6 .20336 7 .17913 8 .16103 9 .14702 10 .13586 11 .12679 12 .11926 13 .10297 14 .10758 15 .10296 16 .09895 17 .09544 18 .09235 19 .08962 20 .08718 21 .08500 22 .08304 23 .08127 24 .07967 25 .07822 26 .07690 27 .07569 28 .07459 29 .07357 30 .07264 The use of this Table. What Annuity to begin presently and to continue 28 years, payable at yearly payments, will 640 l. purchase, accounting compound Interest after the rate of 6 per cent. Look in your Table for 28 years, and right against it in the next column you shall find .07459, which shows that one pound ready money will purchase an annuity worth .07459, and to continue 28 years, wherefore say by the Rule of three. If one pound ready money, will purchase an Annuity worth .07459 to continue 28 years, what Annuity shall I purchase for the same time, paying 640 l. ready money? Answer 47 l. 14 s. 9 d. as by the operation doth appear. FINIS. The fourth Part: BEING AN ABRIDGEMENT OF THE PRECEPTS OF ALGEBRA Written in French By JAMES de BILLY: And now translated into English. With divers Questions added, which were not in the Original. LONDON, Printed Anno, MDCLIX. ALGEBRA BREVIS. I Have been always of opinion, that the practice of Algebra should not be entangled with a great number of precepts: This Science is of itself dark enough, without adding unto it new obscurity, by the confusion of many different operations. You have here an Abridgement which hath pleased many of good judgements, and I hope, such as will with attention read it, shall from thence receive both satisfaction and profit. I shall in the first place set forth a Table of three ranks: In the first of which there is a progression natural, of which the terms are disposed in that order, that immediately under them you have the Cossick Characters, of which they are Exponents, and in the lowest Cell, a progressron Geometrical, which beginning with an Unite, may be doubled, trebled, quadrupled, etc. we have for the greater ease only doubled them. Observe then that R is a Coffick Character signifying a Root, and that the Exponent thereof is marked above in the uppermost rank, and C is the Cossick Character of a Cube, whose Exponent is 3, and so of the rest. I call those terms, which are in the upper Cell, Exponents; because they expound the Cossick Characters, and the Numbers of the Geometrical Progression which are below. Mark well this manner of speaking, consider diligently this Table, Exponents Characters Cossick Progr. Geometr. etc. &c &c 8 QQ Q Square quare quare 256 7 S2 second 128 6 QC. Square Cube 64 5 S Sur 32 4 QQ Blquadiate 16 3 C Cube 8 2 Q Square 4 1 R Roor 2 0 N Numbers absolute 1 and for the present content yourself with this. You may continue this Table, if you please, infinitely in this manner. Take two numbers, which if you multiply, they will produce some Exponent, you shall presently see what a character is to be put under that Exponent. For example, If you would know the character of the Exponent 6, take the numbers 2 and 3 (because these multiplied together make 6) after that add their characters which are Q and C, and you shall have Q C for the character of the Exponent 6. In like manner the character of the Exponent 8 is QQQ, because under the exponents 2 and 4, which multiplied together produce 8, are contained the characters Q and QQ. Also the character of the exponent 12 will be Q Q C, because 12 is a number produced by the multiplication of 2 and 6, or of 3 and 4. But if the exponent be a first number (that is to say, a number not produced by the multiplication of any two other) mark in what order it is after the exponent 5, and call it sursolide second, third, or fourth, etc. according to its rank. The character of 5 is , of 7 second, of 11 3; and so consequently under such expoponents as are first numbers, under which only are found such solides. CHAP. I. The Alegorithm of Cossick Numbers, simple, compounded, or diminished. BY the word Alegorithm, I mean all the operations comprehended under these four kinds, Addition, Subtraction, Multiplication, and Division. By the word Cossick simple, I understand such as have not this + (which signifies plus, nor this − signifying minus, expressed before them. On the other side, by Numbers composed, are meant such as have the sign +, and by diminished, such as have the sign −. Note such numbers as have no sign expressed, are supposed to have this of +. Sect. 1. Addition of simple Cossick numbers. All simple cossick numbers are of the same denomination (that is to say, have the same character) or of different: If they be of the same denomination, the Addition is as in common Arithmetic. Example, 5Q added with 3Q, makes 8Q If they be of different denominations, they must be added by the interposition of the figure +, as 6R added to 4Q makes 6R+4Q, in like manner 3 added to 4R makes 3+4R. Sect. 2. Subtraction of simple Cossick numbers. Either they are of the same, or different denominations: if of the same, you must subtract as in ordinary Arithmetic; for example, 3Q subtracted from 8Q, there rests 5Q. If of different, you must subtract by the interposition of the sign − as 6R subtracted from 4Q there rests 4Q − 6R, so 3C subtracted from 66, there remains 66 − 3C. Sect. 3. Multiplication of simple Cossick numbers. You must here have regard both to the absolute number, and to the Cossick characters: If therefore a Cossick number be to be multiplied by an absolute, you must multiply the absolute numbers, and unto the product give the same character, as 5R multiplied by 12 produce 60R. But if you multiply Cossick numbers by Cossick, you must multiply the absolute numbers together, and to the product give the character of that exponent, which is made by the addition of the exponents belonging unto the foresaid Cossick characters. For example, 2R multiplied by 3Q make 6C, because the exponent of Q is 2 added to one the exponent of R, make 3 the exponent of C, which in this respect ought to be given to the product. In like manner 5R multiplied 4C it makes 20QQ for the reason above given. Sect. 4. Division of simple Cossick numbers. The speculation of this is marvellous. But the practice of it is by putting the Divisor under the number to be divided, drawing between them a little small line in manner of common fractions. For example, 13Q divided by 7R, the quotient is 13q/7R and 6QQ divided by 5C the quotient is 699/5C. Sect. 5. Addition of numbers composed and diminished. Some order is to be observed in this, in which the numbers are to be disposed in such manner, that those that are of the same denomination must be put right under one another. After you have done this, if they have the same sign, they are added as in common Arithmetic, and to the product give the same sign. As for example, 7Q − 4C added to 3Q − 2C, the Sum is 10Q − 6C. But if the numbers be of different signs, the lesser must be subtracted from the greater, and to the residue you must give the sign of the greater number as 6Q+7R added with 7Q − 12R, give for the Sum total 13Q − 5R. 6Q+7R 7Q − 12R 13Q − 5R 6Q − 7R 7Q+12R 13Q+5R Sect. 6. Subtraction of numbers composed and diminished. There is nothing more intricate to beginners than the precepts commonly given for subtraction. You have here an order plain, sure, and very easy to practise. Change the sign of the particulars of that number you desire to subtract, and after this change add them with the number from which the subtraction is to be made, and you shall have the residue: As if from 6Q − 10R you would subtract 18Q − 15R by the 5 Sect. the residue will be 5R − 12Q. So likewise if you would subtract − 8R − 9Q from 16R+6Q, the residue will be 24R+15Q 6Q − 10R Add − 18Q − 15R Add − 12Q+5R Resid. 16R+6 Q Add 8R+9 Q Add 24R+15Q Residue. Sect. 7. Multiplication of numbers composed and diminished. Mark what I said of Cossick simple in the 3 Sect. and remember that the same signs have always the sign + in the product and different − & there is no difficulty in multiplication, so as you multiply every particular of the mulplicand by every particular of the multiplicator, as in common Arithmetic, as if you multiply 3Q − 2R by 8R, multiply 2R by 8R, it makes 16Q, and because the multiplicator and the multiplicand have different signs, the product must have the sign of −; and therefore that shall be 16Q, farther 3Q by 8R make 24C, to which you must give the sign +, because the multiplicator and the multiplicand have the same sign, so that the product of this multiplication will be 24C − 16Q. So 2R+4Q multiplied by 3Q − 5 the product is 6C+12QQ − 10R − 20Q 3Q − 2R multiplicand 8R multiplicator 24C − 16Q Product. 2R+4Q mult. 3Q − 5 multip. − 10R − 20Q 6C+12 QQ/ 6C+12Q Q − 10R − 20Q Sect. 8. Division of numbers composed and diminished. There is no great difficulty in this, only put a line between the Dividend and the Divisor, and you have the quotient, as 4C − 3Q+2R divided by 5R − 4C make for this quotient 4C − 3Q+2R/5R − 4C Sect. 9 Algorithm of Fractions. I shall not give any particular precepts, because if a man understand the Fractions of common Arithmetic, and practise according to what is before said, there will be no need of them. CHAP. II. The rule of Algebra, with the explication thereof. IT was merely necessary by the precedent rules, to trace out insensibly the way to Algebra, which cannot be practised without Addition, Subtraction, Multiplication and Division. Having therefore made plain these difficulties, we will proceed in the proposition of the rule of Algebra, and the explication of every part of it briefly and plainly. Sect. 1. The rule of Algebra. You must first for the number unknown put 1R, and after examine this root according to the tenor of the question, until you come to an Equation. Secondly, this Equation is to be reduced, if need require. Thirdly, you must divide every part of the Equation by the number of the greatest cossick character. After which, either the quotient, or some root of the quotient, will give the root unknown: This is the rule of Algebra, let us now explain it. Sect. 2. How the Equation must be found. The rule saith, this is done by examining the question propounded, according to the tenor of the same. That is to say, you must well observe all the conditions of the question propounded, to the end you may fully accomplish it. For after you have gone thorough it, you shall find an equation between two numbers. As if I search a number, which added with its square shall make 20, I suppose this number unknown to be 1R, the square thereof is 1Q. (because every number multiplied by itself, makes its square) than 1Q+1R is equal to 20. See thus an equation found between 1Q+1R and 20. Sect. 3. How your Equation must be reduced. Your Equation being found, it is reduced by adding the same number to both the terms of the equation, or by subtracting from them the same number. So is it performed by multiplying, or dividing both the terms by the same number. For by this means your equation shall remain the same after these things done. As for example, 1R+1Q = 20, adding throughout 2C, you shall have also an equation between 1R+1Q+2C and 20+2C. So subtracting 1R from the terms, you have 1Q = 20 − 1R. Likewise multiplying, or dividing both your terms by 3, you shall have by the multiplication 3R+3 Qequal to 60, and by the division the equation with be between on ⅓R+⅓Q and 20/3. Now to make your reduction judiciously and profitably, you must take care always that your greatest character remain alone on one side of your equation. As of all the reductions before made, there is none useful but the second, because in that only you find on the one side alone 1Q = 20 − 1R which is the only end of your reduction. I said in the rule of Algebra, that your equation must be reduced, if it be necessary; because it sometimes happens, that there is no need of it: as when your equation falls out between two simple collateral numbers, I call those numbers collateral, whose exponents do not surpass one another by more than an unite. Sect. 4. When you must extract the root. When your cossick numbers are simple and collateral, you must not extract any root; but if you divide by the number of the greatest cossick character, the quotient shall show you the value of the root, which is all you seek for in Algebra. For example, If you find an equation between 2R and 28, dividing simply 28 by 2 the quotient, shall be the value of 1R. But when the terms of your equation are not collateral, you must extract some root; either square, cube, squared square, etc. according to the cossick character which remains after your hypobibasme. Now hypobibasme is nothing else but an abatement, or depression of the character, and is done by subtraction of the letter exponent from the greater. As if you find an equation between 10QC and 90 QQ take notice of the exponent of QC in the Table inserted at the beginning, the which exponent is 6; afterward look to the exponent of QQ which is 4, subtract 4 out of 6 there rests 2, of which the cossick character is Q. From hence I conclude, that 10Q are equal to 90, after dividing 90 by 10, and finding 9 in the quotient, I conclude, that the square root of 9 must be extracted by reason of the character Q. Sect. 5. How to extract the square root of numbers, compound and diminished. No man hath yet perfectly found out the way to extract the root of numbers compound and diminished, unless the exponents of three terms of the equation keep between them in some situation or other an Arithmetical proportion, that is to say, the same distance. As if the equation be between 1Q and 20 − 1R, you may now extract the root of 20 − 1R, because the exponents of the three numbers which make the equation, are 2, 0, 1, which thus place, 0 1, 2, keep the same distance. The greatest cossick character left after the Hypobibasme shows the root to be extracted; as in this example, before the square root is to be extracted, because the greatest character is Q. The method to be followed in the extraction, see here propounded in general terms. Take first the half of the number of roots. Secondly, to the square of this half add, or from it subduct your absolute number according to the signe of + or −. Thirdly, extract the square root of this sum, or of the residue. Fourthly, to this root add, or from it subtract half the number of roots, and this last sum or residue shows you the value of the root unknown. For example, I would find a number, the double whereof added to its square, should be equal to 24, I shall find an equation between 2R+1Q and 24 by the second Section. Moreover; I shall reduce this equation after this manner 1Q equal to 24 − 2R by the third Section. Then if I divide 24 − 2R by 1, the number of the greatest cossick character there still remains 24 − 2R, because an unite doth neither multiply nor divide. Then in as much as the three terms of the equation do keep an Arithmetical proportion, I extract the square root of 24 − 2R in this manner. I take first half the number of roots, which is 1. Secondly, the square of 1 is 1, to which I add the absolute number, which is 24, because of the sign + before it, that makes 25. Thirdly, I extract the square root of 25, which is 5. Fourthly, from this root I extract the moiety of the number of roots which is 1, because of the sign − the residue will be 4; whence I conclude that to be the value of one root, and that the number sought is 4, whose double is 8, added to the square (of 4,) 16, makes 24. Here note that numbers diminished, where the absolute number hath the sign—, have two roots. The greater is extracted, as before we have taught; the lesser is found out by subtracting the square root of the residue from the half sum of the roots. As if I seek a number whose octuple diminished by 12, shall be equal to its square, you will find an equation between 1Q and 8R − 12. The greatest root is 6, the lesser is 2, here both the roots answer the question. But this happens not often. But if you be to extract the biquadrate root: First, extract the square root, as is taught, and again extract the square root of this, and this shall be your biquadrate root. As if the equation be between 1QQ and 2Q+8, you may find the square root to be 4 by the method taught, first, taking half the number of the squares, etc. and afterward you must extract the square root of 4, which is 2, this shall be the value of the root. In like manner, if your equation be between 1QC and 2C+48, first, I extract the square root of 2C+28, which is 8, of which extract the cube root 2, because the root to be extracted is the square cube, as the character QC, which is one of the terms of the equation denoteth. Sect. 6. How to know if the question be impossible, vain, or ill propounded. You may know the question to be impossible, if you come to an equation impossible: As if following the conditions of the Problem, you meet with an equation between 6R and 24R, or between 3Q+5 and 4+●Q. Secondly, the question is vain, when the equation is between two equal numbers of the same denomination, as between 6Q and 6Q. Thirdly, the question is ill propounded, when without any difficulty many numbers will answer the problem propounded. CHAP. III. Algorithme of second roots with their use. ALgebraists sometimes use more than one root to find out divers numbers propounded, and then to the end they may proceed with less confusion they usually help themselves with second roots, which they express thus, 1A, 1B, etc. Sect. 1. Addition of second roots. If your second roots be of the same denomination, add the numbers, and to the sum give the same denomination, as 5A added to 4A make 9A, if they be of different denomination, add them with the sign of + as 5 A added to 6B make 5A+6B. Sect. 2. Subtraction of second roots. If they be of the same denomination, subtract one number from the other, and to the residue give the same denomination, as 5A taken from 9A, there remains 4A, if different, they are subtracted with the sign − as 6B taken from 8A, there rests 8A − 6B. Sect. 3. Multiplication of second roots. If they be of the same denomination, do as you do with the first roots, as 4A multiplied by 7A make 28AQ, if of different, both denominations are retained in the product, as 3R multiplied by 5A make 15RA. Sect. 4. Division of second roots. Division is only performed by the interposition of a little line, as is before taught, notwithstanding it is to be observed, that if for example 3AR be to be divided by the Divisor 1R the quotient shall be 3A, because in such case there needs nothing else but to take from the Dividend the character of the Divisor. Sect. 5. The extraction and use of second roots. After you have found and reduced your equation, according to the manner of working in second roots, you must extract the root after the manner taught in the precedent Chapter. As if 1AQ be equal to 25, I say that 5 is the value of the second root, and if 1AQ be equal to 4A+12, you must take the moiety of the number of roots, etc. As is said in the 5. Sect. of the precedent Chapter, and you shall find 6 to be the value of 1A. Now since the end of the second roots is to be reduced to first, you must not forget after you have found the value to begin again your work, and to put in first roots that which you have found to be the value of the second, as I shall show you in some example in a Chapter following. CHAP. IU. The Algorithme and extraction of the roots of furred & irrational numbers. Furred roots are those that have a radical sign before them, and which in propriety of speech ought to be called absolute numbers, notwithstanding they cannot be expressed by any common number, neither whole, nor broken, we will hereafter express the radical sign by this character ℛ. There are many sorts of furred roots, some are simple, as ℛ Q5, that is to say, the root square of 5, others are compound, as ℛ Q5+ ℛ C6, that is to say, the root square of 5, plus the root cube of 6, some are universal, whose radical character, extends to all the particulars following it, and for that end are enclosed in a Parenthesis in this manner ℛ Q (14+ ℛ Q4) the root universal of 14 joined with the root square of 4, all which number is 4 for 14+, the root square of 4 which is 2, maketh 16 whose root is 4. Sect. 1. Reduction of furred roots simple to the same denomination. First, you must put the radical signs under the numbers to which they belong. Secondly, you must multiply the numbers by the signs across, for to get new ones. Thirdly, you must add the signs together, which is done by multiplying their exponents, & give the character of the product common to the two new products, as if you would reduce to the same denomination ℛ Q5 and ℛ C4, you must first place them as followeth. Secondly, you must multiply the numbers 4 and 5 by their signs across, that is to say, you must take the square of 4, and the cube of 5, which are 16, and 125. Thirdly, the exponents of the signs ℛ Q and ℛ C, which are 2 and 3, aught to be multiplied together, the product is 6. I look then in the table what cossick character is under the exponent 6, and finding QC, I take that for my common denominator, and in stead of my two first furred roots, which were of different denomination, that is to say ℛ Q5, and ℛ C4. I have two new ones of the same denomination, that is to say, ℛ QC of 125, and ℛ QC of 16. Sect. 2. Multiplication and Division of furred simple roots. If the roots be of the same denomination you must only multiply and divide the numbers by themselves, and to the product and quotient giuse the same radical sign as ℛ Q7 by ℛ Q2 give for the product ℛ Q14. In like manner ℛ Q36 divided by ℛ Q12, gives for the quotient ℛ Q3. But if the roots be of different denomination, you must reduce them to the same denomination by the precedent Paragraph, and after multiply and divide as shall be showed, for example, ℛ Q3 multiplied by 2, the product is ℛ Q12, and ℛ Q12 divided by 2, the quotient is ℛ Q3. Sect. 3. How to know whether two furred roots be commensurable or not. You must divide the greatest root by the lesser, if the quotient be rational, the two roots are commensurable: if otherwise, they are not. As because ℛ Q24 divided by ℛ Q6, gives for the quotient ℛ Q4, which is 2 a rational number, I conclude these two root: ℛ Q24 and ℛ Q6 to be commensurable. In like manner, since root square 24 divided by ℛ Q8, the quotient will be 3, a furred number and irrational, you may conclude those two roots, ℛ Q24, and ℛ Q8, to be incommensurable. Sect. 4. Addition of simple irrational roots. If the roots be incommensurable, you must add them only by the signe+as ℛ Q24 added unto ℛ Q8 makes ℛ Q24+ ℛ Q8. But if they be commensurable, you must add a unite to their quotient rational, and you shall have a sum, which being multiplied by the lesser of the two roots to be added, will give a product which shall be the sum sought. As ℛ Q24 added with ℛ Q6 makes ℛ Q54, because ℛ Q24 divided by ℛ Q6, gives 2 for the quotient rational, to which I add a unite, and it is 3 by which (always reducing them to the same denomination) I multiply ℛ Q6, which is the lesser of my two roots, and I find for my sum ℛ Q54. Sect. 5. Subtraction of simple irrational roots. If they be incommensurable, you must subtract them by prefixing the sign − as ℛ Q8 subtracted out of ℛ Q24, the residue shall be ℛ Q24 − ℛ Q8. But if they be commensurable, you must take away a unite out of the quotient rational, and you shall have the residue, the which being multiplied by the lesser of the roots given, shall give a product which shall be the residue fought, as if I be to subtract ℛ Q● from ℛ Q4, dividing the greater by the lesser, the quotient rational is 2, from which if you take 1, there remains 1, by which (reducing them first to one denomination) if you multiply the lesser root, that is to say, ℛ Q6, the residue will be ℛ Q6. Sect. 6. Addition and Subtraction of furred numbers, composed and diminished. I have here no new precepts, only advertise you, that you may remember what I have said before of Cossick numbers, touching the signs of + and − in the fifth and sixth Paragraph of the first Chapter, and what is delivered in the fourth and fifth of this Chapter, touching the Addition and Subduction of simple furred numbers, and these will be no difficulty, as if you be to add 5+ ℛ q. 24 with 3+ ℛ q. 6, you will find 8+ ℛ q. 54. In like manner, if you subduct 3 − ℛ q. 6, from 5+ ℛ q. 24, there rests ℛ q. 54 − 42. Sect. 7. Multiplication of numbers furred, composed and diminished. This multiplication hath no great difficulty, nor needs new precepts. Remember only that the same signs have in the Product+and different − with this, that your Multiplication is not good, if the particulars to be multiplied be not first reduced to the same denomination. For example. 5+ ℛ q. 24 by ● − ℛ q. 6, 'tis to be done after this manner: + ℛ q. ●4 by − ℛ q. 6, maketh − ℛ q. 144, or − 12 after+● by − ℛ q. 6, maketh − ℛ q. 150. Further, + ℛ q. 24 by 3 is + ℛ q. 216. Lastly, +5 by 4● is+15, than the whole Product will be 15+ ℛ q 216 − ℛ q. 150 − ℛ q. 144, or 3+ ℛ q. 216 − ℛ q. 150 because ℛ q. 144 is a rational number, to wit 12, which being subducted out of 15, because of the sign − leaveth 3. Example: Sect. 8. Division of furred numbers, compounded or diminished. If the Divisor be simple, the division is made by the interposition of a line between the Divisor and the compound number to be divided, as if ℛ q. 2+ ℛ q. 5 be divided by 8, the quotient will be ℛ q. 2+ ℛ q. 5,/ 8 and so of others. But because it sometimes may fall out (though very seldom) if the Divisor also will be a Binome, or compound number, that is to say, a furred number compounded of two particulars with the sign +, or a Trinome that is compounded of three particulars, etc. See here the manner to divide in such a case. If the Divisor be a Binome, you must multiply by his Apotome, as well the number to be divided, as the Divisor (and if the Divisor be an Apotome, you must divide by the Binome, as well the Dividend, as the Divisor) by means of this Multiplication you shall have a new Dividend, and a new Divisor. Now this new Divisor will be always rational, and therefore needs only to to be set under the Dividend with a line between. As for example: ℛ q. 6 − 2 by ℛ q. 5+ ℛ q. 3, I take the Apotome of my Divisor (that is) ℛ q. 5 − ℛ q. 3. by which I multiply both my Dividend and my Divisor, by one of the multiplications is produced ℛ q. 30 − ℛ q. 20 − ℛ q. 18 + ℛ q. 18 + ℛ q. 12, for my new Dividend, and by the other is produced 2 for my new Divisor. So that the quotient of my Division will be ℛ q. 30 − ℛ q. 20 − ℛ q. 18+ ℛ q. 12/4, because 2 must be first squared, and then 4 put underneath the Dividend. If your divisor be a Trinome, you must observe the same method, multiplying the dividend and the Divisor by the Apotome of the Divisor, that is to say, by the same Divisor only, changing the sign of the last particular. After this is done, you shall have a new Dividend and a new Divisor, which shall be a Binome. Then you must again seek a new Dividend and Divisor, which now will be simple and rational. Last of all, if you will not take this pains, the Division is good, if under the Dividend you subscribe the Divisor with a line between. Sect. 9 Multiplieation of roots universal. You must reduce the root to be multiplied, and the Multiplicator to their squares or cubes, according to the radical sign prefixed, and afterward perform you multiplication, as is taught in Sect. 7 of this Chapter. Afterward you must affix the sign radical, and enclose all in a Parenthesis. This is better understood by an example, as if you multiply ℛ q. (7+ ℛ q. 3) by 2, the squares of the one and other numbers are 7+ ℛ q. 3 and 4 then the first being multiplied by the last, maketh 28+ ℛ q. 48, and therefore if you close this number within a Parenthesis, and put before it the same radical sign, the product will be ℛ q. (28+ ℛ q. 48.) In like manner, if you would multiply this number 7/2 − ½R+ ℛ q. (49/4 − 3/4 q. − 7/2 R) by itself, to get the square of it, you must call to mind the fourth proposition of the second book of Euclid, which showeth that a line being divided into two parts, the square to the whole is equal to the square of the parts, and to double their Rectangles, you must therefore conceive this number, as divided into two parts, of which the first is 7/2 − ½ R, and the last ℛ q. (29/4 − 3/4 q. − 7/2 R) take then the squares of the parts, which are 49/4+3 4+3/4 q. − 7/2 R, and 49/4 − 3/4 q. − 7/2 R, the double of the Rectangle of the parts is ℛ q. (9604/16+220 16+220/16 q. − 2744/8R+7C+12 8R+7C+12/16 q. q.) than the square of the number proposed is the sum of these three numbers, that is to say 49/2 − ½ q. − 7 R+ ℛ q. (9604/16+220 16+220/16 q. − 2744/8 R+7 C − 11/16 q. q.) The square of the Apotome is the same number, putting only the sign − before the universal root, & the sum of the two squares is 49 − 1 q. − 14 R. Sect. 10. Division of roots universal. You must reduce the roots to be divided, and the Divisor to their Squares, Cubes, etc. And after divide them as is taught in the 8 Sect. and when this is done, enclose all in a Parenthesis, with the same radical sign which was before. As if you divide ℛ q. (13+ ℛ q. 17) by the root square of 5, their squares are 13+ ℛ q. 17 and 5, than the first being divided by the last, the product will be 2 3/5+ ℛ q. 17/25, than the quotient of the division proposed will be ℛ q. (2 3/5+ ℛ q. 17/25.) Sect. 11. Addition and Subduction of roots universal. Many trouble themselves to give precepts intricate enough, the short and most certain is to add them with the signe+, and subduct them with the sign −. As for example, ℛ q. (3+ ℛ q. 2) added with ℛ q. (ℛ q. 5+6) will be ℛ q. (3+ ℛ q. 2) + ℛ q. (ℛ q. 5+6) and the same first root subducted from the last, the residue is ℛ q. (ℛ q. 5+6) − ℛ q. (3+ ℛ q. 2.) Sect. 12. Extraction of the roots of Binomes and Apotomes. 1 Take the difference of the squares of the one and other part of the Binome. 2 Add and subduct the square root of this difference from the greatest part of the Binome. 3 Conjoin the square root of the moiety of the sum, with the square root of the moiety of the residue, by the signe+, if it be a Binome, and by the sign −, if it be an Apotome; and thus the extraction is finished. As if you would extract the square root of this Binome 3/2+ ℛ q. 5/4, you shall first take the square of the first part, which is 9/4, and the square of the second which is 5/4, the difference of these two is 4/4, that is to say 1. Secondly, you must extract the square root of this difference, which is 1, you shall add and take it away from the first part of the Binome, by the addition you shall have 5/2 for your sum, and by subduction ½ for your residue. Thirdly, join the root square of the sum, with the root square of the residue by the sign+& you shall have ℛ q. 5/2+ ℛ q. ½ for the square root of your Binome propounded, and consequently ℛ q. 5/2 − ℛ q. ½ shall be the root square of the Apotome 3/2 − ℛ q. 5/4. CHAP. V The Use of Algebra. 'TIs much to have taken the pains to learn all that we have hitherto shown or taught: but I dare boldly say, that those that shall rest here, do as yet know nothing to the purpose, although they may know all the precepts; it behoveth us then to make a step further, to apply and bring those precepts into use and exercise. 'Tis that which I desire to demonstrate in this Chapter, by some questions, the solution of which will give great light to the attaining of perfection in this Art. Wherefore I entreat thee (Reader) not to omit this Chapter, in which I pretend to yield thee some pleasure and delight, as also an illustration of what hath been before treated of. Sect. 1. Questions resolved by one simple Equation. Question I. A Lexander one day told Ephestion, that he was elder than him by two years; thereupon Clitus tells them, that he was as old as both of them (their ages added together) and four years over and above. The Philosopher calisthenes being present at this discourse (saith he) I well remember that my father, who was 96 years old, had the age of you three. It is demanded here, how old Alexander was when he held this discourse, as also how old Clitus and Ephestion were. I put for the age of Ephestion 1 R of years, whence it follows that Alexander had 1 R+2; therefore Clitus had, R+6, and those three together, according to the condition of the question, aught to be equal to 96: therefore there is an equality between 4 R+8 (which is the sum of the three ages) and 96, take away 8 from both parts of the equation so there will remain on one side 4 R equal to 88, which divide by the number of the greatest Cossick Character, that is to say by 4, the quotient gives 22 for the value of one root, which was supposed for the age of Ephestion. Therefore Ephestion was at that time aged 22 years, Alexander 24, and Clitus 50, which all together make 96 years. Question II. A Hare is 100 Geometrical paces distant from a Dog that swiftly pursues him, and the Dog runneth two times and an half faster than the Hare: It is demanded how many Geometrical paces the Hare will have run when the Dog overtaketh her; I put for these Geometrical paces 1 R; therefore the Dog which runs 100 paces more than the Hare, will have run 100+1 R: and for that the Dog runs twice and an half swifter than the Hare, I take two numbers in like porportion to one another, that is to say 5 and 2, and conclude that there is the same proportion between 100+1 R to 1 R, as between 5 and 2; therefore the product of the first number 100+1 R multiplied by the last number 2 (which is 200+2 R) is equal to the product of the two means 1 R and 5 (which will be 5 R) therefore if you take from both parts 2 R, there will remain 200 equal to 3 R, and therefore I divide 200 by 3, which is the number of the greatest Character, and find in the quotient 66⅔ which will be the value of the root. I say therefore that the Hare will have run 66 Geometrical paces, and ⅔ when the Dog shall have overtaken her, and the Dog will have run 166⅔ paces, which make twice and an half more than 66⅔. Question III. The Architect Vitruvius in his ninth Book Chap. 3. tells us, that Archimedes found the quautity of silver that a Goldsmith had mixed in a golden Crown which he had made for the King Hiero (who was obliged by vow to present it to the gods) weighing 100 pounds. It is demanded by what means Archimedes could arrive to the knowledge of that secret. The common opinion is, that he took two masses, the one of gold, the other of silver, which weighed as much as the Crown; afterward he filled a vessel up to the brim with water, which vessel was placed in some great basin, that the water that should be forced out of the first vessel, might be preserved and not lost. Thirdly, he gently put in the two masses and the Crown, one after the other, into the prepared vessel, taking exact notice of the quantity of water that issued out of the vessel at each time, and concluding from thence, that the Goldsmith had mingled 6 pounds and ⅔ parts of silver. We will suppose then that the mass of gold weighing 100 pounds, did cast out of the vessel 60 pounds of water, & that the mass of silver also weighing 100 pounds, cast forth of the vessel 90 pounds of water, and that the Crown cast forth 65 pounds. I put afterward for the silver mixed in the Crown 1R, and constitute twice the Rule of three after this manner. If 100 pounds of gold give me 60 pounds of water, how much will 100 − 1. R? and I find 6000 − 60R/100 for my fourth number. Secondly, if 100 pounds of silver give me 90 pounds of water, how much shall 1R? and I find 90R/100. Now these pounds of water 6000 − 60R/100 and 90R/100 added together do make 6000+30R/100 pounds of water cast out, which ought to be equal to 65 pounds of water, cast forth by the Crown, and therefore if we reduce them, we shall find 6000+30R equal to 6500 (this reduction is made by multiplying the denominator 100 by 65, for seeing that this fraction 6000+30R/100 is equal to 65 it shall be also equal to 65/1, Mark well this kind of reduction once for all. and therefore there will be the same proportion of the numerator 6000+30R to the denominator 100, as of the second numerator 65 to 1. Therefore the product under the extremes 6000+30R is equal to the product of the means 6500) take away therefore from both parts 6000, and there will remain an equation between 500 and 30 roots, and therefore divide 500 by 30, the number of the greatest character, you shall have the value of the root 16⅔ for the pounds of silver mingled by the Goldsmith in the Crown: Sect. II. Questions resolved by an Equation compounded. Question I. TO divide 8 into two such numbers as their squares being added together, may make 34. I put for the first 1R, therefore the second shall be 8 − 1R, their squares are 1Q, and 64+1Q − 16R, which added together, do give for their sum 64+2Q − 16R, the question imports that the sum of the squares is 34. Therefore there is an Equation between 64+2Q − 16R, and 34, which being reduced by addition and subtraction, there will remain also an equation between 2Q and 16R − 30, and the whole divided by 2, which is the number of the greatest Cossick Character, there will yet remain an Equation between 1Q and 8R − 15, from which I extract the root, as hath been shown in the fifth Section of the second Chapter. The half of the roots is 4, his square 16, from which take the absolute number 15, rest 1, whose square root 1 added to the half of the number of roots, gives for its sum 5, which is the value of the root; therefore the two numbers sought shall be 5 and 3. Question II. To find two numbers whose product may be 12, and the difference of their squares 32, I put for the one of them 1R; therefore seeing that the product is 12, the other number shall be 12/1R (for if the product of two numbers be divided by one of those two numbers, the quotient shall be the other number) their squares are 1Q and 144/1q whose difference is 144/19 − 1Q equal to 32, as appears by the question; therefore there will be an equality between 144/1q and 32+1Q, & therefore if we make the reduction, as in the third Question of the first Section, we shall also find an Equation between 144 and 32Q+1QQ, also between 144 − 3Q and 1QQ; it behoveth then to extract the square root of the number 144 − 32Q. The half of 32 is 16, whose square is 256, to which add 144, makes 400, whose square root is 20, from which take the half of 32, to wit 16, there remains 4. See here the square root but seeing that the squared square root ought to be taken, I take again the root of 4, and I find 2 for the value of the root. Therefore seeing that the second number hath been put 12/2R the same second number shall be 12/2, that is to say 6. Question III. Two Merchants join in company, and together bring in the sum of 165 Crowns; but the first man's money hath been exposed twelve months entire, and the second man's money only eight months: it happens that they gain but 28 Crowns, which added to 165 make 193, which they distribute to one another in such sort, as the first takes 67 Crowns, as well for his principal money as for his profit, and the second takes 126 Crowns, the question is what each of those Merchants brought into stock. I put for the money of the first man 1R, therefore seeing that the sum of both was 165 Crowns, the second man's money is 165 − 1R. Now if you take away 1R, which is the sum the first man brought in, from the sum he received, which was compounded of the principal and profit, you will find that the first man's profit will be 67 − 1R, and by the same argument you shall find, that the second man's profit will be 1R − 39 Now you must find what one root gaineth in eight months, which will be done by the Rule of Three, thus, If in 12 months there be gained 67 − 1R, how much will there be gained in 8 months, and the fourth number shall be 134/3 − 2R/3 for the first man's profit in 8 months; after that I seek what the second man hath gained by another operation of the rule of three, saying; If 1R gain 135/3 − 3R/2, what will 165 − 1R gain, and I find for my fourth number 7370+2/3Q − 464R/3,/ 1R which is equal to the second man's profit, which we have already found to be 1R − 39, and therefore by reduction there will be an equation between 1Q − 39R and 7370+2/3Q − 464/3R, all which 39R being added, and ⅔Q taken away, there will be an equation between 7370 and 1/3Q+347 3Q+347/3R and consequently between 7370 − 347/3R and ⅓Q, and therefore multiplying all by ⅓, which is the number of the greatest character, there will be yet an equation between 1Q and 22110 − 347R out of which the square root must be extracted. The half of the number of roots is 347/2, whose square is 170409/4, which added to 22110, make 208849/4, whose square root is 457/2, from which if you take half the number of roots, there will remain 110/2, that is to say, 55 for the value of one root, and was the money the first man put in bank; and for that we have found in the pursuit of these operations, that the first man's profit was 67 − 1R, that same profit will be 67 − 55, that is to say 12, by the same reason the second man's stock shall be 110, and his profit 16. Sect. III. Questions resolved by furred numbers. Question I. TO divide any given number (as for example 4) according to mean and extreme reason, that is to say, to divide 4 into two numbers, in such manner as that the whole 4 may bear the same proportion to its greatest part, as the greatest part bears to the least. I put for the greatest part 1R, therefore the least shall be 4 − 1R. Therefore there is the same proportion of 4 to 1R, as of 1R to 4 − 1R, and therefore the square of the middle part 1Q is equal to the product of the extremes 16 − 4R, from which I extract the root according to the rule before prescribed. The half of the number of roots is 3, whose square 4, added to 16, makes 20; out of which the square root ought to be extracted according to the precept: but seeing it is no square number, I must content myself by putting the radical sign before it thus, ℛ Q 20, from which I take the half of the number of roots, and I have for Residue ℛ Q 20 − 2, which is the value of the root, by which I shall find with facility, that the other part will be 6 − ℛ Q 30. For the proof of this operation, it behoveth that these two parts added together make 4, and that the lesser 6 − ℛ Q 20, being multiplied by 4, make the product equal to the square of the greater part ℛ Q 20 − 2. Question II. To divide 8 into two parts, between which, 2 may be a mean proportional. I put for the first part 1R, therefore the lesser shall be 8 − 1R, and seeing that 1R and 2, and 8 − 1R ought to be proportional, it behoveth that the square of 2, which is 4, be equal to the product of the extremes, which is 8 R − 1Q, and therefore after the reduction it will be found, that 1Q is equal to 8R − 4, whose square root is ℛ Q 12+4. For the one part of 8, and for the other part 4 − ℛ Q12, both the one and the other root do resolve the question, as you will find, if you take the pains to examine it. From this practice may be framed an universal Canon, which may serve for the resolution of an infinite number of Algebraical Problems, which may be conceived after this manner. The sum given, which containeth the two extremes, aught to be distributed into two equal parts, that the square of the half may be taken, from which the square of the mean proportional given, must be taken, and the square root of the Residue added, and taken from the half of the given sum, will show the two parts sought. As for example, I take the half of 8, which is 4, whose square is 16, from which I take 4, the square of 2, which is the given mean, and there remains 12, whose square root added to the same half, makes 4+ ℛ Q12▪ and taken from the same half, maketh 4 − ℛ Q 12. Question III. To divide any given number (as for example 4) into three numbers continually proportional, in such sort as that the squares of the extremes joined together, may be triple the square of the mean. I put 1R for the number of the middle part, then seeing all three ought to make the same sum of 4, we shall have for the sum of the extremes 4 − 1R. Now seeing that of three numbers continually proportional, the square of the sum of the extremes is equal to the square of the extremes, and to the double of the square of the middle part. I take the square of this sum 4 − 1R, which is 16+1Q − 8R, from which I take 2Q, which is the double of the square of the middle part, and there will remain 16 − 1Q − 8R for the sum of the squares of the extremes: Therefore seeing that the condition of the question requireth a triple proportion, there will be an equation between 16 − 1Q − 8R and 3Q, add therefore 1Q on both sides of the equation, and you shall have 4Q equal to 6 − 8R, and dividing the whole by 4, which is the number of the greatest character, the equation will be 1Q equal to 4 − 2R, whose square root is ℛ Q5 − 1 for the middle number sought, and the sum of the extremes shall be 5 − ℛ Q5, which being divided into two parts, by the Canon of the precedent question, in such sort as ℛ Q5 − 1, be the middle proportional, you will find that the extremes are 2 and 2 − ℛ Q5, therefore the three numbers are 2 and ℛ Q5 − 1, and 3 − ℛ Q5, which all together make 4, and are in continual proportion, and the squares of the extremes are triple the square of the mean. Sect. IU. Geometrical questions resolved by Algebra. Question I. THere is a piece of ground of a greater length than breadth, whose angles are right angles, and in a triple proportion, Before you attempt the resolution of such questions, you must draw their figures. and their squares taken together, are quintuple their sum. The sides, the diameter, and the capacity or superficies of that piece of ground is demanded, I put for the least side 1R, therefore seeing they are in a triple proportion, the other side shall be 3R, their squares shall be 1Q and 9Q, which added together make 10Q, which ought to be quintuple, the sum of the numbers. Now the sum of the numbers is 4R, and its quintuple 20 R, and by consequence, see here an equation between 10Q and 20R, which are two collateral characters, and therefore dividing 20 by 10, which is the number of the greatest character, you shall find 2 for the least side, therefore the greater side shall be 6. Therefore the superficies shall be 12, and the diameter ℛ Q40. Question II. There is an equilateral triangle, whose superficies is ℛ Q243. The side and perpendicular is demanded, supposing that the perpendicular of an equilateral triangle doth always divide the side into two equal parts, I put for the half of the side divided, 1R; therefore the side shall be 2R. Now seeing that in every equilateral triangle; the square of the side is equal to the square of the perpendicular joined to the square of the half of the side, which is 1Q of the square of the whole side, which is 4. I shall have 3Q for the square of the perpendicular, and so ℛ Q3Q shall be perpendicular, which multiplied by the half of the side, which is 1R (reducing it first to its square, because of the radical sign which is in the number multiplied) you shall have ℛ Q3QQ for the superficies of the triangle: therefore there will be an equation between ℛ Q243 and ℛ Q3QQ, and therefore there will be also an equation between their squares, which are 243 and 3QQ, and the whole being divided by 3, the number of the greatest character, there will be yet an equation between 81 and 1QQ. I extract therefore the squared square root of 81, and have 3 for the half of the side, 6 for the side, ℛ Q27 for the perpendicular, and ℛ Q243 for the superficies of the triangle. Question III. There is a Semicircle, whose Diameter is divided according to mean & extreme reason, on which there is raised a perpendicular produced to the circumference, and the lesser line which is drawn from the extremity of the diameter, to this point of the circumference, is ℛ Q20 − 2, the quantity of the diameter, and of its parts, and of this perpendicular is demanded. To resolve this question, it is presupposed that the greater part of the diameter shall be equal to the given line, as may with facility be Geometrically demonstrated. That being done, I put for the lesser part of the diameter 1R, therefore seeing that the other part is given ℛ Q20 − 2, the whole diameter shall be ℛ Q20 − 2+1R, which multiplied by 1R, giveth for the product ℛ Q20 − 2R+1Q, equal to the square of the given quantity, which is 24 − ℛ Q320, and by due transposition you shall find 1Q equal to 24 − ℛ Q320+2R − ℛ Q20, from which the square root ought to be extracted, taking exact notice that the particles which have the Cossick Characters may hold place with the number of roots. I consider therefore in this term of the equation, the number of roots which is 2 − ℛ Q 20, whereof I take the half, which is 1 − ℛ Q 5, to whose square 6 − ℛ Q 20, the absolute number 24 − ℛ Q 320, aught to be added. and the sum will be 30 − ℛ Q 500, whose square root ought to be extracted as from Apotomes, as hath been shown in the last Section of the fourth Chapter, that root is 5 − ℛ Q 5, which added to the half of the number of roots 1 − ℛ Q 5 gives for the sum 6 − ℛ Q 20, which is the value of 1 root, that is to say, of the lesser part of the diameter; and therefore if you add it to the greater part, you shall have 4 for the quantity of the whole diameter, from whence the perpendicular is easily known, provided the rules of Geometry be in any reasonable manner understood. Sect. 5. Questions resolved by the second Roots. Question I. THree men have amongst them a sum of money: The first saith to the second, If you deliver me the half of your money, I shall have 100 Crowns: The second saith to the third, If you deliver me ⅓ of your money, I shall have 100 Crowns: The third saith to the first, If you deliver me ¼ of your money, I shall have 100 Crowns. I demand how much money each one hath. I put for the first man's money 1 R of Crowns, and for the second man's money 1 A, and for the third man's money 1 B: therefore the first which hath 1 R with ½ of the second man's money shall have 1 R+½ A equal to 100, and by consequence ½ A shall be equal to 100 − 1 R, and multiplying the whole by 2, 1 A, shall be equal to 200 − 2 R. I begin again therefore the operation, and in stead of 1 A, I put for my second number 200 − 2 R. Now the question requires that the second man, (with ⅓ of the third man's) shall have 100; therefore there will be an equation between 200 − 2 R+⅓ B, and between 100, add to both parts of the equation 2 R, and take away 200, there will yet remain an equation between ⅓ B and 2 R − 100, and multiplying the whole by 3, you will have 1 B equal to 6 R − 300: That being found, I begin again the work, and in stead of 1 B, I put for the third number 6 R − 300▪ which added to ¼ of the first man's, makes 25/4 R − 300, which ought to be equal to 100, therefore if you add on both parts 300, there will be an equation between 25/4 R and 400, therefore if divide 400 by 25/4, you shall find 64 for one root; therefore the second, which had 200 − 2 R shall have 200 − 128, that is to say 72, and the third shall have 84. Those three numbers do perfectly satisfy all the conditions of the question. Question II. Two men divide between themselves three hundred Crowns in such sort, as that the second man's money divided by that of the first man's, makes ●/●. It is demanded, how much each of them hath. I put for the first man's money 1 R, and for that of the second 1 A, there is therefore an equation between 1 R+1S A, and 300, and therefore 1 A is equal to 300 − 1 R, therefore 300 − 1 R/ 1 R is equal to 3/2, and by consequence in cross multiplying these two fractions, I shall have 600 − 2 R equal to 3 R, therefore 600 shall be also equal to 5 R, and dividing 600 by 5, I shall find 120 for the first man's money, so shall the other have 180. Question III. To find two numbers whose product may be 10, and the sum of their squares 29; I put for the first 1 R, and for the other 1 A, the product is 1 R A, equal to 10; therefore in dividing the whole by 1 R, there will be an equation between 1 A and 10/1 R, and therefore I begin again the operation, and put for the first 1 R, and for the second 10/1 R, their squares are 1 Q+100/19 equal to 29, and after the reductions and extractions of the roots, I find 5 and 2 for my sought numbers. Sect. VI Questions resolved indefinitely. THat question is said to be resolved indefinitely, in which the numbers are demonstrated in such Algebraical terms, as do satisfy all the conditions of the question proposed. Question I. To divide 12 into four numbers Arithmetically and continually proportional, I presuppose that when there is four numbers in Arithmetical proportion, the sum of the extremes is always equal to the sum of the means: whence it follows, that in our question the sum of the extremes shall be 6, and the sum of the means also 6. I put for the second 1 R, therefore the third shall be 6 − 1 R, their difference is 2 R − 6, presupposing 1 R to be the greater number of the two; if therefore I add this difference to 1 R, I shall have 3 R − 6, and if I take it from 6 − 1 R, I shall have 12 − 3 R, and therefore the four numbers in continued Arithmetical proportion, shall be 3 R − 6 ‑ 1 R ‑ 6 − 1 R ‑ 12 − 3 R | and the question is indefinitely resolved, in pursuit of which you may take such a number as you please for the value of 1 R, provided nevertheless that you do admit of feigned numbers less than nothing: However, if you will have no other numbers than what are real, you ought to take the value of 1 R, beneath 4 and above 2, which will be easily understood by a little experience. Take for example 5/2 for the value of 1 R, therefore the first number, which is 3 R − 6, shall be 15/2 − 6, that is to say 3/2, the second shall be 5/2, the third 7/2, and the fourth 9/2, which added together make 12, and are in continued Arithmetical proportion. And so you may take an infinite number of others. Question II. A Vintner hath three sorts of wine, the first is worth 4, the second 6, and the third 10 pence the pint; of these three sorts of wine he desires to fill a vessel which contains 80 pints, which may be worth 8 pence the pint; I demand how many pints he ought to take of each sort. You ought here to consider, that the number 80 must be divided into three such numbers, as the first multiplied by 4, the second by 6, and the third by 10, the sums of the three products added together may make 640 (because that all the wine which shall be in the vessel to be filled will cost 640 pence, seeing that if one pint be worth 8 pence, 80 pints will be worth 640 pence) I put therefore for the third number 1 R, which multiplied by 10, maketh 10 R, which being taken from 640, doth leave for the residue 640 − 10 R, which is a number containing the first 4 times, and the second 6 times. On the contrary, seeing that the third hath been put 1 R, therefore 80 − 1 R shall make the sum of the first and the second, which multiplied by 4, will give 320 − 4 R, which being substracted from 640 − 10 R, will leave 320 − 6 R double to the second number, and therefore the second shall be 160 − 3 R. In like manner the same sum 80 − 1 R multiplied by 6, will produce 480 − 6 R; therefore if you take away 640 − 10 R, there will remain 4 R − 100L, double to the first, and therefore the first shall be 2 R − 80. See here the question resolved indefinitely, the first number is 2 R − 80, the second 160 − 3 R, and the third 1 R. The terms between which you ought to take the value of 1 R are 53⅓, and 40; if therefore you take 46 for the value of 1 R, you shall have 46 pints of wine of 10 pence the pint, 22 of 6, and 12 of that of 4 pence the pint. Here I entreat you to consider, A special Note. that it is impossible perfectly to understand the rule of Aligation, without the knowledge of Algebra: For if you propose this question to one skilful only in Arithmetic, he will give you for the 3 sought numbers, 40, 20, and 20. And if you tell him, that of the sort of wine of 4 pence the pint, you have but 16 pints, he will remain astonished; whereas by your question resolved indefinitely by Algebra, you will be able to give satisfaction to this condition in infinite kinds and manners. Question III. Two numbers are sought, which have 56 for the difference of their Cubes, and which added together make 6. I put for the difference of those numbers 1 R, and if from the difference of the two Cubes you take the Cube of the difference of the sides, dividing this residue by the triple of the difference of the sides, you have for the quotient the product of the sides, it follows that if from 56 you take 1 C, and that the residual 56 − 1 C be divided by the difference of the sides, which is 1 R, the quotient 56 − 1C/1R shall be triple, the product of the sides; and therefore if you divide this quotient by 3, you will have for the product of the sides 56 − 1C/3R, which is equivalent to 56/3R − 19/3, therefore if you divide 6, (which is the sum of your two sought numbers) into two parts, whose product may be 36/3R − 59/3, you will have the question resolved indefinitely. Now to attein this, I have given you a Canon in the second question of the third Section, the half of the sum 6 is 3, whose square is 9, from which if you take the product found, there will remain 9+19/3 − 56/3R, whose square root added, and taken from the half of the sum, gives for the resolution 3+ ℛ (9+19/3R − 56/3R) which shall be the greatest sought number, and 3 − ℛ Q (9+19/3 − 56/3R) which shall be the least. Now these two numbers resolves the question indefinitely, in such sort, that if you take two for the value of the root, you will find that your two sought numbers shall be 4 and 2, and every other number (taken for the value of 1 R) above 2, will resolve the question. Appendix. Questions in Algebra, most of which require the Rule of Three in their Operation. Question. I. A Merchant receives in exchange for 568 Crowns four kinds of money; of the first 7 makes one Crown, of the second 18, of the third 21; and of the fourth 28 make one Crown. Moreover, he received of each sort of money a like number. I demand, How much he received of each kind of money? Put 1 R for the quantity of each sorts of money, and then constitute the Rule of Three after this manner. If 7 Pieces of money be worth Crown 1 what is money 1 R I answer 1/7 R Crowns. If 18 Pieces of money be worth Crown 1 what is money 1 R I answer 1/12 R Crowns. If 21 Pieces of money be worth Crown 1 what is money 1 R I answer 1/21 R Crowns. If 8 Pieces of money be worth Crown 1 what is money 1 R I answer 1/28 R Crowns. For if 7 pieces of money make 1 Crown, 1 R of money will make 1/7 R of Crowns, and so of the rest. Now those Fractions described in the fourth place, do make together 7●/2●2 R of Crowns, equal to the number 568 Crowns. Divide therefore 568 by 71/252, and you have for the value of 1 R 2016, and so many pieces of each kind of money he received: which thus I prove. If 7 Pieces are worth 1 Crown what Pieces 2016 Answer 288 Crowns. If 18 Pieces are worth 1 Crown what Pieces 2016 Answer 112 Crowns. If 21 Pieces are worth 1 Crown what Pieces 2016 Answer 96 Crowns. If 28 Pieces are worth 1 Crown what Pieces 2016 Answer 72 Crowns. For 2016 pieces of the first kind of money do make 288 Crowns, and as many of the second kind make 112, as many of third kind make 96, and of the fourth kind 72, which added together do make 568 Crowns. Question II. A Certain man hath two measures of wine, the one worth 12 Crowns, the other 15. Now he desires of both these wines to fill another equal measure, whose worth may be 13 Crowns. I demand what part of each of those wines he must take, to fill the other to be worth that price? Put 1 R for the part of the measure of the worst wine, and for the part of the measure of the best wine 1 − 1 R, then work by the rule of 3 thus. Meas. Cro. Crowns Worst wine 1 12 1 R? maketh 12 R Best wine. 1 15 1 − 1 R? makes 15 − 15 R For if one measure of the worst wine be worth 12 Crowns, 1 R of one measure of the same wine will be worth 12 R of Crowns, & if one measure of the best wine be worth 15 Crowns 1 = 1 R of a measure of the same wine will be worth 15 − 15 R. Therefore 1 R measure of the worst wine and 1 − 1 R measure of the best will be worth 15 − 3 R Crowns, which ought to be equal to to 13 Crowns. Add therefore 3 R to each part of the equation, and the equation will be between 3 R+13 and 15; take away therefore 13 from both sides, and the equation will be between 3 R and 2. Divide therefore 2 by 3, and you shall have 2/3 for the value of 1 R, and so much aught to be taken of the measure of the worst wine, and ½ part of the measure of the best wine, which thus I prove by the Rule of Three. Meas. Cro. Crowns Worst wine 1 12 ⅔? worth 8 Best wine 1 15 ⅔? 5 13 Cr. For ⅔ of a measure of the worst wine is worth 8 Crowns, and ⅓ of the measure of the best wine is worth 5 Crowns, which added together make three. Question III. I Have a measure of wine worth ten Crowns; How much water must I mix with one measure, that a mixed (like) measure may be worth seven Crowns. Put for the measure of water 1 R, then frame the question by the rule of Three thus. Meas. Meas. Meas. Wine, water Crowns mixed Crowns 1+1 R 10 1? worth 10/1+1 R For if a measure of wine, together with 1 R of a measure of water, be worth 10 Crowns, one measure of wine and water mingled together will be worth 10/1+1 R & so the equation will be between 10/1+1 R and 7, which by cross multiplication is reduced to 10 and 7+7 R, take away 7 from each part of the equation, and it will be between 3 and 7 R; divide 3 by 7, and you have for 1 R, 3/7, and so much water ought to be mingled with a measure of wine, that a measure of the mixture may be worth seven Crowns, and is thus proved. Measure of wine & water Crowns Measure Crowns 1 3/7 10 1? worth 7 For if 1 of wine and 3/7 of water be worth 10 Crowns, 1 of that mixture is worth 7 Crowns. Question IU. THere are in a certain vessel 20 measures of wine, of which each of them are worth 12 Crowns the measure: Now this vessel is filled up with water, and then one measure of this mixture is worth 10 Crowns, I demand the content of the vessel. Put for the measures 20+1 R, then work by the rule of Three thus. Measures Measures Meas. of wine of water Cro. mixed Crowns 20+1 R 240 1? worth 240/20+1 R For if one measure of wine be worth twelve Crowns, a measure of wine 20+1 R measure of water together will be worth 240 Crowns: therefore one measure mixed will be worth 240/20+1 R Crowns, so the equation is found between 240/20+1 R and 10, which by cross multiplication is reduced to 240 and 200+10 R; take away 200 from both parts, and there will remain an equation between 40 and 10 R: Divide 40 by 10, and you have 4 for the price of the root, and so many measures of water were put into the vessel, and therefore the whole vessel contains 24 measures, thus proved: 24 measures wine and water worth 240 Crowns, 1 measure? worth 10 Crowns. Question V. TWo Letter-carriers belonging to two cities distant each from other 140 leagues set forth towards one another, at one and the same time; the one travels eight leagues a day, the other six. I demand on what day they shall meet together. Put 1 R for the day, then work by the rule of Three thus. Day's Leagues Day's Leagues 1 8 1 R? 8 R. 1 6 1 R? 6 R. The first therefore in 1 R of days shall travel 8 R of leagues, and the latter 6 R, and both of them together will have measured 14 R of leagues, that is 140 leagues. There is therefore an equation between 14 R and 140. Divide then 140 by 14, and you shall have 10 for the value of 1 R. So the tenth day finished, they met together, which thus I prove. Day's Leagues Day's Leagues 1 8 10? 80 1 6 10? 60 For the first in 10 days went 80 leagues, and the latter went 60 leagues, which both together make 140, the distance of the two cities from one another. Question VI. A Certain Merchant bought a quantity of wool, and another quantity of wax, which cost him together 124 Crowns. Now 100 pound weight of wool cost him 7 Crowns, and 100 pounds of wax cost him 14 Crowns; but the quantity of wool that he bought was double to the quantity of wax I demand how many pounds of each sort the Merchant bought? Put for the wax 1 R, and for the wool 2 R of pounds, then work by the rule of Three, thus. pounds' Crowns pounds' Crowns 100 wool 7 2 R wool? 14/1●0 R 100 wax 14 1 R wax? 14/106 R Therefore there will be an equation between 28/100 R of Crowns, and 124 Crowns. Divide therefore 124 by 28/100, and you shall have for the value of 1 R 442 6/7, and so many pounds of wax he bought, and of wool 885 5/7, which is thus proved. 100 pounds, wool 7 Crow. 885 pound. 5/7 62 Cr. 100 pounds, wax 14 Crow. 442 pound. 6/7 62 Cr. And so for wool and wax together he expended 124 Crowns. Question VII. A Certain man buys a number of els of Velvet, which he selleth again; he buys 5 els for 7 crowns, and sells 7 els for 11 crowns, and gained on the bargain 100 crowns. I demand how many els of Velvet he bought and sold in all. Put for the quantity of els 1 R, then work by the rule of Three thus. els Crowns els Crowns 5 7 1 R? 7/5 R 7 11 1 R? 1 1/7 R You see if 7/5R of crowns which he laid out be substracted from 1 1/7R of crowns which he receiceived, there will remain 6/35R of crowns for the profit. Therefore the equation will be between 6/35R and 100 R, divide 100 by 6/35, and you have 583⅓ for the value of 1 R, and so many els he bought and sold. Thus proved. els Crowns els Crowns 5 7 583⅓? 816⅔ 7 11 583⅓? 916⅔ By which you may perceive there is 100 crowns gotten. Question VIII. Certain man buys a number of els of Velvet, paying 11 crowns for 7 els: Now he sells the whole again after the rate of 5 els for 7 crowns, and lost 100 crowns by the bargain. I demand how many els he bought and sold in all. Put for the number of els 1 R, and then work by the rule of Three thus. els Crowns els Crowns 7 11 1R? 1 1/7R. 5 7 1R? 7/5R. Then if 7/5R of crowns which he received, be substracted from 1 1/7R of crowns which he expended, the loss will happen to be 6/35R of crowns. So the equation will be between 6/35R of crowns, and 100 crowns. Divide therefore 100 by 6/35, and you shall have for the value of 1 R, 583⅓ els, and so many els were bought and sold, which thus I prove. els Crowns 7 11 583⅓? 916⅔ 5 7 583⅓? 816⅔ Where you see he lost 100 crowns by the bargain. Question IX. A Man buys 100 pounds of wax for 17 crowns. I demand, how many pounds he must sell for one crown, that so on 102 crowns he may gain 18 crowns. Put 1R for the number of pounds, then work by the rule of Three thus framed. Crowns Pounds Crowns Pounds. 17 100 102 600 1 1 R 102+18? 120 R. For 102 crowns do give 600 pounds, and if for one crown there be given 1R of pounds there will be 1●0R of pounds given for 102+18, which said 120R are equal to 600 pounds, the quantity of the wax sold. Therefore the equation shall be between 120R of pounds and 600 pounds. Divide therefore 600 by 120, and so you have for the value of 1 R, 5 pounds, and so many pounds are sold for one crown; so as that in 600 pounds 18 crowns may be gained on 102, thus proved. pounds' Crowns pounds' Crowns 100 17 600? 102 5 1 600? 120 Where in the first example 600 pound made 102, in this it makes 120, that is 102+18. Question X. A Man buys 100 pounds of wax for 17 crowns, in disposing of which he loseth 18 crowns, on 102 crowns. I demand how many pounds he sold for one crown? Put for the number of pounds 1 R, then work by the rule of Three, thus constituted. Crowns Pounds Crowns Pounds 17 100 102? 600 1 1 R 102 − 18? 84 R. So there will be equation between 84R of pounds and 600 pounds. Divide therefore 600 by 84, and the value of 1R will be 7 pounds and 1/7 and so many pounds, he sold for one crown, and lost 18 crowns on 102 by the bargain, which thus I prove. pounds' Crowns pounds' Crowns 100 17 600? 102 7 1/7 1 600? 84 Where you see he laid out 102 crowns for 600 pounds, and in stead thereof received but 84 crowns, that is 102 − 18 crowns. Question XI. A Certain man agrees with a servant for 12 month's service, for ten crowns and a coat; but at the end of seven months, he gives him the coat and two crowns. I demand then at what rate he esteemed the coat. Put for the price of the coat 1R of crowns, and say by the rule of three. If 12 months require 1R+10 crowns, how much will one month require? and you shall find that it will require 1R+10/12 of crowns: again say, If 7 months require 1R+2 crowns, how much will one month require, and you shall find it to be 1R+2/7 as here under appears. Month's Crowns Month Crowns 12 1R+10 1? 1R+10/12 7 1R+2 1? 1R+2/7 Therefore there will be an equation between 1R+10/12 and 1R+2/7 seeing both are the reward of one month, which equation by cross multiplication is reduced to 7R+70 and 12R+24, take away 24 from both parts, and the equation will be between 7R+46 and 12R. Again, take away 7 from both parts, than it will be 46, equal to 5R; divide 46 by 5 and the price of a root, and so of the coat is 9⅕ crowns, as appears hereunder. The reward of 12 months is 19⅕, and of 7, 11⅕. Month's Crowns Month's Crowns 12 19⅕ 7? 11⅕ Question XII. A Certain Citizen agrees with a slothful servant for 30 days, that every day he wrought he would give him 7 Groats; but for every day that he idled, and wrought not, he was to allow his master five Groats: when the 30 days were passed, it happens that the servant was to receive nothing from his master, nor the master from the servant. I demand then, how many days he laboured, and how many he idled? Put 1R for the days of labour, and 30 − 1R for the days of idleness, and then frame the rule of Three thus. Day Groats Days Groats Labour 1 7 1R? 7R Idleness 1 5 30 − 1R? 150 − 5R Now seeing that his work and play came to one reckoning, there will be an equation between 7R and 150 − 5R, add 5R to each part, and the equation will be between 12R and 150. Divide therefore 150 by 12, and you have for the value of 1R, 12½, and so many days he laboured, and 17 day's ½ he idled, which is thus proved. Day Groats Days Groats Labour 1 7 12½ 87½ Idleness 1 5 17½ 87½ Where you see the reward is the same with the mulct. Question XIII. ONe sells 20 pound weight, part Saffron and part Ginger, for 45 crowns; but he sold 1 pound of Saffron for 3 crowns, and 1 pound of Ginger for ½ a crown. The question is, how many pounds of each sort he sold? Put for the Saffron 1R of pounds, and for the Ginger 20 − 1R of pounds, then by the rule of Three work thus. pounds' Crowns pounds' Crowns Saffron 1 3 1R? 3R Ginger. 1 ½ 20 − 1R? 20 − 1R/2 Therefore the equation is between the sum of 3R of crowns, and 20 − 1R/2 of crowns added together, and 45 crowns, now that sum is 10+5/2R of crowns (for 20 − 1R/2 is equal to 10 − ½R, to which if you add 3R of crowns, it makes the sum 10+5/2R) therefore the equation shall be between 10+5/2R of crowns, and 45 crowns. Take away 10 from both parts, and it will be between 5/2R and 35. Divide therefore 35 by 5/2, and you shall have 14 for the value of 1 R. And so many pounds of Saffron were sold, and 6 pound Ginger, which thus I prove. Pound Crowns Pounds Crowns Saffron 1 3 14? 42 Ginger 1 ½ 6? 3 Where you see that the price of 14 pound Saffron, and 6 pound Ginger added make 45 crowns. Question XIV. A Certain Tradesman hath 2 sorts of coin, in number 560 pieces, worth 160 crowns; a certain part thereof is worth each piece ⅓ of a crown, and each piece of the rest ¼ of a crown. I demand the number of the first and latter sort of money? Put 1R for the first, and 560 − 1R for the latter, and then constitute the rule of Three after this manner. Money Crown Money 1 ⅓ 1R? ⅓R 1 ¼ 560 − 1R? 560 − 1R/4 The fourth number found is equal to 160 crowns, and the sum of their numbers makes 140+1/12R (for 560 − 1R/4 is equal to 140+¼R, to which if you add ⅓R, makes the sum 140 +1/12R.) There is therefore an equation between 140+1/12R, and 160 crowns. Take away 140 from both parts, and then the equation is between 1/12 and 20. Divide therefore 20 by 1/12, and you have 240 for the value of 1 R, and so much money there was of the first sort, of which each piece was worth ⅓ of a crown of the latter kind 320, each worth ¼ of a crown. Thus proved. Money Money Crowns 1 ⅓ 240? 80 1 ¼ 320? 80 Where you see the numbers in the fourth place make 160 crowns. Question XV. IN the Army of the Emperor, the number of the Infantry were octuple to the number of the Cavalry, among them there is distributed 392000 crowns, so as that every Foot Soldier had 5 crowns, and every Horseman 16. The question is, Of how many Horsemen the Army consisted, and of how many Footmen. Put 1R for the Horsemen, and 8R for the Foot, according to the condition of the question, and then constitute the rule of Three thus: Crowns Horsemen Crowns Horse 1 16 1R 16R Foot 1 5 Foot 8R 40R 56R Therefore 56R of crowns shall be equal to 392000 crowns: wherefore divide 392000 by 56, and you shall have for 1R 7000 for the number of Horsemen; therefore the Foot shall be 56000, eight times as many, and so there will be distributed to the Horsemen 112000 crowns, and to the Foot 280000, which together make 392000 crowns. Question XVI. A Man hath a certain sum of money in a purse, which a slander by judgeth to be 600 crowns, whose error he thus corrects. If to what I have in this purse, there be added ½ ⅓ and ¼, and from the sum there be substracted 1/12 part of my money, than I should have 600 crowns. The question is, How many crowns he had in the purse! Put 1R for the number of crowns. If the parts ½R, ⅓R, and ¼R, which together make 1 1/12R, be added to 1R, the whole makes 2 1/12R; take away 1/12R, and there will remain 2R equal to 600. Divide therefore 600 by 2, and you have 300 for the value of 1R, and so much money was in the purse. For if you add the ½, ⅓, and ¼, to wit 150, 100, and 75, it will make the sum of 625, and taking away 1/12, to wit 25, the rest will be the number 600, which resolves the question. Question XVII. Certain Traveller goes 9 miles a day, another Traveller after the tenth day past, gins his journey from the same place, and goes every day 14 miles. I demand in how many days the latter will overtake the first. Put 1R for the number of days, therefore the first over and above 90 miles, which he hath gone in ten days, goeth in 1R of days, besides 9R of miles, seeing that every day he goeth nine miles. But the latter going 14 miles each day, goeth in 1R of days 14R of miles, and because the first of necessity in 1R of days goes as many miles together with 90, which he went in ten days as the latter went in 1R of days; sigh that then they are to meet together, the equation shall be between 9R+90 and 14R. Take away 90 from both parts, and it will be between 90 and 5R. Divide therefore 90 by 5, and 1R makes 18. Therefore in 18 days they shall come together. For the first in 18 days went 162 miles, which added to 90, he made the first 10 days, makes 252 miles, which the latter went in 18 days. Question XVIII. A Traveller goes nine miles a day, another Traveller after the end of ten days gins the same journey. I demand, how many miles a day the latter aught to travel, that so in 18 days he may overtake the first. Put 1R for the miles. Therefore in 18 days he will have traveled 18R of miles. Seeing therefore that the first travelling every day nine miles, went in 18 days 162 miles, and adding thereto 90, which he went the first ten days before the second set forth, it is manifest that he traveled 252 miles. Therefore the equation is between 18R and 252. Divide 252 by 18, and you have 14 for the value of 1R, and so many miles the latter aught to travel, to overtake the first in 18 days. Question XIX. A Certain man dying, made his will and testament, leaving 3000 crowns to be distributed between his Wife, Son, and two Daughters, on this condition, that the portion of the Son might be double to that of the Mother, and the portion of the Mother double also to the portion of each of the Daughters. The question is, how much each one's portion was? Put for the portion of one of the Daughter's 1R, for the Mother's portion 2R, and for the Son's portion 4R. So there will be an equation between 8R and 3000 crowns. The portion of one Daughter 1R that is 375 Cro. The portion of th'other Daughter 1R that is 375 Cro. The portion of the Mother 2R that is 750 Cro. The portion of the Son 4R that is 1500 Cro. Divide 3000 by 8, so the value of 1R will be 375, the portion of one of the Daughters, and therefore the Mother's portion will be 750, & the Sons 1500. Question XX. A Certain man receiveth of a Merchant, a quantity of Saffron for 10 crowns; and again he receives of the same man 24 pounds more of Saffron, at length he returns to him 30 pounds thereof again, and the merchant computing the price of the Saffron, restoreth to him 14 crowns. I demand the price of a pound of that Saffron. Here you see 10 crowns +24 pounds to be the whole debt which the buyer owed to the merchant, & in like manner 30 pounds − 14 crowns. Therefore there will be an equation between 10 crowns +24 pounds, and 30 pounds − 14 crowns. Add 14 crowns on both parts, and the equation will be between 24 crowns +24 pounds and 30 pounds, take away 24 pounds from both parts, and it will be between 24 and 6 pounds. Divide 24 by 6, and you have 4 for the root, and so many crowns one pound of Saffron is worth, which I thus prove. 2½ pounds are bought for 10 crowns, and so the buyer received of the merchant 26½, which were worth 106 crowns. If therefore to the merchant there be restored 30 pounds, the merchant oweth to the buyer 3½, seeing he received only 26½, but 3½ pounds are worth 14 crowns, which the merchant rendered to the buyer. Question XXI. TWo men enter into fellowship in trade, now the second brings with him double the money that the first brings, and 5 crowns over and above. They gain by their traffic 960 crowns, of which the first takes to himself 300 crowns, and the second 660. I demand how much each put in bank. Put for the first 1R, and for the second 2R +5, the sum of both together is 3R +5, which have gained 960 crowns. Then work by the rule of three thus: 3R+5 960 1R? 960R/3R+5 You shall find that the first which brought in 1R, hath gained 960R/3R+5 which number is equal to 300 crowns which he received. This equation by cross multiplying will be reduced to 960R, equal to 900R+1500. Take away therefore 900R from both parts, and there will Remain 60R equal to 1500, divide 1500 by 60, and you have 25 for one R, and so much the first put in bank, therefore the second put in 55 crowns, and is thus proved. For both put in 80 crowns. 80. 960 25? 300 80. 960 55? 660 Question XXII. THree Merchants gain together 700 crowns, which thus they distribute amongst themselves (having regard to the sum each one brought into bank) so as that the portion of the second surmounted the portion of the first by 12 crowns, and the portion of the third surpassed that of the second by 16 crowns. I demand how much each man's portion was. Put for the portion of the first man 1R, and then shall the portion of the second be 1R+12, and the portion of the third 1R+28. These 3 portions together make 3R+40, equal to 700. Take away 40 from both parts of the equation, and the equation will be between 3R and 660. Divide therefore 660 by 3, and you shall have 2●0 for the value of 1R, and so much was the first man's portion, so the portion of the second shall be 232, and of the third 248, which all together make 700. Question XXIII. A Caterer buys a number of Fowls, so as that if he had bought the ⅓ ¼, and ⅕ of the number, and over and above 6, he would then have 100 just. I demand the number of Fowls he bought. Put 1R for the number, whose ⅓, ¼, and ⅕ make 47/60R, which with 6, make 47/60R+6 equal to 100 Take away 6 from both parts of the equation, and the remaining equation will be between 47/60R and 94. Divide therefore 94 by 47/60, and the value of 1R will be 120, to wit, the number of Fowls that were bought. For its ⅓ is 40, and its ¼ is 30, and its ⅓, 24; which all together with 6, make the sum 100 Some Examples in Algebra concerning Squares. To find two numbers in a given Excess, so as their Squares may have also a given Excess. LEt there be sought two numbers, whose Excess is 4, and the Excess of their Squares 144. Put for the lesser number 1R, and therefore the greater number must be 1R+4, whose squares are 1Q and 1Q+8R+16, their excess is 8R+16, which ought to be equal to 144. Take away 16 from both parts of the equation, and the equation will be between 8R and 128. Divide therefore 128 by 8, and you shall have for the value of 1R, 16 for the lesser number, the greater therefore will be 20, that the Excess may be 4. The squares of those 2 numbers are 256, and 400, whose difference or excess is 144. Two numbers being given, to find another, with which multiplying both the given numbers, makes the greater number a square, and its lesser the side of that square. LEt the two given numbers be 200 & 5, and let the sought number be put 1R. Now 1R multiplied in 200, produceth 200R, and 1R multiplied in 5, makes 5R, which ought to be the side, and so multiplied in itself, aught to make a number equal to 200. But 5R multiplied in itself, makes 25Q, the equation therefore is between 200R and 25Q. Divide therefore 200 by 25, and the value of 1R is 8, which multiplied in 200, make the square 1600, whose side is the number 40, which is also produced by the multiplication of 8 in 5. Some Examples relating to Cubes. To find a number, which multiplied in itself, and the product multiplied by some given number, may produce a number in a given proportion to the Cube of the found number. LEt the given number be 20, find another number, which being first multiplied in itself, and then the product multiplied by the given number 20, may produce a number in a quintuple proportion to the Cube made of the found number. Put for the sought number 1R, which multiplied in its self makes 1Q, which also multiplied by 20, makes 20Q, the Cube of 1R is 1C, to which 20Q ought to have a quintuple proportion; so the equation is between 20Q and 5C: Divide 20 by 5, and you have for the value of 1R, 4, the number songht. This 4 multiplied in itself makes 16, and 16 multiplied by 20, makes 320, which is quintuple to 64, the Cube of 4. To divide a given number in two parts, so as that their Cubes may make a given sum, which shall be greater than the quarter part of the Cube described of the given number. LEt the given number be 10, to be divided into two parts, whose Cubes may make 370, which number is greater than the ¼ part of the Cube of 10. Put for the first number compounded of 1R, and half of the given number 1R+5, and for the second 5 − 1R. So these two numbers do make the given number. Their Cubes are 1C+15Q+75R+125, and 125 − 75R+15Q − 1C; their sum is 30Q+250: For +1C, and − 1C, as also +75R, and − 75R, do mutually destroy one another: and of 15Q, and 15Q, are made 30Q; also 125, & 125, make 250. Therefore the equation is between 30Q+250, and 370. Take away 250 from both parts, and the equation will be between 30Q and 120. Divide 120 by 30, and you have 4 for the value of 1Q, and the value of 1R, 2. The first part put 1R+5 shall be 7; and the second put 5 − 1R shall be 3. The Cubes of those parts are 343 and 27, which together make 370. FINIS.