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- ELEMENTS OF ALGEBRA
PRELIMINARY TO THE DIFFERENTIAL CALCULUS
AND FIT FOR THE HIGHER CLASSES OF SCHOOLS IN WHICH
THE PRINCIPLES OF ARITHMETIC ARE TAUGHT.
BY
AUGUSTUS DEMORGAN,
OF TRINITY COLLEGE, CAMBRIDGE,
AND PROFESSOR OF MATHEMATICS IN UNIVERSITY COLLEGE, LONDON.
SECOND EDITION.
Ghat a benefite that onelp thong is, to haue the Mitte whettey anv
sharpened, I neave not trauell to Declare, sith all men confesse it to be ag
qreate as mate he. Bxcepte an toitlesse persone thinke he mate bee to wise,
But he that moste feareth that, is Teaste tn Daunger of tt. Gherefore io
conclude, 1 see moare menne to acknowledge the benefite of nomber, than I
can espie willpng to studic, to attaine the benefites of ft. gelann praise it,
but fete Door greatly practise it: onlesse it bee for the bulgare practice,
concernpng MMerchaundes trade. Meiherein the Vesire and hope of gain,
maketh manp willpng to sustaine some trauell. for aide of whom, I viv
Sette forth the firste parte of Arithmetike. Wut tf thet knetwe How farre thts
Seconve parte, Dooeth excell the firste parte, thet would not accoumpte anp
tyme loste, that were imploied in it. Yea thet mouly not thinke any tyme
foell bestowed, till thet had gotten soche habilitic dp tt, that it might be their
aide in al other studics.”— ROBERT RECORDE.
LONDON:
PRINTED FOR TAYLOR AND WALTON,
BOOKSELLERS AND PUBLISHERS TO THE UNIVERSITY OF LONDON,
28 UPPER GOWER STREET.
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PREFACE
TO THE
FIRST EDITION.
In the title-page I have endeavoured to make it clear that
it will be impossible to teach algebra on the usual plan by
means of this work. It is intended only for such students as
have that sort of knowledge of the principles of arithmetic
which comes by demonstration, and whose reasoning faculties
have therefore already undergone some training.
Algebra, as an art, can be of no use to any one in the
ce. of life; certainly not as taught in schools. I appeal
to every man who has been through the school routine
whether this be not the case. | Taught as an art it is of little
use in the higher mathematics, as those are made to feel who
attempt to study the differential calculus without knowing
more of its principles than is contained in books of rules.
The setence of algebra, independently of any of its uses,
has all the advantages which belong to mathematics in
general as an object of study, and which it is not necessary
to enumerate. Viewed either as a science of quantity, or
as a language of symbols, it may be made of the greatest
service to those who are sufficiently acquainted with arith-
metic, and have sufficient power of comprehension, to enter
fairly upon its difficulties. But if, to meet the argument that
boys cannot learn algebra in its widest form, it be proposed
to evade the real and efficient part of the science, whether by
presenting results only in the form of rules, or by omitting
and taking for granted what should be inserted and proved,
557114
4 PREFACE.
for the purpose of making it appear that something called
algebra has been learned: I reply, that it is by no means
necessary, except for show, that the word algebra should find
a place in the list of studies of a school; that, after all, the
only question is, whether what is taught under that name be
worth the learning; and that if real algebra, such as will be
at once an exercise of reasoning, and a useful preliminary to
subsequent studies, be too difficult, it must be deferred.
Of this I am quite sure, that the student who has no more
knowledge of arithmetic —that is, of the reasoning on which
arithmetical notation and processes are built—than usually
falls to the lot of those who begin algebra at school—that is,
I believe, begin to add posztive and negative quantities to-
gether,— will sooner find his way barefoot to Jerusalem than
understand the greater part of this work. And I may say the
same of every work on algebra, containing reasoning and not
rules, which I have ever seen; provided it contained any of
the branches of the subject which are of most usual appli-
cation in the higher parts of mathematics.
fe The special object to which this work is devoted is the
developement of such parts of algebra as are absolutely
requisite for the study of the differential calculus, the most
important of all its applications. / The former science is now
so extensive, that some particu ar line must be marked out
by every writer of a small treatise. The very great difficulty
of the differential calculus has always been a subject of
complaint; and it has frequently been observed that no
one knows exactly what he is doing in that science until
he has made considerable progress in the mechanism of its
operations. I have long believed the reason of this to be
that the fundamental notions of the differential calculus are
conventionally, and with difficulty, excluded from algebra, in
which I think they ought to occupy an early and prominent
place. I have, therefore, without any attention to the agree-
ment by which the theory of limits is never suffered to make
v!
PREFACE. 5
its appearance in form until the commencement of the dif-
ferential calculus, introduced limits throughout my work:
and I can certainly assure the student, that, though I have
perhaps thereby increased the difficulty of the subject, the
additional quantity of thought and trouble is but a small
dividend upon that which he would afterwards have had to
encounter, if he had been permitted to defer the considera-
tions alluded to till a later period of his mathematical course.
On those who offer theoretical objections to the introduction
of limits in a work on algebra lies the onus of shewing that
they are not already introduced, even in arithmetic. What
is V2, supposing geometry and limits both excluded ?
I have been sparing of examples for practice in the
earlier part of the work, and this because I have always
found that manufactured instances do not resemble the
combinations which actually occur. They are but a sort of
parade exercise, which cannot be made to include the means
of meeting the thousand contingencies of actual service.
The only method of furnishing useful cases is to take some
inverse process, and the verification of literal equations (as
in the seventh and following pages of this work) is the most
obvious. With these the student can furnish himself at
pleasure, the test of correctness being the ultimate agreement
of the two sides, after the value of the unknown quantity
has been substituted.
The only remaining caution which he will need is, not
to proceed too quickly, especially in the earlier part of the
work. He must remember that he is engaged upon a very
difficult subject, and that if he does not find it so, it is, most
probably, because he does not understand what he is about.
Wherever an instance or a process occurs, he should take
others as like them as he can, and assure himself, by the
reasoning in the work, that he has obtained a true result.
It was at first my intention to write a second volume on
the higher parts of the subject. But, considering that of
|
i
6 PREFACE.
these there are several distinct branches, it has appeared to
me that the convenience of different classes of students will
be consulted. by publishing them in several distinct tracts,
which may afterwards be bound together, if desired.
AUGUSTUS DE MORGAN.
London,
July 31st, 1835.
*,%* This Second Edition differs from the first only in
verbal amendments. Since the publication of the first edi-
tion, I have carried on the consideration of the principles of
algebra in two other works. In the first, or Elements of
Trigonometry, besides the consideration of periodic mag-
nitudes in general, I have (Chapter IV.) given a view of
that extension of the meaning of symbols which must ac-
company the complete explanation of the negative square
root. In the second, or Connexion of Number and Mag-
nitude, which is an attempt to explain the Fifth Book of
Euclid, I have entered upon what is in reality the most
difficult part of the application of arithmetic to geometry.
Both of the preceding works may be made supplementary to
the present one, though the latter is altogether independent
of it.
University College, London,
October 16, 1837.
!
CONTENTS.
PREFACE.
PeeteeCr UGE LO Ngee ker %, se 6! ~ js (©. -¢ 6° “© te (es © |e «
CHAPTER I.
EQUATIONS OF THE FIRST DEGREE Ad Vat Cie oS SOU ope OAS
CHAPTER II.
ON THE SYMBOLS OF ALGEBRA, AS DISTINGUISHED FROM
ME OR ARITHMETIC .. 6 2 Se 8s 6 8 wes
CHAPTER III.
EQUATIONS OF THE FIRST DEGREE WITH MORE THAN ONE
PNONOONEN GANLEY oe “so ef ie We 0 ‘ese “es 1 © e« e
CHAPTER IV.
ON EXPONENTS, AND ON THE CONTINUITY OF ALGEBRAIC
CPUS STON MMH) 8 @ “0 «je. 6 le wii6 “ss © “#« %© a s
CHAPTER V.
GENERAL THEORY OF EXPRESSIONS OF THE FIRST AND
SECOND DEGREES; INCLUDING THE NUMERICAL SOLU-
TION OF EQUATIONS OF THE SECOND DEGREE"... .
CHAPTER VI.
ON LIMITS AND VARIABLE QUANTITIES . -« «© «© « « «@
CHAPTER VII.
CLASSIFICATION OF ALGEBRAICAL EXPRESSIONS AND CON-
SEQUENCES. RULE OF DIVISION SMe soe fi 5 6. Set foes
44
124
151
166
8 CONTENTS.
CHAPTER VIII.
ON SERIES AND INDETERMINATE COEFFICIENTS .. . .
CHAPTER IX.
ON THE MEANING OF EQUALITY IN ALGEBRA, AS DISTIN-
GUISHED FROM ITS MEANING IN ARITHMETIC EF has
CHAPTER X.
ON THE NOTATION OF FUNCTIONS (6 eee eres
CHAPTER XI.
ON THE BINOMIAL THEOREM . « «© «© “© 6) 6 seemuetGnS
CHAPTER XII.
EXPONENTIAL AND LOGARITHMIC SERIES eC YO te OO
CHAPTER XIII.
ON THE USE OF LOGARITHMS IN FACILITATING COMPU-
TATIONS e . . e e ° e ° e e e e e * e ° e
207
222
ELEMENTS OF ALGEBRA.
INTRODUCTION.
Iv is taken for granted that the student who attempts to read this
work has a good knowledge of arithmetic, particularly of common
_and decimal fractions. Whoever does not know so much had better
begin by acquiring it, as the shortest road to algebra.*
In arithmetic, we use symbols of number. A symbol is any sign
‘for a quantity which is not the quantity itself. Ifa man counted his
sheep by pebbles, the pebbles would be symbols of the sheep. Our
symbols are marks upon paper, of which the meaning of every one is
determined so soon as the meaning of 1 is determined. If we are
speaking of length, we choose a certain length, any we please, and
callit1. It may have any other name in common life, for instance,
a foot or a mile, but in arithmetic, when we are numbering by means
of it, it is 1. We now introduce the sign +, and agree that when
we write + between two symbols of quantity, it shall be the symbol
for the quantity made by putting these two quantities together.
Thus, if
1 stand for the length
1 + 1 stands for the length
1 +1 is abbreviated into 2, a new and arbitrary + symbol. Similarly
2 +1 is abbreviated into 3, 3+ 1 into 4, and so on.
* The references in this work are to the articles (not pages) of my
Treatise on Arithmetic, and serve either for the second or third editions.
Arbitrary, that is, any other would have done as well. It is 2 that
stands for 1+1, and not <, 3, 00, or any thing else, because certain
Hindoos chose that it should be so. See Penny Cyclopedia, Art.
ARITHMETIC.
|
ree INTRODUCTION.
When 1, 2, 3, &c., mean 1 mile, 2 miles, 3 miles, &c., or 1 pint,
2 pints, 3 pints, &c., these are called concrete numbers. But when
we shake off all idea of 1, 2, &c., meaning one, two, &c., of any thing
in particular, as when we say, “‘ six and four make ten,” then the
numbers are called abstract numbers. To the latter the learner is first
introduced, in regular treatises on arithmetic, and does not always
learn to distinguish rightly between the two. How many of the
operations of arithmetic can be performed with concrete numbers,
and without speaking of more than one sort of 1? Only addition and
subtraction. Miles can be added to miles, or taken from miles.
Multiplication involves a new sort of 1, 2, 3, &c., standing for repe-
titions or times, as they are called. Take 6 miles 5 times. Here are
two kinds of units, 1 mile and 1 time. In multiplication, one of the
units must be a number of repetitions or times, and to talk of multi-
plying 6 feet by 3 feet, would be absurd.* What notion can be
formed of 6 feet taken “ 3 feet”’ times ?
But in solving the following question, ‘‘ If 1 yard cost 5 shillings,
how much will 12 yards cost?” do we not multiply the twelve yards
by the five shillings? Certainly not—the process we go through is the
following : Since each yard costs five shillings, the buyer must put
down 5 shillings as often (as many dimes) as the seller uses a one-yard
measure; that is, 5 shillings is taken 12 times.
In division, we must have the idea either of repetition or of
partition, that is, of cutting a quantity into a number of equal parts.
“Divide 18 miles by 3 miles,” means, find out how many times
3 miles must be repeated to give 18 miles: but “ Divide 18 miles
by 3,” means, cut 18 miles into three equal parts, and find how many
miles are in each part.
18 miles divided by 3 miles gives 6; meaning, that 3 miles must
be repeated six times to give 18 miles.
18 miles divided by 3 gives 6 miles ; meaning, that if 18 miles be
cut into three egual parts, each part is 6 miles. The answer in ab-
stract numbers is the same in both cases; 18 divided by 3 gives 6. ,
But now we ask, How many times does 12 feet contain 8 feet?
* In old books the following is sometimes found. ‘‘ What is
£99. 19s. 113d. multiplied by £99. 19s. 11$d.2” The only intelligible ©
meaning of this is as follows: Ifa stock of money is to be increased at
the rate of £99. 19s. 118d. for every £1 in it, how much will that be ©
when the stock itself is £99.19s.113d.? Let the student answer this.
See
INTRODUCTION. iil
The answer is, more than once and less than twice; which is not
complete, because we have not an adequate idea of parts of times,
that is, parts of repetitions. In talking of times, we use a figure of
speech which we may liken to a machine which works by starts, each
start doing, say eight feet of work, and which is so contrived that
nothing less than a whole start can be got from it; either 8 feet or
nothing. It is plain that such a machine cannot execute 12 feet of
work, or any thing between 8 feet and 16 feet. But, let us now sup-
pose the machine to be made to work regularly, at 8 feet a minute.
Let us still continue to call 8 feet a start, then 12 feet must be called
a start and a half. In the same way we say that 12 feet contains
8 feet a time and a half, the notion of half a time being equivalent to
that of repeating not the whole 8, but its half,
When we speak of dividing one fraction by another in arithmetic,
this is what is meant; for instance, < divided by 2 gives coe or e
3 7 15 3
contains >, = ofatime. Let the learner study the following pro-
positions.
If 2 of £1 were gained in a day, then ; of a pound would be
: ahd EE:
gained in 75 of a day.
wh ers : 5.
If the signification of 1 be changed, so that what was 7 is now gs
: 14
then what rid is now —.
3 shea
If ; Sf the line A be the line B, then : Af tha line Avis a of the
line B.
The want of a proper comprehension of such questions as the
preceding is a great source of difficulty to most beginners in algebra.
If the preceding pages be not readily understood, it is a sign that the
reader is not sufficiently acquainted with arithmetic for his purpose in
reading this work.
The symbols of artrumetic have a determinate connexion ; for
instance, 4 is always 2-+2 whatever the things mentioned may be,
miles, feet, acres, &c. &c. In atceBra, we take symbols for num-
bers which have no determinate connexion. As in arithmetic we
draw conclusions about 1, 2, 3, &c., which are equally true of 1 foot,
2 feet, &c., 1 minute, 2 minutes, &c.; so in algebra we reason upon
1V INTRODUCTION.
numbers in general, and draw conclusions which are equally true of
all numbers. This, at least, is one great branch of algebra, and ex-—
hibits it in a view most proper for a beginner.
But this is a definition in a few words, and can only be understood
by those who have already studied the subject. No science can be
defined in a few words to one who is ignorant of it. We shall begin
by giving an instance of a general property of numbers and fractions.
Take (8) units,* and the fraction which, taken (8) times, gives a
unit, that is the (8)th part ofa unit. Add 1 to both; this gives 9 and 1%.
The first contains the second (8) times. Take (=) of a unit, and the
, : 2 on Pare : i.
fraction which taken (=) of a time gives a unit, or 1
5 Add 1 to
both, this gives 15 and 25. The first contains the second (@) of a
time. Try the following, in which it will be found that the blanks
may be filled up with any one number or fraction at pleasure.
Take (_ ) units, and the number or fraction which repeated (_ )
times gives a unit. Addi to both; then the first result will con-
tain the second (_) times, or parts of times.
The following are instances which should be tried.
Numberor fraction Number of times
: 1 added to | 1 added to ;
( ) | which repeated ( ) the first. .| theeenem which the third
times gives a unit. containsthe fourth.
7 : : lz 7
eC 3 1; 4 ‘
yi | |
= 20 1s 21 =
1 1 2 1
The connexion between the first and second columns is this, that
the number or fraction in the second is 1 divided by the number or
* By putting 8 in brackets, we wish to call attention to the circum-
stance of the numbers in the different brackets being the same,
INTRODUCTION. vi
| fraction in the first; or the number of times or parts of times which
1 contains the first. That is, if we call the number in the first column
‘¢ the number,”
then the number in the second column is
1 divided by “ the number.”
And the coincidence of the first and fifth columns (which constitutes
_ the thing we notice) may be thus expressed :
Let one more than “ the number” be divided by “ 1 more than
' the times which 1 contains the number,” and the result must be
“ the number.”
The above must be still further abbreviated for convenience. As
“the number” means any number we please, and as “ any number
we please” will be better expressed by any short symbol which we
may choose to make use of, let one of the letters of the alphabet be
employed, say a. Let the addition of a number be denoted by + as
_ before, and let the division of a number by a number be denoted, as
in arithmetic, by writing the divisor under the dividend with a line
_ between them. Let = be the sign that what goes before is the same
_ number as what comes after. Then the preceding property of num-
bers is thus expressed :
a+1
ee
We shall now proceed to lay down the definitions of the first part
of the science.
I. Atcresra is the European corruption of an Arabic phrase,
which may be thus written, al jebr e al mokabalah, meaning restor-
ation and reduction. The earliest work on the subject is that of
Diophantus, a Greek of Alexandria, who lived between a.p. 100 and
A.D. 400; but when, cannot be well settled, nor whether he invented
the science himself, or borrowed it from some Eastern work. It
was brought among the Mahometans by Mohammed ben Musa
(Mahomet, the son of Moses) between a.p. 800 and a.p. 850, and
was certainly derived by him from the Hindoos. The earliest work
which has yet been found among the latter nation, is called the Vija
Ganita, written in the Sanscrit language, about a.p. 1150. It was
introduced into Italy, from the Arabic work of Mohammed, just
b2
Vi INTRODUCTION.
mentioned, about the beginning of the thirteenth* century, by Leo-
nardo Bonacci, called Leonard of Pisa: and into England by a
physician, named Robert Recorde, in a book called the Whetstone
of Witte, published in the reign of Queen Mary, in 1557. From this
work the motto in the title-page is taken.
II. A letter denotes a number, which may be, according to
circumstances, as will hereafter appear, either any number we please ;
or some particular number which is not known, and which, therefore,
has a sign to represent it till it is known; or some number or fraction
which is known, and is so often used that it becomes worth while to
have an abbreviation for it. Thus the Greek letters 7 and « always
stand for certain results, which cannot be exactly represented, but
which are nearly 3°1415927 and 2°7182818.
III. The alphabets used are 1. The Italic small letters; 2. The
Roman capitals; 3. The Greek small letters; 4. The Roman small
letters ; 5. The Greek capitals. They are here placed in the order of
their importance on the subject; and as many may wish to learn
algebra, who do not know Greek, the Greek alphabet is here given,
with the pronunciation of the letters.
A a alpha N » nu
BBC beta Bé Xl
Ty gamma O o omicron
Ad delta Ila@ pi
E ¢ epsilon P e ro
Zagl 1 Zeta Zocs sigma
H heta To Ptae
© 6 theta Tv upsilon
I « idta ® ¢ phi
K x kappa X x% chi
Aa lambda vd) psi
Me mu 2 ow oméga
1V. Under the word number is always included whole numbers
* The young reader may need to be told that the thirteenth century
does not mean a.p. 1300 and upwards, but a.p. 1200 and upwards. The
first century is from the beginning of a.p. 1 to the end of a.p, 100: the
second from the beginning of a.p. 101 to the end of a.p, 200, and so on.
INTRODUCTION. Vil
and fractions. Thus, 24 is not called a number in common language,
but in algebra it is called a number; or, if it be necessary to dis-
tinguish it from 2, 3, 4, &c., it is called a fractional number, while
the latter are called whole numbers.
Under the word number the symbol 0 may be frequently in-
cluded, which means nothing, or the absence of all magnitude or
quantity. If we ask, what number remains when 6 is taken from a,
and if we afterwards find out that 6 and a must stand for the same
number, then the answer is, “no number whatever remains, or there
is nothing left.” If we say that 0 remains, we make the symbol 0
an answer to a question beginning, ‘‘ What number, &c.?” which is
equivalent to including O under the general word number.
V. The sign + is read plus (Latin for more), means in correct
English* “increased by,” and signifies that the second-mentioned
number is to be added to the first. Thus a+b) is read a plus 6,
and means a increased by 8b, or the number which is made by adding
b to a.
+a by itself can only mean a added to nothing, or 0 +a, which
is a itself.
VI. The sign — is read minus (Latin for /ess), means “ dimi-
nished by,” and implies that the second number is to be taken away
from the first. Thus a—0d is read a minus b, means a diminished by
b, and signifies that 0 is to be taken away from a.
When a is less than 6, the preceding stands for nothing at all, but
a direction to do what cannot be done. Thus 3—6 is impossible.
We shall hereafter have to investigate what such an answer means,
that is, the problem which gave it being of course impossible, what
sort of absurdity gives rise to the impossibility.
VII. The sign x is rendered by the English word into, and
* Those who have attempted to substitute a short English term
have never been able to find one which does not shock the ear. We
have seen ‘‘a add b,” and ‘‘a more b;’’ which should be, ‘To a add
b,” or “b more than a.” Neither of the last would be convenient,
and ‘‘a increased by b” is too long. The same remarks may be made
on ‘fa less b,” and ‘‘a take away b,” for ‘‘a diminished by b:” while
‘‘q save b,” used by our oldest writers, is now too uncommon. There
is no language in which the simplest relations are expressed by the
simplest terms.
Vill INTRODUCTION.
means “is the multiplier of,” meaning that the second number is to
be taken as many times, or parts of times, as there are units, or parts
of units, in the first. Thus a x 6 is read a into b, and means 6 taken
a times. Thus, 13} x6 is 6 taken a time and a half, or 9. The
expression ab is called the product of a and 6; the two latter are
called factors of the product, and coefficients of each other. 0 Xa
and a XO must both mean 0; for a taken no time, nor part of a time
whatsoever, cannot give any quantity, and nothing, however often
repeated, yields nothing. As this is connected with a mistake always
made by beginners, we shall (to impress the results on his memory)
give two problems* the answer to which is self-evident, and put the
answers in an algebraical form; desiring the learner, of course, to
remember that no new information is gained, but only an opportunity
of making certain results occupy a space proportional to their
importance.
There is a number of boxes, none of which contain any thing.
How much do all together contain ?
If a be the number of boxes, then 0 repeated a times, or
a x'0NS 20,
There is a box full of gold, of which no part whatsoever belongs
to A. How much belongs to A?
If p be the number of pounds of gold in the box, then A’s part is
O X p, or 0. id
The reason of the mistake is, that the beginner retains the notion
that “not multiplied” implies “ not diminished,” and ‘ not changed
at all.” But multiplication in the arithmetic of fractions, and in
algebra, means the taking a number of times, or parts of times. A
number noé multiplied at all yields no number, for no part of it is
taken; a number which remains the same is multiplied by 1, or taken
once. Thus a is 1 Xa.
When letters are employed, or numbers and letters together, the
sign X is dropped. Thus ab means b taken a times; 3a means a taken
3 times. It is only when two numbers are employed that x becomes
necessary. Thus 3 x 6 must not be written 36, which already stands,
not for 6-+6-+6, but for 3 x ten+6. It is more common to place
points between numbers (thus, 3.6) to signify multiplication. But
* Problem, any question in which something is required to be found.
INTRODUCTION. 1X
. as this may be confounded with B40, the student should always
write the decimal point higher up, thus 3°6.
VIII. The bar between the numerator and denominator of a
fraction is read “ by,” and this is the word for division. Thus 5
is read “‘a by 6,” and means that it is to be found what number of
times and parts of times a contains b. Thus “ > or 3. by 2; 18-14,”
means that 3 contains 2 a time, and halfa time. That a half is
: 1 ) : ne ;
written yo “<1 by 2,” is consistent, because it is the part ofa time
which 1 contains 2. The division of a by 6 will sometimes be
denoted by a—b.
a . : ; ;
A (which beginners usually confound with a) has no meaning.
How many times does six contain nothing? The answer is, that the
ve. . a
question is not rational. a is * for by a we mean a number of
units and parts of units, so that a itself is the answey to “how gaany
times, and parts of times, does @ contain 1?”
IX. The following are then synonymes for a, which the student
should repeat till he is very familiar with them :
Peels ax toa ete = &e.
core? EG
a—OoO ta 1 Sr x 1 &e.
X. The following abbreviations are used for the connecting words
equals and therefore. By a = 6 we mean that a and bare the same ,
numbers: it is read a@ equals b. By .*. we mean therefore, or then,
or consequently. Thus,a = 6 and b=c .*, a =c is read a equals
b, and b equals c, therefore a equals c.
XI. Every collection of algebraical symbols is called an expresston,
and when two expressions are connected by the sign =, the whole is
called an equation.
An identical equation is one in which the two sides must be
always equal, whatever numbers the letters stand for, such as the
equation already described (page iv),
a+1
+1
= 4
xX INTRODUCTION.
which cannot be untrue for any value of a; or is true, whatever
number we may suppose a to stand for.
The following are very obvious identical equations :
a+b=b+a a+1+1l1=a+3-1
ata = 2a atata= 3a
1 1 1 1 1
5 O54 = a s¢+5a¢+s2@ =a
2a+3a—a+4a = 8a+6a+5a— lla
The following are not obvious, but will be found to be true in
every case in which they are tried.
1 1 2+ 3a
ifaTif2a ~ 1F3a+2ua
a ax
ite +e
An equation of condition is one which is not always true for every
value of the letters, but only for a certain number of values. Thus
a
Reveal eee Oey, Abs sh DF; LS Aa
Ui wea tt MET Derr | ana G™e— 16
cannot be true unless 6 is 6 and a is 15.
Again, a = b+ c cannot be true for every value of a, 6, and c,
but requires or lays down the condition that @ must be the sum of
band c. Whena number may be written for a letter in an equation,
and it remains true, that number is said to satisfy the equation.
Thus, ~—3 = 12 is satisfied by a = 15. .
XII. When an algebraical expression is enclosed in brackets, it
signifies that the whole result of that expression stands in the same
relation to surrounding symbols as if it were one letter only. Thus,
a—(b—c)
means that from a we are to take, not b, or c, but b—c, or what is
left after taking c from 6. It is not, therefore, the same as a—6b—c.
Exampie. What is a—(b— {c—d}) when a = 20, b = 12,
c=10, and d=3. Here c—d is 10—3, or 7; b—(c —d) is
12—7,o0r5; a—(b—{c—d}) is 20 —5, or 15.
Also, (a + 6) (c+ d) means that c + d is to be taken a + b times,
and p(q +7) that g + r is to be taken p times.
These are the first outlines of algebraical notation. Others will
develope themselves as the work proceeds. We shall now lay down
some very evident characters of the four principal processes.
|:
INTRODUCTION. Xi
1. Additions may be made in any order, without affecting the
result. It is evident that all the following six expressions are the
same :
14243 14342
2+3+1 3+2+41
3+1+42 2+1+43
Also a+b+c+d = b+c+d+a = b+a+d+e, &c.
2. The order of additions and subtractions may be changed in
any way which will not produce an attempt to subtract the greater
from the less. Thus, if a man lose £20 and gain £50, he has the
same as ifhe first gained £50 and then lost £20, with this exception,
that if his property be less than £20, he may do the second but cannot
do* the first. Thus 10 — 20+ 50 is impossible; but 10 + 50 — 20
is possible. Also 8 — 6+ 10—11 admits of the following forms :
8— 6410-11 8410—l1— 6 10+8—11~— 6
8+10— 6—11 10+ 8— 6-11 10—6+ 8-1];
but does not admit of the following :
10—11+8—6 —l1+ 8+]0— 6 &c.
10—11—6+8 — 6—11+ 8410 &c.
Question. What are the conditions necessary to the possibility
of a—b-+c—d? Answer. That a be greater than (or at least not
less than) b, and that a—6-++c be not less than d. We shall here-
after return to this point.
3. Multiplications and divisions may be performed in any order.
For instance, abc means that c is to be taken 6 times and the result
a times: the rules of arithmetic shew that this is the same as bac;
that is, as c taken a times, and the result taken 6 times; and this
whether the numbers be whole or fractional. The student is supposed
to have demonstrated these rules before he commences algebra.
It is also shewn in arithmetic that division and multiplication
* According to the language of common life a man may lose more
than he has, that is, lose all that he has and incur a debt besides. But
this is always on the tacit supposition that not only what he has, but
what he may get, is liable for his debts. No such supposition exists in
arithmetic until the meaning of words is altered ; 10—20 is, in every
arithmetical interpretation, impossible.
Xil INTRODUCTION.
may be changed in the order of operation; that is, a divided by 4 and
the quotient multiplied by c, is the same as a multiplied by c and the
product divided by b. Or
a Ca
CS Sa
In fact, owing to the extension of the meaning of words by which
multiplication includes taking parts of times (page iii), every multi-
plication is a division, and every division a multiplication. For
instance, to divide a by six is to take the sixth part of a, or to take
a one sixth of a time, or to multiply a by e Similarly, to divide a
Lys : ‘
by z isto ask how many fourths of a unit a contains; the answer to
which is, four times as many times as a@ contains units, which
multiplies a by 4.
Again, to divide 10 by : is to multiply 10 by To divide
10 by 5 is to find how often 10 contains two-thirds of a unit. Now
a unit is made up of = and > the second of which is half the
, 2 ; ;
first; that 1 contains 3 2 time and a half. Therefore, 10 contains
two-thirds ten times and ten halves of times, or 15 times. That is, 10
divided by 5 is 10 taken once and a half or 10 multiplied by 15,
: 3
that is, by 5"
Similarly, to divide by ; is to multiply by 7 How often does 10
contain the half of 7 units? Twice as often as it contains 7 units;
; AEE 3 20 : 20. 2
that is, twice os of a time or 7 ofatime. But 7 8e of 10, or 10
2 :
taken of a time.
The learner should practice many different cases of the following
general assertions.
a multiplied into : isa divided by ;
a divided by 7 is a multiplied into 5
: P eer ees a
or 9 xa>= q 2p xX a
p q
INTRODUCTION. X11
The first operation to which the beginner must be accustomed is
the conversion of algebraical expressions into numbers, upon different
suppositions as to the value of the letters. For instance, what is
b
oa when q@ =-— and je
a—b ‘ 5
1 2 9 2 1
Sh ems io T= 5 5 = 7
ae a
a oY 10 2G
(ae or
10
1+aa aaa 1
I — — py |
s or aye true when q@ ee
| eae, 9 Oued S
Bea Sig I+aa=14+7=—=
3 5 13 5 13
] = | ——— ] ey (4 | = ee ES
ta=143= 5 (l+aa)+(1ta) = B38
3 3 oO 27 1 3 1 11
ON ioe Se alias Be ay 2+5a= 7
1 ) 7a ie 97
ne
But (1+aa)+(1 +a) =3
therefore the above equation is not true in this case.
Whatis qa+b(a+b) when g—=4 § = 3?
@a+b=7 b(a+6)=3x7=21 aa=I6
aa+b(a+6) = 16+21 = 37
But the most instructive exercise is the verification of equations
which are asserted to be identical. For instance,
aa—bb _ aaa+bbb
a—b ~— aa+tbb—ab
Lleta=4 b6=2,thenaa=16 bb=4.
aa—bb _16—4
| =a =(6) aaa=64 bbd=8
: aaa+bbé tae ok ane 6
a@erti—ab — io+4—8 ~ 12 — ©)
Let a= 5 b= i.
X1V INTRODUCTION.
eats 7
aa bb 9 Tae 7
rey ES TT
Soe 6
a eee 91
aaa+bbb 2% 6 ae = (+)
bb—ab 4: 12 eee
aa+ a ae 3 6
The following is a list of identical equations, which the learner
should verify; first, by giving whole values to the letters, next, by
giving fractional values. It is needless to give instances of each,
because the test of correctness is in the two sides shewing the same
value, no matter what.
letters is that no values must be assumed which will produce in the
The only restriction upon the values of the
equation an expression for the subtraction of the greater from the less,
or which will produce 0 in the denominator ofa fraction (pages vii, ix).
, (atat+y)(a+a—y)=aa+2axr+nu2z—-yy
(a+b)(a+b) = aa+2ab+6b
(a—b) (a—}) = aa—2ab+bb
(a+ b) (a—b) = aa—bb
(mm—nn) (mm—nn)+4mmnn = (mm+nn) (mm+nn)
(a+b) (a+b)+(a—b) (a—b) = 2aa+2bb
(a+b) (a+6)—(a—b) (a—b) = 4ah
(a+b+c) (a+b—c) (6+c—a) (c+a—b)
= 2aabb4+2bbec+2ccaa—aaaa—bbbb—ccec
(a+b) (a+b) (a+b) = aaa+3aab+3abb+bbb
1 1 1 hia 3saa+2—6a
a "e a—1 oe a—2 = aaa+2a—3aa
atb a a—b 2aa+2bb
a—b a-b — ~aa—bb
axx+bxz+ce= Ce ee
EEE YYYY __ rre+-rry+-ryyt+yyy
Rd SY very tyy
INTRODUCTION. XV
er+a_,rr+(a+ec)r+ac
r+b” rat+(b+cor+bc
The reduction of several terms into one can be performed so as to
_ produce a more simple term, when all the terms are alike as to letters.
_ Thus,
3a + 2a = da atta — 4a = 4a
dsax+2a72 = 5xx aab+V7aab—3aab = 5aab
eee —sa—6Gat2a—a = 5a
In the last instance, all the additive terms make up 15a, from which
_@ is to be subtracted three times, six times, and once, or 10 times in
all; that is, 10a is to be subtracted: and 15a—10a = 5a.
Similarly a+ b6—3a+4+4b = 5b—2a. For band 56 added
give 6b, which is added to a, after which 3a is subtracted. But,
subtracting 3a where a had previously been added, is the same as
subtracting 2a without previously adding a, which gives
a+5b—3a = 5b6—2a
EXAMPLES.
atab—2ab+4a+6a = lla—ab
222+62—42x2—7e4+e=2244+27r4+
32—15+42—x2—7 = 244 —22
L+yY+u—yt3u = or
If, upon looking through such an expression as either of the
above, we find the following terms containing the simple product ry *
(with their signs)
+62y, —zy, +42ry, +2ay, —llay, —l2<2y,
the single term which represents the result of all those containing vy,
is found as follows; the whole number of additions of wy to the
rest of the expression amounts to 12, and there are 24 subtractions of
it in all, Let all the additions, and as many of the subtractions, be
abandoned, there will then remain 12 subtractions uncompensated by
any additions, and —12ay must be appended to the rest of the
expression. :
When two terms are not exactly the same as to letters and number
XV1 INTRODUCTION.
of letters which enter them, no such simplification can take place.
For example, aa +a cannot be reduced. It is not* 2a, nor 2aa, nor
aaa; itis a taken a times added to a taken once, and is therefore a
taken a + 1 times, or (a+ 1)a. This is also (a+ 1) taken a times,
or a(a+ 1), page xi.
Thus a + a is algebraically reducible to the single term 2a, but
aa -+ a is not so reducible, nor does it admit of any simplification of
form, except that which is contained in the formation of its arith-
metical value, which cannot be found till we know what number a
stands for.
How many times is a contained ina+ 6? This question admits
of no answer till we know how many times 6 contains a; therefore
es which has been
he
we can only use the algebraical representation +
chosen to signify the number of times which a+ contains a. But
let the student observe that this is not an answer to the question, but
a method chosen to represent the answer.
How many times is a contained in ma—na? Here, though —
we cannot completely answer this, till we know what m and z stand
for, yet the algebraical meaning of ma and na puts our algebraical
answer one step nearer to the arithmetical answer, that is, enables us
es Ma—Nd ;
to answer otherwise than by directly writing down ages a For it
is evident that when z times is taken away fron m times, the re-
mainder is m—~n times; that is, ma—mna contains a, m—~n times.
This must be taken notice of in all algebraical operations. The
question, ‘‘ What is 8a-++ 5a?” cannot be answered until we know
what a stands for; but to ‘“¢ What is the most simple algebraical form
of 8a+5a?” we answer 13a. And by the algebraical operations of
addition, subtraction, multiplication, &. we mean the methods of
changing algebraical expressions into others which are of more simple
character. For instance, “to add a+6toa—b.” The following
(a+ 6) + (a— 6)
is the first form of the result, derived from representing the thing to
be done under algebraical symbols. But its most simple form is 2a;
and in the reduction of the preceding expression to 2a, consists what
* These are all mistakes to which the beginner is liable.
INTRODUCTION. XVIl
- we shall call algebraical addition. We shall now go through several
of these processes.
Appition. We wish to add a+btoc+e. If to a+b we
add c, giving a+b6-+c, we have not added enough, because the
quantity to be added is not c, but e more than c. Therefore
a+b6+c-+e is the result, or*
(a+b)+(c+e) =at+b+c+e.-.-- (1)
Toadd c—etoa-+ bd, we first add c, giving a+6-+c. But this
is adding too much, for e should have been taken from c and the
remainder only added. Correct this by taking e from the result,
which gives a+6+c—e.
(a+ 6) +(c—e) = atbtc—e weee (2)
In the results of (1) and (2) further reduction is impracticable.
We shall now try the following.
To add a+ 6 toa—b,
(a+b6)4+(a—b) =a+b+a—b = 2a
To add 34—a to 2a—z,
(3%—a)+ (2Qa—2) = 3a4—a4+2a—4 = 2r+a
To add ab—b to 2ab+c—6b.
fae eae o—60+ab—b or 3ab+c—7b
From these, and similar operations, we have the following rule of
addition :
Write + before the first terms of all the expressions but one, and *
consider the aggregate of all as one expression. Make the reductions
which similarity of terms (as to letters) will allow.
ExaMPLes. a—b+3c—ab
da6—b> +2¢0—2
4x+6a+ab—7 are to be added.
Answer. 9a—2b+4ab4+3c¢+3a—7
* When we wish to refer to an equation afterwards, we place a
letter or number opposite to it, as is here done.
c 2
XVlll INTRODUCTION.
Add. Add. Add.
a—b a—2b at+ma—4
b—'c Di Ge 6—' a4-+2ma
c—d c—2d 12ma—12— 3a
d— x d—2x pta-—s
G—£#. a—v—C—d-— 20 l5ma—2a—103 4p
The following rule is also derived from (1) and (2):
An expression in brackets preceded by the sign + will not be
altered in value if the brackets be struck out.
a+(b+c—e)=a+b+c—e
Susprraction.—To subtract b +c from a. If we first subtract 6,
which gives a— 6, we do not subtract enough, since & should have
been increased by c, and the whole then subtracted. Hence c must
also be subtracted, giving a—b—c, or
a—(b+c)=a—b—c ---: (3)
To subtract b—c from a. If we now subtract 6, giving a — 6,
we have subtracted c too much; or a—Q6 is less than the result
should be by c. Consequently, a—6-c is the true result, or
a—(b—c) =a—b+e.::-- &)
The following are instances:
a—(ec—ad) = a—c+a = 2a—c
a—(a—c) =a—a+c=ece
3a+b—(2a—b) = 5a+b6—2a4+b)=4a4+2b
a+b—(a—b) =a+b—a+b = 2b
mx—(q—smx) = mx«—qt+3m« = 4mx—¢q
WHEN AN EXPRESSION IN BRACKETS IS PRECEDED BY THE
SIGN —, THE BRACKETS MAY BF STRUCK OUT, IF THE SIGNS OF ALL
THE TERMS WITHIN THE BRACKETS BE CHANGED, NAMELY, + INTO
— anpd — 1nTtTo +. This is evident from (3) and (4).
As the neglect of this rule is the cause of frequent mistakes, not
only to beginners, but to more advanced students, we have printed it
thus to attract attention. Still further to impress it on the memory
of our younger readers, we beg to inform them that to neglect this
INTRODUCTION. X1X
-rule is the same as declaring that all debts are gains, and all property
a loss; that to forgive a debt is to do an injury, and that the more a
_ man is robbed the richer he grows; with a thousand other things of
the same kind.
Another proof of the preceding rule is as follows. If we wish
to find
a—(b+c—p—q) seartre Se abit (A)
“i
we must remember that if two quantities be equally increased, their
_ difference remains the same: thus the difference of a+z and b+ 2
is the same as that of a and 6. Weare told to diminish a by
ny Naas ie
- which is the same as if we diminished a+ (p+ q) by
(b+c—p—q) + (p+q)
or by b+c—p—q + p+q orby b+e
that is, a—(b+c—p—gq) is the same as
atpt+q—(6+c) or a+t+p+q—b—c
or a—b—c+p+q-. Boe ata. (B)
on comparing (A) and (B) the rule is obvious.
The rule for subtraction is as follows:
Consider the first term of the expression to be subtracted as having
the sign + ; then change every sign, annex the expression thus changed
to the expression which is to be diminished, and make all practicable
reductions, as in addition.
From @+60—c—2+22+3ab—14
Take c—2a+2+2—4ab423
Rem’. 3¢+60—2c—22x4+2z—T7ab—163
From a+c x+y—3—a Geb -- odie
Take 2c—a x—y+t3—a a—2b+c+d—e
Rem’. Q2a—c 2y—6 P—i 2a ee
From a@+60+2c+3d+4e—5f—6g
Take 12d+4e—3c+2a+b—g+f
Rem, 5c -9d—6f—5g—a
XX INTRODUCTION.
What is
a—(b—(e +2£)) + (b — (a — 26))
This, by the rules for expunging brackets, is
a—b+(¢+2)+b—(«#—25)
which, by the same rule, is |
a—b+c+a+b—244+26 oO a+c+42b
Shews that
a— ja—(a—(a—2))} man 3
a— \b—(a—(b —«x))} = Qa—2b+72
Since all the rules for addition and subtraction are independent
of the order in which the terms of the expressions are written, we
shall in future not inquire whether an expression is written in a
possible or impossible form (page x1), but consider the impossible
forms as meaning the same thing as the possible ones. Thus, in
3—7-+8, we shall not regard the subtraction as necessarily to be
performed first, and therefore treat the expression as impossible, but
we shall consider the order of the operations as immaterial, and the
above as 3-++8—7; and the same for other expressions. We could
not have done this if the rules for addition and subtraction had
required any preliminary inquiry into the possibility of the order of
the terms. Ifwe have to subtract the preceding, say from 12, we
find that the employment of the impossible form, namely
12—(8 —7 4+ 8) = 12—34+7-8=8
gives the same result as if the possible form had been employed,
as in
12—-(8 + 8—7) = 12—3—84+7=8
Mottiprication and Drvision.—In the multiplication and
division of algebraical quantities, it must always be borne in mind
that the letters may represent either whole numbers or fractions.
We shall first express the rules which have been found in arithmetic
for the addition, &c. of fractions which have whole numerators and
denominators.
- a °
First observe that rR (when a and 6 are whole numbers) is the
answer to all of the following questions, which are in effect the same.
INTRODUCTION. XX1
1. Ifa unit be divided into 6 parts, and a of those parts be taken,
how many units or parts of units result?
2. How many units or parts of units are there in the bth part
of a?
3. How many times, or parts of times, does a contain 6? Thus,
3 : : ;
7 tepresenting the seventh part of unity repeated three times, is the
answer to the questions, “ What is the seventh part of three?” and
_“ How many parts of a time does three contain seven?”
: It is only the method of speaking, or the idiom of our language,
which prevents our explaining in a similar mauner the meaning of
fractions which have fractional numerators and denominators. For
a
b
same way as where a is 3 and Bb is 7, we shall produce a (yet)
unintelligible idiom. Let us suppose some concrete unit, say a
mile of length.
: A . rey,
instance, if we attempt to explain — where a is 23 and 0} is 9 2 the
se ier 3 2.
What is = miles, or 7 of a What is = miles ?
mile? - 4
Cut a mile into 7 equal parts Cut a mile into < equal parts,
and take 3 of them. and take 23 of them.
The words in italics are unintelligible, and not having a meaning,
may have one given to them.* By altering the manner in which the
first of the two preceding is expressed, without changing the meaning,
we may find a mode of speech which shall not become unintelligible
4. :
when 9/8 written for 7, and 23 for 3. As follows:
* The same letter may either stand for a whole number or for a
fraction ; and we might find out how to group those idioms which belong
to whole numbers with those belonging to fractions, so as to be able to
pass from one of the first set to the corresponding one of the second.
But it will be more convenient to take the modes of speaking which
belong to whole numbers, and agree that when fractions are spoken of
| they shall be used for the corresponding expressions. We have already
_ done this in arithmetic in the word muttiplication, which originally
. means, “‘ taking a thing many times,” but which we have also made to
signify ‘‘ taking a thing a part of atime.” Thus we speak of multiplying
by one half.
XXll INTRODUCTION.
Find the length which taken Find the length which taken
7 times gives a mile, and take 4 ity aie gives a mile, and
that length 3 times.
take that length 23 times.
The reduced value of 24 — ; is = of a mile, or 5% miles, and
the student may easily shew by the rules of arithmetic that this coin-
cides with the length, : of which is a mile, repeated 23 times. And
5% is also the answer to the question, “‘ How many times and parts of
: ee
times does 24 contain a
It is usual in algebra to make use of modes of speaking with
regard to letters, drawn from our notions of whole numbers; but as
the letters themselves may stand either for fractions or whole numbers,
these modes of speaking are too confined unless they are understood
to imply the corresponding modes of speaking, with regard to
fractions. For instance, suppose it asked, ‘‘ What is the value of
one acre if a acres cost y pounds?” The answer would always be
given as follows:— Divide the number y into 2 equal parts; then
each of those parts is the number of pounds which an acre costs.
Thus, if 18 acres cost £36, since 2 is the 18th part of 36, £2 is the
in| 1 .
price ofan acre. But if 3 ofan acre cost £2., and if we speak of
eet iB. 1 :
dividing 25 into = equal parts, we must mean the same as if we
: 1 : Mer 1 1 :
said “ take 23 two times,” or divide 25 by 3 (according to the
arithmetical rule). Now, though it may appear at first sight almost
ridiculous to say that dividing a quantity into 1G equal parts is the
same thing as taking it 10 times, we must remember, Ist. that we
have done something of the same sort, when we said that dividing
into 10 equal parts, and taking one part, is multiplying by = ;
2d. that we do say this, when we say that if 2 acres cost y pounds,
the price of one acre is found by dividing y into x equal parts; for
letters may stand either for fractions or whole numbers.
We should recommend the student to solve various simple
INTRODUCTION. XXlil
questions in the rule of three, with fractional data.* 1st. By imitation
of a question with integral data. 2d. By independent reasoning ; as
follows:
I. If 6 yards cost 7 shillings,
how much do 5 yards cost?
As 6 yards to 5 yards, so are
7x5
ge
or 55 shil-
7 shillings to
lings, the answer.
If - of a yard cost 7 of a
shilling, how much do : ee
yard cost?
2 4
As 3 of a yard to5 of a
yard so is : of a shilling to
54
7%5 10
Feb he
ee Qern of a shilling, the
3
answer.
2 5 a
FLAT = of a yard costs rs of a shilling
5 15
2 yards cost > X 3 or Tootteee
15 15
; ts —— wee eee es or 0
1 yard costs > - 2 or =
15 30
4 yards cost = Xx 4 or +e ae
4 30 . 10
7 of a yard costs Tee 9 or Sie aa
We shall now write all the rules of arithmetic relative to fractions
in an algebraical form, these rules being those with which the student
is already familiar as applied to whole numbers. We shall first
suppose the letters to denote whole numbers.
ma
mb
2.
4
Q1~ atom S18
i=}
°
4
a
s~8 oe SHS
ad+be
— Sa hed iy
ad—bc
Syeeres:
_. a—be
b
RIS
Qe
|
i)
|
* Datum. A thing given, that is, a number given as part of a
question. ‘‘2 pints cost 4 shillings, how much, &c.;’ here the data
are 2 pints and 4 shillings.
XXIV INTRODUCTION.
a ac a Cc ac
bispueciay nie) hy eden
a a a be
birdy mas Abe On eae
a Cc ad a d
bk dik, bc a a
The learner must make himself acquainted with each of these
equations, which he will easily do if he have the requisite degree of
familiarity with the operations of fractional arithmetic. For example,
in ; x = > he will recognise the rule, “To multiply one frac-
tion by another, multiply their numerators for a numerator, and their
denominators for a denominator.”
The preceding rules all hold good when the letters stand for frac-
tions. We shall take the demonstration of this in one instance,
namely, ; = —. Let a be the fraction a let b be-, and let m be
- d
mE where p, 7, T, 8, v, and y, are whole numbers.
Pp
a ta @ ae ps
be LUmmuge (By the rule.)
&
ma == xX = mb 22 eee
Bf ff a. y s ys
xp
ma Yq ryps
mb aay? Te rygr
ys
)
But 7YP5 — XPS _ Ps
syorie - (ey) Ggramulyr
ma ¥ a \ where a, b, and m, are frac-
or chi = a ;
mb b tional.
It is shewn in arithmetic that the order in which multiplications
or divisions are made may be changed without affecting the result,
both in questions of whole numbers and fractions. We need not,
when speaking of both, distinguish between multiplication and
division ; for division by a is multiplication by =. The following
summary should be attended to and repeated on several products.
The expression abcd is the product, 1. Of a, b, c, and d, in any
INTRODUCTION. xXXV
order; 2. Of ab and cd; 3. Of ac and bd; 4. Of ad ana bc;
5. Of abc and d; 6. Of abd and c; 7. Of acd and b; 8. Of bed anda.
The product of single terms may be expressed by writing the
letters consecutively in any order which may be convenient, and
actually multiplying the numerical coefficients, ifany. The sign x
becomes necessary between two numerals only. Thus, 2ab x 4cd
might be written in the following ways:
— Qab4cd 2 x 4abcd Sacbhd Sabed, &c.
It is most convenient to write the numerical coefficient first, and
letters of the same sort together, the order of the alphabet being
generally preferred. Thus,
2aabx 3abbe is written Gaaabbbe
:
labax4abax = 48aabbaxx Babex sab = Saabbe
When fractions are written side by side of each other, or of single
letters, multiplication is intended to be denoted. Thus,
© and is zac’
dp of
[Though a, ab, abc, &c. are called integral expressions, and
aac ac
bed’ b’
arithmetical character; for, since letters may stand for fractions, that
2 cs means 2.x =X OX
&e. fractions, this refers to their algebraical, not to their
which is integral considered algebraically, may be fractional con-
‘sidered arithmetically, and vice versd. For example, suppose that
1 1 : :
a stands for ; and 6 for fe then the algebraically integral expression
4 ey : ; :
z is the arithmetical fraction = while the algebraical fraction 7 or
is the arithmetical whole number 2.
In speaking, therefore, of integral or fractional expressions, we
efer always to their algebraical appearance, not to their arithmetical
values. The latter depend on the particular arithmetical value of the
‘etters, on which no supposition is made.* |
* The part of algebra which treats of letters considered as repre-
enting whole numbers only, is generally called the theory of numbers.
t is very rarely, if ever, of.use in the application of algebra to the
ifferential Calculus, and is therefore altogether omitted in this work.
d
XXVI INTRODUCTION.
The multiplication of algebraical quantities depends on the rule
which appears in the following equations :
m(a+b) = ma+mb
m(a—b) = ma—mb
First, it is required to take a+ 6, m times. If a be taken m
times,.it is clear that for every time and part of a time which a has
been taken, b or a part of 6 has been omitted. Consequently, ma is
too little by mb, or ma + mb is the product required.
Secondly, it is required to take a—b, m times. Here, if a be
taken m times, 6 or a part of b too much has been taken for every
time or part ofa time which a@ has been taken. Consequently, ma
is too much by mb, or ma—mb is the product required.
Hence we may prove the following equation :
m(a+b—c—d) = ma+mb—mcec—md
as follows: let a+6 be called p, and let c+d be called gq; then
a+b—c—d is p—q (page xix.), and
m(a+b—c—d) = m(p—q) = mp—mq
But mp = ma+tmb mg = me+md
mp—mg = ma+mb—(me+md)
= matmb—mc—md
The following are applications of the preceding rules :
1
3(a+b) = 34435 35(a—b) = 35a—356
ab(a—b) = aab—abb 2a(a—aa)*= 2aa—2aaa
3abc(ab—ac+4) = 3aabbc—3aabcec+12abe
(s+) aa+8 6(E+2) = 5242.
* The student will observe, that in these examples we do not inquire
whether the expressions are possible or not, but how to apply the rules
when they are possible. For instance, a— aa is impossible when a is
greater than 1, and possible only when a is less than41. And let it be
observed in proceeding, as a matter of great importance in the sub.
sequent part of the work, that the rules investigated will not serve to
distinguish between possible and impossible expressions.
INTRODUCTION. XXVIl
a
ry 4 frys We y
| 1(zz+1) = a A Dia ee
Ses ac ad ae
Bettas Saas tags ass
ih. pg eee. 5.) = qrs+prs+pqs—s
In order to multiply a + 6 by c +d, let us, for a moment, make.
stand fora+6. Then, p multiplied by c+ d, is the same as c-+d
nultiplied by p, which is pe + pd, or
(a+b) (c+d) = (a+b)c+(a+b)d
But (a+b)c =ac+be, (a+b)d = ad+bd
. (a+b) (e+d) = (ac+be)+ (ad+bd)
= ac+tbc+ad+bd
To multiply ¢+6 by c—d, let a+6 =p, and we have
y(c—d) = pe—pd, or
(a+b) (c—d) = (a+b)c—(a+b)d
(ac+bc)—(ad+bd)
= ac+bc—ad—bd
To multiply a—b by c—d, let a—b =p, and we have
»(c—d) = pe—pd, or
(a—b) (c—d) = (a—b)c—(a—b)d
= (ac—bc)—(ad—bd)
= ac—bc—ad+bd
There are two methods of conducting this process, the first of
which is recommended to the learner for the present.
1. To multiply a+ b—2c by d—a—ce.
Here a times and c times the multiplicand are to be successively
aken from d times the multiplicand; that is, a +c times is to be
aken from d times.
XXVIli INTRODUCTION.
d times the multiplicand = ad+ bd—2cd
Add { ee ue = aa+ab—2ac
c times ditto = ac+bc— 2cc
a+ctimes ditto = aa+ac+ab+bc—2ac—2cc
f Subtract fom d times ad+bd+2ac+2cc—2cd
ditto which gives —aa—ac—ab—be
2. From looking at the preceding instances, it appears that the
rule of multiplication is as follows: Consider the first terms as having
the sign +3 multiply every term of the multiplicand by every term of
the multiplier, and put + before the products of terms which have the
same sign, and — before the products of terms which have different
signs. The preceding example is here written in the usual way, with
the proper signs written to every term by the preceding rule.
a+b—2c
d—a—c
from G-s-- a ages ae
from @ «+++ —aa—ab + 2ac ;Product required.
from C eee: Ee ee
Where like terms are found in different lines, so that subsequent
reductions may be made, it is convenient to put like terms under each
other, as in the following example :
Multiply rxa—2e2+1
By r—4
Froomz w«rr—2rr+er
Product required.
From 4 —4v7r+8r—4 \ : a
rax—6ru+9r—4 ditto in simplest form.
But the manner of doing this, which (as yet) we recommend, is
the following :
Multiply xx—2z+1
Ry x—4
, ee Subtract second
From & times multiplict wrv—2xvx+ux
line from first,
Take 4 times ditto 4xn—84+4|
as in third line.
xxx—Oxux+9u—4 Product required.
‘ INTRODUCTION. XXixX
- The three instances which follow are of particular importance, and
ihe learner should be able to write them (and similar ones) at sight.
‘Mult. a+) a—b ath
By a+b a—b a—b
To We ah From aa—ab From psec
Add ab+bb Take ab—bb Take ab+bb
Mee, ca—2ab4bb6%) am 9 bd
[The following definitions may be appropriately introduced here.
A square is a four-sided figure, with sides of equal length, and
-
with contiguous sides perpendicular to each other.
A cube is a solid figure enclosed by six equal squares; or, a box
of the same length, breadth, and thickness.
A square 4 inches long contains 4 x 4 squares of one inch long ;
a square of x inches long contains «x squares of one inch long.
There are 2 rows of square inches, and # squares in each row.
A cube 4 inches long contains 4 x 4 x 4 cubes of one inch long ;
a cube 2 inches long contains xa cubes of one inch long. There
are x layers of cubic inches, and x cubic inches in each layer.
Owing to this connexion of xa with the square on a line of «
units, and of rr with the cube on a line of x units, it has always
‘been customary to call vx the square of x, and xxx the cube of x.
But rrez is called the fourth power of x, rxxaxw the fifth power ;
and so on. So that 2 itself should be called the first power of x,
ax the second power of x, and xxx the third power of x, But the
words square and cube are so conveniently short that they have
never been abandoned. |
The last mentioned products may be thus stated :
i (a+b)(a+6) =aa+2ab+6b
a2
XXX INTRODUCTION.
or, The square of the sum of two quantities is the sum of their squares,
augmented by twice their product.
2. (a—b)(a—b) = aa—2ab+bb
or, The square of the difference of two quantities is the sum of their
squares, diminished by twice their product.
3. (a+b)\(a—b) =aa—bb
or, The sum of two quantities, multiplied by their difference, is the
difference of their squares.
Thus let the two quantities be ab and 2a.
ab x ab = aabb 2a x 2a = 4aa
2(abox 2a) =Agas
(ab+2a) (ab+2a) = aabb+4aab+4aa
(ab—2a) (ab—2a) = aabb—4aab+4aa
(ab+2a) (ab—2a) = aabb—4aa .
The following are examples for the student:
1
: : 1
Square* of (a + —} =aete2 +
a aud
(043) (o-1) mat
Square of (Qax + b) = 4aarx + 4abxr + bb
(Qaxr + b) Qax — b) = 4aarx — bb
Square of (a+ b+ c) = (a+ 6)(a+b)+2a@+4+ dce+cc
=aa+bb+cc + 2ab+ 2bc+2ca
(a+tb+c)\(a+b—c) = «a+ b) (a+ b)—cc
= aa + bb—cc + 2ab
(c+a— b)(b+c—a) = (c +a— b) (c—a—b)
\
* By + occurring several times in an equation, two equations are
combined in one. The upper or under sign is to be taken throughout.
Thus,
a+b = cj 0 3s bith
atb=c—d or a—b=c+d
The student is recommended never to use this double sign himself ;
but as it frequently occurs in books it is here shewn.
| INTRODUCTION. XXXI1
[ = cc—(a—b) (a—b)
2ab+cc—aa—bb
a = 2ab—(aa+bb—cc)
From the last two examples, shew that the product of the four
quantities
atb+c, a+b—c, b+c—a, c+ta—b, is
| 2aabb+2bbcec+2ccaa—aaaa—bbbb—cccc
Shew the preceding by help of the following (which shew also),
Square of (V+9—7) = ppt+qqtrr+2pq—2qr—2rp
The following are miscellaneous examples in multiplication, to be
done without paper :
(a+bxz) (at+ecx) = aat+abxe+acxt+bcxua
(x+a) (a+b) = xx+ax+bu+ab
(wn—a) (c—b) = xx—ax—bu+ab
(+1) (@—3) = rx—2a—3
(a—1) (@—3) = wx—424+3
(2241) (a#—1) = 2xa—z-1
The following theorems are given for exercise :
1. If a and b be two quantities, of which a is the greater, and if
S be the square of their sum, D the square of their difference, and P
the product of their sum and difference, then
S+ D = 2(aa+ bb) S—D = 4ab
S+P = 2a(a+ db) S—P = 2b(a+))
D+P = 2a(a—)) P—D= 2b(a—))
2. If two numbers differ by a unit, the difference of their squares
is their sum; and if two fractions are together equal to a unit
1 3 ae Sa: : é
(such as 7 and a) their difference is also that of their squares.
8. The sum of the squares of xx—yy and 2.vy is the square of
RUT yy.
[According to the rule, the square of a—Qé is the same as the
square of b—a; for that of the first is aa—2ab-+ bb, and that of
the second 6b—2ab-+aa, which two are the same (see page xi).
J
XXXli INTRODUCTION.
But one of the two, a—b or b—a, must be impossible, except only
when 6 == a, in which case both are nothing; for in every other case
either a—b or b—a is an attempt to subtract the greater from the
less. But for the same reason, one of the two, b—a or a—b, must
be possible; therefore aa+bb—2ab, which is the square of that
possible one, must also be possible. That is, if either of the two,
a or 6, exceed the other, aa+bb must be greater than 2ab. And
we also see that i¢ does not follow that an algebraical process is intel-
ligible because it gives an intelligible result; for it appears that the
algebraical rule of multiplication would apply to (3—7) (3—7),
which is absurd, and give the same result as (7 —3) (7 —3), or 16.
This is a defect, the remedy for which we shall afterwards have to
find ; and we see, that so far as we have yet gone, we can never
know that any process is correct which is to lead to the value of an
unknown quantity, until we re-examine the process after the unknown
quantity has been found by it.]
Divisions in algebra we shall for the present divide into two
classes; those which it is obvious how to do, and those which it is
not obvious how to do. For instance, to divide ab by a, that is, to
tell how many times ab contains a, the answer evidently is, that since
ab means the same as ba, or a taken b times, therefore ab must
contain a, b times; or ab divided by a is b. In this case the simplest
rule is as follows: To divide, where there has already been a multi-
plication by the quantity which is made a divisor, suppress all the
symbols of the multiplication.
This will be seen in the following examples :
Dividend.| Divisor. | Quotient. || Dividend.| Divisor. | Quotient.
ab a b l2aax| 6aax pa
abe ab c 2sbyz| 4by 10z
2abxz 22 ab aaaa | aaa a
aabba ab aba aaaa aa aa
Gabcc | sabe 2¢ LYZ LY Z 1
[One of the most common errors of a beginner is a mistake
between 0 and 1, arising from a confusion of subtraction and division.
This is partly a result of the idiom of our language, as follows. Ifa
beginner be asked how many ¢imes does 7 contain 7, the answer is
INTRODUCTION. XXXill
sure to be, no times at all; and in one sense this is correct, for 7 does
not contain 7 a number of ¢imes, but one time. But it must always
‘be understood in algebra that times means time, or times, or parts of
a time, or time and parts of a time, or times and parts of a time.*
Therefore, though
x diminished by x leaves nothing,
x divided by x gives one.|
Closely connected with the preceding is the theorem in fractions
chat ; and — are the same. For “ multiplied by m gives
|
this divided by m gives —: But multiplication, followed by division
Ma
P , and
(Gnultiplier and divisor being the same), leaves any quantity the same
as at first.
_ From the rule of multiplication it follows, that if any quantity
‘contain the same letter or letters in every term, it is the obvious result
of multiplying another expression by that letter or the product of
those letters. Thus, ab+ac is b+c multiplied by a, aab—abe
lis a—c multiplied by ab. Hence, to divide an expression by a
‘letter or a product of letters, strike out those letters from every term
‘of the dividend. But remember to write 1 where all the letters of
any term are thus struck out. For instance, a+ ab divided by a
ives 1 +b, ac—aac-+acc divided by ac gives 1—a-+c.
The following are iustances, arranged as before:
Dividend. Divisor. Quotient.
2ab—2bc+4abe 2b a—c+2ac
aqaa—aa+a a aa—a+l
6ab—3a+3b) See be 2ab—a+b
aab—abb ab a—b
ALXLY—XLXYY LLY a—xY
* An act of parliament, or any other legal instrument, always speaks
of men under the title ‘‘man or men,” &c. If this should happen to be
1eglected, and a single offender should plead that ‘‘men” only were
»rohibited from doing as he (one man) had done, it would be called a
“quibble.” If the student, after this warning, should ever say that x
‘ontains x no times, it would not only be a quibble, but a very useless
juibble, because nothing is to be got by it.
rl
XXXIV INTRODUCTION.
In dividing ab by a we might proceed as follows. The result |
; Pe ivvab ‘ : : Al
of the process is the fraction al which is not changed in value if |
ls Shee. gryi
both numerator and denominator be divided by a. But this gives - |
which is b. :
Such a process is of no use in the preceding case; but suppose |
that ab is to be divided by ac. The complete division is here
impossible until we know what numbers a, b, and c, stand for. But
a the symbol of the result, may be reduced to Ma by the preceding
Cage |
theorem. The division here is not completed, but reduced to a more
simple division.
2b a b aab ks ab Sammn __ mn
eat ee Gee Gaam 2a
Dicsky be 3ab 3 21vww _ 3vw
dt Pemaee anal, aera 282w . “4a
i
The division of an expression of several terms by another of one
term may also admit of reductions. For example, ry + ys—ér
divided by wyy is
v 1
vY EFI SS 5 ee di Rae oe, ©
ayy apy i ded wd x7 ro
2vu—axrx+vewr 2 9
Soa
atb+c 1 1 sy 1
abes je ac ab
b 1 b 1 1
ee) 1 8 6a
ad a aa a a
r+4y—32+2 f 2y z 1
6 Sigs 273
All that precedes contains the obvious cases of division ; of those,
the answer to which requires further process, the following is an
instance: “ How often is 2 + y contained in axx + yyy?” As we
shall not need such a process for some time, we defer it to its proper
place. In some cases, however, the preceding theorems of multi-_
plication (page xxix) furnish an answer at once. For instance, we
know that va —9 must contain «+ 3, c~-3 times.
INTRODUCTION. XXXV
Fractions may frequently be reduced to simpler terms by in-
spection, of which the following are instances:
mob | 1 +b 3r-4+6r2 1+22
a—ab” 1—b y—3e. B8y¥—1
a—aa 1—a aa+3ab_ a+3b
Qa+az° 2+2 vata WA 2
It is frequently necessary to arrange expressions in a different
form, without altering their value, by performing inverse operations
‘upon them with the same data, such as addition followed by sub-
traction, or multiplication followed by division. The four following
methods of writing 2 exhibit this process.
a:
t+a—a r—a+a oT
a
Thus a+2 = 2a+2-—a = (+ *)\a
aa+2ab—c = aa+2ab+bb—(c+bb)
= (a+b) (a+6)—(c+6b)
4ac : b 4
b6—4ac — i Tg eee = abc ——>)
mtn = mn (+ ++ =n = +1) =m(1+2
ee n m
yee vile
wutay 71
The following are instances of those reductions of fractions which
will occur hereafter. The rules with regard to fractions which are
proved in arithmetic are here applied in conjunction with the
algebraical methods of addition, &c. At the head of each section
stands an example without any complicated expressions, containing
the arithmetical process used in those which succeed.
(See page v.)
a ma
1 -=
—a
—S=
—
XXXVI
INTRODUCTION.
AS ca ye wih) dali
y—1 — y—1 a” ax+aa
1 a—b a+b 4
2 5 10 5a
Tr—4 3(7r—4) _ 357—2
10 Fs 410 ee pis
Lette 1 1
MO: ab(; + a5)
b—a++ ab(b—a +3 abb—aab+a
tl Ree _ ayte £ ay—r 2 ee L— Ay
oy. Af ¥ y ¥
[pester tem. {ha
x x
9 9s! _ yt OS
y+ y+ y+1
ad ab ab aa
ar SO = a— Cl
a+b a-+b a+b a+b
aa—2ab 4ab+ bb
ob a
aa—2ab Q2aat+bb
a+b+—7 5° a
yr1 oid ag Yatul
1 Ly pals y—1 y—1
amide hs GY ps: b—a-e x ip EY
a—c a—c y
ab+bc+ca ‘gem ab—cc
a+t+b-+c ~ atbte
ur. @ ec ay+ber a@ 2 _ ay—be
ee ah) yuu by b. Yeas
a+b a—b _ (a+b) (a+b)—(a—b) (a—b) _ _ 4ab
ib ach oee (a—b) (a+6) "aa bb
Neti Pr ty _ wetyy
aly “ay y ta Tay
Beve cans ch—ad a—b a ad—bc
Sol aemcemece oa c—=d 0 9t, mite =o
INTRODUCTION. XXXV11
ees SS Ory — Vy PES Bu OB CCL
ie ry pe oR A TA) FP PPY
By ei as oy a ay
IV. 7 Xx r = cae y -
r—l +2 (@—1t)(r7+2) #42
St r—1 7 (we tt)\(@—1) 7 41
2ab | a—b 6aab Sar yv _ 3av
Ae eer aa—bb yy Qa 2y
m 2m _ 36 pe 399 3pq
an ° 3bn 2a g SES: ce
The student is recommended to make himself well acquainted
‘with every example given in the preceding list, but no more (see the
Preface); as a better method of obtaining examples will be given.
All that has preceded is purely arithmetical, and the letters may
be considered as mere abbreviations of numbers, and all identical
equations as abbreviations of arithmetical propositions. Thus,
(a+ b)(a+ 6) = aa+2ab+ bb
represents the following sentence: If two numbers be added
together, and if the sum be multiplied by itself, the result is the
same as would arise from multiplying each number by itself, and
adding to the sum of these products twice the product of the
numbers.
An arithmetical problem is one in which numbers are given, and
certain operations; and the question asked is, what number will
result from performing the given operations upon the given numbers.
For instance, what is the fiftieth part of the product of 25 and 300.
An algebraical problem is one in which numbers are either given
or supposed to be given (as will presently be further explained),
ind a question is asked of which it is not at once perceptible what
yperations will furnish the answer. Such is the following : —The
aumbers 3 and 17 are given; what number is that, the double of
vhich will fall short of 17 by as much as its half exceeds 3? And
he questions asked are the following. 1. Is there any such number?
'. If there be, by what operations on 3 and 17 may it be found ?
'. What is the result of these operations, or the number required.
“he answers to which (as the student may afterwards find) will be
€
XXXVIll INTRODUCTION.
that there is such a number, that it is found by taking two-fifths of
the sum of 3 and 17, and that in consequence the number is 8.
If we had contented ourselves with the first two questions, it
would have been unnecessary to have specified that the numbers in
question were 3 and 17, for the same problem might have been
proposed about any other numbers, and the process of solution would
(as may afterwards be shewn) have been the same whatever the
numbers might have been. That is, if the following question had
been asked ;—The numbers a and 6 are given; what number is that,
the double of which will fall short of b (the greater), as much as its
half exceeds a (the less)? The answer is = (a+ 6) and the veri-
fication is as follows:
The double of = (a +b) is : (a+b), or . a os
=.
, 4 4 4 4
This falls short of b by 6 — (Fa ++ =b) or b — Pi op zo
or =b— =a; but the half of = (a + b) is=(a +B)
—
oe 1 1 1 1 4
which exceeds a by z (a+ b)—a, or 34 + = b—a, of abies a
the same as that by which the double of = (a +b) falls short of 6.
Which was to be done. Now, observe that the preceding not only
informs us of the general process by which this problem may be
solved, but it also shews in what cases the problem is impossible.
1 4
For the excess or defect above-mentioned turns out to be 6 — se
a : uf
which is an absurdity, unless ee be greater than (or at least not less
4 : :
than) B43 that is, unless 6 be greater than 4a. This was the case
in the first instance where 6 was 17 and a was 3. If it be not so,
we may pronounce that the problem is impossible; for instance, let
the student try to find a number or fraction, the double of which
shall fall short of 11 by as much as its half exceeds 3. In this
problem there must be a contradiction; and when we know there
is one, and set ourselves to find out how it arises, we see it in the
following :— The number sought is presumed to have a half which
exceeds 3 (so that it must be more than 6) and a double which fails
INTRODUCTION. XXXIX
hort of 11 (so that it must be less than 53). But a number which
‘xceeds 6 cannot be less than 53; therefore the clauses of the pre-
.eding problem contradict each other.
We see, then, that we may propose a problem which is impossible
rr contradictory, or has no solution. But, on the other hand, we
aay propose a problem which admits of an innumerable number of
nswers. These we will call unlimited problems, And, as in the
ase of impossible problems, there are some of which the impossibility
3 evident, as “ To find a whole number which shall be the half of
4
:
-even ;”
and others, in which it requires investigation to discover the
mpossibility, as “* To divide 10 into two parts, whole or fractional,
f which the product shall be 30;” so in the case of unlimited
yroblems, there are those in which the unlimited nature of the result
‘hall be evident, and others in which it shall not be so. For instance,
o the question, “‘ To find two odd numbers which added together
hall make an even number?” it is clear that the answer is, “ Any
'wo odd numbers ;” and to the question, “ What two numbers are
hose of which half the sum added to half the difference shall give
he greater number?” the answer is (but not so evidently), “ Any
“wo numbers.” Between these two extremes, we can conceive there
nay be problems which admit of 1000 answers, others of 999
iswers, &c. &c. down to problems which admit only of one answer.
\nd even when we find that a problem is impossible, we may yet
‘hink proper to ask, why is it impossible? what are the two parts of
he problem which contradict each other, and by how much? that is
“0 say, what sort of change, and quantity of change, in the conditions
of the problem, will render it possible ?
_ To all these questions, arithmetic gives no means of answering,
‘and we have therefore to consider algebra as a distinct science, which
sroposes objects of which arithmetic knows nothing, and therefore
is we may suppose, uses language, finds methods, and adopts inter-
‘oretations, of which arithmetic furnishes no examples.
If the student have read* a little of geometry (a science which he
ja" In England, the geometry studied is that of Euclid, and I hope it
‘ever will be any cther; were it only for this reason, that so much has
been written on Euclid, and all the difficulties of geometry have so uni-
‘formly been considered with reference to the form in which they appear
‘in Euclid, that Euclid is a better key to a great quantity of useful
{reading than any other.
“|
ae ll
Oe
x] INTRODUCTION. 3 |
should begin to study at the same time as algebra, if not before), he
knows that all the questions of geometry are made for him, that is,
the reasonings, &c. are put together before his eyes, and all he has to
do is to comprehend and agree to one step of the process after another,
This is called synthesis (cuvdeois, a putting together), or the synthetical
method, in opposition to analysis (aveavors, an unloosing, or bringing
asunder), or the analytical method. The latter consists in taking the
problem to pieces, if the phrase may be used, that is, reasoning upon
the whole problem, reducing it to more and more simple terms, and
so coming at last to those considerations which must be put together
to make a solution and to verify it.
We now proceed to establish the principles of algebra analytically;
and instead of laying down new names or new principles, and putting
the science together, we begin from arithmetic, such as we know it,
and leave all additional considerations till the want of them is felt,
We shall thus see one new result spring up after another, until we
find the necessity of speaking a new language, and giving interpreta-|
tions to symbols which we did not at first contemplate. How this is
done, and what it leads to, we cannot otherwise explain than by di-
recting the student to proceed to the first chapter.
! ELEMENTS OF ALGEBRA. .
CHAPTER I.
i EQUATIONS OF THE FIRST DEGREE.
‘We now proceed to the solution of equations of the first degree.*
This term must be explained.
’ To find the degree of a term generally, count the letters in it.
Thus, abc is of the third degree; aabc is of the fourth; for though
there are only three letters, yet one of them occurs twice. The fol-
lowing are examples:
Of the first degree, a, b, c, x, 2, p, &c.
Of the second degree, aa, bb, cx, be, pz, &c.
Of the third degree, aaa, aab, abb, abc, pac, &c.
and so on.
To find the degree of a term with respect to any letters, count
those letters only. Thus 3aarwy, of the fifth degree, is of the third
Jegree with respect to x and y, of the fourth degree with respect to
z and wz, of the second degree with respect to x only, of the first
degree with respect to y only, and so on. A term which does not
sontain 2 at all, is of no degree with respect to x, or is independent of x.
The degree of an equation with respect to any letter, is the degree
of the highest term with respect to that letter. Thus the equation
LE — ZLLX = YZ — YyYX
's of the third degree with respect to 2, of the second with respect to
y, and of the first with respect to z.
* Commonly called simple equations.
B
2 EQUATIONS OF THE
The solution of an equation of condition is the following problem :
—Given an equation of condition, containing a letter the value of
which is unknown; what is that number for which the unknown letter
must stand, in order that the equation may be true? Are there more
such numbers than one? if so, how many, and what are they? Or is |
there no such number, that is, is the equation impossible? and if
so, how is that to be ascertained ?
The scope of this will be better seen by some instances, which the |
pupil may verify. |
The equation
24—l1=5x—19
is true then, and then only, when z is 6.
The equation
20 — Jet eee
cannot be true, whatever x may stand for.
The equation
162 = 48 4+ ve
is true when @ is 4, and is also true when 2 is 12; but never in any
other case.
The equation
12% = 48 +242
is never true for any value of z.
The equation
axx+lla=—6ar2r +6
is true when 2 is 1, when wv is 2, and when z is 3; and in no other
case.
As an example of verification, let us try the latter equation when
zw == 4. Then
Ct toms 6x2 = 96
lle = 44 622 +6 = 102
car + lla= 108
But 108 is not = 102; therefore rrr + 1124 is not = 6r27+6,
or « = 4 does not satisfy the equation.
We shall now use the following evident truths :
1. If equal numbers be added to equal numbers, the sums are
equal numbers. That is, if a= 6 and c=d, then a +e =>6+ d.
If a=b—c and «=p—gq, thna+ar=(b—c)+(p—q) =
b+p—c—q. Ifa=ua—y,and b==x1+y,thena+b = (r—y)
FIRST DEGREE. 3
+(ety) = r—yterty = 2e. Ifa = b+c,thena+v =
tot.
2. If equal numbers be taken from equal numbers, the remainders
we equal numbers. That is, if a = b and c = d, then a—c = 6b —d.
lf a=p—q and b = p— 2q, then a—b = (p—q)—(p—2q) =
9—gq—pt+2q=q. Ifa=2z+y, thena—m=z+y—™m.
3. If equal numbers be multiplied by equal numbers, the products
‘we equal numbers. That is, ifa =bandc=d,ac=bd. Ifa=
d+c, and z=n, then az=n(b+c)=nb+ne Ifd=l—z,
2d = 2/—2v.
x av
| mY az Pak
| Then 25+25 ee OF t+ = 2
32+3— = 6
o 3824+227 =6
4. If equal numbers be divided by equal numbers, the quotients
are equal numbers. That is, if a= 0b and c=d, then = — x If
@ = n, then > =>. If a = b—c and p+q = 2, then < —
p+”
=< If 7x = 14, then 2 =o, OM a) =a oe
The following abbreviations will be used, on account of the con-
tinual occurrence of the phrases :
(+)a@ means, add a to both of the last-mentioned equal quantities,
which gives ....
(—)a .... subtract a from both of the last-mentioned equal
quantities, which gives ....
(x)a .... multiply both the last-mentioned equal quantities
by a, which gives ....
(+)a .... divide both the last-mentioned equal quantities by
a, which gives ....
(+) (—) (x) and (=) by themselves I use to denote that the
dwo last-mentioned sets of equal quantities are to be added. The
following will explain the use of these abbreviations:
i 1. a = b—c a—b=q+ux
(F)e ate=6b (+)6 a=qt+«x+6
4 EQUATIONS OF THE
2. c—d = l—m 22-3 = 9
(+)d+m ct+tm=Il+4+d (+)3 2a = 12
3. ptq =a—b llz+18 = 100
(—)q p=a—b—q (—)18 llx = 82
4, pt+q-2z = 3at+4
(—)q—z p = 3a+4—q+2
: tao Te 52,3
+ 4 6 12 4
or 6727—427481 = 142—527+4+9
6. Avs & (a+b)x =c
(—)a x= 2 (+)a+b L= es
rf a—b+2c—3d-= z—a+6
64+3c—2d = 4z—a—2b
(+) a+5c—5d = 54—2a—b
8. 2a% = b—z
a=b+z
b—z
(=) 28 = eae
We shall now proceed to the solution of equations of the first
degree, containing one unknown quantity, by means of the principles
in pages 2 and 3, and the preceding operations.
1. What value of 2 will satisfy the equation
3“2—7 = «x+19
(+)7 32 = £+26
(—)x 2% =z 2G
(—)2 ia hh
Verification. If X = 13, 8x2— 7 = 32
z+19 = 32
2. 3x2+16 = 102749
(—)3x 16= 7z2+9
(—)9 ef.
(+)7 ]
B
- 3.
(+)13
(+)x
. (+)21
4.
(x)2
Gx) 2
(x)3
(+)3a
chat is
(=+)13
Verification.
Verification.
Verification.
gn
FIRST DEGREE.
i a= l, 37+16 = 19
20z—13 = 1024—2
20% = 1153—2
21x = 1153 = —
231 :
C= 2x21 (Ar, 123.)
a OR, sil A
= > = 53
if a= 54, 207—13 = 97
1024—2 = 97
& i Hi hy
pi tg Some
2f oe
“An grea lea
Qa+— = 4-27
67+42 = 12—32
67+4244+32 = 12
be geo
12
13
mereet ree Ie SAG
Be ae gt gh et 1S Of Te oY 73°
Hy ; 1 12 3 10
ear ig) sat a 10
And ] 738 I ( of =)» or ] aw ich is also =
4
The same equation might be more easily solved by multiplying
both sides by any common multiple of 2, 3, and 4. The least com-
mon multiple is the most advantageous ; why, will appear on trying
a higher one, as follows :
ri & &
aie nn 4
36 is a common multiple of 2, 3, and 4.
(x )36
or
(+)92
3627 362 _ 364
7 Ur aoe oe CA
18%+122 = 36—9u
1842+127+92 = 36
B2
6 EQUATIONS OF THE ;
|
that is 39.2 ==, 86 |
(—)39 fy > which, reduced to its lowest
: 12
terms, 1S = —
13
Now try 12, the /east common multiple of 2, 3, and 4.
OH x x
gta Mee
127 424 P 124
(x )12 S243! = 12-8
or 62+42 = R—32
Proceed as in the last case but one; and no reduction of the result to
lower terms is necessary.
5. ab+a—b=1
This equation differs from the preceding in having two unknown
quantities. The real answer is, that there is an infinite number of
values of a and 6, which will satisfy this equation. Ifwe choose a
value of b, we can find the value of a, which, with the chosen value
of 6, will satisfy the equation. For instance, I ask, can 6 be = 12?
Substitute 12 for 6 in the above equation, which then becomes
12at+a—12= 1
or 1l3a—12= 1
(+)12 13a es bY,
(+)13 a stl
The answer is, 6 may be 12 provided a be 1. In this case,
Hoo 12,°and 12-4+-1-—Jo¢=74.
Without making any particular assumption about the value of 6,
Jet us suppose it given; in what way must we combine this known §
on the first side with the known unit on the second side, so as to
point out the manner of finding a so soon as a particular value shall
have been assigned to 6?
Resume the equation :
ab+a—6= 1
(+)6 abt+a=1+6
But ab-+-a is a taken one more than 6 times; that is, ab-+a =
(1+ d)a.
UB ae
i FIRST DEGREE. 7
Therefore (1+b)a=1+06
ao. 146 _
Chi +b fom baa l
The answer, then, is the following: 6 may be what we please,
provided a be 1.
Verification. If a = 1,
ab+a—b = 6+1—b= 1.
_ We have given this instance to shew how soon the operations of
algebra lead to unexpected results. We will now take another
‘instance. |
6. sy =xrtyt+l
Knowing the value of y, to find that of «.
Le (—)x sy—x=ytl
sut xy—- is x taken once less than y times, or (y—1)a. Therefore
en eee. 4 ond cist
| Particular case. Let y = 5, then
5-4 6 3
| 5 ae a2
as ES 2s
Verification. “Ly = 5x9 iath a
3 did HOR ye 15
zt+ty+l= Short == Rie as sets
1
General Verification. ea a Ly = " ar ,
Ca 0 tet UG eet,
aa
ere yyy yt
= oes
wee eel)
* ee yt
7. Two labourers can separately mow a field in 4 days and 7
‘lays. They begin to work, and on the second day are joined by a
hird, who alone could mow the field in 10 days. The third remains
8 EQUATIONS OF THE
with the former two for a certain time, after which he leaves them;
and it is then found that exactly four-fifths of the field have been
mowed. How many days is this altogether?
[This question is introduced to shew how very soon algebraical
symbols may be made to simplify complicated arithmetical reasoning. |
The fractions of the field which the first and second could mow in
a day are ; and > Let w be the whole number of days; or, that
number being unknown, let 2 stand for it until itis known. Then
the first man, who does one-fourth in one day, two-fourths in two
days, &c., will in « days do 2-fourths, or the fraction 5 of the field.
xv 5
In the same time the second does 25 but the third, who works one
day less, at the rate of one-tenth a day, does — Therefore, all that
is done of the field is
ec. @«t—t1
AN Th
2 + eS : ,
But by the question this is 3? which gives
Gf &£—i 4
407 Ap ee
The least common multiple of 4, 7, 10, and 5, is 140 (Ar. 103).
(x 140 ata +o (a —ly)= ane
or 352+20¢%+14(¢—1) = 112
or 30x4+20¢%+142—14 = 112
[because 14(7—1) = 147—14]
therefore 69x—14 = 112
(+)14 69 =e
1
(+)69 = P=5=15
Verification.
In 1 day and 5 —— > of a day, there is mowed by the first ; and =
1 42 a
of Oh ier —s of rt or = of the field; the second mows md of i or
12
35’ that is — ag ; and the third, who works one day less than the others,
' FIRST DEGREE. 9
i 1 Lisa tghey at 19
or only =; of a day, does in that time on of apc Gamat that is
(38. .
{60 of the field’ But
| | Pieet 88h. WBr Ay eds
: Beet ang AGO ee
at 7 1
rl 64 5
A common multiple of 24 and 6} (not the least, which is 95,
‘and with which the student should also solve the equation) is 570,
which contains the first 228, and the second 90 times. It is alsoa
‘multiple of 5.
(x)670 =P @—3) —"P @—4) = 1710-F w+)
or 228 (aw —3)—90 (w—4) = 1710—114(@ +1)
But 228 (a@—3) = 2287r— 684, &c.
Therefore
(228 x—684)—(90x—360) = 1710—(1142+4114)
or —«-228.x«—684—90x+360 = 1710—1142—114
(+)6844114r—360 2282—90r+1]42
= 1710—114+684—360 or 2522 = 1920
(+)12 2l2 = 160
160 13
Gel oo festa yeaa
Verification. If ¢,=
O70 197
Sota k) 215,194 4494
eh 5 105.) AK OL
Ge
76
e—4 21. 228-228
eae 195 399 19x 21
wy
ew—3 r—4 194 FBG Ha O54G yi =e 184
Wace) 6h). 7 &K21.=«19xX21" 5xi9x21 , 5x21
181
ool ge OS Nob te:
5 5) soem 50 24)
10 EQUATIONS OF THE
3— art _ sp the same as before.
From these cases we may lay down the following rules for the
solution of equations of the first degree.
1. To clear an equation of fractions, multiply both sides by any
common multiple of all the denominators: generally, the least common
multiple is the most convenient.
The following are some useful applications of this principle:
f= (x)bd Va M8 o ad = be
From the preceding equation the student is left to deduce the
following :
cb ad ad be
tactieds * b= -— Car d a
Aken Wee 1 hae pee
a co'b “adc «eda
From the first of the following equations let the student deduce all
the rest.
ab cd cday abpq a cdx
cee = -— = =
ry =p pqe cdy y pqb
ab dxy b Ce
—=-— gbnq = rycda— |]
c PY a J dy apq
2. Any term of an equation may be removed from one side to the
other if its sign be changed. If this have not already occurred to the
student from the preceding examples, it may be established by the
following :
Let a+b =c+d—e
(—)b a=c+d—e—b
(+)e ate =c+d—b
In applying the rule for clearing an equation of fractions, care
must be taken, when the denominator is removed, to remember that
the sign which was placed before the complete fraction now belongs
to the complete numerator, which should, therefore, be placed in
brackets, or the proper rule for addition or subtraction applied at
once. The following example will shew what is meant.
r—a c+@r r—e
or RENO y Bee da
(x)ab abx+a(a—a)—(c+2) = abd—b(x—e)
or abx+(ax—aa)—(c+2) = abd—(ba—be)
or abx+ax—aa—c—xv = abd—bx+be
FIRST DEGREE. 11
The mistake to which the beginner is liable is, to write —c + x
and — bx — be, instead of —c— vr and — bx + be.
By the second of the preceding rules,
aba t+ax+ba—ax=abd+aa+c+be
or (ab+a+b—l1)x =abd+aa+c+be
a bd+aatctb
(—) ab+a+b—1 C= eral eee. (1)
Verification.
_ abd+aa+c+be
aa S| abta+b—1 aac
abd+aa+c+be—a(ab+a+b—1)
abta+b—1
_ abd+aa+c+be—aab—aa—ab+a
En ial ash b— 1 Too wore
abd +a—ab—aah+c+be
a ab+ta+b—1
a—@ _ abd+a—ab—aab+c+be (2)
b b(ab+a+b—1) er a
And by similar processes,
c+e abc+tac+bc+abd+aa-+ be
Bb ab(ab+a+b—1) 8 8— °°" e
r—e abd+aa+c—abe—ae+e 4
| eat a(ab+a+b—1) —"'" (4)
Reduce (1) (2) (3) and (4) to a common denominator, which can
be done by multiplying the numerator and denominator of (1) by a6,
of (2) by a, and of (4) by b. Then form
oo — <**, or (1)-+(2)—(3)
which will be found to be
aabbd+aabd + abbe + abe —aab —abd—bc—be
ab(ab+a+b—1)
aabd+aad+abe+ae—aa—ad—c—e
a(ab+a+b—1)
the latter of which arises from dividing the numerator and denomi-
nator of the former by 6.
By similar processes,
d—-— or d — (4)
or
will be found to have the same value as (1) + (2) — (3).
The student should not pass the preceding solution until he is
able to repeat the whole on paper without the assistance of the book. °
12 EQUATIONS OF THE
In the preceding equation, let c and e be each equal to nothing,
which reduces the equation to
C—t x x
the value of x to
a abd+aa ;
~ ab+a+5oe
and the value of each side of the equation, as just found, to
aabd+aad—aa—ad
a(ab+a+b—1)
abd +ad—a—d (dividing numerator and denominator by a.)
ab+a+b—1
These the student should find for himself from the equation.
We now look into particular problems to see what explanations
may be necessary. Various unforeseen cases will present themselves ;
and each case will be explained, and a problem will be given for
each, to shew how it may arise.
’
Anomaly 1. Let a=26b6=3 d==
The equation then becomes
xr—2 A i
ire
eoenty
© eae 6
_ 2X3 Fh 2ee ee
~ 2X S+2+3 =f oe
1
x
2
Then x 5
On attempting to verify the equation, we see that a contradiction
ned Fesese :
appears; for x is oy and the operation « — 2, which is therefore im-
possible, appears in the second term. It seems, then, that an eqyua-
tion may give a rational solution ; and on attempting to verify the
equation by this solution, the latter may be found to be impossible. The
question now is, can such an equation arise from a problem? if so.
is it the problem itself which is absurd, or the way of treating it!
If the latter, how is the method of solution to be set right ?
PROBLEM IN ILLUSTRATION. A enters into this bargain with B.
that he is to take B’s property and pay his debts, taking his chanc«
of gain or loss. On examination, it is found that B’s property is
(debts allowed for) exactly the same as that of A, with this excep-
tion only, that B is in partnership with another, and he and his
partner have made a similar bargain with C. On examining C’s
affairs, he is found to be insolvent by £100. The result of the whol«
FIRST DEGREE. 13
is, that this transaction falls short by £75 of making A’s property
twice as great as it was. What was A’s property?
Let x stand for A’s original property, which is, therefore, that of
B and Co., independent of their share in the engagement with C.
From the concluding paragraph we might presume that A is benefited
by the transaction, namely, that the £2 of B and his partner is more
than sufficient to cover the loss arising out of their engagement with
C. Let this be so; then x —100 pounds remains for B and his
partner, of which B’s share, namely, 3 (x — 100) is by the bargain
transferred to A, who has, therefore,
——
aX+ es
+ doubles his property all but Oe and is, therefore, the same
‘hing as 24 — 75. Therefore,
2+ = 2u—75
Coe 22+x2—100 = 4x— 150
4% —22—a2=-150 — 100
z = 50
This introduces into the equation the anomaly we are now con-
i. 50—100.. . ‘ ;
sidering, for >> mame impossible. It also contradicts the suppo-
sition on which it was obtained, namely, that x is greater than 100.
We cannot, therefore, depend upon this solution. Suppose we try
he other supposition, namely, that x is not greater than 100. In that
zase, B and his partner have to pay £100, of which they can only
make good £2; of the remainder, or 100—.z, A must, by the
dargain, make good B’s part, or }(100 —2). This he. loses by the
Tansaction ; and having vz at first, he has now only
east 100 —2z
2
Chis doubles his property all but £75, or rather we must now change
his mode of speaking, which may seem to make one part of the
>»roblem disagree with another, and say simply that A’s property is
£75 less than twice what he had before. Hence,
x — = = 2x — 75
ux.) 2 2x2 —(100— x) = 4x — 150
7 Cc
14 EQUATIONS OF THE
or 2x—100+2 = 4x—150
or 2x+x—100 = 4r—150
which is now the same equation as before, and not absurd in its
present form ; for though (since it yields = 50) — 100 is absurd,
yet 22 -+x2—4100 is not so. Hence we see that the effects of the
wrong supposition, which made us write 2 + 4 (a — 100), where
we should have written « — 3 (100 — 2), disappear in resolving the
equation, and leave the same result as we should have obtained by
proceeding correctly.
The anomaly arises from an error of this sort. If 6 be greater
than c, we know that
a—(b—c) = a—b-+c
but if we have a — b +c, and wish to bracket b and c together, we
cannot do this correctly until we know which is the greater. If it be
b, the preceding is a—(b—c) ;
but if it be c, this should be
a+(c—b).
One or other of the preceding two is absurd, except only when 6 = c,
which makes them a — O and a + 0, or a for both.
From hence we may see, so far as one instance can shew it, that
any mistake which amounts to no more than writing a —(b—c)
instead of a +(c— }), as the representative of a— 6 +c, makes no
difference in the final result. We here write, side by side, the solu-
tion of two equations, which only differ as above.
a—x x—b r—a b—«
Lae rer ty ehh’ SR RET ar gay og
ba—(a—x) = bc+(ax—b) bx+(ex—a) = bc—(b—2z)
bx—at+a = bct+a—b bxt+a—a = bc—b+2z
The remaining part is common to both.
bx = bc+a—b
_. bec +a—b
a
Anomaly (2). Let
ax+b =ca+d
ax—cxz = d—b
(a—c)x« = d—b
d—b
a—cC
M 0
I
FIRST DEGREE. 15
“Let it happen that d is less than b, but a greater than c, as in the case
of 92,+-4=2r+1
| There is then an impossible subtraction in the numerator of the
result ; and it is sufficiently evident, on other grounds, that the equa-
tion is impossible ; for if a be greater than c, ar must be greater than
_ex; from which, if 6 be greater than d, ax + b must be greater than
cx +d, and cannot be equal to it. A similar question may now be
‘proposed to that which arose upon the last anomaly. (See p.-12.)
- Prosiem I.1n rttusrration. In the year 1830, A’s age was
"50 and B’s 35. Give the date at which A is twice as old as B.
This must either be before or after 1830. Try the second case,
and let the required date be 1830 + «.
| Then A’s age will be 50 + 2
RE nae 3 jee 3D + 2
and 50 +2 = 2(35+2)
“or 50+2 = 70+227
2a2—x2 = 50—70
Here we see an impossible subtraction, and it is also evident that
‘2x2-+70 must be greater thana+50. Now, try a date before 1830 ;
| say 1830 — x.
Then A’s age was 00 — x
Lik) aera 35 — x
and 50—a = 2(35— 2)
or 50 —az = 70—2z
| 2x—x = 70— 50
or i AN
|
‘An evidently true answer; for in the year 1830 — 20, or 1810, A’s
age was 30 and B’s was 15. Here then we see that an impossible
subtraction may arise from assuming a date to be after a certain epoch
which is in fact before it, or vice versa.
Prosiem II. A and B have accounts together. The state of
their affairs is this: Give A half as much as will make their dealings
worth £500 to him, and give B £100, and they will then, after settling
their account, have equal sums. How does their account stand ?
The balance is either in B’s favour or in A’s. Take the latter,
and suppose A ought to receive £2. Then 500 —z will make this
transaction worth £500 to him, because
16 EQUATIONS OF THE
a +(500—2x) = 500 |
Give him 3 (500 —~) and he will have (when B pays him)
500 —az
t+ 3
Now B, when he gets the £100, must pay £x to A, and will
therefore have £(100—.). But they have then equal sums; therefore,
500—a
x+ 5 = 100—2z
(x)2 22+ (500—2) = 200—2z
or 22+500—a2 = 200—2z
Qa4+227—2 = 200—500
or 3x2 = 200—500
which is impossible. Try the other case, and suppose the balance to
be in B’s favour, and that he ought to receive £x. Then, to make A
worth £500, his debt to B must be paid, and he must receive £500
besides: that is, he must receive 500+ 2. But of this one-half only
is given, or 3} (500+ x), out-of which, when he pays £2 to B, he will
have
500+ x
2
Now B gets £100 and also £x from A, and will therefore have
100 + x pounds. And, since they have then equal sums,
Ot = 100d
CS 500 + «—22 = 200+ 22x
224+ 24—7r = 500 — 200
or ox = 300 and « = 100
Therefore A owes B £100.
Prostem III. A traveller proceeds along a road on which, at
various intervals, are found direction-posts variously numbered, point-
ing north or south. As soon as he reaches No.1, he proceeds in the
direction pointed out by it till he reaches No. 2, and soon. He finds
the first direction-post after he has travelled 16 miles north, and he
finds also that he changes his direction at every post which he meets
after the first ;* that the distance between every two posts is double
* The problem does not say whether he changes his direction at the
first, or not.
FIRST DEGREE. 14
‘that between the preceding, and that, at the fifth post, he is 86 miles
‘north of his first position. What is the arrangement and character of
‘the posts ?
After travelling 16 miles north he reaches a post, and we are not
told whether he is there directed to go on or turn back. Let us
suppose the former, and that he travels 2 miles further north before
hé reaches the second post. He will then be 16+ 2 miles north of
his first position. At the next post he has to turn back, aud proceed
‘2x miles south. Here two cases arise. If 22 be less than 16 +2,
the third post will be north of his first position by 16-+2— 2
miles; but if 22 be greater than 16+, the third post will be
south of his first position by 27—(16+~ 2) miles. Suppose the
first; then between the third and fourth post there are 42 miles,
‘and he has to go north from the third, therefore he will meet the
_ fourth post at 16-++a2—22-+ 4-2 miles north of his first position.
( At the fourth post he turns south, and after proceeding 82 miles,
meets the fifth post, his position north of his first position being then
(16+2—2x+4r—8r miles. But this by the problem is 86
miles: consequently
164+-2-—2274+427—82 = 86
or 82—42+2r—a2 = 16—86
“or 5x2 = 16—86
in which there is an impossible subtraction. Let us now try the other
hypothesis, and suppose that at the first direction-post he has to turn
south, and that be finds the second after proceeding x miles south,
or at 16 — vz north of his first position, or « — 16 south, according as
x is less or greater than 16. By the same attention to the expressed
conditions of the problem, we find that
16—x7+22—424+82 = 86
i 8x—424+2x—2 = 86—16
l hapa
x= 14
Consequently the positions of the posts are as follows :
South ——————— North
| Ales (2)¢¢ £01) £4(3) (5)
26 fae 16 30 86
} 53
Cee
18 EQUATIONS OF THE
Under each is marked the number of miles at which it is from his
first position.
If we collect together and look at the correct and incorrect equa-
tions which we have found in the three preceding problems, we shall
have the following :
Prosiem I.
Incorrect, 50+ 2 = 2(35+4.2) or a = B0—7U years after 1830
Correct, 50—a2 = 2(35—2z) orz = 70—50 .. before ..
Prosiem II.
O—. o0—500 ;
Incorrect, ae 5 r +z= 100—rorz = cae mE 8 whichBowesA
500 +: 500—200
Correct, aio = 100+z2ore = — = Ree aay
Prosceo III.
Incorrect, 16 +a2—2e+4x—82 = 86 orz = milesnorth
: miles south
Correct, 16—x+227—47 + 82 = 86 or Le
From which, as well as from other instances, the following principle
is clear:
When the value of x, deduced from an equation, contains an im-
possible subtraction, both the equation and the meaning of # have
been misunderstood, and require alteration.
1. To correct the equation, alter the sign of every term which
contains 2 once only as a factor.
[The words in italics are inserted to remind the student that we
cannot draw any conclusion from the preceding as to equations
whicb contain such terms as rz, rxrz,&c. All our equations have
been of the first degree. ]
2. To correct the result, invert the terms of the impossible sub-
traction (that is, change 50 —70 into. 70 — 50), and let the quality of
the answer be the direct reverse of that which was supposed when the
incorrect equation was obtained. Thus, change years after into years
befure; property into debi; distance measured in one direction into
that exactly opposite; and so on. Or, whatever alternatives it may
be possible to choose between in assuming 2, provided one be the
direct reverse of the other, then, if one alternative produce an impos-
sible subtraction in the value of x, the other is the one which should
have been chosen.
i
}
: i
FIRST DEGREE. 19
An equation generally obliges us to take a more extensive view of
‘the question than the words of the problem will bear, and will fre-
. quently shew that the view taken of the problem is not in every part
a consistent whole.
In the preceding questions we have taken eare to leave every
possible case open in the statement of the problem: thus we have
said (Problem I.), “‘ Give the date at which A is twice as old as B,”
not “ How long will it be before A is twice as old as B?” because
the latter would be tacitly assuming that the event is to come,
| ‘whereas it would be found out that the event is past, and the implied
statement is erroneous. We write underneath the correct and incor-
rect mode of enunciating the question.
Correct. INCORRECT.
In the year 1830, A’s age was In the year 1830, A’s age was
50 and B’s 35. Give the date 50 and B’s 35. When will A
at which A is twice as old as B. be twice as old as B?
ANSWER. ANSWER.
_ 20 years before 1830, or in Never; but A was twice as
1810. old as B 20 years ago.
The chance of an impossible subtraction occurs in both; but in
the first it arises from a question being left open to the student, who
may choose the wrong alternative; in the second it arises from a
wrong alternative being already tacitly assumed in the problem.
The alternatives presented by a problem may generally be ascer-
"tained with ease; but if not, the equation itself is frequently a guide.
Anomaly 3. Let 32—10 = 2xz—8
3sx—2x2x = 10—8
gh Fes
On verifying this equation we find each side to contain an
s impossible subtraction, for 3r—10 is 6—10, and 2x—8 is 4—8.
After what has been said on the last case, we need not dwell upon
this; the problem in the next page will furnish an instance.
Somewhat similar to this is a mistake in the process which may
introduce an impossible subtraction into both numerator and deno-
minator of the answer. If, in solving aa+b6 =ca#x-+d, thus,
ar—cx =—d—borr= oe? we afterwards find a less than c, and
t
20 EQUATIONS OF THE
d less than 6, it is a sign that we ought to have chosen the process
cx—ax =b—d. This is a mistake in the order of operations only,
and not in the conception of the problem.
Prosiem. Divide the number 13 into two parts, in such a
manner that three times the first may exceed half the second as much
as the first exceeds 4. ;
It will be found from the resulting equation, namely,
(where «x is the first part) that c = 1, and therefore the parts of 13
required should be 1 and 12. But the problem is then impossible,
for three times the first does not exceed half the second. But if the
words “fall short of” be substituted for “‘exceed” throughout the
problem, the equation becomes
13—2
2
the answer to which is 2 = 1 as before, and the problem is possible ;
—37 = Aes
for 32 or 3 falls short of }(13—~.) or 6, as much as 2 or 1 falls
short of 4,
We now ask how it happens than an equation gives a rational
result, by which, when it is tried, the equation itself is proved to be
irrational; and are we to conclude that no answer to an equation
holds good until it has been tried upon the equation, and found to
satisfy it rationally? Let us examine our first instance. The
equation is
3x—10 = 2x—-8
and the answer # = 2, when applied to the equation, gives
6—10 = 4—8
It so happens that the rules for solving an equation give the same
answer to both of the following:
dsx—l10 = 2x—8
10-32 = 8-22
And the following example will shew how this happens:
az—o = cxz—d b—ax = d—cz
ax—cx = b-—d b—d =ax—cez
Liem 4 in both.
= C
Hence, when an anomaly of the kind treated in this article occurs,
FIRST DEGREE. 21
‘it is the sign of a misconception of the right way of viewing the
problem, but of a misconception which no way affects the result.
Anomaly 4. If we first solve the equation
ax+b =ca+d
which gives t= ace and if it should happen
Ge
I
(
:
(without our observing it during the process) that a =c ora—c=0,
we shall then find an answer of the following form,
d—b
L = ——
0
which is unintelligible; because there can be no answer to the ques-
tion, “ How often is nothing contained in d—b?” or, at least, if there
be any answer, it is, “ Nothing, however often it may be repeated,
yields nothing, and therefore cannot be repeated often enough to
yield d—b.” On returning to the equation, we see that the supposi-
tion of a= c gives ar =cz, so that, as far as the equation only is
concerned, it is always true if ) = d, and never true if b is unequal
to d. But in giving further explanation of problems which produce
such equations, we shall employ the following principle, the propriety
of which is obvious.
When any supposition (such for instance as making @ = c in the
preceding equation), makes the results of ordinary rules unintelligible,
then, instead of making a exactly equal to c, let it be made very
nearly equal to c, and observe the result: afterwards suppose it still
nearer to c, and so on; the succession of results will inform us whether
any rational interpretation can be put upon the result of supposing
a=c, or not. We shall now try a problem in which the preceding
difficulty will be found to occur.
Prosiem. There are three trading companies, of 4000, 5000,
and 9000 shares respectively, and all three would, if broken up, pay
the same dividend upon their shares; but if every shareholder in the
second advanced his company £10 on each share, and every share-
holder of the third advanced £12 in like manner, then the first two
companies together would have the same total assets as the third ;
what is the dividend which each company could now pay ?
Let x pounds per share be that dividend. Then, after the advances
‘supposed in the problem, the three companies could pay z, «+10, and
w +12 pounds per share respectively; which, taking their number of
shares into account, supposes them to be in possession of 4000 z,
22 EQUATIONS OF THE
5000 (a + 10), and 9000 (+12) pounds respectively: between
which the last clause of the problem gives the equation
4000a + 5000(x +10) = 9000(« + 12)
(+)1000 42+45(2+10) = 9(a + 12)
or 9x+50 = 9x2+108
In which we recognise the anomaly which is the subject of this
article. Nor can we in this case, as in page 18, account for the
impossibility by supposing that we have mistaken the problem, and
that the three societies are at first insolvent, and are in debt upon
each share ; for, if we make this supposition, and let x be the amount
they fail for upon each share, we have already seen (and the student
must make himself sure of the same in this particular case) that the
resulting equation will be
50—92 = 108 —9z
which is equally impossible with the former. We shall, therefore,
now try the consequences of a slight change in the conditions, agree-
ably to the preceding principle. For instance, we will suppose the
third society to have only 8999 shares, instead of 9000. The equation
then becomes
4000a + 5000(# +10) = 8999 (a +12)
or = 4000 « + 50002 + 50,000 = 8999 «+ 107,988
4000 x + 500012— 89992 = 57,988
or & = "Ofoe
The answer therefore is, that each society could at first pay £57,988
per share. Let us try the effect of a still smaller change, and suppose
the third society to want only the hundredth part of a share of 9000
>
: 9
shares, that is, to have 8999 =a shares. Then we have
4000 «+5000(4+10) = g999 2 706 > (n+ 12)
40002 +5000x + 50,000 = 8999 «+8999 > x x12
4000.x +5000 z—8999 «x = 8999 x 12—50,000
1 899999,
on = soon 12—50,000
(x )100 x = 899,999 x 12—5,000,000
2 = 5,799,988
FIRST DEGREE. pis ae
or each society can pay £5,799,988 per share. In the same way, if
we take the third society at 8999 mies shares, we shall yet have a still
greater answer, and so on. A similar result would be obtained by
increasing the 4000 or 5000 shares of either of the other societies.
Therefore, the answer to the preceding question is, that no number is
‘great enough to satisfy the conditions of the question; but that if these
conditions be slightly altered, an answer may be found, which answer
is a greater number the slighter the alteration just alluded to. Dis-
missing the problem, which we have only introduced to shew that
such anomalies may arise in the application of algebra, we return to
the consideration of similar equations.
The solution of
| Gr Da Te
. Cc
1S 8 ame
a—b
in which, should it happen that a = 6, the answer is unintelligible,
‘being -, and the equation impossible, being
ax =ax+ec
but if a exceed b by any quantity, however small, the equation and
its answer are both rational. Let a exceed b by the fraction of unity
1 !
ae then the equation becomes
(64+-)a= ba+e
or br+— = br+e
(—)dx —=¢ (x)m x = mc
The same might be obtained from the preceding answer, for
: 1
To make a exceed b by a small quantity, ms must be small, that
is, m must be large; and in this way we may get an equation whose
answer shall be as large as we please. For instance, let c be 1; and
suppose we want an equation of the preceding form whose answer
24 EQUATIONS OF THE
1 ‘
. Let m = ————.. Then such equations as
shall be 1,000,000 et m 7,000,000 q
the following have the answer x = 1,000,000.
1
7 = FE
1,000,000 is u
A number which does not satisfy an equation may, we can easily
conceive, nearly satisfy it. But this word nearly is too indefinite for
our purpose, as we shall now shew. Suppose we have the equation
7x = 2x+ 3, and we try whether 2 = 1 will satisfy this equation.
It will not; for the first side is 7 (a being 1), and the second side is
5; that is, the first side, instead of being equal to the second, is
greater by 2. The same result applies, in the same words, to the
equation 7x = 5x + 1988, if we try = 1000 upon it; for the first
side becomes 7000, and the second 6988. Shall we then say, =1
satisfies the first as nearly as = 1000 does the second? Is 7 as
near to 5 as 7000 to 6988? If we look at the differences only, we
must answer yes; for
7—5=2
7000 — 6998 = 2
but in the common use of the word near* (to which it will be conve-
nient to keep) it would be said that 7000 is nearer to 6998 than 7 is
to 5. In the first case, the difference is 2 out of 7000; in the second
I
it is 2 out of 7. Keeping to this meaning of the term, we shall in
future consider ax and ax-+c as nearer to equality when z is greater
than when it is smaller. And in this sense we say, that when a
problem leads to such an equation as
ax = ax+ec
the result is, no number is great enough to be an answer to the
problem ; but the greater any number, the more nearly is it an answer
to the problem.
Tried by the common rules (which, in this case, lead us too far)
the answer to the preceding equation is
e= or
.
a—d 0
* In the making of a bargain, £6998 would be considered as the
same price, within a trifle, as £7000 ; but any thing at £7 would be con-
sidered dear as compared with the same at £5. We might multiply
instances in which the same quantity would be considered small under
some circumstances, and great under others.
FIRST DEGREE. 25
‘ ; c : i
‘d it is customary to say, that 9 means an infinite number, and that
.e answer is infinitely great. Taken literally, such phrases are un-
‘eaning, because we know of no number which is infinitely great,
at is, greater than can be counted or measured. But the word
‘often used ; and we shall, therefore, adopt it with the following
‘caning :
f
: jee :
_ By saying that — is infinite when g = 0, we are to be understood
q
{ meaning no more than a short way of expressing the following :—
‘Ben aie stall - is great; if g become still smaller, : is still
| 1 ;
‘eater, and so on; so that ; may be made to exceed any given
|
‘mber, however great, if g be taken sufficiently small. And when
2 say the answer to a problem is infinite, we mean that no number
. great enough to satisfy the conditions of the question ; but that
y great number nearly does so, a still greater still more nearly,
d so on; so that the problem may be answered within any degree
nearness (short of positive exactness) by taking a sufficiently great
umber.
, Anomaly 5. The solution of
| /_ ax+b=ca+d
d—b
a—c
oo =
_ Now, if it should happen that, after the solution of this equation,
‘becomes necessary to suppose a equal to ¢ (as in the last case), and
also equal to d, the answer
j d— ; 0
os must be written =
‘ich has no meaning, On returning to the equation, we no longer
d any necessity to give x one value rather than another; for, if a
=c, and b = d, then ax + b is equal to cx + d, whatever « may
| Therefore the answer to the question is, that every possible value
‘w satisfies the conditions. We shall apply this in the following
‘ Propiem. Is there any number such that a times one less
‘nthe number, added to 6 times two more than the number, is
actly c times the number; a, 6, and c, being given numbers or
ctions ?
-
26 EQUATIONS OF THE
Let x be that number, then
a(«—1)+b6(@+2) = cx
ar—at+6x44+26 = cr
ax+bx—cx = a—2b
(a+b—c)x =a—2b
_ a—2b
~ atb—c
Verification.
c—3b — ac— 8ab
a ae atb—c a(t—l)= a+b—c
3a—2c . 3ab—2be
e+e He | Oe
ac—2be __ c(a— 2b)
a(x—1)+ 6(2 +2) = a+b a oes
= iGo
following :—What number is that, one less than which multiplied by
8, added to the product of 2 more and 4, is equal to 12 times the
number? Here we find a—26 = 0, and a+b—c = 0; s0 that the
preceding answer takes the form - trying this case by itself, to find
out the reason, we have the equation, 3
8(a—1)+4(% +2) = 124
8x—8+42+8 = 12a
122 = ie
which being always true, the answer is, that every number and frac-
tion whatsoever, which is greater than 1, satisfies the conditions of the
problem ; a result corresponding to the interpretation we have already
0
seen reason to put upon the form iy
We shall now proceed to the solution of some problems, in whicli
the preceding method and principles will be applied. As instances
will be taken from different parts of natural philosophy, we shal
divide them into sections, and state at the head of each the facts or
which the solution depends.
Exampxes.—- Section I. Specific gravities.— By the specifi
eravity of a body is meant the number of times which its weight i:
FIRST DEGREE. OF
of the weight of an equal bulk of water. Thus, when we say that the
“specific gravity of brick is 2, we mean that a mass, say a cubic foot
of brick, weighs twice as much as a cubic foot of water.* A cubic
foot of water weighs about 1000 ounces avoirdupois.+
- Prosiem I. One pint of water is added to three pints of milk
(specific gravity 1°03): what is the specific gravity of the mixture ?
Ifa pint of water weigh m ounces, then one pint of milk weighs
'1:03 xm ounces, and the whole four pints of the mixture weighs
m+ (1°03 m) xX 3, or m+3:09m ounces. But four pints of water
weigh 4m ounces; therefore the specific gravity of the mixture (see
the preceding definition) is
3°09 40 4:
er Oe OF or 102285
i 4m 4m 4
N.B. Here is an instance of a quantity m introduced for conve-
‘nience, and which disappears in the process. The question itself is
too simple to require an equation.
Prosiem II. A number of cubic feet (m) of a substance whose
7 specific gravity is a, is mixed with m cubic feet of another, having the
| specific gravity 6. What is the specific gravity of the mixture ?
| It is plain that the m-+n cubic feet of mixture will be as heavy
as ma+nb cubic feet of water. Therefore the specific gravity re-
‘ . ma+nb
quired is a ae
ma+t+nb
m+n
Prosiem III. How much ofa specific gravity 2 must be mixed
with 20 cubic feet of specific gravity 10, in order that the specific
gravity of the mixture may be 5?
Let 2» be the number of cubic feet in that quantity. Then the
whole 20+. cubic feet of mixture has the same weight as
20x10 +-2~x 2, or 200+ 2-2 cubic feet of water. Therefore the
Exercise. Try to slew that must lie between a and 6,
specific gravity is Oe and
200 F 27 — 5 or 2004-24 = 5(20+2) «. x = 334
20+ 7
Generalisation of the preceding. How much of a specific gravity
* Atmospheric air is often taken as the standard of gases. Water is
about 800 times as heavy as.-air.
+ Easy to recollect, and remarkably near the truth. Let the student
deduce it from Ar. Art. 217.
28 EQUATIONS OF THE
a must be mixed with m cubic feet of a specific gravity b, in order
that the specific gravity of the mixture may be c?
Let « be the quantity required. Then 2 cubic feet of specific
gravity a weigh as much as az cubic feet of water, and m cubic feet
of specific gravity 6 as much as bm cubic feet of water. Hence the
whole m + x cubic feet of mixture weigh as much as bm + ax cubie
feet of water. Hence, as in the particular instance above,
bm+axr ae bm+axr = c(m +x)
M+
= cm+cx
“. bm—cm =cx—ax or (b—c)m = (c—a)x
b—c
le .m
C— a
This is rational when 6 is greater than c, and ¢ greater than a; that
is, when 6—c and c—aare possible. Jt is also rational when 6 —c
and c—a are both impossible; since, in this case, the apparent irra-
tionality arises from our having converted
bm+ax = cm+cx into bm—cem = cx—ax
instead of cm—bm = ax—cx
Cc—_—
and the rational answer is 7 = .m. In this case a is greater
pai
than c, and ¢ greater than b. That is, this problem is rational when
c lies between a and b. If c do not lie between a and b, then a
rational problem is formed, as in page 18, by supposing 2 the direct
reverse of what it was last supposed to be; that is, by supposing the
m cubic feet of specific gravity 6 to allow of the substance of specific
gravity a being subtracted from it, or to be itself a mixture already
containing that substance. That is, solve this problem: How much
ofa specific gravity a must be taken from m cubic feet of specific
gravity 6, so that the specific gravity of the remainder ay be ct
The answer will be found to be
c—b b—c
i m orr=s= Ay /
C—a ai—c
according as c is greater than both a and 8, or less than both. But
here may arise another of the anomalies previously explained. Sup-
pose, for instance, we ask how much ofa specific gravity a (= 10)
must be taken from m (== 20) cubic feet of a specific gravity b(= 6),
in order that the specific gravity of the remainder may be c (== 12)
FIRST DEGREE. | 29
(Call this problem A.) Here, though the problem is evidently im-
‘possible, the answer will be rational, being
c—b — 12—6
Sm = 5-20 = 5-20 = 60
*and the impossibility is detected, not in the form of the answer, but
‘on looking at the problem, with which the answer is inconsistent ; for
60 cubic feet cannot be taken from 20. The equation from which
this answer results is
OTS = 12 or 120-102 = 240—12u
20—v
‘in which, with the answer x = 60, the anomaly 3 explained in p. 19
occurs. On correcting this equation, as done in p. 20, it becomes
| 10z—120 = 12¢—240 or S—** — 12
L— 20
which is derived from the following problem :——-From how much of
specific gravity a(= 10) must m (= 20) cubic feet of specific gravity
b(=6) be taken, in order that the specific gravity of the remainder
may be c(—= 12)? (Call this problem B.)
| Whether the answer 2 = 60 is to be called possible or impossible
depends upon the answer to the following question. Was problem
B within our meaning or not when problem A was proposed ? that is,
did we mean to take the one of the two, A and B, which should turn
out to be rational ; stating A, because we supposed it, before examina-
tion, to be that one ? or did we mean to confine ourselves within the
limits of the literal meaning of A? Because, in the first case, the
answer is, that we have chosen the wrong alternative, that the other
should have been chosen, and that the answer is 2 = 60; in the
‘second case, the answer is that the problem is impossible.*
Prostem 1V.—The specific gravities of gold and silver are
191 and 10}; a coldsmith offers a mass of } of a cubic foot which
' * Thave here stated this purposely, because the matter of convention
‘is more obvious in this problem than in that of page 16, where there is
am even chance of our choosing the wrong alternative at first. The
problem before us will appear strained, simply because the alternatives
of north and south of a post come more frequently into practice than
those of taking a known from an unknown, and an unknown from a
known, mixture. I state this because some writers on algebra seem
to imply, by making this sort of extension rest only on the most obvious
and usual examples, that they wish the student to consider it as not con-
‘ventional, but necessary.
D 2
30° EQUATIONS OF THE _ ;
he asserts to be gold, and which is found to weigh 260 pounds.
Can it be all gold? if not, may it have been adulterated with silver ?
and, in that case, in what proportion were silver and gold mixed ?
Since a cubic foot of water weighs 1000 ounces, and gold is 194
times as heavy as water, a cubic foot of gold weighs 19,250 ounces,
and 4 ofa cubic foot weighs 48123 ounces, or 300 pounds 123 ounces,
Therefore the mass cannot be all gold. Again, a cubic foot of silver
weighs 10,500 ounces, and } of a cubic foot weighs 164 pounds
1 ounce. Therefore the mass is heavier than its bulk of silver, and
lighter than its bulk of gold, and consequently may be a mixture of
the two. In this case, let a be the quantity of gold, 2 being a fraction
ofa cubic foot; then }—~ 2 is the remainder of silver, and 19250 x is
the number of ounces the gold weighs, and 10500 (4—~) the same
for the silver. But the whole weight is 260 pounds, or 4160 ounces ;
therefore
19250 2-+10500(,—2) = 4160
or Hut SA SBS pre 2 ee ih nearly, or LL of a
v 8750 © 1740 aaa J, ON Teer ace
., about 12 parts out of 17 are gold, and the rest silver.
This is a case of the celebrated problem first solved by Archi-
medes,* and may be generalised as follows:—In what proportions
must substances of specific gravities a and b he mixed, so that the
specific gravity of the whole may be c? To one cubic foot of the
first let there be # cubic feet of the second, to produce the mixture
required. Then the 1 cubic foot of the first, weighing @ cubic feet of
water, and « cubic feet of the second, weighing bx cubic feet of
water, the whole 1+ a cubic feet weigh a+b. cubic feet of water.
But since its specific gravity is to be ¢, it weighs c(1+ 2) cubic feet
of water; therefore, |
atbz=c(l+z) o #@=
to which the remarks in pages 28 and 29 apply.
Exampes. Section II. The Lever. Generally speaking, a bar,
when suspended by any point in it, will only rest in one position,
namely, hanging downwards ; but if properly loaded with weights, it
may be made to rest in any position, and is then said to be a lever in
equilibrium, that is, evenly balanced. The rules are, 1. Treat the
* See the Penny Cyclopedia, vol. li. p. 277.
FIRST DEGREE. a
‘weight of the bar itself as if it were all collected in its midddle point.
2. Call the number of pounds* in any weight, multiplied by the
number of feet by which it is removed from the point of suspension,
(the moment of that weight; then a bar will be in equilibrium when
the sum of the moments of the weights on one side of the pivot is the
same as that of the weights on the other side. If the bar be not
suspended by its middle point, the weight of the bar itself must be
taken into account as if it were all collected in the middle point.
If the sum of the moments on one side be not equal to that on the
- other, the side which has the greater sum will preponderate.
. Prositem I. A bar 18 feet long, weighing 40 pounds, has
| weights of 12 and 20 pounds at the two ends. Where must the pivot
| be placed, so that the bar may rest upon it?
I~ le a
Wea hey 40 lb. 20 lb.
Let the weight of the bar (40 lbs.) be collected at the middle
| point C. Then AC=CB= 9 feet. We do not know on which
"side of C to place the pivot, which may produce an incorrectness in
_ the equation similar to that in Anomaly 1. page 12. This, however,
_ will not affect the result. Let the pivot be at D between B and C;
that is, let 12 Ib. at A and 40 lb. at C balance 20 Ib. at B. Let
AD=z feet. Then CD =(x—9) feet, DB=(18—.) feet. The
_ moments of the weights are 12, 40(7—9) and 20(18—2); and by
. the preceding principle,
1224+40(a—9) = 20(18—z), or = 10;
' therefore the pivot is 1 foot to the right of the middle point, and the
~ problem has been rightly interpreted, because x — 9 and 18 —2, tried
by the result, are both possible. If we had imagined D to be on the
left of C, and AD =~ as before, we should have supposed the weights
at C and B conspiring to balance that at A, and DB =18—z as before,
but CD=9—-z instead of r—9. The equation would have been
| 122 = 40(9—2) + 20(18—2)
or 122 — 40(99—x) = 20(18—2) or x = 10
[ Any other units may be substituted for pounds and feet; but care
must be taken to use the same units throughout the whole of each
| problem.
D2 EQUATIONS OF THE
‘differing from the first in having
—40(9—2) instead of +40(a2—9).
The solution « =10, would have shewn us the Anomaly 1, page 12.
To generalise this problem, let the length of the bar be J, its
weight* W, the weights at the left and right extremities P and Q.
Then let AD =a, which gives (supposing the pivot right of C),
CD=a2—3/, DB=/—xz. The equation becomes
Px+W(2#—3l) = Q(l—2z)
; _ —Wi+201 wae
ment. * =2P+2Qq0W = (Pong saa
Exercise. Prove from the value of xv just found that 2 is greater
than, equal to, or less than, $/, according as Q is greater than, equal
to, or less than P,
The preceding problem contains the principle of the steelyard.
Prosiem II. Ifa bar 20 feet long, weighing 6 pounds, be sup-
ported at 9 feet from the left extremity, how must we place upon it~
weights of 16 and 17 pounds, 7 feet apart, so that the whole may be
balanced (the 16 Ibs. being supposed on the left) ?
16 lb. . 6b. 6 17 ie
‘a RRC in a tak a Pee Te eA
A E DC ¥F B
Let C be the middle point, D the pivot, and E and F the places
of the weights. Then AD=Q feet, BD=11 feet, AC =10 feet,
EF=7 feet, DC=1 foot. Let AE=z; then ED=9—2, DF=
AF—AD>=AE+EF—AD=24+7—9=27—2. The system
therefore consists in ,
16 Ib. at dist. 9—x from pivot; moment 16 (9—-2),
which balances
6 lb. at dist. 1 foot from pivot; moment 6 x1 or 6
17 lb. at dist. c—2 feet from pivot; moment 17 (a— 2).
Hence I16(9—x) = 64+17(@—-2) «= 52, feet.
The equation has been rightly formed, for 9—w and 7—2 are
both possible. If we had supposed E and F to fall on the same
side of D, the resulting equation would have been
* That is, W is the number of pounds, or other unit, in its bt hoe
and ( the number of feet, or other unit, in its length.
FIRST DEGREE. 3a
. 16(9—x) +17(2—2) = 6
sor 16 (9—2) = 6—17 (2—2) L= OGG feet:
| from which the same value of x is obtained, but 2—~ is impossible.
Even if we had made the evidently impossible supposition that E and
F are both on the same side of D as C (amounting to supposing that
the weights at E, F, and C are counterbalanced by no weight at all
on the opposite side), the equation which results, treated by ordinary
rules, would present no direct sign of impossibility until we came to
compare the result with the equation. In this case,
[Oo ae a aS a a a ae Ua Gee
A Dew) wk: F 7.5
ifA E=xz, we have DC =1, DE=2—9, DF =(4—9)+-7=27—2;
‘and the moments of the weights at C, E, and F, are therefore 6,
16(v—9) and 17(~w—2). To trace the effect of the preceding
‘supposition, namely, that there is no weight on the left side of the
pivot D, we must see what will follow from supposing
) 6 +16(e—9) +17(2—2) = 0
as if such an equation were possible. This gives
| 6 +162—144 +17x—34 = 0
332—172 = 0 33a = 172 2 = Bal,
the same as before. But the preceding equation is impossible, since
the addition of three quantities must give more than nothing. At the
same time, we see that s—9 is also impossible.
___ We must here make a remark similar to that in page 14. The
equation
x—(c—b)=0, o x+b—c=0
‘is possible, since it merely indicates that r==c—b. But the equation
x+(b—c)=0
‘is impossible. Nevertheless, when 5 is greater than c, x-+(b—c) is
the same thing as x+b—c; and by attempting this conversion
when 0 is less than c, we might be led to the impossible form
{ r+(b— c) = 0,
where we should have adopted the rational form
x—(c—b) = 0.
From this we conclude, that if we meet with
tCt+p = 0,
34 EQUATIONS OF THE
it is a sign that, in forming the quantity p, we have inverted the order
of the terms in a subtraction; that is, we have supposed p was b—c
when it should have been c—b. Let us call the latter g; then if
we had proceeded correctly, we should have had
x—-q=0, o x+=q.
The problem in page 32, generalised, is as follows. A bar / feet
in length, weighing W pounds, is supported at a feet from its left
extremity. How must we place P and Q pounds (P being on the
left) m feet asunder, so that the bar may be balanced? The equation is
P(a—x) = W (3l—a) + Q(a+m—a)
a Pa+Qa+Wa—i4{W/—Qm
=~ Or
Exercisr. Supposing the bar to be continued to the left of A
and the right of B, but the continuation on either side to have no weight,
explain, as in page 33, the case where W=20 pounds, /=50 feet,
a=5 feet, P=4 pounds, Q=7 pounds, m=10 feet.
Examptes. Section III. Miscellaneous. Prositem I.—A straight
line AB, 10 inches long (continued both ways), is cut by the point
C at 7 inches to the right of A. Where must the point D be placed,
so that AC may bear the same proportion* to CB which AD bears
to DB?
A Cee D
Here AC =7 inches, CB=3 inches. The point D must be
either between A and B, or to the right of B, or to the left of A. Or
there may be (for any thing we have shewn to the contrary), more
than one such point; for instance, one such point to the right of A,
and another to the left. But if we consider the conditions of the
problem, it will appear impossible that D can lie any where but to
the right of B. For, suppose D to be placed between A and B, say
between C and B; then, according to the problem, AD (greater than
7 inches) contains BD (less than 3 inches), the same times and parts
ofa time which AC (7 inches) contains BC (3 inches). This, the
* To those who have never used this term mathematically (see
Ar. 177, 178), we may state, that a bears to b the same proportion
: : : c
which ¢ bears to d, when the fraction - is the same as 7
-_——
FIRST DEGREE. 35
‘least consideration will shew, cannot be. Fora similar reason (this
-let the student explain), D cannot lie between A and C. Neither can
| D lie on the left of A; for then, of AD and DB, AD the less must
contain DB the greater, as AC (7 inches) contains CB (3 inches),
or 24 times, which is also impossible. Let us then suppose D on
the right of B, and let AD =~, inches; then BD=a—10 inches.
By the problem, x bears to x—10 the same proportion as 7 bears
to 3; that is,
x 7
( 2—10 > 3
(x )3 (4% —10) D2, ==, KL — 1 OF
: : 35 :
which gives « = on 17% inches.
If we had supposed the point D to lie between A and B (say
' between C and B), then AD being x, DB would have been 10—z,
and the equation would have been
L a : :
se a 3 which gives x =7;
‘that is, D coincides with C. This is a case which we have not put
among what we have called anomalies, because the result, though not
expected, is intelligible without further explanation. It implies, that
if we would place D between A and B, so that AD should be to
‘DB in the same proportion as AC to CB, D must occupy the
same place as C. But if we suppose D to lie on the left of A,
and let x stand for AD, we have DB =10-+2, and the equation
_ becomes
> aE
= 7 pa
) ——_ = =, or CN ies |
10427 3? see 0,
| which presents the anomaly in pages 14..19. This shews that we
have measured x or AD in the wrong direction, and that if it had
) been made .to fall to the right of A, the equation would have been
7 7z—3x2 = 70, or 2% = 173 inches,
' which we have found before.
Now let the problem be altered so that C shall stand in the middle
} between A and B. For instance, let AB =12 inches, and AC =6
‘inches. Is there a point D, not coinciding with C, such that AD
_ dears to DB the same proportion as AC to CB? Certainly not ; for
AC contains CB once exactly, but if D be placed right of B or left
of A, AD is greater or less than DB. But ifa point be placed at a
4
36 EQUATIONS OF THE
great distance from A or B on either side, AD contains DB very
nearly once (see p. 24); and by removing D to a sufficient distance,
AD may be made to contain D as nearly * once as we please. We
might therefore expect, if we attempt to find D in this case by an
equation, such an anomaly as that in page 21, explained in page 24.
If possible, let the point D, on the right of B, satisfy the conditions
of this problem. Let AD=vx. Then DB=2r—12 and AC=6,
CB=6. Therefore the equation of the problem is
& 6 : .
—a =G= 1 or w= a@—12 (See page 21.)
The generalisation of the preceding problem, in supposing AB=a,
AC =), and AD=za, and placing D on the right of B, gives the
equation
eee b Pe oy ab
2 lonelt, teat beel) mae ee,
In order to meet all the cases which may occur in the application
of algebra, we will now take a problem in which the answer will go
beyond the notion which was formed when the problem was proposed,
not because the thing proposed to be done is impossible, but because
the answer is not within the limits of what is usually necessary or
convenient. For instance, in ordinary arithmetic, a figure placed on
the right of another means that it is to be multiplied by ten before
adding it to the other. Thus 24 is 2x10+4. We do not commonly
use fractions in the same way : thus 23 4 never stands for 23% 10-+4,
or 29; but it might do so if we pleased ; similarly 31 23 might stand.
for 33x 10-+23, or 343. We now propose the following
Prositem II. What is that number, consisting of two digits the
sum of which is 10, and which is doubled by inverting the digits ?
We see that 91 is not double of 19, nor 82 of 28, nor 73 of 37,
nor 64 of 46. Therefore, with the restrictions on the decimal system
usually adhered to in practice, the problem is impossible. What sort
of answer then are we to expect if we reduce the problem to an
equation? Let x and y be the digits in question: then
x+y =10, or y=10—z2.
* For instance, place D on the right at a thousand times the distance
of B from A. Then AD is to DB as 1001 to 1000, or in the proportion
il een .
of 1+ ooo '° 1) which is very nearly the proportion of 1 to 1.
FIRST DEGREE. i
The number formed by placing the digit # before y, is 102 + y
‘ust as 24 is 2X 10+ 4, 58 is 5 x 10 + 8, &c.); and when the
‘igits are reversed, the number is 107+ 2 (just as 42 is 4x10 +2,
|5 is 8X10+5, &c.). By the problem, the second is double of the
rst; that is,
l0y+a2 = 2(10r+y), or 20x%+4+2y.
herefore 10y—2y = 20x—az, or 8y = 192.
ut y = 10—2, or 8(110—2z) = 192;
hierefore, z= >= 2, y = 10—27 = 7.
The answer therefore is, that if it be understood that none but the
: sual single digits shall be placed in the columns of units and tens,
ie problem is impossible ; but that, if the method of writing numbers
e extended, so that a fraction placed before a fraction shall be con-
dered as meaning 10 times as much as when it stands alone, then
ie problem is possible, and the direct and inverted numbers are,
haere
2° 75a and eats
ik i a means 22 x104+75 and is 36 =
71> 2— means 7 x 10+2=° and is 73
,1e second of which is double of the first.
We may then lay down the following: When a problem has a
‘actional answer, that answer can only be used on the supposition
iat any usual method of combining whole numbers, which is neces-
wy in forming the equation, shall also be applied to fractions.
This principle, when introduced into common life, often induces
'3 to suppose fractional parts of things which, in the strict and ori-
nal meaning of the terms, have no fractional parts. Thus there is,
roperly speaking, no such thing as half a horse: there may be half
\\e body of a horse (as to bulk or weight), half the power of a horse
‘vhich is but the half of a certain pressure), or there may be a horse
“half the size, half the power, &c. of another horse—that is, we may
\ilve any quality of a horse which can be represented by numbers ;
/at not the complete idea which we attach to the word, because it
|mtains notions which have no reference to number or quantity.
az ;
38 EQUATIONS OF THE
Nevertheless, we do not speak absurdly when we talk of a stean
engine of the power of 203 horses, because it is there only the horses
power that we speak of, which can be numbered in pounds of weight
or of wolves eating half a horse, because we then speak only of
weight of flesh. Thus the problem—A horse can draw two tons
certain horses drew 5 tons, how many were there ?—is absurd in th
strictest sense, but not so if we confine ourselves to the only qualit
of a horse which is concerned in the problem, namely, power ¢
drawing: and we may either say, there was 23 times as much powc
as a horse possesses, or the power of 23 horses. The same remark
would apply to the following: The reckoning came to £5, and th
share of each person is £2, how many persons were there ?— whic
cannot be solved in the strictest meaning of the words, but in whic
we may say that the whole reckoning is 23 as much as that of on
person, or that of 2} persons.*
ProsiEM III. There are two pieces of cloth of a and a’ yards i
length. The owner sells the same number of yards of both sorts |
b and b’ shillings per yard. Ifthe remainders were then sold, of tl
first at c shillings a yard, of the second at c’ shillings a yard, the tot:
prices of the two pieces would be the same. What number of yar
was first sold of each ?
N.B. To avoid using too many letters, it is usual to employ tl
same letter with one or more accents, to signify different numbe
which have some common point of meaning. Thus a and a’ are tl
lengths of two pieces of cloth, b and 0’ the prices per yard of the fir
pieces taken from each, and ¢ and c’ the prices per yard of t
remainders. But a and a’ differ as much in meaning as to the nur
bers they may stand for, as a and 0; either may stand for any numb
named.
Let x be the number of yards first cut off from each piece; th
the remainders are a— wz and a’—vw yards. The sums received |
the first are therefore bw and b’2 shillings; for the remainde
¢(a—2) and c’ (@—2) shillings. Consequently, by the conditions
the question,
* So when we say that the yearly mortality of a country is 1 out
403 persons, we mean 2 out of 81.
+ The symbol a’ may be read a accented, a’, a twice accented, and
on. Buta dash, a two dash, &c. are shorter, though not quite so corm
in grammar,
hs FIRST DEGREE. 39
} be+c(a—x) = Va+c(a'—2z)
bxa+ac—ca = bVat+acd—cxz
ba—cxt+cr—J'xr = dc —ac
| (6+¢—b'—c)x = ac —ac
ae —adac
> py aa ay Rag ar
_ Suppose we try to apply this to the following case: Let the num-
ver of yards be 60 and 80; let the prices of the number of yards
aken from each at first be 10 and 9 shillings a yard; and let the
omices of the remainders be 4 and 3 shillings a yard. We have then
w=60 a =80 bB=10 B=9 c=4 ¢=3
; atgatl ‘ac—ac _ 80xX3—60x4 __ 0
b+c—b'—c” 104+3—9—4 #0
im anomaly already discussed in page 25. We have there seen that
t implies that any value of x will solve the equation, and this we
shall find to be the case in the present instance. For if we return to
the equation, we find it becomes
107 +4(60—r) = 9r+3(80—27)
or 107+240 —42 = 974+240—32
or 6x2+240 = 6%+240
which is true for all values of z. Hence the answer is, that in this
particular case the total prices of the two pieces are the same whatever
{
number of yards be first cut off.
Let us now try another case. Let the pieces be 60 and 80 yards,
as in the preceding ; but let the first pieces cut off be sold at 5 and
4 shillings a yard, and the second at 2 and 3 shillings a yard; then
meeouea = 50 b=5 b=4 c=2 c=3
dé —ac_ __ 80X3—60X2 __ 120 _ gg
aero b—c , 5+3—4—2 2
the number of yards cut from both is 60 ; that is, the whole of the first
piece is taken, and 60 yards of the second, which are sold at 5 and
4 shillings a yard (giving 300 and 240 shillings). Then the re-
mainder of the second (we need not mention the remainder of the
first, which being nothing, brings nothing), 20 yards, sold at 3 shillings
a yard, brings 60 shillings. The produce of the first is 300, of the
second 240 + 60 shillings, both the same, as the problem requires.
;
AQ EQUATIONS OF THE
The third case we will take is as follows: Let the pieces be 60
and 80 yards; let the pieces cut off be sold for 7 and 3 shillings a
yard, and the remainders for 5 and 2 shillings a yard,
a=60 a =80 6=7 B=3 ec=6 C=2
80Xx2—60x5 160—300
TFS 3—5 ee
and contains an impossible subtraction. From the conclusions in
page 18, we must suppose some wrong alternative has been employed
in choosing «. But, on looking at the problem, no such thing
appears; « yards are to be first sold of each. But the problems in
page 18 were purposely* put in a form in which the alternatives were
obvious ; how then are we to widen the expressions used in stating
this problem so that the problem, as we have given it, may be only
one case out of two or more? Remember that we alter no number,
but only the quality of the result. The seller begins with 60 and 80
yards of the two cloths in his possession, and ends with none of
either, having in his pocket the same receipts from both pieces. Our
problem says he first sells a certain number of yards from both, and
the answer upon this supposition shews the problem to be impossible.
We have been previously directed in such cases to alter the quality of
the result; let us do this, and suppose he begins by buying the same
quantity of both. We must preserve the condition that he begins
with 60 yards of the first; and ends with none; therefore, if he begin
by buying 10 yards more, he must sell the whole 70. If so, he also
buys 10 yards more of the second sort, and sells the whole 90. But
as we are to alter none of the numbers, but only change their names,
if he buy more he buys at 7 (0) and 3 (0’) shillings a yard; and
when he gets rid of all he has (not all he has eft, for that belongs to
one particular alternative), he sells at 5 and 2 shillings a yard.
Therefore the wider problem, of which the one proposed is one of the
alternatives, is as follows:
* We may here remark that the extension of a problem appears
natural or not according to the idiom of the language in which it is
expressed. Thus, comparing together the case here given, and that in
page 19, the former appears forced, because we have no very common
word to denote either buying or selling as the case may be; the latter
appears natural, because the words “ give the date,” implying asking for
a time, either before or after a given epoch, are perfectly consistent with
“
FIRST DEGREE.
ExTENDED ProBLeEm.
There are two pieces of cloth
' aand d’ yards in length. The
a
wt &
owner concludes a bargain re-
specting the same number of
yards of both sorts at b and 0’
shillings per yard. If his stock
of both were then sold, of the
_ first at c shillings a yard, of the
-second at c’ shillings a yard,
4]
CasE FIRST PROPOSED.
There are two pieces of cloth
a and a’ yards in length. The
owner sel/s the same number of
yards of both sorts at 5 and 0’
shillings per yard. If the re-
mainders were then sold, of the
first at c shillings a yard, of the
second at c’ shillings a yard, the
total prices of the two pieces
the total results of the transac- would be the same.
- tions in each sort of cloth would
or x
be the same.
The case first proposed leads, as before, to the equation
x)+0'x
a'c! — ac
c(a—x)+b4 = c(a—
|
_ but the general case leads to this equation only when the owner is the
seller in the bargain mentioned. If he be the buyer, he first pays bx
and b’x for what he buys of each sort, and then sells the stocks a+ x
and a’+ atc andc’ shillings per yard. Therefore c(a+«)—bwx
_ is the balance in his favour from the first, ay e'(a’+x)—b'x from
the second. The equation is
c(a+2)—ba = c(a+2)—bx
i yy ace—a'c
¥ meri lg a
the idiom of our language. But if we translated these two problems into
a language, in which there was a word in common use, such as trafficking,
_ of the science, though some regard it as a defect.
_ decide this point for himself, when he has had sufficient experience of
to denote either buying or selling, and in which there was no usual way
of asking for a date, without implying either before or after some other
date, then the present extension would Ege natural, and the one in
page 19 forced.
The equations of algebra of course take no cognisance of such differ-
ences of idiom, which is generally considered one of the great advantages
The student must
the advantages and disadvantages arising from such extensions.
E2
42 EQUATIONS OF THE
which differs from the former, consistently with the rules in page 18,
by an alteration of the sign of every term which contains 2, and an
inversion of the subtraction a’c’— ac in the result. When
a=60 a@=80 b6=7 VW=3 c=5 c=2
we have already tried the alternative of supposing the owner to be
seller in the first transaction, and have found an impossible subtrac-
tion 160— 300 in the result. If we now try the second alternative,
and suppose him to be buyer, we shall get the rational answer
300—160 or 140; which will be found to satisfy the problem. We
might enter upon various other cases of the same question, but we
shall leave them for the present to the student, and state a problem
which is in all respects analogous to the preceding, and presents
similar alternatives in a different form. The equations of the two
problems will be the same.
Prosiem IV. The paper is a map of a country, of which AB is
a level frontier. All the roads rise from left to right, and fall from
right to left, and all the miles mentioned are meant to be measured
perpendicularly to the frontier on a level with it. CD isa parallel
to the frontier (whether right or left of it is not stated), and T and
V are towns on, above, or below (as the case may be), the line BD
perpendicular to the frontier. P and Q are two frontier towns;
R and S two towns, on, above, or below the points R and S, according
to the position of CD. The roads rise or sink, as the case may be,
a number of inches per mile for every level mile from the frontier,
as follows :
From P to R b inches per mile.
From Q to S§ D’ inches per mile.
From R to V c inches per mile.
From § to T c’ inches per mile.
Now T and V are on the same level, BV is a level miles, and BT
a’ of the same. Required the distance BD and its direction.
FIRST DEGREE. 43
We have given this as an exercise, and shall merely point out the
‘esulting equations on all the different suppositions. In all of them
e stands for BD in miles.
1. If D lie to the left of B,
‘ither c(a+az)—ba = c'(a+2)—b'x according as TV is
Dy bx—c(a +2) =Ua— c(a’ + x) \ above or below PQ.
2. If D lie between B and T (as in the figure),
c(a—x)+bx4 = c'(a'— 2) 402.
3. If D lie between T and V,
ba—c(a—a) = b'x+¢'(a'— x) when a’ is greater than a,
w bx+e(a—x) = b'a—c' (x — a’) when a is greater than a’.
In the figure a is greater than a’.
4, When D lies to the right both of T and V,
bx—c(x—a) = b'x—c (x—a’)
Ir c(a—a) —bx = c(a—a)—b'x
he first when TV is above, the second when TV is below, the level
if PQ. Only one of these seven equations can be altogether true;
nd as only the alternative in the set marked 3 is directly given in
he conditions of the problem, six equations may need examination.
3ut after the explanation of the anomalies (1), (2), and (3), any one
if the six equations may be made to give the true answer. The
nomaly (4) will be found to arise when b + c’=b’+ c, and (5) when
n addition to this, a’c’ = ac.
The student may make this problem an illustration of every case
vhich we have found to arise.
Before proceeding to other forms of equations, we shall now,
aving found impossible subtractions arise in the solutions of problems,
nd having seen the method of interpreting them, proceed to the
nvestigation of rules, hy which the interpretation may be deferred to
ny stage we please of the processes, and by which the symbols of
mpossibility may be used as if they were real numbers, without
reating error,
44 ON THE SYMBOLS OF ALGEBRA.
CHAPTER Il.
ON THE SYMBOLS OF ALGEBRA, AS DISTINGUISHED FROM
THOSE OF ARITHMETIC.
Sympots such as 50—70 (page 15), 200—500 (page 16), have
received the name of negative quantities. This expression is not a
correct one, because 50 —70 is not a quantity, but the contradiction
which arises out of writing down directions to do that which is
impossible to be done. But we have seen, pages 14..19, that
50—70, when it is the answer to a problem, implies 20 things of
some sort, as much as 70—50, but 20 things of a nature directly
the contrary of the things first supposed; so that in forming the
correct equations, these things diminish that which they increased in
the incorrect one, and increase that which they diminished.
Conceive a problem consisting of various steps, such as would
arise from joining two or more problems together, in such a way that
the result of the first must be known before we can proceed with the
second. If we solve the first problem, and find by an impossible
result that we have chosen a wrong alternative, we should first retrace
our steps, set the matter right, and when we have obtained an
intelligible result, proceed to the second problem. But, we may ask,
are there no rules by which this may be avoided, and by which we
may continue the process, as if the result obtained had been rational ?
To try this question, we must examine the consequences of proceeding
with the symbols of impossible subtraction, to see what will come of
applying those processes which have been demonstrated to be true of
absolute numbers.
If we look at a problem which presents alternatives, we shall see
that not only will the answers be of different kinds, according as one
or other alternative is the true one, but the methods in which the
unknown answer (v) must be treated, in order to form the correct
equation, are different in the two different cases. Thus, in page 15,
Problem I., when x was years after 1830, the processes into which #
entered were 50+ and 35+.2; but when 2 was years before 1830,
ON THE SYMBOLS OF ALGEBRA. 45
hese processes were 50—az and 35—a. And this brings us to
iefine what we mean by different problems as distinguished from
lifferent alternatives of the same problem.
This distinction has been already tacitly laid down; for, when we
ame to an impossible subtraction a—b, we always looked for that
aodification of the problem, which, not changing absolute numbers,
ut only the sense in which they were taken, would have given b—a
s the answer. Or when we came to Anomaly I., page 12, in which
ot the answer, but the verification of it, was impossible, we chose
hat modification which, without altering absolute numbers, inverted
ae impossible subtractions in the equation. Hence, without per-
eiving it, we have been led to make use of the following definition,
in which, however, it must be recollected, that we have obtained it
ntirely from equations of the first degree, and that it must therefore
considered as limited to such equations, and possibly capable of
xtension for equations of higher orders.| It is convenient to con-
ider those problems only as different alternatives of the same problem,
a which, 1. the absolute numbers employed are the same; 2. the
quations only differ in having inverted terms with different signs
ee TE)
4 instead of + 2 ;
aving the signs of the unknown quantity altered throughout (as
O—«r=2(35—z2) instead of 50+2—=2(35+4-~2) in page 15);
the answers are either the same in both, or else only differ in the
aversion of a subtraction (such as 70—50 instead of 50—70, in p. 15).
We might propose a problem appearing to have alternatives, but
vhich, in this view, is two different problems combined together.
‘or instance, A has £60, and is to receive the absolute balance that
ppears in B’s books, whether for or against B; but C, who has
1200, is to take B’s property, and pay his debts. After doing this
t is found that C’s property is three times that of A. What is the
bsolute balance for or against B?
| 100
such as —
in page 13), or else in
_ IfB have x pounds, the equation is
3(60 +2) = 200+2 or x =10
If Bowe x pounds, the equation is
3(60+2) = 200—2% o «a =5
We have here two different equations not reducible one to the
‘ther by any of the changes allowed in the preceding definition ; and
46 ON THE SYMBOLS OF ALGEBRA.
therefore, so far as the preceding is considered as reducible to equa-
tions of the first degree, the two cases are distinct problems.
The step we now make is to apply the rules of algebra to the
symbols of impossible subtractions, and afterwards to proceed cor-
rectly with the inverted subtractions, that we may see, by comparing
the results, whether the errors committed may be corrected or not by
simple and general rules. And since we know that such a subtraction
as 3—7 will, when set right, be 7—3 or 4, let us denote it for the
present by 4, in which the bar written above 4 is not* a sign of sub-
traction, but a warning that we are using the inverted form 3—7
instead of 7—3. Thus we might say, that 10—14 should also be
denoted by 4; but here we must stop until we have some further
assurance upon this point; for we cannot as yet reason upon such
symbols as 3—7 and 10—14, since they represent no quantity
imaginable, and we have not yet deduced rules. All we can do is
to go to the source from whence they came, and see whether, by the
same means which gave 3—7, we might have got 10—14 in its
place.
Suppose a problem, wrongly expressed or understood, gives
2r—xr=50—70 (as in page 15). Ifwe take the correct equation
50—ax =2(35—.2), we may add any quantity to both sides. Say
we add a to both sides, which gives
(50+a)—2 = (70+a)—22
the solution of which is
22—x = (70+a)—(50+a) = 20
If we had taken the incorrect form 50+a—=2 (35+), and added
a to both sides, we should, by what we suppose to be strict reasoning,
till the result undeceives us, arrive at the conclusion # = (50 +a)
—(70+a). And this is what we want to ascertain, namely, that
just as in the rational form of an equation, we may by previous legiti-
mate alterations obtain the answer in the form 70—50, 71—51,
72 —52, &c.: so, in the irrational form, we may, by the same sort
of process, obtain 51—71, 52—72, &c., instead of 50—70. And
we will not say, that 51—71 is equal to 50—70, because our
* The sign of subtraction is a bar written before the quantity to be
subtracted. The present is not a sign of subtraction any more than the |
bar between the numerator and denominator of a fraction.
ON THE SYMBOLS OF ALGEBRA. 47
‘notion of equal (as far as we have yet gone) applies only to mag-
‘nitude, number, bulk, &c. &c.; but because any equation which
. gives 50 — 70, might also have been made to give 51 — 71, &c., we
_ will call these equivalent to 50 — 70, meaning by the word equivalent,
that the first may stand in the place of the second, or be substituted
for it, without producing any error when we come to correct the
vesult, or any set of operations for which correct substitutes cannot
be found. Thus, then, we say, that O—1,1—2, 2—3, &c., are all
equivalent, and may be represented by 1; a—(a-+6) and (a+z) —
(a+c-+2) are equivalent, and are represented by c; the rule always
being ;—invert the subtraction, and place a bar over the result.
Thus, if we obtain such an equation as
| z+a+b=0
_which is the form most obviously impossible of all, we shall, dy
rules only, obtain the expression
x =0—(a+d)
' which we signify by (a + 0)
| In considering equations of the first degree, we may confine our-
selves to the rational form r—a = 0, and the irrational form «+a=0.
For to these all others can be reduced. For instance (page 5)
equation 4 is reduced to
12
and the first equation of Prob. I., in page 15, is reduced to
Ete = 0
We now find equivalent forms for
- = _- = - = a
a+b a—b axb a
The first will arise from such an equation as the following :
z+(pt+a)—p+q+6)=¢
from which, if we solve it before observing that it is impossible, we
|
have
a= p—(pta)t+q—(¢+b)=a+b
But the same rules of solution will also give
x=p+q—(p+qg+atb) = (a+b)
’
48 ON THE SYMBOLS OF ALGEBRA.
In which the bar over a+ signifies that we have attempted to
subtract from a quantity another which is greater than itself by
a+b. Or we have .
a + 6 is equivalent to (a + b)
Similarly the equation
v+(p+a)—p+q = (q+?)
gives eS p—(pt+a)—(q—(q¢+8)) = a—b
following rules only. But this equation is not always impossible,
for it is equivalent to
x+p+a—p+q = qtb
or x+az=b whichgivese x = b—a
Therefore, a—b is correctly b—a when b is greater than a; when
b is less than a, itis (a—b). We may also give the equivalent form
b + a, which follows from the attempt to solve correctly.
t+(p+a)—p = 6b
in the form t= b+ (p—(p+a))
We have as yet obtained no such expressions as
ab or
oi] 81
but we shall now shew that these arise from inattention to the equa-
tion, not to the problem. That is, we shall deduce ad and ab,
a a sha. :
=a and 5? oF similar forms, from the same equation, whether that
}
equation be true or false.
If p be greater than g, and c greater than d, the multiplication of
the rational expressions p—g and c—d, and the attempt at multipli-
cation of the irrational expressions g—p and d—c, give the same
result, as follows :
P-4d Y—Pp
c—d d—c
pe—qe qd—pd
pd—qd qce—pe
Subtract pce—qce—pd+qd qd—pd—qce+pe
ON THE SYMBOLS OF ALGEBRA. 49
‘shich are the same in every thing but the order of their terms.
“onsequently, the equation
x+qce+pd = pe+qd
aight, by inattention, be thus solved
| a= pe+gd—qce—pd = (q—p)(d—c)
vhere the latter should have been
a (p—7)(¢— a)
vhence, if p—g be a and c—d be 5, the expression ab may be
jbtained instead of the quantity ab.
We have already seen, in page 19, how it may happen that —" ig
b
a
ntroduced in the place of 5
‘han d; then
Let p be greater than g, and c greater
cr+q=dx+p
sorrectly solved, gives
. (c—d)x = p—gq roma
ncorrectly solved, gives
al, eg wy LP
(d—c)x pe b? 7 Ears
whence, if p—g be a, and d—c be b, F is equivalent to ;
We have thus determined either equivalent forms or corrections,
n all the cases which arise from the equation, considered without
eference to the problem from which it was derived. We shall
‘now consider both, taking as a basis what we have ascertained by
sxamples, namely, that the misconception which produces a instead
‘of a, also causes us to add where we should subtract, and subtract
where we should add, in forming our equations.
Suppose we have obtained a as the result of an equation, and
that the next step (if all before had been correct) would have been to
add a toc, or to findc +a.
Correction and Process. The true result is a, but the same mis-
conception by which a@ was produced, has made us suppose this
quantity was to be added where it was in reality to be subtracted,
and vice versd (pages 15-18); therefore the next step corrected is
é—a or c+a is equivalent to c—a.
F
50 ON THE SYMBOLS OF ALGEBRA.
Trial of the incorrect Process. a is the representative of z—(z-+-a),
and c +a (applying rules only) is c + z—(z +4), orc +z—2—a,
or ¢—da, which has been shewn to be correct.
Similarly, where c —a occurs, we know that the true answer is a,
but that the misconception which produced a leads us to suppose it
should be subtracted where it should be added; therefore c + a is
the true result. The incorrect process gives c— (z—(2 _ a)), or
e—2z+(z¢+a), orc—z+2+a, orc+a. We have, then, these
two rules:
C +d. is equivalent to C—4
C— GZ eoeereercer even cta
By similar reasoning (p— a) x (q — 6) is equivalent to (p + a)
(g+b). The incorrect process gives pg—qa—pb-+ab. If we con-
sider a and b as resulting from the misunderstanding ofa problem, we
must take some such problem as a guide. Let it be the following :
A and B have respectively 4 and 5 pounds. One of them loses a
bet to the other, after which the product of the number of pounds
belonging to each is 18. Which lost, and how much?
If we suppose A lost 2 pounds, the equation evidently is
(4—2)(5+2) = 18
But if we suppose A gained wv pounds, the equation will be
(4+2)(5—2z) = 18
First ALTERNATIVE. SEcOND ALTERNATIVE.
5 — 4+
5b+a4 | 5—w
20— 5a 20+ 5%
Av —~-2XX 47+ 72
Add 20—a«—wwu Subtract 20 +a7—vwer
20—xr—ar = 18 20+a—rxr=18
vx + r= 2 VLI—L = 2
From this we see, that in this instance (and trial proves the same
in others of the same kind) the correct equations belonging to different
alternatives have different signs in the terms which are multiplied by
x only once (or contain the first power of x), or (page 1) are of the
first degree with respect to 2; but that the terms which contain 22, or
~
'
ON THE SYMBOLS OF ALGEBRA. ok
(we of the second degree with respect to x, have not different signs.
‘And the same may be proved of terms containing the product of two
wknown quantities x and y, which are not the same; for instance,
_n the equations
and (4+ 2)(5—y) = 18
If we take a term such as abc, we shall find that, on cor-
‘ection, the sign preceding it must be changed; that in such a term
‘
is abed, the preceding sign is not changed: the rule being,
change the sign where there is an odd number of factors to be
sorrected. It is indifferent whether the factors to be corrected be in
| 1 : ; ab
‘he numerator or denominator: if, for example, we take —, we
c
pe Be: A
may, by deducing ss instead of x from the equation which gave c,
ae i
find 7 instead of 3? as follows. Suppose the equation which
‘ives c to be
2e+(c+2) =x#+2:
his, following only rules, may be finally represented either by
2a—2% = 2—(Cc+2) or = C
x by xa—2x or (l—2)e4 = c+2z-—2z =C
: = se
hat is, a = (—)cx =;
[racing the consequences of this result, we find that the uncorrected
: a 1 ab.
orm * is equivalent to-—. Hence the term — is equivalent to
c c c
ib x =, or hah which contains all the uncorrected factors in the
jumerator.
If we now resume (p —a) (g —), of which the continuation of
he incorrect process is pg — p 6 —qa+ab, we must conclude, that,
a the terms of the first degree with respect to a and 6, the signs
‘aust be altered ; but that in the term aé the sign preceding it must
sot be altered ; so that the correction gives
patpbt+qatab
52 ON THE SYMBOLS OF ALGEBRA.
But this is the same as would have arisen if the process had been
corrected one step earlier, as in page 50; that is, if (p+ a) {q+ 5)
had been written at once for (p—a) (q—6)-
It is not necessary to go through all the individual cases that
might arise. We have found in all that the common rules of algebra
may be applied without error to the expressions of impossible sub-
tractions ; that is to say, the correction may be deferred as long as we
please without introducing error, provided that, when at last the cor-
rection is made, the following rule be observed :—In correcting any
term, change the symbols of impossible subtraction by substituting
the absolute number resulting from the real subtraction [thus, put
3 for 3, or (¢+3)—z for z—(z+3)]; change the sign preceding all
such terms as require an odd number of such corrections; keep the
sign of such as require an even number. Repeat this as often as
may be necessary. If the final result be then rational, the problem
has been rightly understood, and the mistakes have arisen from
inattention to the processes which come between the statement and
the result; but if the result after correction be still an impossible
subtraction, then the problem has been misunderstood if there be
alternatives in it, or is itself a wrong alternative of some more general
statement.
As an example, let the final result of a set of operations be
abc+dd ab
ac—d a cd ae
Let it be then discovered that a, b, and e, are not rational expres.
sions, and let them be found to be made, a by attempting to subtrac
a number from 1 less than itself, 6 by attempting to subtract <
number from 3 less than itself, and e by attempting to subtract <
number from 5 less than itself, That is, let a, b, and e, be representec
by 1, 3, and 5. Let c and d, which are rational, be 2 and 6. Then
the preceding expression, before correction, is
1X3x64+2x2 , 1x3
= + = — 6x5
1x6—2 6x2
the first stage in the correction of which is
1844
3
=e ge te + 30
or
ON THE SYMBOLS OF ALGEBRA.
out, — 6—2 (page 47) is signified by 8; and the necessary correc-
ion, according to the preceding rule, gives
|
3 22
| = +30—= or 273
which is the result which would have been obtained, had each
sorrection been made in its proper place.
7 But the preceding form a is not made use of by algebraical
writers, and is introduced here not to remain permanently, buf to
wwoid using the sign of subtraction, to appearance at least, in two
lifferent senses. If we follow rules, without observing where they
ead us, we should obtain processes of the following sort:
| Memes 45) = 3-3-5 — 0—5
and, as in 0+ 5, it is not necessary to the meaning to retain 0,
we might, by imitation, write —5. This is what we have hitherto
written 5. And we shall find, in all the intelligible properties of the
sign —, a close similarity between —5 (properly placed in an ex-
oression) and the legitimate rules by which it is treated, and 5, an
ancorrected misconception, with the rules for obtaining a correct result.
In fact, if we apply rules to + and — as they are applied when
he quantities concerned are rational, we find no distinction between
t and —a, though we have made one until we could establish the
noints of similarity. For instance, )—a when corrected is 6 +«.
And b—(z—a), when 2 is greater than a, gives b—z-+a; so that a,
ationally used with two negative signs before it, gives -+ a in the
vesult. Apply this rule to b—(—da) and it gives b+a; so that
')— and <. Thus,
For a is greater than 6, write a > b.
For a is less than 6, write a < 6
The angle is turned towards the greater quantity. In the sign of
equality = there is no angle towards either quantity.
arithmetical meaning, except when arithmetical quantities are mentioned.
Those who object to it in its present shape, will of course object to the
zoological supposition in page 58.
64 ON THE SYMBOLS OF ALGEBRA.
5. Multiplication and Division. These rules are, so far as the nu-
merical quantities are concerned, the same as in arithmetic. The rule
of signs, as we have seen, is, like signs produce +, unlike signs —
+a6 is both +ax +b and —ax et,
—ab is both —aX +060 and +ax —O
a +a
137 is both ep and —
a +a
ta is both —— = and ay
The terms greater and less cannot always be applied to pro-
ducts as in arithmetic. For instance, 3>2, 5>4, and therefore,
3x 5>2x4; but from 3>—2, —3>—4, it does not follow
that 3x —3>—2x —4, or —9>8, but the contrary —9<8.
But it is seldom necessary to deduce the algebraical magnitude ofa
product from that of its factors ; we therefore leave to the student the
collection of the different cases.
6. Proportion. Four quantities are said to be proportional when
the first divided by the second is equal to the third divided by the
fourth. This definition is the same in words as the definition of
proportion in arithmetic, but the words quantity, divided by, and equal,
have their extended signification. The words greater and less cannot
3 Le ary,
always be applied as in arithmetic. Thus ma being = we have
3:—4::—6:8, where 3 is greater than —4, but —6 is less
than 8.
We shall now apply our definitions to a problem, and shall choose
the cases already noticed in page 45, as being two different problems
when only equations of the first degree are to be used.
A has £60, and is to receive the absolute balance that appears in
B’s books, whether for or against B ; but C, who has £200, is to take
B's property and pay his debts. After doing this it is found that C’s
property is 3 times that of A. What is the absolute balance for or
against B?
Let x be this balance, positive or negative according as it is for or
against B ; then A has 60 a, the positive sign being used* when
* For A’s property is to be increased on either supposition ; hence, if
the balance be +3, he must have 60+( +3); if it be —3, he must have
60—(—3).
ON THE SYMBOLS OF ALGEBRA. 65
' is positive, the negative when w is negative. But C has 200+ 2;
therefore,
3(60+ 2x) = 2000+2
or +32 = 20+2
This contains two equations, one for the positive, one for the negative
sign. But from it follows that
+3ax+32 = (2042) (204+ 2)
' in which the first side is + 922 in both cases, for —3x X —3er
and + 3x xX +32, are the same, namely, +9. Therefore,
+9va = 4004+407+4+ 242
or 8x2x—40r—400 = 0
(+)8 xx—5x—50 = 0
When we come to the solution of equations of the second degree,
we shall find that this equation can be true only for two values of x ;
either r = 10, or x = —5. That is, the balance is either £10 for B,
or £5 against him; which are the solutions already found in page 45.
That either 10 or —5 will satisfy the preceding, may be shewn as
follows :
z= 10 “2s —)
zx = 100 ax = 25
=—57 = —50 Oe =. 195
—50 = —50 —50 = —50
zx—52z—50 = 0 ZL—OxL—ov = 0
Since the extensions of algebra have been so laid down that the |
rules for managing algebraical quantities when they are not arith-
metical are the same as those which must be employed when they
are arithmetical, it follows that the arithmetical case of a problem may
be taken as a guide; for, to say that certain operations follow the same
rules as in arithmetic, or that we must proceed as if the operation was
arithmetical, is only the same thing in different words.
Up to page 56 we have considered the symbols of algebra, which
are not arithmetical, as results of misconception, and have called the
rules by which they are treated corrections. In page 59, &c., by pro-
perly laying down definitions, these same symbols are recognised and
expected, so that the term erroneous no longer applies to them, The
student should not immediately give over the first method of con-
G2
66 ON THE SYMBOLS OF ALGEBRA.
sidering them, but should frequently, while employed upon the rules
(pages 49, &c.), make himself sure that he understands the con-
nexion of the preceding method with those rules ; and in future we
may accordingly employ both methods, it being always understood
that when the first is used, the extensions required by the second are
dropped for the moment.
The following examples of the use of the rules are added for
practice. Such symbols as a and —a are used indiscriminately, it
being remembered that they mean the same thing in practice, but are
referred to two different methods of considering the subject in theory.
8x425=102 8x4i3 = — 102 o 10
644—12 = 2.9 (—6) +28) aye
ab+ed cd—ab (—a)x(—6)xe _ _abe
m aa m —d. raid
obo = pes ae
a+b
ab—ba = 0 a(—b)+6(—a) = —2ab
abc = abe =abc = abec= —abe
EQUATIONS OF THE FIRST DEGREE, 67
CHAPTER III.
EQUATIONS OF THE FIRST DEGREE WITH MORE THAN ONE
UNKNOWN QUANTITY.
Let there be such an equation as
w+y = jee
resulting from a problem which involves two unknown quantities,
z and y. Such a one is the following:
—_ —_ -—_ | —_
ee a aa ae ne
A C B
Prostem. A is a given point ina straight line, and B and C
are two other points. From A to half-way between B and C it is six
feet. What are the distances of B and C from A?
Let us take as our principal case that in which B and C are both
on the same side of A. Let AB = «z feet, AC = y feet; then
BC = x—y, and from B to half-way between B and C the distance
is 3(a—y); whence from A to the same middle point the distance is
xrL— ; (« — y) which is therefore = 6
Here the problem is what is called indeterminate, that is, ad-
mitting of an infinite number of solutions. All that is laid down
relative to « and y is found to do no more than require that their sum
shall be 12, which can be satisfied in an infinite number of ways; for
all the following cases are solutions, and others might be made at
pleasure.
faite —.11 c= 5 y= ll;
m= 2) y = 10 x= 2 a 9
tz=3 y= 9 n= 8 = -
ge
°
ge
re)
Re
re)
ze
°
68 EQUATIONS OF THE
Again, when either x or y is taken negatively, we have solutions
such as the following :
z= —!] y= +13 e-=15 yu
Ee beste. | Of ee er z=16 9 ="—4
1 ‘ 1 3 oS
t= —l5 y = +13; a= 12, y= -5
&e. &e. &e. &e.
the first column corresponding, with the explanations before given at
pages 14-19, to the case in which B only is at the left of A; and the
second to the case in which C only lies to the left of A. Thus, if B
be 1 foot to the left of A, and C 13 feet to the right, the middle point
between C and B is at 6 feet distance from A to the right. Observe
that we cannot in this equation of the first degree, x+y = 12,
include the cases in which the middle point falls to the left of A;
for since this quantity 6, given in the problem, has been treated
throughout as an arithmetical (or algebraical positive, see page 59)
quantity, the equation formed from it cannot include those cases of
the problem in which the corresponding line is so measured that its
symbol ought to be negative.* )
From the preceding and similar cases, we deduce the following
principles :
"1, One equation between two unknown quantities admits of an
infinite number of solutions; either of the unknown quantities may
be what we please, and the equation can be satisfied by giving a
proper value to the other.
2. A problem which gives rise to such an equation is indetermi-
nate, or admitting of an indefinite number of solutions.
Let us now suppose two equations, each containing the same two
unknown quantities. For instance,
x+y = 12 d«—2y = 31
* On reading the problem again, therefore, we perceive either that
we have not sufficiently defined it, by omitting to state whether the six
feet is to the right or the left of A, or else that there are two problems
involved in it, as in page 45, or that these two must be represented in
one equation of the second degree, as in page 64. For the student who
is disposed to try to represent the two cases in one equation, we give the
result, namely,
rrr yyt2ry = 144.
FIRST DEGREE. 69
“the first (considered by itself) has an infinite number of solutions ; so
also has the second. As follows:
Solutions of the Solutions of the
first equation. ~ second equation.
1
z=10 y=2 Ta aA fies
1 x 1 1
ag 105 \ phere oa 10, ey
=ll y=1 z=11 y=1
1 3 1 11
&e. &c. &e. &e.
We have taken the same set of values for 2 in both; and we find
the corresponding values of y different, generally speaking, but the
same in one particular case: that is, we find a set of values 7 = 11,
y = 1, which satisfies both equations. The question now is, among
- all the infinite number of sets of values which satisfy one or the other
equation, how many are there which satisfy both? There is only one,
as we shall find from the following process of solution.
Ifc+y = 12, it follows that =12—y. Substitute this value
of x in the second equation (which may be done, since the solutions of
the second which we wish to obtain are only those which are also solu-
tions of the first), and we find 3(12— y)—2y = 31, or 36—5y = 31,
or y= 1. It appears then, that the supposition of both equations
being true at once consists with no other value of y except 1, or
(Since + +y = 12) of x except 11.
Let the equations proposed be
ax+by =c petqy=r
where a, b, c, p, g, and 7, are certain known quantities.
First Method. Obtain a value of one of the unknown quantities
from one equation and substitute it in the other. The resulting equa-
tion will then have only one unknown quantity,
From the first equation,
water Oy
a
which value substituted in the second gives
per agy=r y=
a
ar—cp
uq—bp
70 EQUATIONS OF THE
To obtain z, find y from the first equation, and repeat the process,
which gives
_ C—axr qe—qar eh GGaar er
pes RRS gee b ate ~ ag—bp
Or substitute the value of y first obtained in the previous expression
for x; thus,
_ o— by _ __ bar—bep
Be wi °Y = “ugly
i a _ tag—cbp— (bar — bap yas cag—bar
‘ by ay ag—bp ~ ag—bp
S attg—or) eae c—by — cq—br
ag—bp ‘°° a ag—bp
cq—br _. ar—cp
ag—bp’ and ae ag—bp
acg—abr abr—bcp
Verification. If x =
earths ag— bp aq— bp
ers acg— bcp as c(aqg— bp) ae
~Tag—op ~ Saye
cpg —bpr aqr—c
pirq = GLa zi ay he sath
Second Method. Multiply both the equations in such a way that
the terms which contain the same unknown quantities may have the
same co-efficient ; then add or subtract the two results, whichever
will cause the similar terms to disappear. Generally, the shortest
method is: Multiply each equation by the co-efficient which the
quantity not wanted has in the other equation.
To find y.
ax+by =c (x)p pax+pby = pe
pet+qy=r (x)a@ pax+qay = ar
Some aqy—bpy = ar—cp
(aeip ies
To find x.
ax+by =c (x)qg agqz«u+bqy = qe
prt+qgy =r (x)b bpx+bqy = br
i) aqzx—bpx =cq—br |
_ ¢g—br
ag—bp
(+)ag—bp
FIRST DEGREE. 71
Third Method. Obtain a value of one of the unknown quantities
from each of the equations, and equate the values so obtained.
; C—axr c—by
he fi psa, oh Nt _ S
From the first equation y ; x ge
From the second equation i rm 2 ey eee
q P
C— ar > me 7) stadia >
: fans tre ited cess br
b q ag— bp
c— by ae alee 1) mee ar—cp
ey P aq—bp
Let the student now repeat all the three processes with the fol-
lowing equations (see page 38).
cb’— bc’
ax+by=c fe ae sae
/ j , ¢ which give : '
ax+b'y =c | Li _ ac—ca
DET ab’— ba’
@) 82-2y=14 (x)2 6r—4y = 28
22+3y=100 (x)3 6x+9y = 300
a) 13y = 272 y = 07
(x)3 9x—6y = 42 (x)2 42%+6y = 200
a 13a = 242 x = 18—
(2.) Z+y =a x—y =b
(+) 2x = ath Gis
(—) 2y =a—b y= >
(3.) pe ty =i] xr—py = 2
The first is Do +4) = |
The second, (X)p px—ppy = 2p
Re EP PY Il Pe
2; 1—y
J = 4 pp Pp t+pp
4a EQUATIONS OF THE
4) 38a—-7=4+4+(@+y) o 2a-y= 11
2y+79 = 5a or 5a—2y = 79
The first, (x)5 10%—5y = 55
The second, (X)2 JO0x—4y = 158
From the first, z= ae = 57
a 3a+4y = 13 4x+5y = 10
(x)4 12”74+16y = 52 (x)3 12x2+1l5y = 30
(—) y = 52—30 = 22
(x)5 1524207 = 65 (x)4 16”+20y = 40
the problem producing these equations must therefore be treated as
before described.
axt+by =c AS _@
Having given ey, = de y= ees |
petqy=r ag— bp ag — bp
we may find the following:
ax—by =C es cq—obr y= cp—ar
pxe—qy =r aq— bp ag— op
ax—by =c ee cq+or __ arto
qy—pxr =7T ig—bp 4 ~ ag—bp
On looking at the two first sets of equations, we see that they
differ in this, that +by and +qy in the first are replaced by —by
and —qy in the second. But we know that av —by is the corrected
form of ax + by (see page 46, &c.), and pxr—qy of pr+yqy; and
since we have proved that the corrections may be deferred to any
stage of the process, it follows that we may take the solutions of
ax+ by = C ie cqg—br 5 ar—cp
pet+qy=r aq—bp aqg—bp
and conclude, that the corrected solutions will be the true result of
the corrected equations. This if done will give results as above; for
the value of x, corrected by the rules in page 52, gives
_ b br— —b
cq+or , rat ae :
—aq-+bp : bp—uagq aq—obp (Pagges)-
| FIRST DEGREE. 73
and a similar process must be followed for y.* In this way, any
‘results derived from an expression such as aa + by + cz, may be
‘made to furnish the corresponding results which would have been
derived from ar —by—cz, cz +by—az, or any other variation
which is only iz sign. We write underneath the different cases which
“may arise, and the manner of referring them to the one which is
(faeces as the representative of all.
The ollowing May be considered as Or as
| ax+by—cz ax+by+cz ax+by+cz
| ax—by—cz ax+by+cz ax+by+cz, &e.
— by—ax—cz ax+by+cz ax+by+cz, &e.
i &e. &e. &e.
Any other case may be taken as the representative of the whole:
thus, if it were ax —by—cz, then ax + by + cz must be considered
as ax — by —cz, &e.
_ Prosiem.t On the intersection of straight lines.
| Principle. Itis proved in geometry, that if any angle BOA (for
Heeeplicity, we use a right angle) be drawn, and a straight line AB
cutting the sides which contain it in A and B, and if from P, any
‘point between A and Bt parallels be drawn, PN to OA, and PM
\toOB, and if OA, OB, PM, and PN be measured in inches, or
‘tenths of inches, or any other convenient unit (provided it be the
‘same for all); then (PM meaning, not the line itself, but the number
of units in it) the product of PM and OA, added to the product of
PN and OB, will be equal to the product of OA and OB, or
PMxOA+PNxOB=OAxOB
‘i:
* Attend here to the remarks in the Preface, on the necessity of
working more examples than are given in the book.
: t If the beginner have no know ledge of the most common terms of
geometry, he must either acquire it, or omit this problem altogether.
¢ From this supposition we start; we shall afterwards make use of
H
74 EQUATIONS OF THE
Let there be two such lines, AB and A’B’ (draw a large figure
and insert A’B’), cutting one another in the angle BOA in P; having
given OA, OB, OA’, and OB’, required the value of PM and PN.
Let OA = 10 units OA’ = 7 units
OB= 8 .. OB =15 ..-
PN = = units
PM=y e
Then, because P is on the line AB (preceding principle),
OAxPM+OBxPN = OAxOB or 107482 = 80
Again, because P is also on A’B’ (similar reason),
OA’'x PM+OB’x PN = OA’x OB’ or 7y+ 152 = 105
Here then are two equations which y and x must satisfy. Solving
them by either of the preceding methods, we have
2 10 ; : 39 ;
xo PN is bz units; y or PM is oy units.
General Case. Let OA = a units OA’ = d units.
OB=0 .. OB’=0' ..
PN = 2 units
PN
The equations to be solved, which are followed by the solution at
length, as at page 70, will then be
ay+bx = ab
adytba = ab’
ady+abxe = aab aby+bb'«e = abl’
ady+abe = aab aby+bb« = abb'
(vab—ab')x =ada(b—b) (ab—ab)y = bU(a—a)
ers , b—bd me) a—a
ee mean y Raker ae YS 22 eae
the preceding theory of algebraical symbols to extend our results to the
case where P is not between A and B,. All words in italics point out
what is peculiar to the first case.
FIRST DEGREE. 75
Verification,
Sst
ab—ab! .
alo) (emer bf— ab’ sae
—ab’ Thi pig i
ay + bx = abb'-2—* + aba
?
ay + Dy —_ abhb’ : is
‘th 1g mela a sus,
ah ad a’ b
+ ae Oe aw ow nee a ow ow ot
oer eee eee -- >
In the preceding figure are four lines, AB, CD, EF, and GH,
cutting the axes* in as many different ways as is possible, that is, no
tw intersecting both OA and OB on the same side of O. There
are therefore, six points of intersection (as in the figure), and the
perpendiculars drawn from them to the axes OA and OB are
inserted ; but no letters are put to them, it being understood that P is
* Axes, the principal lines O A and OB containing the angle first
mentioned, and lengthened in both directions.
76 EQUATIONS OF THE
the point of intersection under consideration,* while PM and PN
are the perpendiculars, as in the first figure.
Let the distances at which the four lines cut the axes be as
follows, in numbers of the unit chosen :
OA = OC’ = 8 OE =4 OG
OB =6 OLD r= Oe OH =
Let us first take the intersection of AB and CD. Place the point
P at this intersection, and let P M = y and PN = a, as in the first
figure. Then, because P is on the line AB, we have, from the prin-
ciple stated at the outset,
3y+6z2 = 18
But on looking at the line CD, we observe, 1. That the principle
does not apply directly to it, because C D is not contained in the
angle BOA, but in the contiguous angle BOC. 2. That OC (in
length 8 units) is not measured in the direction OA, but in the
contrary direction. If, therefore, we write 8 instead of 8, and with
this form the same equation as we should have formed if OC had
been measured towards A, we shall then, by correcting the equation
as in page 66, obtain another equation with which to proceed.t That
is, because the point P is on C D, we have
8y +3a = 8x3 or (page 66) —8y+3x = 24 = —24
or Sy—3ax = 24
[For, since 347—8 y = — 24 denotes that 8y exceeds 32 by 24,
this may be written 8y — 3x = 24. ]
Solving the two equations thus obtained, namely,
3y+6x = 18] foals
we get
By — Su = 24 ly =32
which, with the succeeding results, may be verified by measurement,
as well as from the equations.
* Let the student draw this figure on a large scale in ink, and mark
the letters belonging to the point under consideration in pencil, to be
rubbed out on passing to a new point.
+ We cannot refer to any single page kere. This principle is the
total result of Chapter IT.
‘FIRST DEGREE. © 77
To apply the same method to every case, we must distinguish
all the lines OA, OB, &c. by the negative sign, which do not fall in
the directions of OA or OB. That is,
OA=3 OC=8 OE=4 OG=
OB=6 OD=3 OF=2 OH=4
Proceeding with these as in the last case, we have as follows:
pe creection of Equations. Corrected equations.
1. aBwa op} yt T 3446" = 18
8y+3c=24 8y—-3r=24
2. AB ad BE) aes 3y+62 = 18
4yi2e¢=4x2 4y+22= —8
3. AB and one ie _ 8y+6a = 18
Qy+4a=2x4 Y®—4xr=—8
LODad ERI SYtRT a Sy—3x = 24
5. CD and an} Byt3e= 24 8y—3a = 24
6. EF and GH} SY +27 = 3x2 scion
Qyt+4a=2x4 2y-—4x=—8
These corrected equations are some of them not arithmetically
true ; for instance, 4y+2xv = —8. But remember that they are made
on the supposition that P M(y) and PN (2) are measured in such a
direction that P falls within the angle AO B; which may not be the
case. We may, however, apply the rules in page 70 to their solution ;
of which we shall give one case (the second) complete, as follows ;
4y+2x = —8
Multiply the second equation by 3.
12y4+6x = —8x3 o 12y+6u = —24
Subtract this from the first, which gives
H 2
78 EQUATIONS OF THE
(3—12)y = 18—(—24) = 18424 = 42
18—3 —3(—42) _ 18414 1
ai y _ 18—3(—4%) _ 18+14 _ pi
On placing the point P at the intersection of AB and EF, we
see that PM (y) is measured in a contrary direction to the one
supposed, and PN (wr) in the same direction, as would be pre-
sumed from the negative sign of the first, and the positive sign of
the second.
Proceeding in the same way, we have the following values of x
and y in the six cases :
Intersection of, Value of « (PN). Value of y (PM).
1. AB and CD = 3=
2. AB and EF 5. —45
3. AB and GH 2° big
4.CDadiEF - ~—62 :
5. CD and GH 4, an
6. EF and GH ; —22
[Remember that in such an expression as —13, the — refers to
the whole ; that is, it is —(1+3) or —1—4, not —1+4.]
Upon examining the six lines by a well-constructed figure, it will
appear that whenever an answer is negative, it is measured in a
direction contrary to that which was supposed in the principle at
the. beginning.
The corrections might have been deferred to the end of the process,
in the following manner :
The equations (see page 74)
ay+bx = ab ady+ba = ab
, b—0 ye bb’ a'—a
ive = 10 -—=S] cass hat
6 ab—ab’ ab—ab’
FIRST DEGREE. 79
Take the two following sets (the second being case 6),
ay+bx=ab 4y+2e24=4x2
ay+bz = ab’ Qy +42 = 2x4
which agree if
= 4 b= 2 a= 2 = 4
Substitute these values in the expressions for x and y, which give,
Fi 2—4 —2+4
Pete x _ = (—8) x ————
2X2—4%X4 a patie
2 16 4
= (—8) x (—3) = 59 = 5
ir ie 6 2
ER A. jet OX Hae, DE
4 2X2—4%X4 20 5
the same as before. We leave this problem, and proceed.
We have shewn, page 67, that one equation between two unknown
quantities admits of an infinite number of solutions ; and that a second
“equation must be given by the problem, or it is not reducible to a
Single answer. But this second equation must be independent of the
first, that is, must not be one of those which can be reduced to the
first. For instance, r+y == 12 admits of an infinite number of
solutions; and if the second equation be either of the following,
1 1
22+2y = 24 32+3y = 36 gt tsy = 6
3a—18 = 18—3y 2a+y = 24-Y, Ke.
the same infinite number of solutions will still exist; for if 7+y = 12,
all the equations just given must be true. That is, instead of giving
two equations, we have only given the same equation in two different
forms.
Now, we have already found (page 25), that in one case the index
ofan infinite number of results was the appearance of the result in the
: 0 favre
form = We proceed to see whether this will be the case here.
If ax+by =c 7 eee cq—br ie a
pzt+qy =r aq—bp ag—bp
To try a case in which the second equation is dependent on the
first, suppose
p=ma gq = mb r = me
80 EQUATIONS OF THE
in which case the second equation becomes
max+mby = mc (+)m axrt+by=c
the same as the first. Substitute the above values of p, g, and r, in
the values of x and y, which gives
__cmb—bmc __ 0 amc—cma
~ amb—bma” 0 Y = amb—bma
in which the same anomaly appears as in page 25, and with the same
interpretation,
If there be ¢hree unknown quantities, by reasoning similar to that
in page 67, we may shew that there must be as many as three inde-
pendent equations, or else the problem admits an infinite number of
solutions. We will shew in one instance the method of proceeding.
Let |
2x+4y—32z =O tee see (1)
5x—3y+2z = 20 «--.-- (2)
3x+2y+5z2 = 50 .....- (3)
Multiply both sides of (1) by 2, and of (2) by 3, in the results of
both of which z will have the same coefficient.
Equ. (1) x 2 4x+8y—62 = 20
Equ.(2)x 3 15x—9y+6z = 60
(+) 19z—y = 80 seeceeee (4)
Repeat a similar process with equations (2) and (3),
Equ.(2)x5 25x2—15y+102 = 100
Equ. (3) x 2 6z+ 4y+102 = 100
(~) 19x—19y =a)
(+)19 x—y =Oort =y eevee (5)
Two equations are thus found (4) and (5), containing x and y
only, not z. These solved, give
_ 40 ee
are "¢ ae
Substitute these values in either of the three given equations, the
second, for example. Then
40 40
5x 3x D4+2¢ = 20 nee
FIRST DEGREE. 8]
There is an artifice* which is useful when one only of the three
“unknown quantities is required. Suppose, for example, that in the
preceding equations only the value of z is wanted. Take two new
“quantities, m and n, not yet known, but afterwards to be determined
in any manner that may be convenient.t Since the two sides of an
equation may be multiplied by any quantity, multiply (2) by m, and
(3) by n, and add the results to (1). This will give
(2+5m+3n)x+(4—3m+2n)y
+(2m+5n—3)z = 10+20m+50n.-(A)
Since m and n may be taken at pleasure, and since the value of
3 i iad is wanted, let m and n be such that
294+5m+3n = 0 or 5bm+3n = —2
4—3m+2n =0 or 3m—2n = 4
then m and n must be the solutions of the preceding equations, which
8 2 i : :
aa =. But in (A) the terms which contain v and
y disappear, being multiplied by 0. Therefore,
(2m+5n—3)z = 10+20m+50n
give m =
10+20m+50n 10+ 20x = +50 (-3
mont br—3 eS a ee
(Multiply numerator and denominator of this fraction by 19, and
reduce it),
_ 10X19-++20x8—50x26 _ —950 _ 950 __ 50
2X8—5xX26—3x19 —171 5 WY 9
Anomalies. A problem may give rise to two eaten: which are
absolutely incompatible with each other, such as the following:
z+y =12 x+y =13
or it may happen, that if there be three unknown quantities and three
equations, one of the latter may be impossible if both the others be
true, though it be not inconsistent with either of the others singly.
For instance, take
x—y = 10 y—z = 11 x—z = 12
* An artifice is a name given to any process by which either the
principle or practice of any method is shortened, either generally, or in
any particular case.
+ Such quantities are usually called arbitrary.
82 EQUATIONS OF THE FIRST DEGREE.
The ist and 2d aretrueif » — 20 afd za—-—-l
The istand 3d .... x = 20 y = 10 z=8
Whe 20 and Sd’ a. x = 20 y = 19 z=8
But no values of x, y, and zg can satisfy all three together, as will be
evident by adding the first two, For if r—y = 10, and y—z = 11,
then
(a—y)+(y—z) = 21 8 L—2= 21
which is inconsistent with the third equation.
It is left to the student to examine the general solution in page
70, and to shew that when the two equations become incompatible,
the values of x and y take the form discussed in page 21, namely,
that of a fraction which has 0 for its denominator. It might also be
shewn, that the problems which give rise to incompatible equations
admit of an interpretation similar to that already derived from results
of the form ; in the page last quoted.
It will be found, that two equations such as are treated in this
chapter will in no case become inconsistent except when they can be
so reduced as to have their first sides identical, and their second sides
different ; such as
Q2+3y = 10 24+3y = 12
The subtraction of the first from the second would give 0 = 2,
an absurdity of the same character as av = a4" +c (page 23), from
which the form = was first derived,
ON EXPONENTS. 83
CHAPTER IV.
ON EXPONENTS, AND ON THE CONTINUITY OF ALGEBRAIC
EXPRESSIONS.
Tue continual occurrence of the multiplication of the same quantity
two, three, or more times by itself, has rendered some abbreviations
necessary, which we proceed to explain.
x multiplied by 7, or 2X, is called the second power of «.
LL seseeeeees Ly or LXX, ++++++ the third power of x.
LLL oeeeeeeees XL, or LXKXA, ++++e+ the fourth power of z.
and so on. Or, xes* multiplied together give what is called the
mth power of x. The second and third powers are usually called the
square and cube. Thus rz is called the square of x, and is read
# square; and xara is called the cube of z, and is read x cube: and
a number multiplied by itself is said to be squared; multiplied twice
by itself, it is said to be cubed, &c.
By an extension, v itself is called the first power of z.
The abbreviated representation of a power is as follows: over the
letter raised to any power, on the right, place the number of times the
letter is seen in that power. Thus,
XL is written x*
erie cs tances EY
EEE ce ccct cs He
and soon. Here 2, 3, 4, &c. are called exponents of x. Similarly,
(a+b) x (a+b) is written (a+ 6)?
(a+b) x (a+b) x (a+b) «--.+++» (a+6)8
and soon. The following results may now be easily found:
* The beginner’s common mistake is, that x multiplied n times by x is
the nth power. This is not correct; x multiplied once by x (a2) is the
second power ; x multiplied n times by z is the (1 +1)th power.
84. ON EXPONENTS.
XXXL = x? DORE ae CSTE es Me RO
(a+xz)y = a®+2ax+2°
(a—x)? = @—2azr+2*
(a+2)(a—2) = @—2
(a 4+ax+27)(a—2) = a&—2*
(a+by = a&+30°b43ab?+6°
(a—byY = &—3a2b+3ab*—b
To multiply together any two powers of the same letter, let the
exponent of the product be the sum of the exponents of the factors.
Thus, to multiply 2° and 2%,
uP is LHX
ot jeoa wD ese
a xX aU is HXHXLHULAE or LX or ut4
By the extension previously made of calling @ the first power of a,
and considering it as having the exponent 1, or as being 2', this rule
includes the case where 7 itself is one of the factors. Thus,
axa = 8 o a! xc) =e ee oe
Examples. %* X xv? = gl4 atx gh — zl
BKK Kat aa gl aS sei 8 angie
3ab’x4a?b = 12a? 2abxab = 2a°%?
Co ce yl? x abe? — anes
the latter term, written at full length, would be
aaabbbbbbbcceeeeeeeeeeeeef fff ff
To divide a power by another power of a less exponent, subtract the
exponent of the divisor from that of the dividend. For instance, what
is z'° divided by x3? Since 10 is 7 + 3, or since a! = 27 x 2
10
x
° 3 oe 10—3 TY) .
(—)a@ ~ 2 aa zor az%, Similarly, 23 — x (or w!) == 27; 22 gil
=zlorr; e4—— 2 = a; ath — ya = gb; g2h2? — abc = abc’.
Anomaly 1. If we apply this rule to the division of 2@ by x°, we
have c¢—— 2? = 44-5, If it should be afterwards found that a = b,
the preceding result becomes x2¢-¢ or x, a symbol which as yet has
no meaning. We return, therefore, to the original operation, which
is, to divide «* by 2 where b = a, or, which is the same thing, te
divide 2* by a*. The answer is evidently 1.
ON EXPONENTS. 85
Now why, instead of 1, the rational answer, did we obtain x,
which has no meaning? Because we applied the preceding rule toa
case which does not fall under it. That rule was, “to divide a
power by another power of a /ess exponent,” &c., and was derived
from the preceding rule of multiplication, which rule of multiplication
did not apply to any cases except where both factors were powers
of x, and, consequently, where the exponent of the product was greater
than the exponent of either factor; that is, where 2@ is the product,
and 2” one of the factors, that rule does not apply unless 5 be less
than a.
__ When we come to this symbol (2°) we must do one of two things,
either, 1. Consider 2° as shewing that a rule has been used in a case
to which it does not apply, strike it out, and write 1 in its place ;
or, 2. Let 2° (which as yet has no meaning) stand for 1; in which
ease the rule does apply, and gives the true result. Therefore, we
lay down the following definition.
By any letter with the exponent 0, such as a°, we mean 1; or every
quantity raised to the power whose expenent is 0, is 1.
Anomaly 2. If we apply the equation 24 x = x4-? to a case
in which 6 is greater than a, say b = a+ 6, the mere rule gives
result which has no meaning. The reason is as before; a rule has
deen applied to a case to which it was not meant to apply. To find
the rational result, remember that 2¢+6 = x@ x a; and
“et ae 1
a =
‘he last result being obtained by dividing both numerator and deno-
minator of the preceding fraction by a. The following are similar
nstances ; in the first column is the rational process, in the second
he (yet) improper extension of the rule :
3 3 3
at Fi 1 xv xt
2 2 2
maaemy ee bla eae,
x? a?. x6 x® ae
17 17 17 P
yall ep NY Rad ape Soh
86 ON EXPONENTS.
To make the rule applicable, we must agree that \
(which as yet have no meaning) shall stand for
4 Sat 1
yao ns
3
that is, instead of a—! being the sign that the result is to be abandoned,
and 12 written in its place, we agree that 2-1 shall mean the
same as ix. And we are at liberty to give a—! any meaning we
please, because as yet it is without meaning. We lay down, therefore,
the following definition :
A letter with a negative exponent means unity divided by the same
letter with the same numerical exponent taken positively ; or,
i
Ci paid means —
ve
Our two rules for multiplication and division will now be found
to be universal. The following are instances, arranged as before:
1 eh | Lhe ay ee 5
aN ee eo? eg 8 8 a 8) Sr e
evar eae #4 ad
1 2° ae
a“ x xv
A 1 1 1 ;
ey, ene Sy Oe Rey oe 3.74 — yp 3-4 —S
Richa spree eo pt gl |) =a-7
We have now added to our first definition of an exponent in suck
a way that two fundamental rules remain true in all cases where the
exponents are any whole numbers, positive or negative. These rules
are,
Ce ee idee ll hin
But we have no meaning for a number or letter with a fractional
exponent, such as
xe xs as y%
3 Oe
Instead of waiting until some improper extension of the preceding:
rules shall force us to the consideration of the manner in which it will
be most convenient to give meaning to the preceding symbols, we
shall endeavour to anticipate this step. And first we shall ask what
should aw? mean, in order that the preceding rule may apply to it!
In this case, since ++} == 1, we must so interpret v® that
- 1 yee 1
2X TAS eee Orie
ON EXPONENTS. 87
“consequently, x® is that quantity which, multiplied by itself, gives x,
or is what is called in arithmetic the square root of x. Similarly,
j ”
since }+43+3 = 1, we must so interpret x* that
1 1 1
atets
1 1 1
Cox ee = =z! or x
that is, x must be the cube root of 2. By a root of x we mean the
“inverse term to power; that is, if m is the third power of n, n is called
-the third root of m. As follows:
Equation implied in the fore-
Name of zn. going name.
Square root of m nn = m
Cube root of m nnn = m
Fourth root of m nnnn = m
Fifth root of m AnNNnn = mM
&e. &e.
Thus, 4096 is a number which has exact square, cube, fourth, and
sixth, and twelfth roots.
Its square root is 64 because 64 x 64 = 4096
Its cube root is 16 .... 16x16x16 = 4096
Ms fourth rootis 8 .... 8x8x8x8 = 4096
Its sixth root is 4 .... 4x%4x%4x4x%4x4 = 4096
Mea twellth root is 2 .... ee = 4096
pbs. ee eB ay:
These results should, if the preceding interpretation can be relied
on be thus expressed :
2% fy
64 = (4096)% 8 = (4096)4
2 1
16 = (4096)3 4 = (4096)!
2 = (4096)%
But, according to the notation best known in arithmetic, they would *
_ be thus expressed :
* ./ is derived from the letter 7, the initial of radix, or root. This
symbol is now generally used for the square root, which, in ninety-nine
out of a hundred of the applications of algebra, is the highest root which
will occur.
&8 ON EXPONENTS.
posit abet 4
64 = 4096 8 = V 4096
aba 6 »-_—_—
16 = 4096 4 = 4096
2 = '/ 4096
Proceeding with the interpretation of the fractional exponents,
x8 ought to signify the cube root of x; for, if the preceding rules are
to remain true, we must have
2 2 2
axuexxn? = 2 =
m
and, by the same sort of reasoning, we may conclude that x* should
stand for the nth root of «”. But here we may shew that we cannot
decide upon the propriety of the preceding interpretation without
some further acquaintance with the connexion between roots and
powers. For instance,
xt or 22+} ought to be x? x at or eV x
But 23 is 3; therefore,
2 5
x * ought to be x? or Vx?
Consequently, we Va ought to be Vx
where, by “ ought to be,” we mean that if it be not so, we cannot,*
under the preceding interpretation, apply the common rules of arith-
metic to our assumed fractional exponents. Again, since } +} = 3,
I aes 6;—
x x 23 or Vx x Vx ought to be a or Va?
but as yet we have neither proved that
as a i Bien 6
a Va = Vx or that VexVa = Vo
For the purpose of shewing that the conjectural interpretation as
yet given leads to no erroneous results, we premise the following
arithmetical theorems. ,
Theorem I, If a be greater than 8}, than a? is greater than b?,
a than b3, &c. For aa is then the result of a multiplication in which
more than b is taken more times than there are units in b; therefore
aa is greater than 60: in a® or a’a, more than 6? is taken more times
* The student must remember that we are perfectly free to make a4
and 22 mean different things, if we only take care not to confound the
two. But it would be inconvenient that 24 should any where have a
meaning different from that of §.
ON EXPONENTS, 8&9
than there are units in 6, and so on. By changing the order in which
the letters are named, the theorem may be differently worded, thus :
if 6 be less than a, then 6? is less than a?, &c.
Theorem II. If abe greater than b, a—! is less than b—', a? is
1
7 ie ‘
_fessthan b—?. For if a be greater than b, a8 less than —; and since,
b
‘a Se 1
in that case, a is greater than 6°, therefore is less than py and
“soon. Similarly, if a be less than b, a—! is greater than b—!, &c.
Theorem IIT. If a be equal to b, then a? is equal to b?, a? to 6°,
-and soon. This is evident from page 3.
Theorem IV. Ifa be equai to b, the square root of a is equal to
the square root of b, the cube root of a to the cube root of b, and
soon. Let mand n be, for example, the fifth roots of a and b (which
last are equal), then a and dare the fifth powers of m and n; if m
_were the greater of the two, its fifth power a (Theorem I.) would be
greater than 6, which is not the case. Similarly, if m were the
_ greater of the two, 6 would be greater than a; therefore m must be
equal ton. In the same way any other case may be proved.
Theorem V. If a be greater than 6, the square root of a is
greater than the square root of b, &c. (We put the preceding argu-
ment* in different words), Since a is the square of its square root,
and 6 the same ; if the square root of a were equal to the square root
of b (by Theorem III.), the square of the first (or a) would be equal
to the square of the second (or b), which is not the case. If the
Square root of a were less than the square root of b (by Theorem I.),
the square of the first (or a) would be /ess than the square root of the
second (or 6); which is not the case. The only remaining possibility
is, that when a is greater than b, then the square root of ais greater
than the square root of b. Similarly, if a be less than b, the square
root of a is less than the square root of b: and so on.
Theorem VI. An arithmetical quantity has but one arithmetical
Square, cube, or any other root. For suppose a, if possible, to have
two different cube roots, m and n; one of these two is the greater, let
itbe m. Then (Theorem I.) the cube of m (or a) is greater than the
cube of n (also a); but the last two assertions are contradictory,
therefore a cannot have two different cube roots, &c,
* The nature of the argument is supposed to be well understood from
the last; it is practice in the use of terms which is here given.
90 ON EXPONENTS.
All whole numbers have not whole square roots, or cube roots ;
and the higher the order of the root, the fewer are the whole numbers
lying under any given limit which have a root of the kind. The
following table will illustrate this.
Numbers which have a whole
Value of
Square Cube Fourth Fifth Sixth the Root.
root. Root. Root. Root. Root.
] 1 1 1 | 1
4 8 16 32 64 2
27 8l 243 729 3
16 _ 64 256 1024 4096 4
25 125 625 3125 15628 5
36 216 1296 7776 46656 6
49 343 2401 16807 117649 7
64 912 4096 32768 262144 8
81 729 6561 59049 531441 9
100 1000 10000 100000 1000000 10
&e. &c. &c. &C. &c. &e.
A number which has not a whole root has not an exact fractional
root. At present we shall only enunciate the following proposition,
without proving it, leaving the student to try if he can produce any
instance to the contrary.
No power or root of a fraction* can be a whole number.
Consequenily, all those problems of arithmetic or algebra are
misconceptions, which require the extraction of any root of a number,
unless that number be one of those specified in the preceding table
(continued ad infinitum) as having such a root. But though we may
not look for the exact solution of such problems, we shall shew that
solutions may be found which are as nearly true answers as we
please ; that is, we shall prove the following theorem.
Though there is no fraction whose nth power is exactly any given
whole number, we may assign a fraction whose nth power shall differ
; 4 64
* That is, of a real fraction ; eee. &c. are whole numbers in a
fractional form. It is not necessary to prove this strictly here; because,
were it not true, what follows would not be incorrect, but only useless.
ON EXPONENTS. 9]
from that whole number by less than any quantity named, say *0001,
or -0000001, or any other small fraction.
N.B. With the most convenient method of finding such a fraction
we have here nothing to do, but only with the proof that it can be
found. This is shewn in arithmetic as to the square and cube root
only at most. We must enter into the proof of this at some length,
and shall lay down the following Lemmas.*
Lemmai. The powers of 2 are formed by one addition: thus,
Bee ==) 2% OP 1 De see D3 934.93 = 24
or generally, Qr4 or — Wx Q = Qrtl
Lemma 2. The powers of a fraction are formed by forming the
powers of the numerator and denominator: thus,
(4) = a ¥. a v G@ - aaa a
ie ered b F ie,
b bbb b3
Lemma 3. If p be less than ¢ If p_ be less than g
Then ap is less than aq GUC” Gave sho oak b
ap is less than bg
Lemma 4. If v be less than unity its powers decrease con-
tinually. For instance, if v be one half, its square (3 x 4) is one half
_ of one half, which is less than v; its cube is one half of one fourth,
which is less than v2; and so on.
Lemma 5. If one of the positive terms of an expression be
increased the expression itself is increased, &c. Thus a— 6 is
increased by increasing a, and decreased by decreasing a; but it is
decreased by increasing b, and increased by decreasing 6.
Lemma 6. If v be less than 1, then
(1+v)? is less than 1430 oF 1+(4—l)v
Bere ess. 1+7e, or, 1+(8—1)v
Merpy sess. 1+15v or 14(016—1)v
or (1+ v)" is less than ] + (27— 1)v
Firstly, (1+)? or (1+v) (1+) is 1420+, which, since
(Lemma 4) v is greater than v*, is (Lemma 5) increased by writing
* A Lemma is a proposition which is only used as subservient to the
proof of another proposition.
92 ON EXPONENTS,
v for v*. But it then becomes 1+4+2vu-+v or 1+30. Therefore’
1+2v-+v7 is less than 1+3v; that is, (1+)? is less than 1+3v.
Again, (1+ v) islessthan 1439
(Lemma 3) (1+v)*?(1l+v) «++. (1+3v)(14 v)
or CR ee my ee 1+4v+32?
Still more, then, (Lemmas 4 and 5 as before) is (1+ )° less than
1+4v+3v, or 1+ 7.
Again, (l+v) islessthan ]+7y
Therefore (Ll4tv)* -see. » Cl e7 v) (1 +17)
or 148v+7v?
Still more is (1+v)4 less than 148vH+7v
, or] + 168
We might thus proceed through any number of steps, but the
following is a species of proof which embraces all. Suppose that
one of the preceding is true, say that containing the mth power; that
is, let
(1+v)y" be less than 1 +(2"—1)v
Then (Lem. 3)(1 +v)"(1 +1) is less than ‘1 +(2"— Dra! +0)
which product may be found as follows:
1+ (2% —1)v
1+ v
1-1 (2% —1)v
v + (2% —1)v?
Add 1 +2" + (2% — 1)?
or (1+ v)"*? is less than 1+ 2" + (2"—1)e*
Still more (Lemmas 4 and 5) is * 142" + (2—1)v
less than A ig
or 1 +(2" bee
or (Lemma 1) 14+ (2°t'—1)v
We have proved, then, that
if’ (1+ v)" be less than 1+ (2"—1)0
it follows that (1+ ¥)"*! is less than 14+(2"t1—1)v
or in the series of propositions contained in this lemma, each one
ON EXPONENTS. 93
“must be true if the preceding be true. But the first has been proved,
therefore all have been proved.
Lemma 7. If x be greater than a, then
(@ +a)? is less than x? +3ax or 2°+(4—l)ax
(T+a)> --0-eee, B+7ax? or 23+(8—l)ax®
(T+a)* .--.eee, x*+15aa° o xt+(16—1)axs
or (%+a)” is less than 2” +(2"—1)aa"-!
, a.
Because @ is greater than a, als less than 1; therefore,
(Lemma 6),
G +<)" is less than ] + (2"—1)-
But 1+- es ate ", (Lemma 2) (1 ~ 2)" = eee
(w + ayn
gn
Therefore, is less than 1] + (2” — 1)-
Multiply both sides by x” (Lemma 3), which gives
arr
(+a)” is less than a” +(2"— De
or 2"? +(2"°—l)ax"-}
We proceed to shew the proposition in pages 90 and 91 in a par-
ticular case. Say the number is 10, the power mentioned is the cube.
Can a fraction be found whose cube shall be within, say 0001 of 10?
Since (2)? = 8, and (3)? = 27, 2 is too small and 3 too great.
Examine the cubes of the following fractions falling between 2 and 3;
21, 2:2, 23, &c. We have (2:1) = 9°261, and (2:2)? = 10°648;
whence 2°1 is too small, and 2:2 too great. Examine the fractions
2°11, 2°12, 2°13, &c. lying between 2:1 and 2:2. We find,
1b) = 9°938375 (2) 16)* = 10'077696
Therefore 215 is too small, and 2°16 too great.
Proceeding in this way, we shall find,
(2:154)° less than 10 (2:155)> greater than 10
(2:1544)° .: 10 (2°1545)3 . 10
(2°15443)3.. 10 (2°15444) 5 10
&e. &e.
94. ON EXPONENTS.
The only question now is, shall we thus arrive at two fractions,
one having a cube less than 10, and the other greater, but both cubes
so near to 10 as not to differ from it by*0001? Observe that in the
preceding list,
2:2 exceeds 2°] by only *]
2°16 ey dD os ‘Ol
2°155 . 2°154 Ss. ‘001
2: 1LH45 as 5. . Qa ‘0001
&e. &e. &e.
and from Lemma 7, if a be less than a.
(+a) is less than 2° +7 a2?
or (t+a)?—2? is less than 7a2?
Let x be the lower of one of the preceding sets of fractions; then,
since x? is less than 10, 2 must be less than 3, and its square less
than 9. Therefore, 7ax? must be less than 7a x9, or than 63a.
Still more, then, will (v + a)?—.° (which is less than 7a2?), be less
than 63a. Let x-+ = 100000? We mean that we can obtain two fractions
of which the square and cube are within any degree of nearness
(say *0001) of 10, which we call approximate values of 10 and
N 10, and that, on multiplying these two fractions together, we find a
_ product which, being raised to the sixth power, gives a result within
the same degree of nearness to 10°, or is an approximate value of
10%. We shall anticipate the proof of both propositions, as one
specimen of the method of passing from the strict to the approximative
proposition will serve for all.
Let «@ be a number which has both a square and a cube root
(such as 64, or =): Let x be the square root, and y the cube root.
Then
:
y = a therefore (y*)® = aor ¥ YWoao y =a
eee = 2° a?) or Gey)’ = a”
@ therefore (v*)? = a? or x?.2%.a% = a3 or 26
I
2
©
for 2°76 is wrxrrryy: y, in which the multiplications may be
¥y YYYYYY Pp hy
performed in the order
LY LY. LY .LY.LY.LY giving (wy)?
Consequently, vy is the sixth root of a; but x is the square root of a,
and y the cube root, that is,
Veer 5 484 6/—
iH ge Va> or VaxvVa i
Now, suppose a to be a number which has neither square or cube
‘root, such as 10. Wecan find fractions 2 and y, such that 2? and y®
shall be as near as we please to a. Say that 2? = a+p and
y® = a-+q where p and g may be as small* as we please. We will
begin by supposing p and g smaller than a. Hence (Lemma 7),
* As small as we please does not mean that we can choose them
exactly, but that, name any fraction we may, however small, they may
be made (we need not inquire how much) smaller,
96 [LAW OF CONTINUITY. |
(a+ ~p)° is less than a+7paz
(a+q)* See eee a’+3qa
But (22)? or a = (atp)> (yt or > = (a+ 9
therefore,
x is less than a? + 7 pa?
y® seeeee a + 3g
(Lemma 3) 2°y® «----- (a3 + 7pa’) (a2 +3qa)
or (xy)® is less than a + (7p + 3q)at + 21 pqa3
But since x? (being a+>p) is greater than a, 2° is greater than a’;
and since y® (being a+q) is greater than a, ¥° is greater than a’;
hence a°y° or (xy)® is greater than a®.a? or a’. Hence,
(xy)° lies between a and a + (7p + 3q)a* + 21 pqa’
and therefore does not differ from a® by so much as
(7p + 3q)a*+2lpqa’ |
Now, since p and g may be as small as we please, 7p+3g and
21pq may be as small as we please, and, therefore (however great* a‘
and a® may be), may be so taken that the preceding expression shall
be as small as we please. That is, (vy)° may be made as near an
approximation as we please to a®, or vy is an approximate sixth root
of a’.
The preceding demonstration is not frequently given in books on”
algebra, but the result is assumed in what is called the law of con-
tinuity. This term we shall proceed to explain, as it involves con-
siderations which will be useful to the student; but as it may be
omitted without breaking the series of results, we inclose it in
brackets, and also the heading of the pages which contain it.
[The word continuous is synonymous with gradual or without sudden
changes. For instance, suppose a large square, with two of its oppo-
site sides running north and south. A person who walks round this
square will make a quarter-face at each corner, that is, will at once
proceed east or west where before he was moving north or south,
and vice versd, without moving in any of the intermediate directions.
* The product mn, if n be given, and m as small as we please, may
be made as small as we please; only the greater n is, the smaller must m
be taken, in order to give the product the required degree of smallness.
[LAW OF CONTINUITY. | 97
In this case he changes his direction discontinuously. If the figure
had had eight sides he would still have made discontinuous changes
of direction, but each change of less amount; still less would the
changes have been if the figure had had sixteen sides, and so on.
But if he walk round a circle, or other oval curve, there is no
liscontinuous change of direction. Ifa geometrical* point move
‘ound a geometrical circle, there is no conceivable direction in
which it will not be moving at some point or other of its course;
ind between every two points, it will move in every direction
utermediate to the directions which it has at those two points.
The preceding illustration is drawn from geometry, in which
sontinuous change is supposed; and there is nothing repugnant
© our ideas in the supposition, but the contrary, at least, when we
Magine lines to be created by the motion of a point. But in the
ipplication of arithmetic, and eventually of algebra, to geometry, we
lave this question to ask, Can all geometrical magnitudes be repre-
ented arithmetically? For instance, suppose B to start from A,
nd to move in a straight line till it is 100 feet from A, can we, by
neans of feet and fractions of feet, represent the distance to which
3has moved from A, in every one of the infinite number of points
vhich B passes through ?
|" Itis clear that between one foot and two feet we can interpose the
factions 1:1, 1°2, 1:3, &c. feet; between 1:1 and 1°2 feet we can
aterpose 1°11, 1°12, 1:13, &c. feet ; between 1°11 and 1°12 we can
“ Geometrical, formed with the accuracy which the reason supposes
1 geometrical figures,
K
98 [LAW OF CONTINUITY. |
interpose 1111, 1°112, 1:113, &c. feet; and so on for ever.* But
assigning B any geometrical position between 1 and 2 feet distance
from A, we cannot be prepared to say that we shall thus come at last
to a foot and a fraction ofa foot, which will exactly represent that
position. And we shall now proceed to shew one position, at least,
assignable geometrically, but not arithmetically.
1 foot
2 feet
3 feet
4 feet
5 feet
&c.
It is shewn in geometry+ how to assign (by geometrical construe-
tion, not arithmetically) a position to B, in which the square described
on AB shall be twice as great as the square described on 1 foot (as
in the figure). We now proceed to inquire whether the line AB
has, in such a case, any assignable arithmetical magnitude (a foot
being represented by 1). If so, since every fraction of a foot can
be reduced to a fraction with a whole numerator and denominator
(Ar. 114,121), let AB be — feet, where m and n are whole numbers.
That is, let AB be formed by dividing one or more feet each into 7
equal parts, and putting together m of those parts. Let this mth part
of a foot be called for convenience a ‘“ subdivision;” then a foot
contains n subdivisions, and AB contains m subdivisions. Then,
from Ar. 234, it appears that the square on the foot contains n x”
of the squares described on a subdivision, and the square on ABB
contains m xm of the same. Hence, since the square on AB is
double of the square on one foot, we must have
mm =2nn
We now proceed to shew that this equation is not possible undet
the stipulation that m and m are whole numbers. Because n is 4
* The student must not imagine this phrase, or its corresponding
Latin ad infinitum, to mean more than as long as we please, or as far as
we please.
+ Euclid, book ii. last proposition.
[LAW OF CONTINUITY. | 99
- whole number, mz is a whole number, and 277 is twice a whole
_ number, and therefore an even number; but mm equals twice nn,
therefore mm is even. Therefore m is even, for an odd number
multiplied by itself gives an odd number. But if m be even, its half
is a whole number; let that half be m’, then m = 2m’. Substitute
this value of m in the equation, which gives
2m’ x 2m' = 2nn 4mm =2nn or 2m'm = nn
_ which last equation 2» = 2m'm' may be used in a manner precisely
similar, to shew that m must be an even number. Let its half be
n’ (a whole number), then n = 2n’, and substitution gives
an xen = 2mm 4An'n’ = 2m'm' or 2n'n' = m'm
which proves as before that m’ is even. In this way we shew that in
order that the equation mm = 2nn may be true (m and n being
whole numbers), we must have
m (m orhalfof m) (m” orhalfof m’) &c.
n (mn orhalfof n) (n” orhalfof n’) &e.
all even whole numbers for ever. But this cannot be; for if any
number be halved, if its half be halved, and so on, we shall at last
come to a fraction less than1. Consequently, the equation mm =2nn
cannot be true of any whole numbers, and therefore AB cannot be
represented by any fraction —.
The preceding equation (if it could exist) would give
mm mm
— = 2 or —x— = 2 or ev = 2 where 2
nn nn
ne
nr
and we have seen (page 94) that we can admit the equation rr = 2
only in this sense, that, naming any fraction, however small, we can
find a value for 2, which shall give xz differing from 2 by less than
that fraction. That is, instead of satisfying the equation
xx—2 = 0
we can only satisfy the equation
xa—2 = a quantity arithmetically less than ( )
where we may fill up the blank with any fraction we please, however
small. :
This is sufficient for all practical purposes ; because no applica-
tion of algebra which has any reference to the purposes of life can
100 [LAW OF CONTINUITY. |
require a degree of accuracy beyond the limits of our sight, when
assisted by the most accurate means of measurement. If we take the
least possible visible line to be that of a ten thousandth of an inch,
it will certainly be sufficiently near the truth to solve the equation
“2—2 = arithmetically less than one millionth of an inch,
in cases where perfect exactness demands that r~—2 = 0)
The solution which nearly satisfies such an equation as rr —2 = 0
may be either too small or too great; that is, vx may be a little less
or a little greater than 2. See page 94, where sets of solutions of
both kinds are given for the equation rvr—10 = 0. Hence, though
x or V'10 has no existence, yet, since we can find two fractions, say
a and b, as near to one another as we please, of which the first is
too small (or aaa less than 10) and the second too great (or bbd
greater than 10); and since we can carry this process to any degree
of accuracy short of positive exactness, it is usual to make use of
such forms of speech as the following: 10 has a cube root, but that
cube root is an incommensurable* quantity, not expressible by any
number or fraction, except approximately ; that is, we can find a
fraction as near as we please to V 10.
But still the following question remains: though we can, quam
proximé,t solve the equation »r—2 = 0, may there not be processes
to which it may be necessary to subject that approximate solution,
and may not those processes have this property, that any error, how-
ever small, in the quantity to which they are applied, creates an error
which cannot be diminished beyond a certain extent, however small
the original error may be? For instance, when the student comes to
know what is meant by the logarithm of a number (for our present
purposes it suffices to say, that it is the result of a complicated
process), he might happen to meet with a problem the answer to
which is “ the logarithm of «2, where w is to be found from the
equation ra—2 = 0.” Which of the two following propositions
will he take? one of them must be true.
1. “ In taking the logarithm of 2, the error committed in finding
* Having no common measure with 1; not expressible by adding
together any of the subdivisions of 1. ;
+ A Latin phrase frequently applied to this subject, and therefore
introduced here, Translate it, as near as we please.
[LAW OF CONTINUITY. | 101
x may be made so small, that the error in the logarithm shall be less
than any fraction named.”
2. “ If an error committed in finding a number be ever so small,
the error of the logarithm must be greater than (some given fraction,
say) °001.”
To this no answer can be given except the following caution :
Whenever any new process is introduced, or any new expression,
‘it must be proved, and not assumed, that problems involving that process
“admit of quam proxime, if not of exact, solutions.
The student should now apply himself to prove this of the pro-
cesses already described. We will take one case at length.
a+b
Let the result of a problem be =
where an error, which we are at liberty to suppose as small as we
please, has been committed in determining, say 6 and e. Or (which
‘is a way of speaking more consistent with what has preceded), let
'b and e be involved in some equations (such as bb—2 = 0,
eee —3 = 0) which only admit quam prozimé solutions. Let 6’ and
b’ be approximate values of 6, the first too small, the second too
great ; let e’ and e” be approximate values of e. Then the substitution
of the approximate values of the preceding expression gives
= a+JD’ a+b”
ee and ce" el
To compare these expressions, subtract the second from the first,
which gives
a+ a+b" — (actae’+cb’ + Ue”) —(ac + ae’ + cb" +b")
mee c+ co+ce +ce’ +e”
.: a(e’ — e')—c(b” — b’) + (Ve — eb”)
es co +(e’ +e)c + ee”
And since, page 94, e” may be brought as near to e’ as we please, and
b’ to b’, it follows that e” — e’ and b” — b’ may be made as small as
we please ; whence a(e” —e’) and c(b” —b’) may be made as small
as we please, page 96, note. And so may b’e’ —e’b”, for it will be
found to be the same as
b’ (e’—e)—e' (6° — b’)
Hence the numerator of the preceding fraction can be made as small
as we please, because each of its terms can be made as small as we
K 2
102 [LAW OF CONTINUITY. |
please. But since e” is always greater than e’, the denominator is
(page 91, Lemma 5) always greater than
cc+(é+é)ce+ee
Here then is a fraction of which the numerator can be made as
small as we please, but not the denominator; consequently, the
fraction can be made as small as we please. That is, the fractions
a+b’ a+b” .,. 5 a+b
Aae and che” different values attributed to *eray
can be brought as near as we please, or be made to differ as little as
we please. Here, by the same extension of language as before, the
last mentioned fraction is said to have a definite value, to which we
approximate by substituting approximate values of } and e.
The daw of continuity which is assumed to exist in algebraical
expressions, and which must be proved by the student in a sufficient
number of particular cases, consists in the following theorem :
GENERAL THEOREM. PARTICULAR CASE.
Let there be an algebraical P= eee
expression P, which contains a, Dp = eae
and let the substitution of a
instead of a give to that ex-
pression the value p.
Let the substitution of @ +m g = (a+m)+(a+m)
instead of « give to the expres- = a+a?+(1+2a)m+m?
=p +(1+24a)m+m’?
sion P the value gq.
Then if a and a-+-m may be
made as nearly equal as we
please; that is, if m may be
made as small as we please, it
will always follow that p and q
may be made to differ as little
From the last,
g—p = (1+2a)m + m’
each term of which, if m may
be made as small as we please,
may be made as small as we
please.
as we please.
The only question about which any doubt can arise, as regards
expressions hitherto obtained, relates to the values, real or approxi-
mate, as the case may be, of Jt x, &c. If m can be made as
small as we please, can the approximate values of Va and Va+imn
be made as near as we please? To answer this question, we must
premise a theorem which will also be useful in many other places.
[LAW OF CONTINUITY. | 103
It is as follows :
w—y? = (4—y)(@+Y)
—ys = (a—y) (ei +ayty’)
at—yt = (e—y)(@+a°y tary? +y*)
x” —y" — (a—y) (a +a" y + eae a +ay"? + y"~*)
Multiplication will make any one of these obvious; for instance,
B+ uy +cy’+y?
os Seth J
x*+ xy +2?y*? + ry
—ay—aiy?—xy—y*
a+0+0+4 0 -y7
(Observe that in this, the first multiplication which has been given
‘at length since Chapter II., we have employed the second of the
‘methods in the Introduction, and shall do so in future.)
If we examine the series of expressions
aty, w+aryt+y?, r+x°y+ry? +y%, &e.
we shall see that each of them may be made by multiplying the pre-
ceding by y, and adding a new power of x. Thus,
wpayty? = e+y(x+y)
+eeytacy? ty = 8+y(e+ayty’) &e.
So that, if we call* these expressions P,, P,, P,, &c. (we shall seldom
use large letters except as the abbreviations of other expressions), we
have
A a arty PS Py = a+y Ps Py = a* +yPs, Ke.
or generally P, = a t+y Pai
But we shall also find that the same expressions may be made by
multiplying by v, and adding powers of y, as follows :
atay ty? = yi tery)
o+eyteyty = Prue +uy +y2) &e.
* The figures underwritten must not be confounded with exponents.
They are used as the accents in page 38, and are read P one, P two,
_P three, &c.
104 [LAW OF CONTINUITY. |
o Po = y2+2P, Ps = 73+euPy Pree ra Ps ke
or generally, P, = y°+2Pa-i
The theorem we are now upon may be thus expressed :
x—y" = (x—y)Pr-i
and it may be proved as follows:
GENERAL THEOREM. PaRTICULAR Case.
P, = 2*-+-y Pa-1 P= a
P93 tena i — a +zP,
(=) 0 = my" —(a—y P= (—) 0 = att) P,
an—yn=(2—y) Pn 1 a'—y'=(1—y) P,
Let us now examine “10 and V 10 -++m, where m may be made
as small as we please. Let y and x be approximate values of the
first and second, so that (pages 94, 95),
x’ = (10 +m)+v ) where v and w may be made
y= 10 bi
e—y=m+v—w
or (c—y) (@?+ary+y?) = m+v0—e
as small as we please.
m+vu—w
*—Y = Bh ayty
Now z and y are both greater than 2, since 2? = 8 (less than 10);
therefore the denominator of the preceding fraction must be greater
than 12, while the numerator can be made as small as we please.
Hence the fraction (which is = r—y) can be made as small as we
please, or x and y as nearly equal as we please. But w and y are the
approximate values of V 10 +m and V 10.
The student may try to prove the following theorem :
EA a Cia BE, om
a — yf = (a+ y) (ee — ty +2 y?—y’)
wD — yh = (u+y) (@— sty + ay? — ays + ryt yf)
&e. Ke.
r+ y? = (a+ y) (a —ary 4+ y*)
eP+y =(wty) (at—wy + ary? — xy> + y')
&e. Sc,
We shall now proceed with the general theory of exponents. |
ON EXPONENTS. 105
To raise a power of x to any power, multiply the exponents of the
two powers together for an exponent ; for instance,
(x3)*# = x°X* = gl? for (23)* = 23, 23, oy. 73. = 73434343
In a similar manner,
Eig e = 74 (23)6 = 7's Cry = tb
aa pe a gv? ey oe — gee
(x")” = Ceol = pen
A power of a product is the product of the powers of the factors.
Thus,
(abc) = abc.abc.abe = aaabbbece = ak bic?
(a ed i a — ao) ) cc): — a* b® 12
(a™ b" cP) — (a7 )°¢0")" (c?)* = qm" b"7 cer
3 3
The same rule may be applied to division. Thus, @) = a :
. a a a aaa : :
for the first is BX EX RO bby’ which is the second (page 91).
A root of a root is that root whose index is the product of the
indices of the first mentioned roots. Thus, the fourth root of the cube
(third) root is the twelfth root. To prove this, let the fourth root of
the third root of x be called y. Then,
4 an —
Beye ey x a= y' a= (y*)8 = y”
a Ny poencenso
oF Pee 7 a, but y is also Ge:
We have shewn, page 89, that x can have but one arithmetical
cube root, or twelfth root; and that the cube root can have but one
arithmetical fourth root. Hence the above process is conclusive; it
shews that a fourth root of a cube root of x is a twelfth root of x;
and there is but one arithmetical root of each kind. But when we
come to consider all the algebraical symbols which are roots of 2,
both those which have arithmetical meaning and those which have
not, the student must remember that the preceding does not prove
‘that every fourth root of every third root of x is a twelfth root of «x.
This may be the case, but it is not yet proved.
In a similar way it may be proved that
TERS PR Te WE
eae = Va; (V2 =
I
106 ON EXPONENTS.
The root of a product is found by multiplying together the roots of
the factors. ‘Thus,
/abe = /ax /bx Ve < aees Se (A)
for these two have the same fourth power, namely, abc. For by
definition (page 87),
Ny ere)
((/abc) = ube
and by page 105,
(Jax/bxVe) ae (A/a) x G/6)' x @/e) = axbx-
That is, each side of the equation (A) is a fourth root of abc. But
abc has but one arithmetical fourth root; consequently, each side of
(A) must be that root; and therefore the two sides are equal.
Similarly it may be proved that
JSabe = fax /bx fe Ja = fax /P = by
/ a2bict = \/ a? x V/ b? x / ct /32 = */16 x 1/2 = 24
The same rule may be applied to division. Thus, pecan
for both of these will be found to have the same cube, namely, *:
If a power of x be raised, and a root of that power extracted, the
result is not altered if the order of the operations be changed. That is,
47x is the ptatheeg {v/ a}
To prove this, observe that (page 105),
/ to Jaane = PS EXSEXSEXSE = ial
Simin, °/F = @/ah; as = @/ayt
In the expression Vx, if'a and b be both multiplied or divided by
the same number, the value of the expression is not altered. That is,
b= Vf?
For it has been shewn that
WAL gine =e, "/ ame which is Var J (a = o/ x?
because "(ay =25,
Similarly, J at = Va = \/ a 6 = Nf
ON EXPONENTS. 107
To extract a root of a power, divide the exponent of the power by
the index of the root, if that division be possible without fractions.
se Ly CW recy An
Thus, V2? = 24 = 2°, For x!? = («)*; therefore Vz = x3. Simi-
larly, Vx = a?, 2% = x; and so on.
When the exponent of the power is not divisible by the index of
the root, as in the case of V x7, we have (at least as yet) no alge-
braical mode of operation by which to reduce V a7 to any form in
which the sign’¥ does not appear.* It only remains, therefore,
to find what number or fraction x stands for, and then + to calculate
WaT or a2 V x by the rules of arithmetic.
The only question that remains is about a mode of representing
V 27; and this question we have anticipated in page 86, where we
have found that it would be highly convenient in one respect to let
x3, 73, &c. stand for V2; V2, &ce. But we stopped our course there,
because we had no direct reason to know that all the complicated
relations of roots might be obtained from that notation, without the
necessity of applying rules to fractional exponents different from those
which are applied to common fractions, or of treating the fractional
exponents by rules different from those which apply to whole exponents.
We write in opposite columns instances of the rules which we have
proved in the case of whole exponents, and those about which we
* We may proceed as follows. Since a7 = 2%, we have V al =
Box = 2/26 x nin <4 a2n/'x ; in which, however, the symbol Vea
still remains.
+ There is a certain distinction to be drawn between the processes of
algebra and those of arithmetic. We cannot be properly said to find
results in algebra, but only to put them into the form in which they can
be most easily found by arithmetic. For instance, ‘‘ What is the sum of
aand}?”’ Of this question a+b is not properly a solution, but a repre-
sentation, and the proper answer to the question must be deferred till
we know what numbers a and b stand for. But in the question, ‘‘ what is
the sum of 8 a and 5a?” we can go one step further ; for though 8a+5a
is an algebraical- representation, it is not the most simple one which the
language of the science admits. The latter is 13a; but we are still
without the answer until we know what a stands for. When we come
to the step at which we must pass to arithmetic to get any nearer the
answer, we shall therefore say we have arrived at the ultimate algebraical
; i. : . 37>
form. Thus a+bis an ultimate form; 8a+5a is not. Again V a7, or
3:/ a2» ° afar
at most a2V z, is an ultimate form; V 2” is not.
108 ON EXPONENTS.
wish to inquire as to fractional exponents; the question being, are
the theorems in the second column true, upon the supposition that
m
se n TS
x” represents VV am
ax gt = wSt4 — yl0
OX? — x2
I
(2552
/ 26 =a= x
In the first place we observe that we come by the extension from
the last rule in the same way as by others. We have found that when
b
. . . . a . . a is
6 is divisible by a, * is a correct representation of /x®.
b
But when
b is not divisible by a, a* has no meaning. We give it a meaning;
that is, we say, let it still represent “/ 2, whatever that may be. We
now proceed to investigate the rules which this new symbol requires.
The first column is the general case, the second a particular case.
The references (_) are to the pages in which the rules are contained.
2 Pid
What is gic! 2
m =
= n
7) mM
Xx” means Vx
nq
(106) which = Vv 2m
p
— qd —e,
x* means V x?
nq
(106) which = Van?
a P
Therefore 2” xX x?
— na ont y OL gn
(106) = "4/ a"? x ar?
(64) = "9/anatee
Which is represented by
mq+np
|
Bat 724 +np m dade
ng Neg
oo) ee Ket 3
Therefore 2" x a? = a 2
What is x! x 2%?
ai means a/ x2
(106) which = °/a#
a means Sa
(106) which = Vx
1
Therefore xx x?
(106) — 41 +ax +a= +aa
The question now is, can there be more than two square roots
to +1? Let x be any square root of +1, then #2 must (by the
definition of the term square root) =1, or vr—1=0. But
rx—1 = (rv +1)(x —1); therefore (v +1)(a—1)=0. Therefore,*
either x +1 or r—1 is =O. To x+1 a 0, the only answer is
x= —1; tor—1 = 0, the only answer is x = +1; therefore +1
and —1 are the only square roots of 1. The same process may be
applied to rr = aa, or (wx—a)(4+a) = 0.
Therefore a negative quantity can have no square root which is
either a positive or a negative quantity ; for either of these, multiplied
by itself, is a positive quantity. We will not therefore say that / 49
is no “ quantity,” because we have agreed to give that name to every
symbol which results from the rules of calculation. But /—F
(whenever it occurs) will be what —1 was in page 47, the evidence
of some misconception of a problem, for which the problem must
be examined, and altered, extended, or abandoned, as may be found
necessary. But if we look at the steps by which we established the
meaning of —1, we shall find them to be as follows :+
1. We met with such combinations of symbols as 3—4, &c.
proposing operations which contradicted the meaning which the
symbols 3, —, 4, then had.
2. We examined the problems which gave rise to such combina-
tions, and found out how to make the correction without repeating
the process; so ascertaining what was to be understood by such
expressions as 3—4, &c. that we could either predict their appear-
ance, or explain them when they appeared.
3. We examined what would arise from applying common rules
* When a product ab = 0, either a or b = 0, for if both have a value,
the product, by common rules, has a value.
+ The student may make his understanding what immediately follows
a test of his having understood all that precedes.
| ON EXPONENTS. 111
‘to 3—4, &c., and what would be the effect of deferring the correction
of the misconception till any later stage of the process.
4. From all that preceded, we extended the meaning of the terms
‘commonly used, and in such a manner, that what were till then simply
the results of misconception, became recognised symbols with a definite
meaning, and used with demonstrated rules, not differing in practice
from those with which they were used in their limited signification.
5. And we found that in all cases in which the result produced
‘was simply arithmetical (that is, consistent and intelligible when the
terms had their limited significations), that no error was left in the
‘result by the unintelligible character (with the limited meanings) of
the preceding steps ; but that the result was the same as it would have
‘been if we had retraced our steps and made each step arithmetical.
Thus we observed that +, which before a number, 3, means
addition, before —3 is equivalent to a direction to annex —3 to the
preceding part of the expression; and that though we could not dis-
pense with the extended meaning of + and — in +(— 6)—(+ 4),
yet that +6-+3 admitted of arithmetical interpretation, even though
it were the result of several of the more extended forms of operation.
From this we are at liberty to conjecture, that a further extension
of the meaning of + and — might, by precisely the same train of
operations, give a rational method of using and interpreting Nats
/—2, &c., which at present are wholly unmeaning and contradic-
tory. We say conjecture, because it by no means follows that a
‘method of the success of which we have only one instance, will be
universally applicable. Such a method has been given, but it is not
our intention to explain it here. We shall simply shew that a field
has been left in which the explanation may be looked for.
| of
ce A C B C’
If we suppose a person to set out from A, and stop at B, stopping
first at some other point, and always keeping in the line AB or its
continuation ; and if we suppose distance measured towards the right
to be positive, and towards the left negative, we can by our rules
ascertain the distance +AB through which he goes altogether from
“his first position, as follows :
.
112 ON EXPONENTS.
1. Ifhestopat C (+AC) +(+CB) +AB
2. Ifhestopat C’ (+AC’) +(—CB) +AB
3. Ifhestop at OC” (—AC”)+(+C’B) = +AB
Now observe that if he should leave the line AB or its con-
l
tinuation, if, for instance, he should go through AP, PB, we have no
symbols so connected with AP, &c. that (let {[ simply denote that
some yet uninvented symbol is to be in its place)
F{@GAP)TAPB) = +AB
Therefore we find that in the application of algebra we may yet
have new symbols to employ, and we also fall upon unexplained
symbols such as /—1, &c. May not such extensions be made as
will make V—1, &c , with an extended use of + and —, supply the
place of the yet-to-be-invented symbols? This is for the student of
this work a point for conjecture only, but it will make the following
course advisable.
1. Apply the rules of algebra to such expressions as /—1, &e.
merely to see what will come of using them, without placing any
confidence in the results, or at least more than the experience of a
great number shall render unavoidable.
2. Whenever, in the course of a process, it appears that such
expressions as /—1, &c. have disappeared, examine the result and
see whether it is true.
We now pass to the cube root. Let 2 be any one of the cube
roots of 1. Then #?=1 or ®—1=0 a°—1 = («—1) (a?+2+1) =0
(see page 103, and make y = 1);
therefore either w—1 =O or a?+a4+1=0
The first gives « = 1, and 1 is evidently a cube root of 1. The second
cannot be solved till the student has read the next chapter; but its
solutions (for it has two) both contain the yet unexplained symbol
/—a. They are ,
ahibless ee metre et
and we shall shew of the first of these (leaving the second to the
student), 1. That it does satisfy the equation 2°+ 42-1, if common
rules be considered as applicable to /—3; 2. That it is a cube
root of 1, on the same supposition. ‘
ON EXPONENTS. 1138
ee eee
ey (8) 2a 8 Hiv 3
a 4 = £T ae Ce Cy aren
Therefore
eee? NY 8 tt 10
nj pSLt pe eae /—3
Again 5 oily a
g 4 He A iad > Cae >) )
_ (-1~y—(V¥=3) _ 1-(- 3) __ 4]
4 4 4
Then the three cube roots of aaa or a® must be a, pa, and gu.
‘Hor first axaxa= a’.
paxpaxpa = pia = a’ because p? = |
gaxqaxqa = qa' =a? because g? = |
and the same might be deduced from the equation x*—a?= 0, so
soon as we know how to solve #7 ++ar+a?> = 0.
Let «x be one of the fourth roots of 1. Then we have x* = 1, or
2*—1 = 0; that is,
(@?—1)(a2+1) = 0 andeither 227-1 = 0 or a2?+1 = 0
‘the solutions of 22—1 = 0 are —1 and +1, as before; and the solu-
tions of x?-++1 = 0 are either + = ENG it or == —W —1. There-
fore 1 has four fourth roots, +1, —1, + Monatt BT an Me oe Oe Bia
will be found true upon the application of common rules ; for instance,
(J=1)'= —1 (V1)? = (V1). V1 = —V
(J/—1)' = (V1). Vat = —V 1. VT =
: —(V—1)? = —(—1) =1
The only use which we can logically make of such expressions as
——- : Sein S gy
V—z, previous to any further inquiry, is the following: Let a, 6, c,
and d, be positive or negative quantities. Then
)
| at+bV—ac =c+dV—«
Lae
114 ON EXPONENTS.
cannot be true unless a==candb=d. For suppose a=c+e, that
is, is different from c, then
eet bp Vie =ct+dVvV —z Ve = ar
that is, W—zr is a positive or negative quantity, which is absurd.
Consequently (- e) — (d—b) cannot be a common algebraical quan-
tity. Now if e = 0, and d is not equal to b, we have Var =
0 — (d—b) =0, which is not true; if e be finite and d = b, we have
/—x = -e—0, which does not agree with page 25, since no
quantity multiplied by itself is negative because it is great, or nearer
to negative the greater the quantity becomes. The only supposition
remaining is thate = 0 and d—b = 0, that is,a==c andd= 6;
and the only form at which we have here arrived for J —x is
. But this simply indicates that the equation from which it is
derived is always true; which is the case (so far as such an equation
can yet be said to be true at all), when a = c and b=d. Observe
that we do not lay much stress on the preceding; it only proves that
the equation cannot be true unless a =c and b =d; but it may be
matter of dispute as yet whether the above is true if a—=c and b=d.
For we may as yet reasonably refuse our assent even to the equation
x2 = wv unless x represent magnitude of some sort; we may say that
ideas are contained in the meaning of the sign = which do not apply
except to magnitudes. But if any one should say this, we refer him
to the extended definition of =, page 62, though the beginner must
recollect that he never will comprehend the force of that definition as
applied to /—x, &c. until he has more experience of such symbols.
But there are algebraical quantities analogous to the preceding to
which we must now direct attention. They are such as V3, V5,
2, &c. which cannot be exactly found, but only approximately,
pages 92, &c. We say that
at+bV3=c+dV3B
cannot be true (if a, b, c, and d be numbers or fractions) unless a = ¢
and b =d. The same reasoning applies, word for word, substituting
/3 for V—z, the absurdity deduced being that we cannot have
5 eee
V3= 5
where d, b, e, are whole numbers or fractions.
ON EXPONENTS. 115
In a similar way it may be proved that if all the letters stand for
definite numbers or fractions, and (x and y not being square numbers
or fractions)
a+bVz=c+dVy ae Ria (A)
Then a must =c, and therefore b Vr==dVy. For if not, let
a=cte; substitute, and (—) c
te+bVx=dvy
Square both sides, which gives
(+ 8 + 2A(+tebVae4+(bVayY = dvy)y
or e+2%bVac+b9x = d*y
ae d?y—b?x—e?
therefore z Serer 7 em
that is, Vx, which cannot be expressed in a definite fraction, is so
expressed, which is absurd. The only way of making the equation
A possible is, therefore, by making a = c and b Vr =dvy.
This principle may, in certain cases, be applied to the extraction
of the square roots of such quantities as 4+2 V/3, 21 +4 V5, &e.
Take such a quantity, say 2 + /7, and square it.
QEvVv7y = 242x2V74 (V7)
Be 47 7, SS 7
Now, suppose such a quantity as 11-F 4/7 to be given; how
are we to find out, 1. That it has a square root of the same form ;*
2. That that square root is 2+ /77 As follows: if 11 +4 /7 have p
a square root of the same form, let it be # + Vy, so that
ot) +4 ay a Vy (square both sides)
W44V7 = a+ 2eVy4 (Vy)
=a2a?*+y+22 Vv y
Hence v+y = ll and QaVy — 4V7
Therefore (—) x —2uVy +y= 11-4 V7
* Observe that there is no essential difference of form between
414477 and 2+ V y, fora T=N (4)? x7 =/ 112; whence 1144V7
=11+ 112.
116 ON EXPONENTS.
but the first side is the square of r—V y, or
(c—Vy)* = 11-47 thatis, e—Vy = M11—4V7
But wt+-Vy = V1144V7
Multiply the last two equations together, which gives
(e+ Vya—Vy) = VI 44VIV 1147
or at@—y=V (1144V7)(11—4V7) = V121 112 =3
But 2? +y= 1]
() Yely = ss y=4 Vy=2
Therefore V14 +477 or # + Vy is V7 + 2, as we saw in the
method by which 11 + 4/7 was obtained.
We shall apply the same process to the formation of MG + bv c..
Let this be x + Vy.
Then a+6V c= (a + Vy)
— w+y+2aVy
Therefore a= x?+y and bVe = 2aVy
Therefore q— bVe = x+y—2x Vy = (2— Vy)
or a—Vy = Vane
But a+Vy= N/a
Cap ah J (a—bV ca + bVe) — oi— Bo
But e+y = a
(+) 202 = at+Va—le
p= /ha+aVar—bee
(—) 2y a= a—V a&— bc
Vy = V Pana Vee
4+ Vy = /ka+4Ve—be+V ja— Vee
and this is the square root of a+ bV c.
ON EXPONENTS. aw.
Verification. Let 34+ 1V a?—bic be called p
Let La—iVa—Be b2¢ be called ¥
pg = (4ayr— Bisa, b2c) = 3a?—1(a?—b?c)
= ta°—hat+hb%c = 4b%c; therefore Vpg = LoVe.
According to the preceding theorem,
VatbVe=VprVg (atbVc) = (Vp+Vqy
= (Vp) +2V pV qt(Vq) = pt2V pq+4
But pt¢ =sat+ga=a 2V nq = bVe
Therefore p+-g +2pq is a+bNc; which shews the preceding
theorem to be true.
This theorem is of little practical value, but is very good exercise
in the use of such expressions as b/c, &c. It is a simplification
only when a?— 6?c has a real square root ; otherwise, it is the reverse,
for a square root of a square root occurs only once in Va +bVc, but
twice in the value found for it. Thus it simplifies the first of the
succeeding expressions, but not the second, though both are equally
true.
J 134230 = V10+ V3
J 1342V381 HV 44454 V 34
Anomaly. Apply the preceding result to a case in which 6?c is
greater than a’, or a?—6*c a negative quantity. For example, to
24+/8 (a = 2 iat) G’s=7 8),
S24V8 = V 145 —44-V 1-4 —4 (A)
Is the square root of 2 + /8, therefore, of the same unexplained
character as “~—1, &c.? Certainly not: for it may be found quam
proximé, by the rules of arithmetic, lying somewhere between the
square roots of 2 + /4 and 2+ 79, or between the square roots of
4 and 5.
Are we then to conclude that the expression (A) is in reality
arithmetical? On this we must observe, that a really arithmetical
expression may, by rules only, be made to appear impossible. For
instance,
118 ON EXPONENTS.
ae i ae (e+eV —1)+(y—ev =
xity? = a—(—y) = (wp—(yV = 1)
==. (a+y Vol) ee
To investigate the expression (A) further, extract the square root
of each term by the rule.
Let a=1l16=3 c=-4
/144V—4 = V443V84V3-1V2
Vv pec ai WY
The second sides of the two preceding equations still contain the
square root of a negative quantity; because, since 1 is less than
/2, 4 is less than V2, or 11/2 is negative. Add the two last:
J 144V 44 V1 aod
But this is only the quantity with which we started in another form ;
for it is
VS 4(B45V2) or S242V2 or S24V4x2
We have, then, by following an applicable rule, simply committed
the inadvertence corresponding to the intentional alteration made in
x+y above; and V9 + /8, or generally ve +b c, if a? be less
than b’c, cannot be expressed in a result of the form
Vp+Vq+Vp-Vq
unless g be a negative quantity.
The extraction of Wa +b c is of greater difficulty and less use.
We shall, therefore, omit it.
Exercises. Verify the following theorems :
1. The three algebraical cube roots of —1, are —1 and the
solutions of the equation z2—x +1 = 0, which are }++37—3 and
4— 1/3,
2. The four algebraical fourth roots of —1 are
LdCQ Lee) 1V2(1— Vee
AV2A—-V V1) 44 oe eee
3. The eight eighth roots of 1 are
+1,2V—I1, +4V2(14V—]), £4720=
ON EXPONENTS. 119
Give the reason why this set includes all the values of V1
Having given a quantity which contains radical* terms of the
second degree (that is, square roots) to find a multiplier such that
the products shall be free from radicals.
1. A simple radical term, such as V3. Here V3 is the multi-
plier, or aN 3, where a is rational;+ for V3xaVv3= 3a, which is
rational.
_ 2. A binomial, one or both terms of which are radical, such as
—V3+V2. Since (a+6)(a—b) = a?—2, if a and b are simple
radical terms, a?— b? is rational; therefore a—6 is the multiplier for
a+b, and vice versé. For instance, V3+V2 multiplied by
V3—N2 gives 3—2, or 1; a/3— 31/7 multiplied by 2 J3+3V7
gives 4X 3—3%X7, or 103.
3. A trinomial, containing two or three radical terms; such as
V3+V75—NV7 or Vat+Vb4+N c. Multiply by Va+Vb6—V c,
which gives (Va+ Vb)?—(Ve)? or at+2Vab +b—c or
atb—c+2%Vab. Multiply now by atb—c—2NVab; which
‘gives (a + b—c)2?—(2 ab) or (a+b—c)?—4ab. Therefore
the multiplier required is the product of VatWVb—Vc and
at+b+c—2V ab.
(V8 4V5—V7) (V34V54V7) = 142V15
(142715) A—2V15) = 1—60 = —59
It will seldom or never be requisite to consider more than three
terms.
The preceding can be applied to find the value of a fraction which
has radicals in the denominator. Thus to find 1—(V3 +1) instead
of extracting the root of three and forming the denominator, multiply
both numerator and denominator by VW 3—1, which gives
1 vot 2 Re Soe
Beery 31) (V3.—1)).. 8—1, 4) a(V3—1
the second side of the equation is evidently the more easily found :
* Radix, Latin for root ; radical quantities, those which contain roots.
+ Rational, a term used in algebra, meaning, free from radicals ; for
instance, a common number or fraction. Thus 2 is rational, and also V4,
' though in a radical form. But V 2 is irrational.
120 ON EXPONENTS.
64V7 (V64NV7)(V6EVS — — a
V64+V7_(V6+V DV 6+" 5) _ 64 4/49 4.1/30 4 V 35
V6E—V5 (V6—NV5)(V 64" 5)
The only case of higher roots which is worth consideration is
where there is a simple radical term, ney as V8, a’ b. The mul-
tiplier in the first case is (V8), or 3° for 8° x 5° — 98 = 8. Thus
the following results are obtained :
Va VIVES Vi
Wiig We A) G/N 3 Ohi ee
1 aes 1 nol
Va qt XK Uae
ve 0° tn 7
bt b” a0
nm—1 w—-1
2 een Qe we 2 _ 2V9v64
cae a aon VSN 4 eens
The following miscellaneous examples on all the matters con-
tained in this chapter should be carefully verified.
1
2
axb2x ax B38 = a-2073 =
ah?
= ga! Xi = ee
a” x b” co De aa" = inte b-” —_— Peat
VaVwa=d Vavava=a
m2—m m
iat bt” = oe nas
P_\ vat erent
lary X dpef = (ea) x gee
eee iE 19 4 14
— 5 5 75 3 3 —_— 15 —"s 1b
PT XP oT
: pers pi le
= qi aes / pio Vv (2) =
J/abc = /a2b?.bc = abVbe = ap
be
Serereary bak
Vac’ = V/ abc? abe = ab?ct®/abec
4 8 13 :
abc? = atsptictt) = gbhictadhict
ON EXPONENTS. 12]
i 2 —
Va =a Ve = a° (never used, but might be, why?)
V160=V16x10=4V 10; Vi60 = V3x20 = 220
V3332=14V17 -V3348 = 3V10 V32 0/9
=) oi 7: el les TE
Nl) be oeig Vaio Aan
—— = -(Vari+1)
ci ae
1 1 — vue ae
Va+trt+Vva—r 5 (Vate—Va—z)
»/ /a— PLES Fa eae
eee 2a FOV a balay aay
Va-e—Va—@ 22 ¥
= NF Pe
Java = Jay 142 = Sif 241 +1 = Jays! +*
av b’—4ac dae = 4 pi i meecoiaie/ Beaker
Prana = oy 221 = / 23am = a 22?
GEE AD cok] aa ax+by+(ba—ay)V—1
atyV—1 aiue. eek
Bere eat (iV 1) V4 7 V7. 7
/—k = 4V2.V=1
Rend to 94/ —1
Zs = 4 x \/ 5 aS
b+) (a? +b?) = (ai — b5) (ai+atos + b')
M
(a—b) = (a-
122 ON EXPONENTS.
— (OA Bhat 4a b rae b° = p*)
ag = (taba ee ee
(3 3 yee 3 §
at+ta4+l1) =a 4+2a+3a +2a +1
2\3 } ogy eth el Tha
(,3) = a Cie ) * re ees
[N.B. It is usual to call quantities of the forms aN —1 or V —a?,
a+ V—b, or a+VbV —1, impossible quantities. This they are
at present, as not having received any interpretation; in the same
manner 10—14 was impossible in Chapter I. But, considering
that they will in due time (if the student proceed so far) receive
their interpretation, though not in this work, we shall call them
purely symbolical. All+the phrases of algebra are symbolical ;
but all which contain a letter or numeral, which we have yet met
with, have an interpretation connected with numbers, making them
representatives of magnitude. But /—1 has received no such in-
terpretation ; it is therefore a pure symbol, as much so as + or —,
or more so, inasmuch as yet it indicates neither magnitude nor
operation. Hence, in performing operations with pure symbols, we
can be guided only by experience, or where that fails, we have new
conventions to make. Since it is understood that such symbols will
finally be rejected altogether, unless an interpretation present itself
which brings them under the dominion of common rules, these rules
only should be applied to them. The only case which requires
notice is that of forming the symbol which is to represent V —aX
/ —b. This, by common rules, may be either /—ax—be
V ab, or Jax Sie) multiplied by / 6 SON ea or Vab x —-
that is —Wab. For reasons hereafter to appear, let the student
always take the latter, that is, let ;
VaaxV ab be — Va bison
Ilaving two symbols to indicate the mth root of a, namely,
n iF
1
a anda, we shall employ the first in the simple arithmetical
sense, and the second to denote any one of the algebraical roots,
that is, any one we please, unless some particular root be specified.
ON EXPONENTS. 123
' Thus V 4 is 2, without any reference to sign; but (4) may be either
+2or—2. Thus Vais the cube root found in arithmetic, while
(a)} is either Va, aaideey a or sae Va
(a)% stands for Va, 9 —Va, V—1-Va, and —V—1Va
Similarly a+ 6! has two values, namely, either a+b or a—Vb.
Hence Vb (when b is positive) always stands for a positive
arithmetical quantity. Suppose we wish to represent merely the
numerical value of b, without reference to its sign; we may abbre-
viate the following sentence: “ the number contained in 6, taken
positively, whether 6 be positive or negative,” by /b?, which* is
the same thing. Thus 6 = +V 6? according as b is positive or
negative.
We shail now proceed to the general consideration of expressions
of the second degree. ae.
* I have seen a1, or what is here signified by a+(a?)3, denoted
_ by a+ 4a, in which the ambiguity of sign was referred to/ . But
this was only in one place, and though the want of some express
stipulation as to the distinction between radical signs and fractional
exponents, has led to some variety of usage, I think the conventions in
the text will best agree with the majority of writers. At least, though
I do not pretend to have made any research expressly on this point,
+27 is very familiar to my eye, and = a! not at all so.
124 EXPRESSIONS OF THE
CHAPTER V.
GENERAL THEORY OF EXPRESSIONS OF THE FIRST AND
SECOND DEGREES; INCLUDING THE NUMERICAL SOLU-
TION OF EQUATIONS OF THE SECOND DEGREE,
THE view taken in the First Chapter of equations of the first degree
simply amounted to their numerical solution; that is, having given
two expressions not higher than the first degree with respect to 2,
required that value of 2 which will make the two expressions equal.
We there saw that all equations of the first degree could be reduced
to others of the form az = 6; thus, in page 3, we reduced
a H x
=f i=1-4 t 132=12
It is most convenient to bring all the terms of an equation on one
side ; thus av —b = 0 is in a more convenient form for investigation
thanaz = b. In this manner the whole theory of equations is
considered as involving two fundamental inquiries. The first is ;——
Having given an algebraical expression which contains x, required
that value, or those values of x, which make the expression vanish, that
1s, become equal to 0.
We here (in compliance with common custom) use the term root,
in a manner different from its use in the last chapter. Every value
of x which makes an expression containing x equal to 0, or, as the
phrase is, makes it vanish, is called* a root of that expression. Thus,
* This might be considered an extension of the former meaning, as
follows. Let root have the meaning above assigned, then the square
root of 3 (of last chapter) is the root (as here defined) of 123, the cube
root of 3 is the root of 23—3; andsoon. But we do not make this as
an extension, because there is no corresponding extension for the
correlative term power. And we must confess, that we use this new
meaning of the term root with some repugnance, in spite of its shortness
and convenience. Would not the nullifier do as well? If we were
inventing algebraical terms anew, we should certainly say that 7 is the
nullifier of 24—14, and that the latter is nullified when rx =7. But it
is not advisable to introduce words or sybmols which the student will
not afterwards ‘see in the best writers on the applications of mathematics.
FIRST AND SECOND DEGREES. 125
in page 2, we used language which we may now modify as follows:
“The root of 2r—1—5x2+19 is 6;” “the roots of 16r—2°— 48
are 4 and 12;” “ the roots of «®—62?+ 117—6 are 1, 2, and 3.”
The second fundamental inquiry is as follows: — Having given
an algebraical expression which contains x, what values of x make that
expression po itive, what values make it negative, what values make it
_ purely symbolical (See page 122)? We shall proceed to answer these
"questions for expressions of the first degree. We first make the
following remarks.
1. As we wish the student to keep in mind that we consider
_ various values of x, and the consequences deduced from them as to
the sign of the expression, we shall (whenever we may think it
necessary) use some modification of this latter to denote the root.
Thus, instead of saying that 24—14 vanishes when x» = 7, or that
x = 7 is the root, we shall call the root x,; the second root, if there
be one, v,, and so on.
w
2. We shall always suppose that the co-efficients, a, 6, &c. in such
expressions as ax +6, ax?+bx +c, &c. are either positive or negative
algebraical quantities; unless the contrary be specially mentioned.
Thus, we shall treat of cases where a is 1, 2, —1, — 1/3, &e. but
never, without special mention, of the case where a is Vv —1, or
14+V —3, &c. But we make no such limitation with respect to x.
3. We shall suppose the student to be familiar with the use of
the transformations in page 73 ; for instance, that he can make ax +6
coincide with 21—8 by writing the latter-as 2a -+(—3), and sup-
posing a = 2,6 =—3.
The general form of the expression of the frst degree containing
rvisaxr+b. For instance
r—5 r—2 on AB 5 Ke
2 3
which is G-t+1)2-(G-3)
which coincides with ar+b, by supposing
ee my (27)
i 2m S
The root of ax +6 is readily found ; for let
b , :
ax+b= 0, iben x = ae call this a .
M 2
126 EXPRESSIONS OF THE
When a+b vanishes, it may be written ax,+b, and x, denotes
b b
aa From x,=—-— we get ax, = —b or b=—az, Write
a
this value instead of 6 in ax +b, which then becomes ar—avz,
ora(a—z2x,). Consequently we have this TrHEoREM. If z, be the
root of ax + b, then
ax+b = a(x—2) for all values of «x.
The last is not an equation of condition (See introduction) but an
identical equation. It implies that the two sides are absolutely the
same, but in different forms: indeed it may be thus immediately
traced from aa -+ b without any alteration except of form,
b b
ax+b = a(r+= =a}c—(-* | = aw—a)
because it has been laid down that x, stands for — os
The preceding will seem an unnecessary repetition, but, on
coming to expressions of the second degree, the reason of it will
be seen.
Exercise. Point out, by inspection, the roots of the following
expressions Oy + 4, —42—3, sa—3
ae meet
—-
roots
reduced
roots
‘
Bae
altered ex- 1 *)
pressions 3fe—(— 6 te ~4)2—( tee )}, at -+|
TuroreM. The expressionax + b is of the same sign as a, when x
is greater than the root, and of a different sign from a when x its
Ol oolre
less than the root.
For, ax +b is a(w—zx,); if x be greater than x, 7 — 2, is positive,
(page 63), and a X positive quantity, retains the sign of a (page
64); but if a be ‘less than 2, x—vw, is negative, and a x negative
quantity changes the sign of a.
is =
EXAMPLES. 32 + 9 iS positive for every value of # greater than
1 ; | .
ae ; for instance for ise this we shall try. Ifa = —;
FIRST AND SECOND DEGREES. 127
See.
= +5.
OO} &
mata or
Flan meee, ec ckr
: ‘ 1
The same is negative for every value of x less than — 6? for example,
1
for rs = —-—. For then
J
1 1 1 1 3 i
ee ees 501 = 5 10"
Similarly —4x7—3 is negative (that is, the same as — 4) for every
3 : i.
value of x greater than —7Z and positive for every value of 2 less
3 m es es 4
than a but 97 — 3 '8 positive for every value of x greater than 33
; 4
and negative for every value of a less than 3"
This is sufficient to render our notions of expressions of the first
degree consistent with what we are now going to lay down concerning
. those of the second.
Lemma 1. Ifv+y=p+q, and ry = pq, then x is = one of
the two, p or q, and y is = the other.
Square the first equation, and from the result subtract the second
multiplied by 4, as follows,
a+ cy ty = p?+2pqry”
4xy = 4pq
(=) #—2ey+y = p’—2pqt+ ¢
The square root of the first side is either of the two, v—y or y—v;
that of the second p—y or g—p. Extract the square root, which
gives therefore one of the four following equations :
eg (ly ya = p—g-:- ()
Bay —=g—p----(3) y—r=9—p---- ()
But «+y =p-+4q; which last combined with the four preceding,
separately, gives as follows:
with (1) or (4) 2®=p y =] \ which was to
with (yore) «r= q “yy =p be shewn.
(This lemma will certainly seem most superfluous to the student :
but he must recollect that though, “ if a = p or g, and y = q orp, it
is most evident that r-+y =p+q, and vy = pq,” yet that the con-
verse, namely, that “if~+y =p+q and ry = pq, then x cannot
128 EXPRESSIONS OF THE
be any thing but p or g, and y cannot be any thing but q or p,” is not
equally evident. Thus if
w2—2ax = b?—2ab
it is most evident that « = 6 satisfies this equation, but by no means
evident that nothing but v = 0 satisfies it. In fact r = 2a—6 will
be also found to satisfy it.)
Lemma 2. The product of two expressions of the first degree
[say ax-+b and a’x+b’] cannot be always equal to the product of
any other two expressions of the first degree [say cx-e and c’x+e']
unless the two latter be made from the two former by multiplying
and dividing by a quantity independent of x [that is, unless ax +b =
m(cx+e) and a’x+b'= = (c’x + e’) where m is independent
of x].0, For
(ax +b)(d2+0) = aad'a?+(ab'+a'b)x+b0 .... (A)
(cx +e)(cate) =ccx+(ce+ce)x+ee .---- (B)
If possible, let these two developements* be always equal, what-
ever value is given tow. That is, let
pe+qutr = pat+qde+r
where p stands for aa’, p’ for ce’, &c. (This is merely for abbreviation.)
Now, these two cannot be always equal unless they are absolutely
identical, that is, unless p = p' g = q' and r=r’. This we prove
as follows:—Ifthe two sides of the preceding equation be always
equal, they are equal when a = 1, and also when x = 2, and also
when «x = 3. Let ¢, ¢, fs+ be the values of the first side of the
equation when a is successively made 1, 2, and 3. ‘Then the other
sides will have the same values, according to our supposition ;
that is,
when 21s) ptqtr =4 pv +g tr ts
when X= 2 4p+2qt+r = 4p’ +2 +r sth,
when © = 3 Op+3qtr=t Op'+3q +r = b
Now, apply the first set of equations to find p, g, and r, supposing
* Developement, any expression formed by giving another expression
a more expanded form.
+ These are distinct quantities; for the like abbreviation see
page 103.
FIRST AND SECOND DEGREES. 129
#, t, and ¢, known, as is done in page 80. Then apply the second
set to find p’, q’, andr’. It is clear that, from the perfect likeness of
the equations, p, g, and r, are found by exactly the same operations
on the same quantities which give p’, g’, and 7’. Consequently, the
results will be the same, or we shall have p=p’ g =q’ r=r’.
As an exercise, we give the three results of the first set, which are
t,—2¢,+¢ 8t,—3t,—5¢ P
p = 3 ~ ] q — Sadie SITS OS eS = L fo t; —~3t, + 32,
[The following proof is more simple; but it involves the suppo-
sition of a letter being made equal to 0, which we do not wish to use
till after the discussion in a succeeding chapter.
If pet+gqrtr=px2+qxt+r always,
it is true, among other cases, when + = 0; but it then is reduced to
04+0+7 =04+047 or r=r
‘Therefore pa? +qu+r=pa°+qx+r always
(—)r pe tom ) = px? +q'x
(+)e pe +9 aes ey
This must also be true when + = 0, or
/
0O+g=0+ 9 that is 9 = q
Therefore pr+g =prt+q (—)g pe =p'x orp =p']
It has been proved, then, that the expressions (A) and (B), in
‘page 128, cannot be always equal, whatever « may be, unless
aa =cc, ab'+db=cé+ce, ad bb = ee
Divide the second and third by the first, which gives
ao” *-a’b ce’ Sl éeé bb’ ee’
og ae ae ts anu. cet
Bo. hb Cam & b's vb @Nes @
or — - = - — —~xX-=>-x-
a’ + a c 15 c’ Goce 0 c C
; b’ e’ b € b’ e€ b e’
whence, by Lemma, either = = ~, —- = -, or = = -,and— = -.
a c’ a c a c tb c
Suppose the first.
Bu azv+b = a(x +2) — a (z+)
ny. te = — (cr +e)
c
130 EXPRESSIONS OF THE
Similaly aa+b' = = (ca +e)
/ / /
: au a a sph a 1
But since — = 1 or -xX — =1, iff-=m — = —
cc oe c c m
Therefore ax+b = m(cxuz+e)
a
axth = — (cx +e)
If the second assumption be taken, let the student shew that a similar
result will be obtained.
To proceed with the numerical solution of equations of the second
degree, we shall take the most simple algebraical form, ax?+ bx +c,
to which any other expression of the second degree may be reduced
(page 125). Thus —}2°+2a2—83 agrees with it, if a—=—} b=2
c= —3.
Definition. An expression is said to be a perfect square with
respect to x, when its square root can be extracted in a form which
does not shew x under the sign V- Thus, as may be found by
trial,
V axt+ 2e0r¢+a3 = Va(a+a)
Vatx+2Qaxr2+23 = Va(atz)
Hence a2*+2a*x-+ and we find that 2”?-+ 34+ 5 is 8 complete square, its
roots being
3 = 3
either V2«a+ Fe or —V2x— 75:
2
Generally, if gq? = 4pr, r= ah so that px?+qv is made a complete
square by the addition of
(co-efficient of x)?
4 (co-efficient of x*)
2
bess :
Thus, az?+ ba aie is a perfect square, and su is 4a?x1*?+4abr+ 0%,
the roots of the latter being + (2ar-+ b).
We now proceed to distinguish the peculiarities of different forms
of the expression
aat+ba+e
If b? = 4ac, ‘we have seen that the expression is a complete square.
We shall then look separately at the cases in which 6? is greater
than 4ac, and in which 0? is less than 4ac.
1. Let 5? be greater than 4ac, or let*
b2 = 4ac+e? or 4ac = b*—e
* Why 4ac+e* rather than 4ac+e? Because we wish to signify
that 4ac is really increased. In 4ac+e we do not know whether there
is increase or decrease, till we know whether e is positive or negative
(page 63). But e? is positive, whether e be positive or negative (purely
symbolical quantities being out of the question). Hence the form of a
square is a convenient method by which the student may bear in mind
that a quantity is positive.
FIRST AND SECOND DEGREES. 133
) Pat ya
Mena ba tom te tebe tsa _ 4a%s+4abr+ he
} 4a Aa
| _ @ar+by—e — Qar+b+te\Qar+b—e)
: 7 Gh tga
; or, we have this theorem :
We .& = &—4ac = Vb?—4ac
H SE
the two being identically equal.
2. Let b°= 4ac, then az*+bxr-+c is a perfect square, and so
, is 4a?x? + 4abx + 4ac, which is 4a?x?+ 4abe+b?; and
| 4@x?+4abe+b? — (2ax+b)?
4a a 4a
I axt+bet+te=
| 3. Let 5? be less than 4ac, that is, let
|
[
6? = 4ac—e? or 4ac = 2+ &
4atr2 : Ae ; ‘
re ak Oh acne AaletbAabat Ite!
4a 4a
| = Gar tote
er 4a
Previously to proceeding further, we shall apply the preceding
| expressions to particular cases.
1. Let the expression be 3a°—72-+4. Herea=3,b = —7,
“e=4. And l?= 49, 4ac = 48, whence 0? is greater than 4ac, and
’—4ac=1. This is e?; therefore e =-+1,or—1. Lete =-+1
(6a—7+1)(6r—7—1) i (64 —6) (62 — 8)
f
| ao owe
72 +4— 4x3 : : 742
6(7# —1 3”4—4
+2 RE tke = (x—1)(32—4)
/ as may easily be verified by multiplication. Let the student shew
| that the supposition of e = —1 gives the same result.
3x —4
| r—1
| - 3a°—42
—3r+4
30°— Tx +4
Now, we ask, what are the roots of this expression, or the values
of x which make it vanish. A product becomes = 0 if either of its
N
134 EXPRESSIONS OF THE
factors becomes = 0; that is, let r—1=0O, or, let 31 —4 = 0,
and in the first case
32°—7%#+4 = 0x (3—4) = 0:
in the second, 322—7x+4 = (<1) x0 = 0
; : 4
But if c—1=0, r=—1, and if 3r1—4 = 0, re therefore 1
4 : : :
and 3 are the values of « which make 3a2—7a#-+4 vanish, or its
roots (page 124).
We now inquire what values of x will make
3u°—7x+4 or its equal (*~—1l)GBx—4)
positive or negative. It appears that «—41 is positive when @ is
4 ;
greater than 1, and 34 — 4 when g is greater than ae, while the first
is negative when # is less than 1, and the second when 7 is less
than > (page 126).
Value of a. Sign of e—1. | Sign of 3r—4. | Sign of the product
(w@—1)(37—4)
Less than 1 os _— >
( Greater than 1
less ee “e es he
3 ‘
Greater tl E 7
yan : + | ae a
It appears, then, that the preceding expression is always positive,
except when « lies between the roots 1 and = In this manner we
have determined the following points with regard to 3a°—7xr+4:
er 4 4 :
it is + when « is greater than ae O when « ae — when z is less
than > and greater than 1; 0 when 2 is 1; + when z is less than
1. Follow a similar process with the following expressions.
22°+3¢+1 = («+1)(224+1)
32°+4¢—7 = (x—1)382+4+7)
—22°+62%2—4 = (2—2)(22—2)
Hitherto we have chosen expressions containing no irrational
results: let us now try 32°-+5r—1. Here we have a = 3,b)=5,
c=—1; 6 or 25 is greater (page 62) than 4ac or —12, and
FIRST AND SECOND DEGREES. 135
'b—4ac = 37 = e?; thereforee = + /37. Let e be + /37, then,
page 133,
Beeps t = (2x3a+5+%V37)(2x327+5—¥V 37)
4x3
= 5 (62+5+ V37)(62+5— V37)
Hence the roots of the expression, which call x, and «,, are
be ee V37 +5 _ M375
Ser eat Ore
= —1°8471271 = -1804604 very nearly.
It will be found, as before, that the preceding expression is never
_ negative, except when z lies between 2, and 1,,.
2. Let us take 327—6x+3. Here 6? or (—6)? is equal to
4ac, or 4X3X3. Hence the expression is a perfect square, and
we have (page 131)
32°—624+3 = (V3c2— V3) = 3(a—1)?
This expression vanishes only when 3x2 —%3 vanishes, or
when x =1. But, because there are two equal factors, each of
> a we —_— > =
which is “32 — V3, and to preserve analogy with the preceding
case, it is said to have two roots, which are equal. Thus this ex-
pression has two roots, each = 1.
This expression is never negative, for (r—1)? is positive in all
cases. We can only make it negative by giving a purely symbolical
value to »: for example, 1+ /—1. Then (4—1)? (by rules only,
see page 122) will be —1.
3. In no case hitherto taken has 0? been less than 4ac. Now
try 22°—x+4. Here a=2, b== —1, c=4; and 6? is 1, 4ac
is 32, greater than b%. Here, as in page 133, let 4Aac—b? =e’,
which is therefore 31.
Hence, page 133, we have
(2x2e—1P%+31 — 4r—1)+31
222@—2+4 = ng 5
This expression has no positive or negative root, for (42—1)* being
always positive, so long as @ is positive or negative, must increase
31, and, therefore, (4r—1)?+ 31 can never =0, but is always
positive.. We see, then, that 22°—«-+4 is always positive, for every
positive or negative value of x The least value of the expression
136 EXPRESSIONS OF THE
under this limitation is > for the least value of (4a—1)? is found
P 1
by making 4r—1 = 0, or « =a Consequently, the above expression
has the following property: its least value is > found by making
Fes 2 for every other value of « it is greater. °
The following cases may also be tried by the student.
1
e+eg+l = {Qxr+12+43 \
i
w—ctl = 44(Qe—1%43 ;
1
2284225 = —* {ae 4 36}
We can give the preceding expressions purely symbolical roots;
for instance, to make 27?—x#-+44 = 0, let us make
(4e—1%4+31 =0 (41) = +3]
4¢—l= +V—31 of 42— [eee
Call the roots derived from these x, and z,,
_ 1+V—31 ,
: 4 ‘ 4
which will be found to be roots, by rules only, as in page 122.
We shall now take the more general cases.
1. ax®?+ba+c=0, where b?—4ac = e& (page 132)
and awv’?+ba+c= ~ Qautb+e)(Qax+b—e)
The expression axz?-++6x-+c contains eight different forms, as
follows, which we shall distinguish by the eight letters A, B, C, D,
SCF a MBs ae ae
Sign of a. | Sign of b. Sign of c.
(A) 22a°+52+1 “e + +
(A’) —22*—5a—1 — _ —
((B) 2a%—52+1 + +
| (B)—22*+52-1 _ + —
(O)y 222 O7 + aa _—
(C)—22?—52+1 _ _ +
( (D) Dat) Fel a = : ae
( (D)—22a2+5241 _ a a
FIRST AND SECOND DEGREES. 1gG
We shall first consider that which is common to all the forms;
and then the peculiarities of each.
The roots of az?+bxa-+c are found by solving the following
equations (call x, and #,, their roots)
2ax+b—e = 0 Q2ax+b+e=0
ee eaea! ct ¢ . eer Pe
, 2a os Qa
' But e = Vie 4ac, therefore
= —b+ /i2—Aac re fh uN, Bae
as, 2a PIG Qa
_ And, page 126, the expression 2ax + b—e is the same as 2a(a—x,)
| eat i 2a(a+— = 2a(a— +)
and 2ar-+b—e as 2a(v—z,). This may also be shewn again,
thus,
2a (« — x,)
20
debe = 2(<4 482) — 20(e—=!o4) = 20(e~s,)
2a
4a>(a—ax,)(@—2,,)
~. 46
Hence, when the two roots of an expression of the second degree
are known, and the coefficient of its first term, the expression itself
is known. For instance, what is the expression whose roots are 2
= a(r—xz)\(@— x,
saxe+batec =
1 ; ;
and — a2 and the coefficient of whose first term is 4? By the pre-
ceding formula, this expression must be
4(a—2)(a —(—34)) or 4(u—2)(@ + 8) or 4a*°— 6a —4
If we develope the preceding expression, we find
a(a—a)(e— x,y = ax?—a(a,+4,)X + az,x,
which is identical with aw?+ ba +c; therefore, page 128, we
have
b= —a(x,+ x,) OF Li + x, Saat le %
alo slo
C= 22,2 or 2%, =
Sieh
Coefficient of x
Coefficient of a?
Term independent of
Coefficient of x?
Sum of the roots = —
Product of the roots =
nN 2
138 EXPRESSIONS OF THE
When a = 1, or the expression is 2?-+ bx +c, we have
Sum of the roots —= —§ Product of the roots = ¢
Verify this theorem on all the preceding examples.
We shall denote the preceding forms by placing the signs of
the terms in brackets: thus A will be denoted by (+ + +), A’ by
(— — —), &c. This first case, in which 6?— 4ac is positive, in-
cludes all possible varieties of
(++-—-) (—--—-+) (4+--) (-—++FH)
for in all these, a and ¢ have different signs; ae is negative, and,
therefore, b2—4 ac is positive and greater than b?. Here then, b?>—4ae
is positive without reference to the numerical value of a, b, and c.
This same case may or may not include the following,
(ht). Cea) 0 ee
in all of which ae is positive and therefore the sign of l2— 4ac
depends upon the simple arithmetical magnitudes of 6? and ac.
We shall now examine the cases which have roots; and re-
mark that either of the expressions in any one pair may be reduced
to the other, by simple change of sign. Thus —#—ar+1=
—(x?+x2—1) or (—— +) becomes (+ + —) by entire change of
signs only. And, since, when A = 0, then — A =0, the expressions
in the first of the following columns have roots similar to the corre-
sponding expressions in the second, in every circumstance which
depends only upon the signs of the terms.
Css al» Grea (eee
(+-+) | (@+-) | G--) | C++)
I. Expressions (4+ -+-+)(———); roots not necessarily existing ;
and b°—4ac less than b?, The roots of this expression (when it
has them) are both negative. For, since b?—4ac is less than 62,
/b?—4ac is less than VB, or, page 123, the numerical value of 6.
Therefore — b + \/b?—Aacé and —}— /btan4ae have® the same
sign as —b, or a contrary sign to 6. But the roots are
—b+iV/R— 4ac —}—/ fee
Qa Qa
* For instance, both values of —3-+-2 must be negative; —3-46
has one positive and one negative value.
A
FIRST AND SECOND DEGREES. 139
both of which, as to sign, are therefore negative, because b and a
have the same signs in both, and the numerator is of a contrary sign
to, and the denominator of the same sign as, a and 6.
Il. Evpressions (+ — +) (— + —); roots not necessarily exist-
ing; and 6?—4ac less than b?. By a process exactly similar to the
preceding, remembering that « and 6 have now different signs, we
prove that both the roots, when they exist, are positive.
III. Evpressions (+ + —) (— — +); roots always existing ;
b?—4ac greater than 6’. These two expressions have one positive
and one negative root, the negative root being numerically the greater.
For in this case,* since “l?—4ac is numerically greater than 6,
—b+ /#—4ac and —b— Vb?— 4ac have different signs; namely,
the first is +, and the second —. ‘Therefore,
—b+Vb—4ac (which is +)
2a
es 5/78 4 ich is —
b—vb sia ACSI Seed NET A sign from a and 6.
agrees in sign with a and 6.
If a and b be positive, the second (which is then —) is numeri-
cally the greater (by the preceding note): if a and b be negative, the
first (which is then —) is numerically the greater. Therefore in both
cases the negative root is numerically the greater.
IV. Expressions (+ — —)(— + +); roots always existing ;
b?—4ac greater than &®. Here, by reasoning precisely similar, it
may be proved that there must be one positive and one negative root ;
but that the positive root is numerically the greater. Observe that
a and b have here different signs.
In all these cases we have also the following theorem. The
expression ax?+ bx +c, when it has different roots, never differs in
sign from a, except when the value of x lies between that of the roots.
(Read page 134 over again, with attention.) For we have
axt+bxa+c always = a(“w—x)(«—2,)
One of the two roots x, and z,, must be the greater; let it be x.
Then, if # be greater than «,, it is greater than x,; and e—a2, and
* Remember that in p+q itis the sign of that which is numerically
the greater, which determines the sign of the expression; and that in
pit q that one, either p+q or p—q, is numerically the greater, in which
both terms +p and -+q have the same sign.
140 EXPRESSIONS OF THE
«—x,, are both positive. Therefore a(w—z2,)(#—z,,) has the same
sign asa. Let # be less than x, but greater than «,, (that is, let x lie
between the two roots), then «—~a, is negative, r—w,, is positive;
and a(w#—ax,)(«—z2,,) differs in sign from a. Let @ be less than z,,
then it is less than v,; and r—a, and «—w, are both negative;
therefore a(a— x,)(«—za,,) has the same signasa. A recapitulation
of these three cases gives the theorem in question.
9, ax?+bx+c = 0 where 6? = 4ac or DF? —4ac = O
This case requires that a and c should have the same sign, because
4ac must be positive.
2ax-+b)
Here aa*+ba+e= ee
The two equal roots are derived from
b
2a2+6:= 0 or Ly Ey ae
which are positive when 6 and a differ in sign, that is, in (+ — +)
and (— + —); and negative when 6 and a agree in sign, that is in
(+ +4) and (———). The other cases are entirely excluded,
since a and c must have the same sign. ;
The expression a2?+bx-+c being always a square (a positive
quantity) divided by 4a, always has the same sign as a; observe that
# cannot now lie between the roots.
3. azv?+ba+e = 0 4ac—? = & (page 133)
Here a and c must have the same sign, because 4ac is positive,
being 4?-+e?, the sum of two positive quantities.
0 1
(Page 133) a¢+bzx2+¢c= 1 { Qar+dy +e |
and being a positive quantity divided by 4a, always has the same
sign as a.
The purely symbolical roots (see page 136) are derived from the
equation.
(Qax+bY+e =O o (2ar+b2 =e&x —1
or* 2ax+h = +eV—] = + V 4ace ee
* Observe that p? = q? or + p = +4, gives only two distinct forms ;
for +p = +qand —p = -—qare the same, as also are +p = —q and
—p = +4.
FIRST AND SECOND DEGREES. 141
Bh / tee PY —1 Ge Dat \/ dace 02 oof
eS —— CY a a es
; 24 4 2a
These roots, using rules only, will be found to satisfy the equation,
and also the equations
b c
Lx, wae. Ia
but we cannot at present make an extension of the theorem in
_ page 139, because we can attach no notion of greater or less to x,
and z,.
The numerical solution of equations of the second degree is
_ usually performed by a process instead of a es each case by
itself, as follows:
Let 22a°—7T24+-3 = 0
(—)3 22°—Tx = —3
(+)2 Say? — —
: Complete* the square, 7* — ca -+ ¢)*= (aes re
5
: 7
Extract the root, or a =i ry
- iso a Fayd
ay ie a oe or @ = 7—-G or a
But we should, by all means, desire the student to commit the
following theorem to memory :
If ax?*+ba+ec= 0
: —b+VP—4ac —b—VP—4ac
x = either ~ oC
Qa 2a
Exampues. 1. What are the solutions of
or eae hip +9°)z—p* = 0
Here @ = p—q b= p*+¢? c= —p3
The roots are the two values of the expression
rPEMEVG + PP = 4 P—N CP)
2(p—9)
: b2
* See page 132, where it appears that 27+ bx he is a perfect square,
b
namely, that of a+>-
142 EXPRESSIONS OF THE
(pty = pir 2per +g |
—1420)-9) (—- PP) =e
Therefore the roots are contained in
—(P+9)+ Vop— 4p 9 + pe +a
2(p—q)
2. Let axt—ab«e = b*a—b3
az—(ab+b%)<+b0 = 0
The roots are contained in
ab+P4V@b+RP—4ab
2a
(ab + b?)?—4ab? = a®b? + 2ab? + b*— 4ad
= a?b?—2ab? + bt = (ab—b*)
Therefore the roots are contained in
ab+b?+(ab—b*)
my ya aie a
2 ae
But ab+b?+ab—b ih 2ab —' Fee
2a Qa
— 2 2
Bvt Mee a = au: — a the other root.
2a Qa a é
; < —(ab+b?
Verification, 5 dee a a ee oe s
b2 53 (page 137.)
bx 7 = —
The student should now proceed as follows :
1. To form examples of numerical equations ;— choose two roots
and a coefficient for the first term, and construct the expression which
should have those roots, as in page 137; then find the roots of the
resulting expression by the preceding formula, which should be, of
course, the roots first chosen. Afterwards take any expressions at
hazard ; find their roots, and verify them by actual substitution.
2. To construct literal expressions which shall afford solutions of
more interest than those taken at hazard, choose any expression which
is identically = 0, in which one letter has no higher power than the
second ; such as
ab’—abc+abc—alb? = 0
write 2 instead of 6 in such places as will create an expression, of
FIRST AND SECOND DEGREES. 143
which it could not be known at first sight that it is made to vanish by
x =. For instance, suppose
az’—abc+acz—abac = 0
one of the roots of this should be b. Find the roots by the formula.
Or take the following method: Choose any two simple expres-
sions, one of which only has a denominator; such as = and p. Then
‘the roots of
na—(m+np)x+mp = 0
1—da
Then
should be < and p. For instance, take 6 and
fea —O0. Nn = a~- p=: b
m+tnp = 1l—ab+ab=1; mp = b—ab®
Therefore the roots of
ax’—xz+b—ab? = 0
—ab
should be eo = Flank 7 = besa 2
Anomaly. In the expression az?+bxr+c =0 leta=0. It
then becomes ba +c = 0, giving « = —. But if we examine the
roots of az? + bxr+c = 0 upon the supposition that a = 0, we find
— b?— 4
= igh cee assumes the form (page 25.)
2a 0
—b—V/b?—4ac
—2b
; h ve iF.
a assumes the form 5 (page 21.)
Are we then to say,in conformity to the pages cited, that one root
is infinite, and the other what we please? Apparently not, in the pre-
sent case; we must therefore examine it further. Instead of supposing
a = 0, let us (page 21), suppose it as small as may hereafter be neces-
sary. The Lemma which we here lay down will be useful in every
part of algebra.
Lemma. The expression r/ P+, may, by supposing v suffi-
ciently small, be made to differ from b by as small a quantity as we
please: and, moreover, the same expression may, under the same
circumstances, be made to differ from 6 + =) not only by as small a
quantity as we please, but by as small a fraction of v as we please.
144 EXPRESSIONS OF THE
[In explanation, “i-+-v, however small v may be taken, exceeds
1 by something near the half of v; but Ji+v may be made to differ
from 1+ 3v by less than the ten-millionth part of v, if necessary. |
The first part of this lemma is evident enough: the second we
prove as follows:
(B45 7+ VE +e) (b+ 2—VeR Fo )
= (045) ee
= b+ 2bx = Wap I
4 b?
Q be v
= a a) go a ee
b ++ a5 bial =
ve
462
Therefore Toa VEY +v = en
b+—+VP +0
2b
But the last-mentioned fraction has a denominator, which, when
w is diminished, approaches continually to 6+? or 2b. Let it
be called 26-+w, where, by making v as small as may be necessary,
we can make w as small as we please. Then will
v iar ae v? om v
ice oie V+ = 4P(Qb+w) — 4PQb+4w) *”
that is, b + = differs from V 6?--v by a certain fraction of v, namely,
VU
4b°(2b+ w)
and thence w (see what comes before), that is, since 462(2b + w) can
of v. But since v can be made as small as we please,
be brought as near as we please to 46x26 or 86%, the fraction of v,
1 ee
by which 6 + He differs from 6? v, may be thus represented :
v (a quantity as small as we please)
8 6° (a given quantity) + a quantity as small as we please
and may therefore be made as small as we please.
Precisely the same sort of demonstration may be given of the
following; namely, that 6 — 5 may be made to differ from Vb?—v
by as small a fraction of v as we please.
4
- /b?—4ac may be made to differ from 6 —
FIRST AND SECOND DEGREES. 145
We shall now proceed to apply this theorem to the consideration
of the roots
—b+ivVve—4ac —b—VP?—4ac
a) and
24 24
on the supposition that a may be as small as we please. Hence,
c being a given quantity, 4ac may be as small as we please; and if
_v =4ace (by the last lemma)
4
sr by as small a
fraction of 4ac as we please.
Let, therefore, Vb? —4ac = b — — — px 4ac, in which we
_ may make p (with a) as small as we please.
Then the roots are
ere pe —4pac See pee OA one
2b 2b
: and
2a ; 2a
Sah aoe +4pac
or eee) nc and —— s
b P 2a
Now, diminish a more and more, in which case p is diminished
: c
in the same way. The first root continually approximates to — 7»
2b
and the second to the form — ror But the first is the root derived
from the earlier view of the equation ax?+bx2+c=0 in the
é
ease where a= 0, namely, br +c = 0, which gives ¢ = — 3
The second is yet unexplained.
Propiem in Inzusrratron. a, b, c, and e, are four numbers,
the last three of which are increased by a certain number, and the
first by m times that number. The results are then found to be
proportionals. What is the number?
Let x stand for the number. Then mxr+a, +b, e+c and +e
are proportionals.
mrz+a @f +c
rt+b ~~ v+e
Perform these multiplications, and reduce the result to an equa-
tion of the form P =0, which gives
(m—1) a? + (me +a—b—c)a+ae—be = 0
O
or
or (mx+a)(a+e) = (x+6)(4+0)
146 EXPRESSIONS OF THE
The values of w are contained in
—(me+ta—b—c)+t / (me +a—b—cy—4(m—1)(ae—bc)
2(m—1)
and therefore, generally speaking, there are two solutions of the pro-
blem. But if m= 1, that is, if 2 must be so chosen that «+a, r+6,
x-+c, and «+d are proportionals, the case we wish to consider
arises: for m—1==0; the equation is reduced to
(e+a—b—c)x+ae—bce = 0
which gives only one root; and one of the roots just given takes the
form
—2(e+a—b—c)
Se
The interpretation of this form in page 25 was, that any very great
number would nearly satisfy the conditions of the problem, a still
greater number still more nearly, and so on. Now, the question
becomes, will x +a, x+6,«-+c,«#-+e, approach more and more
nearly to proportionals as « is increased ; that is, will
x+a e+e
= = _ approach to truth in that case ?
Divide both numerator and denominator of both fractions by 2 ;
c
{oe
. ° ax x .
this gives p= 2 Which may be made as near the truth as we
1 =e
x x
; SONGS 5oeec d
please, by taking 2 sufficiently great; for, by so doing, Sel and =
may be made as small as we please, and the preceding equatio
1 1
brought as near to hes please.
Hence it appears, that when a problem which, generally speaking,
has two solutions, has a particular case in which there is only one, we
may say that there is another solution corresponding to an infinite
value of the unknown quantity, in the sense explained in page 25.
But, though we see a confirmation of the interpretation put upon
: 0 ig ee
in page 25, we also see that 7 which is the form of the other
root, does not admit the interpretation of page 25, namely, that any
value of x will satisfy the equation; but it indicates that the rational
: c . Pe
root 1s — >. We shall return to this point in the next chapter.
FIRST AND SECOND DEGREES. 147
The case of a =0 presents new circumstances: let us now sup-
pose onlyc=0. We have then az?+bxr =0, or r(ax+b)=0;
which is satisfied either by x = 0 or by aw+b=0. That is, the
b
roots are 0 and a This would also appear from the general ex-
pressions for the roots.
Similarly, if 6 = 0, we have az?+c = 0.
Cc CG
Capa sss
“6S — cient
a a
which pair consists of a positive and negative quantity when c and a
a
have different signs, and is purely symbolical whenc and a have similar
signs. This also would follow directly from the general expressions.
We choose one from among many instances of the use to which
the preceding theory may be put. Suppose we know the sum of two
quantities (s), and their product (p). Required expressions involving
nothing but this sum and product, which shall give the sum of the
squares, or cubes, or fourth powers, &c. of the two quantities.
By page 138, these two quantities are the roots of the expression
2—sz+p=O0 (x)a® a?—sa%th+pa*= 0
Represent the roots by #, and x,,; we have then
gt? a + p2” oe 0
gt? ghtl 4 + pa = 0
(+) a, x.) + pla +2.) =
Let the sum of the mth powers of x, and 2, be called An; thie,
preceding then becomes
Anis — SAn41 a pAn = 0
or Anse = SAns) — pAn
Now Ao = 229+ 2° = 141 = 2 (page 85)
Av=z2z,+27, =s
Therefore, by the preceding equation,
A,= sA,—pA,= s*—2p
A3= sA,—pA,= s(s¢—2p)—ps = 8—dps
A,= sA,—pAo=s(s*—3ps)— p(s? — 2p) = s*— 4ps* + 2p*
and so on.
148 EPRESSIONS OF THE
There are many equations which may be solved by the assistance
of the preceding theory: the reason being, that though they are not
strictly of the second degree with respect to the unknown quantity,
they are so with respect to some expression containing it. For in-
stance, we wish to find the values of # which satisfy,
e—32a4+1 =>2— Vx2t— 3x +1
In order to clear this equation of the radical sign, we should proceed .
as follows.
V x2 —3x +1=1+32—2?; square both sides,
e—8r4+1] = 14+62472*°—62°4+24
or a*—62°+62°+927 = 0
an equation of the fourth degree, for which no method of solution
has preceded. But, on looking at the original equation, we see
immediately that it is of the form v? = 2—v; for if Ve—3r+1
be v, then a®—32+1 is v2. Letv = /7?— 32 +1, then
oty—2 = 0 vo=1 or —2
First, let v == 1
Ve—B8r+1l=1 oO e~38r4+1l=1..2 is 0 or 3
Next, let v = —2
Ve—3e.1—=—2 324-4 ao = ew
Therefore the preceding equation is satisfied by the following
values of a;
0 3 3+ /21 3—V21
2 2
Again, suppose 22°— 3 = a°. Here 2° is (x°)?; let v = 2°, and
the equation becomes 2v°— 3 = v, the roots of which are —1 and 3.
Hence, «* = —1 or «= $; that is, any values, real or purely sym-
bolical, of “ —1 and V3 are roots of 27°—3 = 2°.
We shall close this chapter with some instances of the process
of clearing an equation of the radical sign. Let it be the following
Vext+Vat1+Var4+2 = 2
.. Va+t+Vatrl = 2—Vr42
FIRST AND SECOND DEGREES. 149
Square both sides
a+2QVaVrt+1l+r4+1 — 4~4V% 74244742
or 2Va(@ +1)4+4Ve4+2 = 5-2
Square both sides again,
a(a +1)4+16V2(e+1)V a +24 16(7 +2) = 25—10x +22
16Vx(v +1)(@ +2) = —(7+302 +32)
Square both sides again,
2562(2 +1)(4 + 2) = (74+ 302 +322)"
which contains no radical sign, and may be developed.
The following are more simple instances, which we leave to
the student.
The equation apt Ib Vogie wo = 4
gives xz—4=0
The equation VetatVatb=c
gives ~ 42xr+4ab—(ce—a—b) = 0
But we must observe that
(x + a)? 25 (x + b)2 = c (see page 123).
gives the same result as the last, and admits of the four following
forms :
VYeatat+Vetb=c —~Vartat+Vart+b=c
Vaet+ —Va+b=c SAGE hd sre fe
The value of x above obtained, namely
(?—a—b)’—4ab
4 ¢?
will only satisfy one of these. Consequently, when we obtain one
of the preceding equations, we cannot be sure but that the problem
has been misunderstood and requires an extension of form which
will give another of the preceding.
We give the following as exercises :
1. Shew that a er cannot be numerically less than 2. Prove
a
1
this by shewing that the roots of a + i= 2—p are purely symbolical
02
150 EXPRESSIONS OF DEGREES.
when p lies between 4 and 0. It may also be shewn from (a — 1)?
being always positive.
2. Shew that a?-+- 6? must be greater than 2ab.
; 1
3. Prove that if a Nee i
1 , 4 1
af >= cD G+ = s—3s G+ = = st— 48° +2
4. If x, and wx, be the roots of the expression a2?-+bx-+c, then
will
LssosS b?>—2Qac b-,./
L— X= + -Vb'—4ac¢ -_~= = + =-Vb—4ae
it 2ac ~—— 2a
a x, _ b?—2ac t vit Pew b
r re eves x as RCA
5. In the expression axz* + bx +c, supposing it previously known
that one root exceeds the other by m, find the roots without the
assistance of the formula. Do the same on the supposition that one
root is m times the other,
CHAPTER VI.
ON LIMITS AND VARIABLE QUANTITIES.
We have already had occasion to observe the effects of particular
suppositions, which make what in other cases are intelligible quan-
- c 0
tities, assume the forms rte a°®, &c. To these we shall now add the
form 0 itself, as requiring investigation on account of the circum-
stances under which we may have to use it; for since we have found
it convenient to reduce every equation to the form P = 0, we might,
without proper caution, be led to such inferences as the following:
If ab = 0 and ac = 0, then ab =ac or b=c. [We® shall give
a striking instance of this, as follows: If « —2 = 0, it follows that
2*—4 = 0, and that 22—2z7—=0. Are we then to equate 2°—4
and z*— 2.x, and proceed in any manner, previously explained,
with the results? If we do so, since 2?— 4 = (4 — 2) (x 4+ 2)
and #?—2¢ = «(7— 2), we have
(c@—2)(w@+2) =(@-2)@ (+) (@-2) at 2 =a
But x—2 = 0 fea or Anon
an absurd result, which indicates some absurdity in the process. The
suspicious step is the division of both sides of an equation by r— 2,
which is 0. If we go through the preceding process without the con-
cealment of 0 which takes place by making the supposition r—2 = 0
and then using w—2 instead of 0, we shall find a most evident
fallacy, amounting to the following :
ax0=0 6x0=0 »ax0=0x0 (+)0a=6b
that is, we have used O as a quantity, have asserted 0 = 0, and have
divided by 0. Returning now to the principle in page 21, we shall
suppose « — 2 instead of being = 0, to be very small, and shall put in
* All that comes between the brackets [ ] is vaguely stated ; the
object being nothing more than to shew the student how liable he is to
error in using such terms as nothing, small, great, nearly equal, &c.
i ON LIMITS AND
opposite columns the two analogous processes, each containing its
own error.
Let zr—2 = 0 Let 2—2 be as small as we please
i: z?—4 = 0 “. £22—4 may be as small as
we please
and « a?—2z7 = 0 and 2*—22z may beas small as
we please
a. v—2e = x?—4 “. 2—22X and x?—4 may be
as nearly equal as we please
or x(x—2) = (x—2) (a +2) | and the same of 7(a~—2) and
(t—2)(e+2)
+(t—2) x= 2+2 (+)(x—2), then x and +2
may be as nearly equal as
we please.
Batwa a= Ute But Z may be as near to 2 as we
Dy please; therefore 2 and 4
may be as nearly equal as
we please.
In the second column there is an error, whichever of the senses
in page 24 is put upon the word equal. If we call quantities nearly
equal whose difference is small, then we do not know that because
w?—2ze and «*—4 are nearly equal, they will still be so after divi-
: : 1 -
sion by r—2. For if r—2 were, for instance, are then divi-
sion by e—2 is multiplication by 1000, or the difference between the
quotients « and «+2 is 1000 times the difference of x2—42x and
a*—4. And the less r—2 is, the greater is the real multiplication
to which division by x—2 is equivalent. If it be said that the quo-
tients x and «+2 do not differ by a larger proportion of themselves
than 2*—2yz and 2*— 4, and that, agreeably to the sense preferred in
page 24, a is as nearly equal to b, as c is to d, when the differences
of the first two and of the last two are in the same proportion as
a and c: the answer is, that c?7—2z and #?—4 must not then be
called nearly equal, because they are small, and because their differ-
ence is therefore small; for both may be small, and, nevertheless,
one may be many times the other. An elephant and a gnat are both
small fractions, if the whole earth be called 1, but they are not
nearly equal in any sense.
VARIABLE QUANTITIES. 153
From the above we gather, that, calling a and b nearly equal when
they only differ by a small fraction of either, we are not at liberty to
say that two small quantities are therefore (because they are small)
nearly equal. |
There are two obvious tests of the absolute equality of a and b:
a—b =0 and eg 1h
We lay down the following definition. Approach towards equality
is measured not by the diminution of the difference, but by the approach
of the quotient towards 1. Thus 3 is not more nearly equal to 2 than
20 to 25 because 3 — 2 is less than 25 — 20; but 3 is not so nearly
equal to 2 as 25 is to 20, because
3 al 25 al 1’
. eo... deci be, & (1 =)
5 exceeds y 5 50 exceeds 1 by P ess than 5
See page 24, for an anticipation of this use of the words “ nearly
equal,” and what precedes in [ | for reasons.
TueorEM. The value of a fraction depends entirely on the
relative, not on the absolute, value of the terms. The following ex-
amples will shew that this is contained in the theorem — = -
1. Find a fraction whose numerator is 583, and which is as
1 583
Pa eee eS A Se re
small as 1000 sean TPT TVs
: F 1
2. Find two fractions, a and 8b, each less than Tite that ; may
be a million.
Answer a= i = :
&” @ = 3000 ~ — 2000,000,000°
a b 1
Here a = 1000,000 3 = 1000,000
3. Find two fractions, a and 6b, each less than 2, so that 5 may
be = m.
Answer. Take any two numbers, p and q (only let p be less than
2q), and let
a=? b=
q mq
Exercises. (All the letters are positive) p is not so nearly
equal top+g as p+mistop+q+m. Ifa be more nearly equal
to 6 than c is to e, then a+c is not so nearly equal to b +e as a is
154 ON LIMITS AND
to b, but more nearly equal than c is to e. Again, mz is as nearly
equal to nz as my is to ny.
We must now say something upon definite and indefinite terms.
A word is called definite when there can be no question as to
whether it is proper or not to use it in any particular case which may
be proposed. A word is called indefinite when it may be matter of
opinion as to whether it is proper or not to use it in any. proposed
case. Thus, equal is a definite term. No two opinions can exist
upon the answer to the question “ Is 4+ 4 equalto9?” But great
is an indefinite term. ‘Is 1000 a great number?” The answer
to this is, there is no way of answering the question in any case
upon which all must agree. We give some examples of each sort.
Definite — equal, exact, larger, nearer, smaller, greater, less,
largest, smallest, &c. as large as, as much as, as great as, &c., quite.
Indefinite — near, small, large, great, much, nearly, hardly, &c.
large enough, small enough, &c.
With indefinite terms we can have nothing to do, unless by
addition to their meaning, so as to give them a signification which
will allow us to use them without presenting as mathematical theorems
propositions which contain matters of opinion. We shall take the
terms near, small, and great as instances. Observe that the term
smaller is not to be considered as altered in the same manner as
less in page 62. It keeps its arithmetical meaning. “If x be
small, 7+ 2 is nearly equal to 7?” This is a proposition in which
all will agree: and the reason is that “ small” and “ near” have a
connexion which is independent of what fraction the speaker may
choose to think entitled to the term “ small.” AB may be a line
which one may call small, and another not small; but all will agree
that in the meaning of the words, “ small” and “ near” is implied
“If AB be small, A is near to B.” But if we come to ask—
What fraction is small, is it a a0 &c.?— The answer must
depend on circumstances. We reject, therefore, the terms small and
near in their common meaning. But the preceding proposition can
be put in a form which will never render it necessary to inquire
what is small or near. “ If 2 may be as small as we please, then
7 + 2x may be made as near as we please to 7;” or, ‘ Let me make
« as small as I please, and I can make 7+ 2 as near to 7 as you
please ; or, “ Name any fraction you please, and let it be such a
VARIABLE QUANTITIES. 155
“one as you choose to call small, I do not ask why: then, if I may
make x as small as I please, I can make 7+. differ from 7 by a
fraction less than the one named by you;” and so on.
Having rejected the terms small, great, and near, in their common
signification, we shall revive them for our own use in algebra, simply
as convenient abbreviations of “ as small as we please,” ‘as great
as we please,” “as near as we please,’ or, ‘as small as may be
necessary,” ‘‘ as great as” &c. &c. In this sense we have various
demonstrable and definite propositions. For instance, “if x be small
? ?
7. aes
— is great;’’ that is, if we may make z@ as small as we please, we
x
1
may make ; a8 great as we please.
When, under certain circumstances, or by certain suppositions, we
can make A as near as we please to P (A being a quantity which
changes its value as we alter our suppositions, and P a fixed quantity,
which does not change when we alter our suppositions) then P is called
the Limit of A. A quantity which we are supposing as great as we
please is said to increase without limit ; one which we are supposing
as small as we please, is said to decrease without limit. The follow-
ing theorems will be evidently true.
If x decrease without limit, the limit of a+wz is a; if x diminish.
without limit, then ; increases without limit; if x approach without
limit towards 6, then the limit of a+ is a+b.
The first and third may appear but a complicated method of
; saying that if r=0, a+a—=a; and if r=6, at+xz =a-+b, which
eb pe
are perfectly intelligible. But, “if x =0 shes’ gy ’ has no intel-
ligible meaning ; in fact, in page 25, we have already anticipated
}
the construction we here put upon that proposition.
One object of this chapter is, to put interpretations upon those
‘ ; a P 1 O
forms which would otherwise offer difficulties ; such as 0, Bite: and
(had we not otherwise found a rational interpretation) a°. But we
| still have the forms
Gp Pn Go. 6 it,
all of which might occur, if we stumbled upon such expressions as
156 ON LIMITS AND
(adh ay (= Ja Ke.
r—d
without observing (what might happen) that 7 = a.
In all these cases, that is, when we get a form which is not a
direct representation of quantity, we shall not ask “‘ What is the value:
of that form?” or in any way enter into the question whether it is
demonstrable that it has a value or not. But the question we shall
always ask is this: “ As we approach the supposition which gives the
unintelligible form, to what value does the expression which gives
the unintelligible form approach?” For instance,
v?—a? 0
when c= a =i
L—a O
But if we examine what sort of change of value takes place in the
above fraction when x approaches towards a, we find that value to
approach towards 2a, as will be afterwards shewn. And it will be
found that we have the following proposition: “ If w may be made
2
as near as we please to a, then can be made as near as we
?
2 a’
: —s ih
lease to 2a.” or, “If x approach without limit to a, then
Pp . ’ pp :
approaches without limit to 2a.”
Shall we then say that
2. 7 O d
when 2 = gL = Se eae Yo 2a (inthis case)?
L— a 0 0
Whether it will be proper to say so, in the common meaning of all
the terms, we leave to the student. But we shall not, in this work,
use such a form, except as an abbreviation of one of the preceding
propositions.
It is usual to make the symbol @ stand for x and this is called
infinity. From what has preceded, and page 25, we shall regard
x = o asan abbreviation of the following: “‘ Let x increase without
limit.”
Again, “let x = 0,” it will often be most safe to regard as an
abbreviation of “let « diminish without limit.” We shall hereafter
return to this. But in equations of the form P —Q = 0, where P and
Q are certainly finite quantities, this alteration will not be necessary.
Turorem I. If A and B be two expressions which are always
equal, so long as they preserve an intelligible form, then the limits
of A and B are also equal.
VARIABLE QUANTITIES. 157
To take a case, suppose that when 2 increases without limit, A
has the limit P, and B has the limit Q; then P must = Q. To
prove this, suppose that
A= P+a B= O65
then, by increasing 2 without limit, a and 6 are diminished without
limit. For, if not, and if @ had the limit «, then the limit of A or
_P+awould be P+. But it is P; therefore a has not a limit,
_ but diminishes without limit.
- But because A = B we have P+a=Q+b. If P be not=Q
it is greater or less. If possible, let P be greater than Q, and let
P=Q+R. Then, since P and Q are quantities not containing x,
R is the same, and P, Q, and R, do not change when « changes.
We have then
Q+R+a=Q+0 or R=d—a
Here, then, is the following absurdity: R (a fixed quantity) is always
_ equal to b—a, which can be made as small as we please by in-
creasing x, because b and a diminish without limit as a is increased
without limit. Therefore P = Q-+R is absurd. In a similar way
it may be proved that P = Q—R isabsurd. Therefore P= Q.
TurorEM II. When « diminishes without limit, the way of
_ finding the limit of an expression is to make + =0, provided, 1st, all
‘the results be intelligible; 2d, that the number of operations be not
unlimited.
For instance, it is clear enough that 1+ 2v + 32, when x dimi-
nishes without limit, has the limit 1+0-+0or1. And, perhaps, the :
student may think it clear that the limit of
142742724 2734 24+ &c. continued for ever
is Pere ay i) Oi 7 Sere eens
or 1, when 2 diminishes without limit. But here we must make him
observe, that when we take w small, though each of the terms a, x, x3,
a4, &c., may be small, yet their number is unlimited. And though
we know that when a certain number of terms is added together, each
of which may be made as small as we please, that their sum can be
made as small as we please, yet we do not know the same of an un-
limited number of terms.
TueoreM III. When @ increases without limit, it is clear that
such expressions as
158 ON LIMITS AND
a+bx, a+bx+c2x’, &c. increase without limit:
b b c raat:
and thata + -, a+- + 72? &e. have the limit a.
x HM vu
But the easiest way, in general, to examine expressions, will be to
: : Ae tes | ,
remember that when « increases without limit, - decreases without
limit.
1 1 ; : ;
Let, then, v =— orx =—: substitute this value of x, and, if
x v
possible, reduce the expression to a form in which its limit will be
evident when v diminishes without limit, that is, when x increases
without limit.
+1
For instance, what is the limit of in such a case.
t—
1 t
oa (5+1)e 1+0
EEN
2.) G2
v Vv
Let & =
© li
Uae,
When v diminishes without limit, the preceding has the limit rs
Let the student now prove the following cases, which we express
in the abbreviated form.
axve+bextec axr?+be+e a
1 — oo a le
pet+q pe+qr+r p
az*+ba+e
petqr+r
Turorem IV. If @ be greater than 1, the terms of the series,
a, a, a®, a, &c., increase without limit; or (abbreviated) a~= «
For a=ata—a=ata(a—l)
or a becomes a? by adding a@ (a —— 1)
Similarly a? becomes a? by adding a? (a — 1)
Generally a” becomes a”*} by adding a” (a —1)
But because a is greater than 1, a—1 is positive, and the addition
of a(a—1) is therefore arithmetical increase. And a? is greater than
a; therefore a*(a—1) is greater than a(a—1), or the third power of a
exceeds the second by more than the second exceeds the first. Simi-
larly, the fourth exceeds the third by more than the third exceeds the
second; and so on. But if to a the same quantity be added as many
a.
VARIABLE QUANTITIES. 159
times as we please, the result may be made as great as we please ;
still more if the same number of additions be made, with a greater
quantity each time than the last. Whence follows the theorem.
TuroreM V. If b be less than 1, the terms of the series 0, b?, b°,
b4, &c. decrease without limit, or (abbreviated) b* = 0.
1 1 ; th:
Let b= 7 then 5» = ae But because Bb is less than 1, a or 38
_ greater than 1; therefore a” may be made (Theorem IV.) as great as
1
we please. Hence, — or b” may be made as small as we please.
9 :
: : — whi } RPT
Find out the first of the powers of tna which is less than 1000000.
Ans. The sixth.
It is hardly necessary to notice, that if a = 1 the terms of the
series a, a, a’, &c. neither increase nor decrease.
Turorem VI. If x be positive, and less than 1, the series of
_ terms
(+2) (4+a4a*%) (l+2+4+2°+2'5) &.
increases, but not without limit. The limit is 5 ; that is, no term
of the preceding, how many powers soever it may contain, can be as
1 F
_ great as rca but may come as near to it as we please. The abbre-
viation is as follows :
1 oc
—— = 1+4+4°4+ 29+ 1.0. +2
1—2z
more generally written
1
m= l+aeta*t2+ &c. ad infinitum.
As this series is a most important part of the groundwork of all
that follows, we shall try to establish the proposition from the method
of its formation. We remark that when z is positive, the terms 1,
d+27,1+7 + a, &c. evidently increase; and that each term is
formed by multiplying the preceding by x, and then adding 1. Thus,
(1+2+2? is 1+2(1+.2), and 1+2r+2°+2° isi+zr7(i+t«¢+42”),
and soon. IfA stand for any term, and B for the next; then
B=1+Az
Now, B is greater than A; therefore, adding 1 more than com-
pensates the diminution which A undergoes by being multiplied by x
160 ON LIMITS AND
(Remember that v is less than 1). But Ax =>A+Ar—A>A—
(1—#)A; or multiplication into # diminishes A by (1—a)A. This
the addition of 1 more than compensates; that is, 1 is greater than
(1—#)A. Divide both of these by 1—w#; and is greater
1—27
than A. But A is any term we please of 1,1+2,1+2+27%, &e.;
; 1
therefore every one of these, how far soever we go, is less than :
Now we have to prove that, though we cannot find A as great as
nf ,
[uz We can come as near to this as we please. Remember that the
way of forming the next term is always, ‘ Multiply by # and add 1.”
Let A differ from a by p; so that
j : x
A=>——p, the next term is] +Aazv =1]+ maar ig
Le ees rity
1 ;
fet a PE Se the next term is
1—w
= 2 é 2.
heh rarer O sooo pe; the next term is
ae
Ne at é'
para ha and so on.
J . 1
Hence we can find a term which differs from ce,
nm is as great as we please. But p is a given quantity, and a”
(Theorem V.) diminishes without limit when m increases without
by pa”, where
P aids 1
limit; therefore pa” may be made as small as we please, or = —pat
—wT
1 obs :
as near to + as we please. But, by continuing the preceding terms
— TL
1
from A, we shall at last come to myst a”, Therefore, by con-
wer 1
tinuing the terms, we come as near to Toop as we please.
We shall now try to find whether (# being less than 1),
l—x# +2°— 2#°+ 2*— &c. continued ad infinitum,
has a limit; or what is the nature of the increase or decrease of 1,
1—w, 1—x+2?, &c. Here we see alternate increase and decrease :
but still under a simple law. To find the next term, multiply the
last term by 2, and take the result from 1, Thus, .
VARIABLE QUANTITIES. 161
l—x+22 = l—a2(l—z)
l—a4+22—2 = 1—2(1—2+4+2°%) and so on.
Or if A and B be two consecutive terms,
B=1—Aa o 14+A—A(14+2)
That is B=A+1—A(1+4+2)
then the next term C = B+1—B(1 +2) &c. &c.
But the results alternately increase and decrease; that is, suppose
B greater than A, then C is less than B. Or, suppose 1 greater than
A(1+72), then 1 is less than B(1 + 2): or
is greater than A,
1+2
1 ,
and Eee is less than B. So that the results are alternately less
1
and greater than :
e 1+c2
: 1
Thus we have 1 is greater than
1+¢2
: 1
1—2z__is less than
1+ 2
: 1
l1—2-+4.4% is greater than
a 6 1+7
&e. &e. &e.
Now, since 2” diminishes without limit as is increased, we can
take n so great that two consecutive terms
l—x+2?— &e. eee hg!
and l—av7+ x72— &e. sees one oe a a
shall differ by a quantity as small as we please (for they differ by
= .”). But we have just proved that one of these terms is greater
than
7 = and the other less. And they differ less from any quan-
tity which falls between them, than they do from each other; con-
~ sequently, either may be (if » be taken sufficiently great) as near to
as we please.
1+02
These two results we express as follows:
= | +a+a?+23+ &e. ad infinitum.
peeP l—a74+2x2?—23 4+ &c. ad infinitum.
1+72
P 2
162 - ON LIMITS AND
\
and is called the swm of the first infinite series, meaning, the
1
1—2
limit towards which we may come as near as we please by continual
addition of the terms 1, x, 2*, 2°, &c.
We shall proceed with this subject in chapter VIII.
Turorem VII. Ifthe numerator and denominator of a fraction
diminish without limit, the limit of that fraction may be nothing,
finite or infinite: that is, the fraction may diminish without limit,
may have a finite limit, or may increase without limit. Neither of
these suppositions is inconsistent with the unlimited diminution of
both the numerator and denominator. |
Take the three fractions
x? —a? v—a? (7 —a)?
(7 —a)? L—a x? —a?
By supposing v« = a, all the three assume the form ~ By sup-
posing 2 to approach as near as may be necessary to a, we may
diminish the numerators and denominators without limit. The reason
of this is, that a—a is a factor of every numerator and every deno-
minator, and «—a diminishes without limit as 2 approaches to a.
For the three fractions are
(w—a)(a+a) (w—a)(x+a) (7 —a)(«#—a)
(@—a) (@—a) (7 —da) (a@—a)(#-+
‘
and when B increases without limit, Pp diminishes without limit.
Hence, as in the last case, AB, under a different form, approaches to
0
the form -.
0)
Thus we see that the three forms
0 ora
— — Ox «
0 CL
are so connected, that any expression which gives one, may be made
to give either of the others.
We now take the form a°, considered, not in the absolute and
defined sense of page 85, but as the representative of
The limit of a* when x diminishes without limit.
164. ON LIMITS AND
AD hes ae then when wz diminishes without limit, y increases
without limit. Let y be a whole number, then
a
C= tie V/a
Firstly, if a@ be greater than 1, all its roots are greater than 1 (for
all the powers of less than 1 are less than 1). Let
V/a=1+v or a = (1+v)
Now, y can be taken so great that v shall be less than any fraction
we may name, however small. For, if not, say it is impossible that y
should be so great as to make v less than &. Then, v being always
greater than k, whatever may be the value of y, 1+ is always
greater than 1+. But, Theorem IV., y may be taken so great that
(1+k)¥ shall exceed any quantity we may name, and shall there-
fore exceed a. Still more, then, will (1+ v)¥ exceed a (v is greater
than k). But (1-++v)y equals a: here then is a contradiction. Con-
sequently, the supposition that » can never be made less than a given
fraction k is not true: that is, » can be made less than any given
fraction, or 1-+-v can be brought as near to 1 as we please. But
if-v= a, therefore, if y increase without limit, we see that
1
Sa or dae or @” hasthe limit ]
Or a? =1 where 0 is used in the sense given in page 156.
Secondly, let a be less that 1, whence - is greater thani1. By
y
1
the last case Af — can be brought as near to 1 as we please; but
a
this is 1+ a, therefore “a can be brought as near to 1 as we
please.
Tueorem X. In any rational* integral expression with respect
to 2, if x may be increased without limit, the term which has the
highest power of 2 may be made to contain the sum of all the rest
times without limit. That is, in the expression
ax+ba2*+cx+e, for example,
let a be any given quantity, however small, and 4, c, and e, any given
* Look at the beginning of the next chapter.
ia i |
VARIABLE QUANTITIES. 165
quantities, however great, yet may be taken so great that ax®
shall contain bx? + cx +e as many times as we please.
The uumber of times and parts of a time which a2 contains
bx? -+ca-+e is expressed by the fraction
aa ar? — x? ax
Rae ep eee
bx? +car+e (ba? +cr+e)—2 Cytine
Cat a
c € ; ; Ai
Let =as “i P- Then, by increasing x without limit, p is
diminished without limit, and will, if « be taken sufficiently great,
E ° Lx
become less than 1. That is, ax* contains ba’*+cu+e, :
op
i hae ax f
times. But —— o xX # Increases
a
T ——
b+1 Pye “apie
without limit, when « increases without limit. The same then does
times, or more than
the number of times which av* contains b2?+cxr+e..
For example, how great must 2 be, in order that we may be
certain that the millionth part of 2° contains 10002?-++ 500 x +1000
more than a hundred thousand times?
one millionth of «% Pi one millionth of x
1 2 1000 ~ 50 1000
000 2? + 500 7 + 0 1000. :
@ x
. 500 ooo .
Now, if x be 1000 or more, or + : = is less than 1. There-
xv
fore, in this case, the preceding fraction is greater than one millionth
of 1001, or than
by
1001000,000 ,
If we take 2 =1001000,000 x 100,000 or 100100,000,000,000,
the preceding fraction becomes 100,000. Hence, the millionth part
of x° is greater than 100,000 times (1000 2?-++ 5007%-+41000). We
do not say that this is the least value of x which will answer the
conditions, but that this, or any thing greater, will do so.
Turorem XI. In any integral and rational expression with
respect to 2, if « may be diminished without limit, the term con-
taining the lowest power of x may be made to contain the rest of
the expression as many times as we please. . For instance, in
a shall
a +10002?-+10023 we may take # so small that
1 q
1000 1000
contain 10002?+1002* as many times as we please: or in
166 ON LIMITS AND
ax?+ba2®+cx-+e, x may be taken so small that e (or ex°, which
is the term containing the lowest power* of «#) shall contain.
ax>+bxz?+cxr as often as we please. This last is evident in this
particular case, because e remains the same when 2 diminishes with-.
out limit, and ax?+ bx2?+ cz diminishes without limit. Therefore,
the second may become less than any given fraction of the first.
Now, take a case in which there is no power of z so low as 2°; such
as aa°+bx?+cxr. Here a may be taken so small that cz shall
contain axz*+ bx? as many times as we please. For the number of
times and parts of times which cz contains az’ + bx? is
cx c fixed quantity
aa®+ba® © aa*+be one which diminishes without limit
which latter increases without limit when @ is diminished without
limit.
Hence it follows, that when 2x increases without limit, 2°, 2°, x‘,
&c. not only increase without limit, but each of them increases without
limit with respect to the preceding; by which is meant that 2° in-
creases so much faster than x”, that x® will come at last to contain x?
as many times as we please. Similarly, when x diminishes without
limit, we find that 2”, #3, 24, &c. not only diminish without limit, but
each of them diminishes without limit with respect to the preceding ;
by which is meant that 2° diminishes so much faster than 2”, that 2°
will come at last to be as small a fraction of 2? as we please. These
notions are sometimes abbreviated into the following phrases, which
it must be remembered are not intelligible, except as abbreviations.
ABBREVIATED PHRASES. Rationat Meaninc.
1. Of two infinitely great 1. Of two quantities which
quantities, one may be infinitely increase without limit, one may
greater than the other. increase so much faster than
the other, as not only to increase
without limit absolutely speak-
ing, but to increase without
limit in the number of times
which it contains the other.
* The algebraical series of whole powers of x is
voce &T9S -g—R gel ao gh ee
answering to
1 1
1
Te: alt = 1 2 wa a ...... (see page 85.)
VARIABLE QUANTITIES. 167
ABBREVIATED PHRASES. RatTionaL MEANING.
2. Of two infinitely small 2. Of two quantities which
quantities, one may be infinitely diminish without limit, one may
less than the other. diminish so much faster than
the other, as not only to dimi-
nish without limit absolutely
speaking, but to diminish with-
out limit in the fraction which
it is of the other.
Let the student now try if he can explain the following ProBLeM.
Be
oT ed
B A ID
iD)
If A and B move together towards the line CD, and if A move
in such a way that AQ is 4 of an inch when BQ is 3 an inch;
that AQ is } of an inch when BQ is } of an inch; or generally that
AQ is x? inches* when BQ is g¢ inches: and if a microscope be
placed over the figure which grows in magnifying power as A moves
towards Q, in such a way that the increase of magnifying power just
compensates the real diminution of AQ, so that AQ always appears
of the same length; then B, instead of appearing to move toward Q,
will appear to move away from Q.
* It is usual to say a inches, when « is less than 1, or when x inches
is really a fraction of an inch ; in which case x of an inch would be more
agreeable to analogy. Returning to the consideration discussed in the
note to page 40, it will be useful to observe that the idiom of our
language makes the connexion between multiplication by 4 and multi-
plication by 3, less obvious than it might have been. We only say 4 of
before an article or pronoun, “ four of the men,” ‘ four of them.” But
we always say, ‘“ one fourth of six,’”’ ‘‘ one fourth of an inch.” If it had
been idiomatic to say “ four of six,” or ‘‘ a four of sixes,” for 24; and
“four of an inch” for “ four inches,” the propriety of extending the
term multiplication to fractions would have been much more obvious,
168 CLASSIFICATION OF ALGEBRAICAL
CHAPTER VII.
CLASSIFICATION OF ALGEBRAICAL EXPRESSIONS AND
CONSEQUENCES. RULE OF DIVISION.
PreviousLy to commencing this subject, we shall make a classi-
fication of the expressions we have obtained. All the terms in use
are usually relative to some particular letter; in what follows we
shall suppose this letter to be a. We shall proceed to explain the
following table.
Functions.
tT we = ae ae sia"
Common Algebraical. Transcendental.
c we - aa a
Rational. Irrational. Exponential,
far EN ee Logarithmic,
Integral. Frac- Integral. Frac- Circular,
_ i ——— fi :
> tional. ~~ tional. Inverse Circular,
Monomial. Monomial, a
. . . . Ce
Binomial, Binomial,
Trinomial, Trinomial,
Quadrinomial, Quadrinomial,
&e. &e.
Any expression which contains a in any way is called a function
ofa: thus a+a, at+b2z’, &e. are functions of 2; they are also
functions of a and b, but may be considered only with regard to 2.
All expressions which contain only a finite number of such ex-
pressions as are treated in the preceding part of this work, are called
common algebruic*® functions, except only where x is an exponent.
Thns Va+22, av®+6, &c. are common algebraic functions, but
a® is not. And we do not know whether 1+ 2+ a?+ &c. ad in-
finitum (page 159) is a common algebraic function of # or not, until
we have found that it is the same as 1+(1—w). All other functions
of x are called transcendental functions; such are a” and all functions —
containing it; such will be (when we come to define them) the
* Usually, algebraic functions; but transcendental functions are
certainly also algebraic, that is, considered in algebra,
EXPRESSIONS AND CONSEQUENCES. 169
logarithm of x, the sine and cosine of 2 in trigonometry, and many
others. A function which contains a? is called an exponential *
function of x, one which contains a logarithm, a logarithmic func-
tion, &c.
Common algebraical functions are divided into rational, which
contain only whole powers of v, as a+a?, ax-2+ 6, &c.; and irra-
tional, which contain roots or fractional powers of xv, as ax?+b,
Ve +2, &e.
Rational and irrational functions of x are both divided into
integral, which contain x only in numerators, as
| a+ Nae
—~ and ———_ ;
Po a a+b?
and fractional, which contain x in denominators,
a+e Vo — Vy
as and’ = ——-—
ba +a? c+ Vi
Integral functions are divided into monomials, which contain only
one power of 2, as 23, ax®, Vba, (a+b) xt; binomials, which con-
tain two distinct powers of w (2° included) as a+b or aa®+ba,
cae V2, mx? -+ na’, &c.; trinomials, containing three distinct
powers; quadrinomials containing four, &c. The two latter terms
are little used: all expressions of more than one term are called
polynomials.
Integral and rational functions are divided into those of the first,
second, third, &c. degrees, according to the exponent of the highest
power which is found inthem. Thus
a+ba is a rational integral function of wx, of the first degree.
a+bxa+cxz? EEGE REST near elics CUare: srehe cPele o aleseetete 6 of the second degree.
&e. &e.
The term a, if written ax°, is of no degree with respect to wx.
a+ b log c+arc
mtvVn
The expression +
* The letter principally considered is an exponent,
+ The meaning of log. c, or logarithm of c, will be afterwards
explained.
) Q
170 CLASSIFICATION OF ALGEBRAICAL
is a rational integral trinomial function of @
.... rational fractional he. Hees
. +.» irrational integral Ue
.... irrational fractional oes ste t ee
. exponential © ao a os
.... logarithmic > 2 sien eee
Rational and integral functions are generally arranged, so that
the powers of x may rise or fall continually in going from left to
right. Thus, az + b—ca? is never so written, but either
—c2z?+axr+6 or bog
in the first case it is said to be arranged in descending, in the second
in ascending, powers of x.
Thus — a— bx 4+¢4%— 2° — 2°
should be a+cx—(b6+1)2?—2°
or —x°—(b+1)e+cx+a
The most important class of functions is the rational and integral,
containing those which appear rational and integral, but of which it
cannot be known whether they are algebraical or transcendental,
owing to their containing an infinite number of terms. Such are
the forms
aha ean) ans + px" + qa"
where a, b, c are not functions of 2, and n is a whole number; and —
Atbatex+eae+ -+++-+ 4 &e. ad infinitum.
The reduction of expressions to such forms is one of the principal
branches of the subject. We shall call the first generally a polynomial,
the second an tnfinite series.
Derinirions. In multiplication of polynomials, the several pro-
ducts formed in the process may be called subordinate products. Thus,
in multiplying @+ 2 by 6+, the subordinate products are ab, az,
bx, and x. A term of a polynomial is all that contains any one
power of x; thus, the preceding product ab+ax+bxr-+f 7, P, can be made
by multiplying another polynomial Q by any third polynomial. For
instance, to divide 82°-+1 by 24+, is the following: To reduce
8x8 +1
2241
. question will also give that of treating any other.
If possible, let 823+ 1 be made by multiplying 27+1 by
a+br+cu?+e2x°+&c. Now, firstly, this latter expression cannot
go higher than ca?; for, ifit did, say to ex, we should have, by the
last theorem, ex? X22 or 2ea* in the product. But that product is
8a°-+1, in which 2‘ does not appear; consequently, ex? and higher
terms are not in the polynomial required, which is therefore of the
form a-+ bx-+-ca®. We have then, if our question be possible,
82+] = (244+1)\(c2?+b2+a)
We have proved that 24x cx? must be a term* of this product; but
it can only be 8 2°, therefore 24 x ca? = 82°, or ca? = 8 a°——2x = 42",
Consequently, 8a3+1 = (24+1)(42?+ bx +a)
= (27 +1)42°+ (Qe +41) (bx+a)
8a +1—(2441)42? or —4a?4+1 = (2441) (ba+a)
of the latter product, 2~xbwx must be a term; but it can only be
if possible, to a simple polynomial. The way of treating this
—42x°, therefore br = —42? +27 = —2z, or
—42? 41 = (24+1)(—22 +4)
= —227(244+1)+(224+l)a
—4772+14+227(2%41) o 241) = 2a4+lya
* This being a subordinate product, which cannot be altered or
destroyed by any other subordinate product.
172 CLASSIFICATION OF ALGEBRAICAL
This last equation is made identical by a=1; therefore 477—2r+1
is the polynomial by which 22-++-1 being multiplied, the product is
8x°-+1: as will be found by trial.
The steps of the preceding process may be arranged after the
manner of division in arithmetic ; the only difference being, that in-
stead of finding a new term in the quotient by trial of the left hand
figures of the divisor and dividend, it is found by dividing the left
hand term of the dividend or remainder by that in the divisor, As
follows, in which the same question is solved in the two different
arrangements.
22%+1)823+ 1(42?-—24+1 1427)14+8275(1—22 4 42?
S23 +47? 1+22
—4774 ] —2274+82°
—47°—2r —2x7—4 72
Cr 1 4277+8 25
A a 422+823
0 0
Great care must be taken to preserve the same order of arrange-
ment throughout, either in ascending or descending powers of x.
The following is the general theory of this process:
First, itis evident that the sum, difference, and product of rational
polynomials, are rational polynomials. Let P and Q be two rational
polynomials, from which it is desired to obtain V in such a way that
P= QV. Here P is the dividend, Q the divisor, and V the quotient
to be found. Assume any convenient polynomial or monomial A,
multiply it by Q, and subtract the product from P, which gives
P—AQ. Call this R, so that
P—AQ —— R .stetuee (1)
Assume any other polynomial A’; repeat the process with R (instead
of P) and Q. Call the result R’.
R—A‘Q\= R° -).o.3e (2)
Assume a third polynomial A”, and let
Ri— AQ, = Rie ee (3)
EXPRESSIONS AND CONSEQUENCES. ive
The object is to simplify the remainder at every step, so as to
reduce it at last to a form in which we may see one of these two
things ; either how to find a new polynomial which shall reduce the
next remainder to 0, or that this is impossible. Suppose, first, that a
new polynomial or monomial A” can be found, which will reduce the
remainder to 0.
ne A“ OQ = 0 eovoeee ee (4)
We have, then, from the different equations
P=AQS+R=AQ+4A/Q4 PR =AQtAQ+A’Q4+R’
= AQ+A’/Q+A”"Q+A”Q _= Q(A + A’+ A”+ A”)
so that A+ A’+ A”+ A” is the polynomial required. Suppose that
instead of (4), we have
Ra A’ Q —_— Ru
and suppose it to be evidently useless to attempt to continue the
process further. We have then
P feos AQ + A’‘Q + A’Q + A’Q + RY
é be uw m1 Lay
es A At AN A ar (aie
a more simple
fraction than ©
= rational polynomial + }
Q
5
v+i1 :
Examp.Le. To reduce —; : to a more simple form
x x
P = 2°+1 Q = 2°4+22
xe + 22) +(A=% She
Peeose = AC. ~~
-20+1=R, A= —3* = 0x
—2e*—4 23 = A'Q
4e9+1=R, AX = “> = 4e
403+ 822= A’Q
—8e2+1=R", Av =—-F = —8
yaa Sra)
16z+1=R”
Q 2
174 CLASSIFICATION OF ALGEBRAICAL
It is useless to carry the process further, and we have
ey 8 i
Saag ae” 22° +44 S hips
From the preceding, we may deduce the following theorem, which
is useful in many parts of mathematics.
If P and Q be two rational polynomials of which P is of the
; as P
higher degree (or dimension, as it is frequently called), then ra)
can be reduced to the form G ite where G and H are rational
Q
polynomials, and H of at least one dimension less than Q.
Exercise. If, in the preceding process, the remainders be
severally multiplied by B, B’, B”, &c. before using them, then
a AN A” it
q-4+3 +35 + ppp’ t BBB
The preceding process may be used in an infinite number of
different ways; for though it is only convenient to employ it as in
the preceding example, yet in the reasoning P, Q, A, A’, &c. may be
any quantities whatsoever.. As in the following example,
P= 1 Q = 1l+z2
1+2a) 1 (de A =
w+ 2
l—z—22= R, le A’ = 2
x? + x3
1—7¢— 237-7
P 1 1—#r—22°— 2x3
—- = —_ = 2 2
Q ite ae 1+2
But this would in most cases amount to no more than an arbitrary
method of adding or substracting fractions. When, however, the
process of dividing the left-hand term of the remainder by that of the
divisor is followed, the result will generally be a symmetrical, and
often a useful, developement. For instance, we thus obtain
Y
ae YS 4
= |]—74+ 2 —7° 4 Les
=l+arda?t+ao+
J
[|= +
& 8
1—~ev
EXPRESSIONS AND CONSEQUENCES. 175
ey ag aa Ge ae Ge ee ea
oy sealer aie aire Se i pe Sema
Lak
1—2v +4?
= |] CASE sh ee mr ee
By this method we can tell whether either of the preceding series
of terms continued without limit will approach a limit or not. For
instance, we see that
Le Oe ade br ae Bi A (A)
antl \ A
when raed added to it. Ifthen wz be less than 1,
rt
becomes
1—~2z7
since 2”+1 diminishes without limit (page 159) when n is increased
without limit, the sum of the terms in (A) continually approaches to
1 ;
rz 28 was shewn in page 160.
Let us, for the present, denote by (P) that P isa rational poly-
nomial; and by (P)+(Q) = (P+ Q), that P and Q are rational
polynomials, whence their sum is a rational polynomial. We have
then, always,
(P)+(Q) = (P+Q), (P)—(Q) = (P—Q),
(P) x (Q) = (PQ); and e2) — & in certain cases.
(Q) Q
Every polynomial which divides (P) without remainder is called
a factor of P; thus v7—1 = (#+1)(w—1) and x+1 and r—1
are factors of x7—1. It is evident, from page 171,
1. That no polynomial can have a factor of a higher dimension
than its own dimension.
2. That if the polynomial be of m dimensions, and one of its
two factors of p dimensions, the remaining factor must be of m—p
_ dimensions.
Thus in
x¢—1 = («—1) (a? 4+224+27+1)
= (x?— 1) (a*# +1)
the polynomial being of the fourth degree, its factors are in one case
_ of the first and third (1 +3 = 4), and in the other of the second and
second (2-2 = 4).
In what follows we speak only of rational polynomials, and by
rational division we mean that the preceding process leaves no
176 CLASSIFICATION OF ALGEBRAICAL
remainder. The following theorems are evident consequences of |
what goes before.
1. Rational division is impossible unless the dividend be at least
as high in degree as the divisor.
2. Where the dividend is higher than the divisor, and rational
division is still impossible, the remainder is of a lower degree than
either the divisor or dividend.
3. If the dividend be of the mth and the divisor of the nth degree,
the quotient is of the (m—n)th, and the remainder not higher than
of the (n—1)th degree. (For so long as the remainder is as high or
higher than the divisor, the process can be continued).
4. The dividend being P, the divisor Q, the quotient* A, and the
remainder R, then
P. R
P = AQ+R or mie Q
5. Every quantity which rationally divides M and N rationally
A+
divides their sum, difference, and product. Let Z be the divisor ;
then
M Nee MN
M—N MN
6. Every divisor of P and Q (in 4.) divides R, and every divisor
of Q and R divides P, &c. so that no two of the three has any
rational divisor which the third has not. For instance, let Z divide
P and Q rationally, then
thd : aa e Q (ee
vA is rational or = TZ ZHNe
Ax (>) = eS therefore Ee Se (sea
—A
— (- Z 2); but this is a which is therefore rational, or Z
also rationally divides R. By similar reasoning the other cases
follow.
7. The highest common divisor of P and Q is therefore the
highest common divisor of Q and R.
* Not strictly a quotient, unless the remainder be nothing. It is what
comes in the quotient-part of the process in trying this point.
EXPRESSIONS AND CONSEQUENCES. ee
8. If one factor of a product be not divisible by x, then those
powers of x, and those only, which rationally divide the other factor,
will rationally divide the product. For instance, in («* + a) (x? + 6x),
since the lowest term is 6a@.x?, and since no power of x higher than
that in the lowest term will rationally divide an expression, it follows
that «? is the highest power of x which rationally divides the product.
But 2x? is the highest which rationally divides «3+ 62°.
9. If an expression be rationally divisible by a power of a, the
quotient is divisible by every divisor of the first, which is itself
indivisible by any power of x. For example, 23—w divided by x
gives z°—1, «—1 is a divisor of the first, and, therefore (were it not
known otherwise), is a divisor of the second.
To prove this theorem, let 2*P (for example) be an expression,
which is divisible, say by x +1; that is, let
ay
a = (A) #8(P) = (A) (e+1)
consequently (8.), (A) is divisible by 2°;
or z = (B) that is (A) = 23(B)
Therefore, 2°(P) = 23(B)(2+1)
(P)= (B)(e@+1) o — = (B)
that is, P (as well as x*P) is divisible by x +1, and the same may be
shewn of any other divisor of xP.
The method of finding the highest common divisor of two rational
polynomials is now exactly similar to that of finding the greatest
common measure of two whole numbers in arithmetic. For example,
required the highest common divisor of a®—wv and 3a°—3z.*. First
separate the monomial factors ; that is, put the expressions in the form
a(z°—1) and 32*(2*—1)
Neglect the monomial factors for the present, and proceed to find the
highest common divisor of
Paw bY and 2t—l = Q
xt—1)x°—1(a
Rem. a—l)at—l(ite?tatl
Rem. OQ
178 CLASSIFICATION OF EXPRESSIONS, ETC.
By (7.) the highest divisor of #5—1 and «4—1 is also that of
a2*—1 and r—1; and, since y—1 divides the former, and also is
the highest divisor of itself, it is the highest divisor common to the
two, and, therefore, the highest divisor of «®—1 and at—1. The
original expressions have also the divisor x; consequently, 2(#— 1) is
their highest common divisor.
Any power of « may be thrown out of a remainder by division,
from theorem 9. For instance, in finding the highest common divisor
of 1—.23 and 1—:
For, since the preceding series is
afl +204 oe 28 + = 5 298 4+ Bo.
and since P is greater than = 4 &c., the preceding series will be
increased by writing P instead of S > &c. But it then becomes
a\1+Pa+Pa*+ Piae+ &e. '
or a}1+(Px)+(P2y+(Px)+ &e.t
which (page 159) is convergent if Pw be less than 1, or # less
1 ; ; Bs .
than Pp: Still more is the original series convergent under the same
circumstances, because its terms are severally less than those of the
last series.
If P be greater than any one of the ratios after some given ratio,
the series converges from and after the term which gives that ratio,
whenever Pw is less than 1. Suppose, for instance, that the thou-
sandth and following terms of the series (A) are
A 2999 + B 20 4 © 1001 &e.
B CB \
999 oer 5 2
or Ag {l+zr+y qe + &c.
Then, by the preceding reasoning, if P be greater than any of the
< : ‘ 4 1
ratios &c. the preceding series converges if x be less than Pp:
B C
A’ B!
For instance, take
1422743277 +425+ &e.
yee Ey ee
rratl ---- Cc.
ratios ent 44
2 is greater than any of these ratios after the first; consequently,
: 1
this series converges from the second term if « be less than a The
hundredth and following terms are
100 799 + 101 7 + 10229" + &e.
Bee
186 ON SERIES AND
ie 101 102 103 Pe
100 101 102 ;
pe, LAU :
of which Too iS greater than any except the first. Consequently,
this series converges from the hundredth term, if « be less than
101 100
Sano aT
vergent whenever x is less than 1, though the term at which con-
vergency begins may be made as distant as we please, by making
x sufficiently near to 1.
Similarly it may be shewn that this series is con-
As a second example, take
l+o+> oe
; 1 1 1
ratios ] 5 3 b &e.
Since these ratios continually diminish, and without limit, a point
of the series will come, after which they will all be less than any
given fraction m, however small it may be. But if m may be made
1
as small as we please, —- may be made as great as we please.
m
Therefore this series is convergent for every value of x, however
great; though the greater x is taken the more distant will be the
term at which convergency begins.
As a third example, take
1+24+4+2.32°+2.3.4434 &e.
ratios a 3, 4, 5, &e.
and as these ratios increase without limit, there is no quantity which
is greater than them all. Consequently, no value can be assigned to
x for which this series must necessarily be convergent. The following
theorem may easily be proved in the same manner as the last.
Turorem. If P be less than any one of the ratios fe 5 Seu;
then the series
At batcxr+ &e.
must be divergent for every value of 2 greater than >
In this way the series in the last example may be shewn to
: 1
diverge from the second term for every value of x greater than =)
INDETERMINATE COEFFICIENTS. 187
: 1
from the third for every value greater than 7 and so on; so that
there is no fraction so small that the series shall not diverge from and
after some term by giving w that value.
[Such necessarily divergent series never will be found in practice :
they are introduced here as a warning against applying to all series
general conclusions drawn from series which may be made con-
vergent. |
In future, unless the contrary be specially mentioned, in speaking of
the series a-+bx-+cx?+ &c., we only mean to speak of series which
may be made convergent. We suppose all the terms positive.
Turorem. Every series of the form a+ ba+ca?+ &c. has this
property, that « may be taken so small, that any one term shall
contain the aggregate of all the following terms as often as we please.
For instance, by taking x sufficiently small, we may make ca?
more than ten thousand times ea? + fat+&c. Let x, be the greatest
value of x which makes e +.fx-+ &c. convergent, and let the sum in
that case be S. Then, for every value of «x less than «,, e +fe+ &e.
is less than S. Now, ca? contains ex + fuxt+ &c.
ca? Cc Cc
eu? + fart + &e. lh as + fu? + &e. EY a(e+fa+ Ke.)
_ times or parts of times. Take x less than 2,, so that S is greater
than e+ fx-+ &c., or
ene ee ey ae stewie
Bille onesies a(e+fu-+ &c.) ex? + fut + &e.
Now, c and § being fixed quantities, 2 may be taken so small
that c— «2S shall be as great as we please; and still more
ca* — (ex? + fxt+ &ce.), which is greater than c— aS. Whence
the theorem is proved.
Exampxe. How small must « be taken, so that we may be sure
the fourth term of
1422743272442? +57*+ &.
contains the sum of all that follow 1000 times at least.
The whole of the series after the fourth term may be written thus:
ABE
Bat (1450+ 5-50 t+ Geb eee (A)
Ok. A :
and z is greater than any of the succeeding ratios ; consequently, we
increase the preceding by altering it to
188 ON SERIES AND
pie of 5a
Satll+ sat 5-50°+ &e.} or ———— ....-. (B)
Ree
re
(see page 182.) We have then to take «, so that
ae OF
4zx* is greater than 1000
4 ones
=o
6
1—-~.
(x) ta 1— ot must be greater than 250 x 52 or 12502,
4x3 /?
which is certainly true if 1—2z be greater than 12502, or 1 greater
1 : ;
than 12522, or 7959 Sreater than x. In this case 42° is greater than
1000 times (B); still more then is it greater than 1000 times (A).
Turorem. If the two series
Ay tay +dea?+ &e. and b)+b,4+b,x? + &e.
be always equal for every finite value of 2, then it must follow that
a, = b,, a, = b,, a, = b,, &c. or the series are identically the same.
Let these series be called a,+ A and 6,+-B, in which, by what
has just been proved, we can make A and B less than the mth parts
of a, and 4,. If possible, let a, and b, be different numbers, and let
a, = b,+¢. Then, since the series are always equal, we have
aatA=b,4+Borb,+f+A =b,4+B; that is, t= B—A. But
because a, and 6, are fixed quantities, their difference is the same;
and we have ¢, a fixed quantity, equal to the difference of two quan-
tities, each of which may be made as small as we please, which is
absurd. Hence a, = 6,+¢ cannot be; and by the same reasoning,
a, = b,—t cannot be; therefore a, = b,. Take away these equal
terms from the two equal series, and divide the equal remainders by
x, which gives
A, +detX+a,%x*+ &c. always equal to bi; +bea+b3a24+ &e.
from which the same species of proof gives a, =, ; repeat the process
of subtracting equal terms and dividing by a, and the repetition of
the proof gives a, = b,; and soon. Hence, if one or more terms be
wanting in either series, the same must be wanting in the other; for
instance, ifa—w be always equal to a,+a,x-+a,2°-+ &c., we must
have’ —“a,;\—1 —=7@,; 0 =a), 10 = a weer
' [The preceding process amounts simply to shewing that we may
INDETERMINATE COEFFICIENTS. 189
make « =O in a series, and take the ordinary algebraical con-
sequences, in the same manner as if the expression were finite in
its number of terms. For instance, if a,+a,2-+ &c. be always
=b,+6,«2+ &c., we have proved that the consequence of making
x = 0, namely, a, = b,, is true. But, as we have sufficiently seen,
it is not safe to say that when 4 = 0, P = Q, except in cases where
we may say that by making ~ sufficiently small (or near to nothing),
P may be brought as near to Q as we please. The difficulty which
we have avoided is as follows: In the series 1 +2” 4+ 2.32? + &c.
we have seen that, take x as small as we may, the sum of the terms
can be made as great as we please. Are we, then, entitled to say,
that when x = 0, the preceding becomes 1+0+0+ &c.or1? If
the number of terms were finite, there could be no doubt of the pro-
priety of answering in the affirmative; but when the number of terms
is infinite, nothing that has preceded will enable us to give an answer.
The student will remember that we have confined the demonstration
entirely to series which admit of being made convergent.
It is usual to prove the preceding* by saying, that when the two
series are always equal, they are equal when x = 0, and consequently
a, = 6,, and so on. This is avoided in the present case; and we
may say that we have proved the following theorem. If two series
(which can be made convergent) are always equal when = is finite,
then they are also equal when 2 = 0.|
* On this point the student, when he is more advanced, may consult
Professor Woodhouse, Analytical Calculations, &c. Preface, p. vill. note.
It is there objected that to make a =0 and thence to deduce a, = b,, is
the same as arbitrarily making a,)=6,. This I conceive to be not the
true point of difficulty. All mathematical consequences are necessarily
contained in the hypotheses from which they spring: so that to invent
any hypothesis is necessarily to invent all its consequences, some of
which may be so near as to appear nothing more than the hypothesis
itself, others so little perceptible as not to seem necessary attendants of
the hypothesis. The real objection to the proof on which this note is
written, I conceive, is this, that having frequently found the passage
from x as a symbol of magnitude, to x as the symbol not of magnitude
but of the absence of all magnitude, to be attended with consequences
which require a special examination, it is not allowable to enter upon any
new ground, without either establishing the accordance of the conse-
quences of x = 0, with those of x = some magnitude, or distinguishing
and explaining the discrepancy, if any.
190 ON SERIES AND
The following are a few instances of the method by which we can
obtain the limits of the sums of many series. First, let us take
P = l+ar4a72?4+23 +2'+ &e.
in which we wish to determine a finite algebraical expression for P.
It is plain that
1+a4+2?+ &c. ad inf. =1 +o}1 +xr+2?+ &e. ad inf.
1
1—w
that is, P= 1lta2P or) BS
a result previously ascertained. Now, let us take
P= 1427432774+4234+ &e.
= = 24+327+4277+5273+ &e.
P—1 1
—— — Pe ae ee ee
1 1 1 1 1
whence pie-il = +5 =F
1
or P — G—ay?
Next, let P = 1+327+527+723+ &e.
-—t = 3+524+72°+925+ &e.
aes = 2427420242273 4+ &. = -
on 1+ 2
whence P = (ey
Next, let P = 14+-474927+162°+ &e.
P=! = 4492416242529 + ke.
St _P = 345x24722+492° + Ke.
é P—1 es MiP 2
ik —-—P)x+1 = 143274+527?+ &. = Goa
ys iter
whence P — G—ay
The same method might be applied to finding a more simple
algebraical expression for the sum of any finite number of terms of
the preceding series. For example, let
INDETERMINATE COEFFICIENTS. 191
P = 142xe43a2+ .-.--) $(n—]l) a"? +na"-!
| Pot = 2432440%+ PsN es Sach ed BAA ed
P—1 n—2 -1
- ah ttre tw..... + xe"-*—n x"
1—a"-1 4 1—(n+1)a"-1+n2"
(page 103) —= eae —nx” 1 = a
(eed ae
cla
__ The student may endeavour to prove the following:
Qn—1a"t1—Qn+ 14" +4+a41
143274+5a2+..+(2n—l)a* = a
} The inquiry which we have most frequently to make is the in-
verse of the preceding; not, having given the series to find its sum,
but, having given an expression, to find the series of which it is the
sum, or to develope it in powers of some one of the letters contained
init. Let us suppose, for example, that we want a series of powers
of x with coefficients, &c. which shall be, in all cases in which it is
convergent, equal to (1+ 7)—(1—v)’. Suppose the series to
be a, +a,7 +aga*+ &c., so that we have
1+27
(1—2)?
Multiply both sides by (1—-)? or 1—2a2-+ 2°
l+a= fagta, +a, x*?+a3 2+ &e. |
— Qa) x—2aya?—2aexv?— Ke.
+d x2 +a, x2 + &e. |
_— Ay + (di —2 do) + (de—2 ay +o) 2? + &e.
The two sides of this equation being equal for every value of x,
the theorem in page 188 gives
= Ag td £+ dex? +0323 + Ke.
: Agra a4 —2 uy a or a= 3
dg—2a,+d) = 0 or Cag nO Me
| a3 —2dg+a = 0 or az = 7
' Xe, &e.
So that the series is 1+ 37-+52?+ &c. as already determined.
The first term of the preceding might have been found imme-
diately : for, since we have proved that the results of « = 0 may be
192 ON SERIES AND
employed, and since (1 +2) + (1— 2)? becomes 1, and the series
is reduced to a), when x = 0, we have a, =1.
We shall now ask what is (1—a#*) + (1—z2) expanded in a
series of powers of x. The first term is 1, found as in the last
entence; let us suppose
1—a‘*
1—2
l—a*t = 14+ a,04 aca? +a,x°+ atat*+asa°+ &e.
— £2 — 04, X*?§ — 9 22> — 3 x*—a42* + &c.
= ltajrvtacr?+asx>+ayz*+ &e.
As there is no first power of x on the first side, we must have
ad,—1=0, ora,=1. Similarly, a,—a,=0, or a,=a,=1; a,—a,=0,
a,==1. But the coefficient of x* on the first side being —1, we must
have a,—a,—= —1, ora,—1 = —1, that is, a, =0; again, a,—a,=0,
ora, = 0, a,—a, = 0, ora, = 0, and soon. Hence the series is
Ll+a+2?+23+0 x 24+0x 2+ &e.
that is l+a+ae°+2°
as might be found by simple division, or from page 103. Thus, we
see that when we assume.an infinite series to represent a quantity
which is in fact a finite expression, the method of determining the
coefficients of the series will shew the coefficient O for every term of
the series which does not exist in the finite expression.
Again, to develope 1 — (1 + 2?) assume
1
Tae = Ay ta xtacx?+asxu?+ &e.
the preceding process will give
A =1 adetao=90 ordg=—1 ag+ae=0 or ag=1
@y =O! ds ay=0' or az:=.0 a3+a3=90 or ag =O
so that the series is
140 x a—a*°+0x 23 +at+0x 2° &e.
or J ~—a*+ 24 — 2° + &e.
If it happen that, by the preceding process, an equation is
produced of which the two sides cannot be made identical by any
suppositions as to the value of the coefficients, it is a sign that the
expression cannot be developed in a series of the form proposed.
If we try to develope 1 (1-+2)x by assuming
& INDETERMINATE COEFFICIENTS. 193
ks ttaox*+asx° + &e.
(1 +2) 0 1 2 3
we shall find
L = do t4+ (a, +0) #2 + (dg +01) 22+ &e.
the two sides of which cannot on any supposition be made to agree;
for there is no term independent of x on the second side which may
be made =1, In fact, we should find
i 1
= 2
. or ae —z3+ &e.
A! 1 ghd a 2
(+)x a(1 +2) gobs eects 5
so that the fraction proposed cannot be developed entirely in whole
and positive powers of x, without the introduction of a negative
power (in this case 2—}),
The following are given as exercises:
4. If P = ata," +ou°+ &c. then
P
oD a Ay + (ay +44) % + (Ay +1 +e) x? + Ke.
iB
1+2 = Ay + (Q— A) & + (Ae— A, +p) 2? + Ke.
oe ae Tay > RE
2. pe 1—w tai —at+ 8—al + &e.
Be (iat) ot
1) 24 (et) se
4 m+ne m mq af es
p+qr- p mE ee EN
_ ) xv? — &e.
ea Ss
194 ON EQUALITY IN ALGEBRA
CHAPTER IX.
ON THE MEANING OF EQUALITY IN ALGEBRA, AS DISTIN-
GUISHED FROM ITS MEANING IN ARITHMETIC.
In page 62, among the extensions of terms, we notified that the word
equal was to be considered as applicable to any two expressions of
which one could be substituted for the other without error. Hitherto
we have only applied this extension to the case of definite algebraical
quantities, either positive or negative: the numerical value of the
quantity determining its magnitude, the sign determining only which
of two opposite relations is intended to be expressed. We now pro-
ceed to consider the word equal, or its sign =, not in a sense wider
than any which the definition will bear, but wider than any in which
we have yet had occasion to use it.
Two expressions are said to be equal when one can be substituted
for the other without error. The whole force of this definition lies
in the answer to the question, What is error? The answer is, any
thing which leads to contradictory results, or which may in any
legitimate way be made to lead to contradictory results.
Results are contradictory when, both being intelligible, or capable
of interpretation, the two do not agree; but they are not necessarily
contradictory merely because one or both are unintelligible, for where
the meaning of any part of an assertion is unknown, we cannot say
whether it be true or false. For instance, in page 23, the expression
ro - was no contradiction of any thing which had preceded, for ;
had no meaning. Having ascertained this, our object was to give
it a meaning which should not contradict any thing deducible from
preceding principles.
We have shewn that if 2 be less than unity, the summation of
» which
: : : 1
i+a-+a2°-+ &c. will continually give results nearer to eas
however can never be absolutely reached by that process. Hence
we used the sign = in the equation
AND IN ARITHMETIC, 195
1
1—wv
= l+a+a?+a3+ &e. ad infinitum.
Arithmetically speaking, we can make the preceding as nearly
true as we please, by taking a sufficient number of terms from the
second side. Algebraically speaking, the above may be considered
as absolutely true ; but it must be remembered that this is only on
a supposition which is arithmetically impossible, namely, that all the
terms on the second side, expressed and implied, shall be considered
as included. For instance, we know that the multiplication of the
first side by 1—2* gives 1+; the multiplication of the second
side by the same gives
l+a+a%+a34+2°+2°+ &e. ad inf.
— 2? — 43 — x*— 2° — &e. ad inf.
' 1+0+0+0+0+0+ &e. ad inf.
in which we can prove, if necessary, that every succeeding term
shall be 0; not by actually looking at every term, which is im-
possible, but by a deduction from what we know to be the law of
the “series: ~Simiiatiy, we shew that 2™ x7” = am+n; not by
looking at every case, which is impossible, but by what we know of
the meaning of 2”.
If we proceed arithmetically, taking any number of terms, how-
ever great, say as far as 2”, we then have
(Lh4a-4a%+ 1... +a") +2") x (1 —22)
= florta?+et-.... +27 +2"
= 1]4+r¢—g"tl—grt?
in which, since z is less than 1, m can be taken so great that a”+1
and 2"+2 shall be as small as we please. Here is algebraical equa-
lity which can be made arithmetical equality, quam proxime.
Let us now suppose that x is greater than 1; say 4 = 23; the
_ series
lt+a+2?+27°+ &e. ad infinitum
is then, arithmetically speaking, infinite, since there is no limit to
the magnitude we may obtain by summing 1+2+4-+48-+ &c.
Should we then assert algebraical equality between = and the pre-
196 ON, EQUALITY IN ALGEBRA
ceding series? We have noright to do so from pages 21, &c. for
k 7 Af.
though it was there shewn how to make it clear that 5 38 above all
‘hrithmetical numeration, the converse does not therefore follow, that
whatever is above arithmetical numeration is properly represented
by m What, then, is the proper algebraical representative of the
‘preceding series? Ifit have one, let us call it P; then, since the
series itself is the same as
] +a} +ar+r°+ &e. ad infinitum}
P cannot be used for the above unless 1+ xP may be cbbatitated
for P. Or we must have.
1
Oe
P = 1+2P which gives P =
ihe same result as when 2 is less than unity. And we may shew,
in the same way as in the last page, that any algebraical operation
performed either upon ; , or upon 1--+ a -+ a? +. &e.- gives -but
one result: indeed, since there is in algebraical multiplication no
stipulation that the quantities employed must be positive, the process
there employed. is equally applicable in the case where x is greater
than 1. But in this case we cannot obtain (as in last. page) ‘any
approach to arithmetical equality, but the direct reverse; for 2+!
and a#”+2 increase, instead of decreasing, as n increases. 2
The method of expanding algebraical quantities in pages 191, &el
does not require that 2 shall lie within the limits of lei
but the process is equally conclusive in all cases. To expand
ae
for instance, we ask what expression of the form dg tac-+ &e,
iil be the same as = All -we know. of the latter lies in -its
definition, that, multiplied by 1-+ 2, ‘it gives 1.- The series Fj
a*— 23+ &e. has this property, independently of the value of xii
We shall now ascertain, 1st, that so far as instances can shew it,
we may prove that this general use of the sign = will not lead to
inconsistent results: 2d, that we are not obliged to depend upon
such a species of proof, but that the result is one which necessarily
follows from the nature of our primary assumptions.
AND IN ARITHMETIC. 197
To try the first, let us assume that, in all cases,
1
1+ 727
We have then, if 2 = 1,
I
5= 1—1+1—1+4+1— &c. ad inf.
= l—x+22—23 +2*+ &e. ad inf.
a result which is in no sense the expression of an arithmetical
equality ; for the above series continued to an even number of terms
must be 0, and to an odd number of terms, 1. We shall now try
whether alternate addition and subtraction of a quantity to and from
itself, ad infinitum, will or will not, when treated algebraically, give
the half of the quantity in question, Let us suppose
P = 1l+a4+a°+23+ &e.
ees lee | Fe 2 ee,
A
+P == Es Aaa
ay ee — 1 — &.
- P—P+ P—&e. = 14+ (@—1) + 2°?9—~a +14 &e.
+1 X21 e+
(page 103) Spl cee i bearere ia ere e
tba t+e* +.&e. 1—1+1— ce.
ah r+1 ui r+
But P = ——, and rs? +294 Ke. = a(1t2ta"+ &e.) =o
if, therefore, the preceding process be algebraically equivalent to
1
1—w
have the following equation :
halving and if1—1-+1—1-+ &c. may be changed into > we
1 1 on Peed 1
ete eats 27-4
2 Ne eas § x 1 us ot |
‘e eT ae eet
which will be found to be true.
The above contains the algebraical artifice of writing the series (A)
in the form ,
)+xu+a?+a°+ &e.
—l—x—2*— &e.
&e. &e.
$2
198 ON EQUALITY IN ALGEBRA
instead of te : tag
lta+224+73+ &e.,
—1l—r— 22 — 2° + &e.
Ke. &e..
or (1—1+1+ &e.) + (@—a2 +44 &e.)+4+ &e.
Bat it is to be observed, that we do not say that every algebraical
use of = will produce arithmetical equalities, but only that when-
ever an algebraical.use of = does produce an arithmetical equation,
we shall find that equation to be arithmetically true: an assertion as
yet uncontradicted. )
We now proceed to the second point.
We have previously so constructed the meaning of the funda-
mental symbols of algebra, that algebra, in certain cases, coincides
entirely with arithmetic; and, more than this, the rules which follow
from the definitions are so constructed, that when the result only
is arithmetical, and preceded by algebraical steps, the alterations
necessary to make these steps arithmetical, produce no alteration in
the result. This being the case at the outset, and it being shewn that
the number of steps through which we pass by algebraical process
S. Le Ae; Oe Coos s; Wy, prnsndias atatamanwet, sro than know, 1st. that all
arithmetical results so deduced may be depended upon, as much as if
they were arithmetically deduced ; 2d, that all results which are not
explicable arithmetically, are such as are perfectly consistent with the
definitions laid down; and, if not always arithmetically true, cannot
produce a result which shall be arithmetical and false. _
The reason why we have appealed to instances is, that the pre-
ceding argument, being general and abstruse, will not be thoroughly
understood by the student until he has a degree of exact compre-
hension of the words employed, which can only be gained by
familiarity with the use of algebraical language. There are also
principles of reasoning,* independent of algebra, which are difficult, —
* Remember that mathematical reasoning means reasoning applied té
mathematics, and is not a different kind of reasoning from any other.
The art of reasoning is exercised by mathematics, not taught by it; noris
the mathematician obliged to use one single principle which is not
employed in every other branch of reasoning. In fact, an opposite of
this is the case ; there are principles in other branches of reasoning which
are not employed in most branches of mathematics. Those who are not
2” AND IN ARITHMETIC. » 199
and which the beginner will therefore not conquer by algebra alone.
Such is the following: that if assertions which are not inconsistent
with each other are rationally and logically used, the conclusions
cannot be inconsistent with each other. But though this, in its full
extent, be a very difficult proposition for a beginner to understand, the
difficulty is not an algebraical one. | |
We have said, or at least implied, that when an_ arithmetical
result is produced, the steps by which it was produced, if not already
arithmetical, may be made so. Under arithmetical equality we in-
clude not only the absolute notion of equality which we recognise in
4+5 = 9, but what we have called the quam proxvimé equality,*
which we. see in
1 Een ee
or Selo tygts + ke. ad inf.
{<= —
2
Let us now suppose that a is less than 1, but that we may sup-
pose it as near to 1 as we please. Also, let x be less than 1 (which
aware of this talk of mathematical demonstration as if it were distinct
from all other kinds. |
But this is not the case; mathematical demonstration is, so far as it
goes, the same thing as any other demonstration ; the superior safety of
mathematics lying, not in the method of arguing, but in the reasoner
knowing more exactly what he is talking about than is usual in other
discussions. ‘The mathematical sciences are concerned in whatever can be
counted or measured ; and whoever talks of mathematical demonstration,
as applied to any thing else, either means merely logical or certain
demonstration, or does not know what he is talking about. |
The opponents of mathematics are, of all men, those who pay this
undue respect to mathematical reasoning ; which they invariably do, by
aSserting as a discovery and a triumph, that those who are only mathe-
maticians frequently reason ill on other subjects. We recommend the
student to believe this whenever he meets with it, and to act accordingly ;
for it is an important truth, though not either a great or recent discovery—
having been, in point of fact, ascertained immediately after the fall by our
common ancestor, who, having till then been nothing but a gardener,
must have found himself but an indifferent tailor.
_*® We are not here stepping even beyond the bounds of ordinary
— <3 »—
arithmetic. For “10,11, &c. have no other but a quam proximé
existence; we can find fractions which, multiplied by themselves, shall
ib 4
oe &e. until we come
1
be as near as we please to 10; we can sum 1, ?
as near as we please to 2.
200 ON EQUALITY IN ALGEBRA
is necessary to the arithmetical existence of the final equation), and
we have
y= lata —at + &. (1)
; arithmetically
pap tet a2 +23 + &e. (2)
1 1
lt2v+22% + 23 + &e.
—aA—axr—ax?*— &e.
+a? +a?x+ &e.
— a— &e.
= ]—(a—2)+(a*—ax + a*) — &e. supposing @ > x
a+ez e—r ata &
atc a+az a+er ae
_ a—@?+a—Ke, r+a?+2°+4+ Ke.
= ate +r ate
a 1 x 1
1+aa+te! 1—2#'a+er
——
——
So long as we suppose a less than 1 (no matter how little), the
preceding process is arithmetical; but the moment we suppose a = 1,
the equation (1) loses all arithmetical character. But still, the last
equation reduces itself to the one given in page 197.
We have chosen a very simple case, but we might, with operations
of sufficient length, and by various artifices, reduce any algebraical
equality to an arithmetical one. But methods of doing this, at once
short and general, cannot yet be understood by the student.
Let x be first supposed positive; let it then become 0, and after-
wards negative. The following and similar tables may then be
made.
Sign of & + O —
. 1
Sign of = a CK —
Aye x? + 0 =
1
. ej + od on
cooee§ % + O ae
1
= + x “f=
AND IN ARITHMETIC. SOOT
These, and other instances, give the following principle: + If,
when x changes from a to b, passing through all intermediate values,
the sign of a function of x change from positive to, negative, or vice
versa, the point at which the change takes place is marked by its value
_ being either nothing or infinite; but the converse is not true, that a
_ function always changes its sign when its value becomes nothing or
infinite.
The infinite here spoken of is the form a3 but, as we have seen,
all methods of obtaining an arithmetical infinite (we should rather
say all arithmetical methods of increasing number without limit), are
not properly represented by =. If we take the equation
—— = lta tarpas4 Ke. ad inf’
we see that when z is greater than 1, the second side is, arithmetically
speaking, only an indication of a method of obtaining number with-
out limit, while the first side is negative. There is then a change of
; : 1 1
sign when @ passes, say from F to 2, and the change takes place when
x = 1, giving,
11.1
= = 1414141 + &e.
—; = 14294448 4+ &e.
We therefore see that a divergent series may be the algebraical
representative of a negative quantity: but the series
1
ile):
which is divergent when 2 is greater than 1, gives a positive result in
all cases. From this we see that when an equality specified ts purely
algebraical, we are not at. liberty to compare magnitudes by any arith-
= 1424+ 32? +423 + &e.
metical comparison, if infinite series be in question. For instance,
if a be greater than a’, b greater than Db’, &c., we may say that
a+b-+ &c. is greater than a’+ b’+ &c.: 1st, so long as the number
of both is finite; 2d, if a, b, &c. be so related that a+b-+ &c. can
never exceed a given limit. But we may not draw this conclusion
202 EQUALITY IN ALGEBRA AND IN ARITHMETIC,
whan @+b-+ &c. increases without limit; nor may we say that the
algebraical representative of 1+2+4-+4 &c. is greater than that of
pee |
1+ 5 ao a + &e.
There is no liability to trangress this rule, because the quantity
in question must cease to be the object of arithmetic. before the
reasoning upon it can fail. Let it then be observed, that with regard
to divergent series, we admit no results of comparison except those
which are derived from their equivalent finite algebraical expressions.
ON THE NOTATION OF FUNCTIONS, 203
CHAPTER X.
ON THE NOTATION OF FUNCTIONS.
We have already defined what is meant by a Junction of x (see
page 168): the present chapter is intended to exemplify the method
of denoting a function, either for abbreviation, or because we wish to
allot a specific symbol to some unknown function.
When we have to consider an expression, such as x? + ax, only
with reference to the manner in which it contains 2, that is, to reason
upon properties which depend upon the manner in which « enters,
and the value of «, and not upon the manner of containing 4a, or its
value, we call the expression a function of x, and denote it by a
letter placed before z, in the manner of a coefficient. But to avoid
confounding this functional symbol with that of a coefficient, certain
letters are set apart always to denote functional symbols, and never
coefficients. These letters will be, in the present work, Let at's
Thus, by Fz, fx, ¢2, ~x, we mean simply functions of 2, given or
not, as the case may be. By F@x we mean the same function of ga
which Fx is of x: for instance, if Fr = «+ a", Foe is or+ (oa).
Examprss, Let 02 = +a? Fo = 1—z?
Fox zs: l1—(l+2*°¥ = —22°— 2x4
oFa = 14+(1—2*)? = 2—2277+24
Let 9% = X* then P(1+2) = (1+27)* 9(2x) = (22)
(a) = at ob = BD &e.
A functional equation is an equation which is necessarily true of
a function or functions for every value of the letter which it contains.
Thus, if ¢7 = az, we have o(br) = abr = b x Oa, or
o(ar) — Vor
is always true when ¢x means az.
Thus the following equations may be deduced :
Lila Oi? ore 2" PEX oY = play)
ae SPATE Serre i)
204 ON THE NOTATION OF FUNCTIONS?
= ot—OY Saray
px = ax+b or—o2 L—Z
Or = ax eut+oy = o(tt+y)
We can often, from a functional equation, deduce the algebraical
form which will satisfy it: for instarice, if we know that ¢(«y)=a2x oy,
supposing this always true, it is true when y = 1, which gives
o(7) = 2x ¢(1). But 9(1) is an independent quantity, made by
writing 1 instead of w in g(w). Let us call it c: it only remains to
ascertain whether any value of c will satisfy the equation. Let
gx =cax; then ¢(wy) =cxry, and «x¢y=x2Xcy =cxry; whence
o(vy) = «oy for all values of c, and gx being ca, o(1) is cx1 ~
or c, as was supposed. Similarly, if
(vy) = (ex), by making = | we have
ey=j}el) =o pr = c
o(ey) = ct = (&! = (pay |
and $(1) =c!=c, as was supposed. This, with the following theorem,
will be sufficient for the investigations connected with functions which
we shall need in this work.
We have seen that if ¢2 = c* we have grax oy = o(x+y); but
we do not know whether there may not be other functions of « which
have the same property. We shall now, however, prove this con-
verse, namely, that the equation gx x ¢y = ¢(«+y) can be satisfied
by no other function of #, except those of the form c’.
Let us suppose @2-to be a function of such a character that
whatever may be the values of x and y, thé equation
PLX OY = O(L+Y) eeeees (A)
is true. For y write a+b, which gives
exxo(atb) = o(a+a+b)
But equation (A), as described, gives (a+b) = gax@b, whence ~
xx paxob = o(e+atb)
For either letter, a, for instance, write ce, which gives
px x o(cte)xob = o(a+ce+e+b)
but OC 4-6) = OC ime
whence Xx 9CX pEX pb = O(a+c+e+b)
ON THE NOTATION OF FUNCTIONS, 205
and so on: that is, if there be quantities of any value whatsoever,
MIRINCLY, Gj Gay By .2000. an—1, Gn, we have
DA, X Pag ++ X PAn-1X PAn = Pat Get ++ +An-1+4n)
Let these n quantities be all equal to each other and to a, which
gives
Nal sees ert = aati iitin Eas
there being x factors. there being 2 terms.
or (pa)" = g(na)
where 7 is any whole number.”
Similarly, if we had supposed m quantities each equal to 5, we
should have had
(pb)" = p(mb)
Now, m and n being whole numbers, let us suppose mb = na,
which gives +
g(mb) = e(na) or (9b)" = (pa)"
n
or ob =(pa)™ but b= —a, whence
e(2a) = (ga)
or the equation ¢(pa) = (¢a)” is true when p is a whole number,
or a commensurable fraction (page 98).
In the original equation (A), let r= 0 and y = 0, whence
r+y=0. Let ¢(0) be called c; we have then exXc= ce, orc = I.
Then let y = —xz, or r+ y = 0, which gives
exxo(—z) = 90) =1 o g(—2) ==
which being true for every value of 2, is true if for 2 we write pa, p
being a whole number or commensurable fraction. This gives
1 1
—p)a) = —-— as -————- = a —p
* We cannot conceive a fraction of the preceding process. (See
page 37).
+ Similar operations performed upon equal quantities must give
equal results.
T
206 ON THE NOTATION OF FUNCTIONS.
or the equation
(pa) = (pa)?
is true if p be a negative whole number or commensurable fraction.
Hence, as in page 204, by making a = 1, we determine
o(p) =e?
where (provided p be commensurable) c may be any quantity
whatever.
The preceding equation will also be true when p is an incom-
mensurable quantity, such as V2, “4, &c.; but the method by
which we shall treat the consequences of this equation will render a
proof unnecessary.
Exercise. Shew that the equation
P(Ery) + o(G—Yy) = 2pux py
is satisfied by
1
aed es (a? +a~)
for every value of a: and also that
o(at+y) = pxu+oy
can have no other solution than
Px = ax
ON THE BINOMIAL THEOREM: 207
CHAPTER XI.
ON THE BINOMIAL THEOREM.
Tue binomial theorem is a name given to the method of expanding
(a+b)" into a series (finite or infinite as the case may be) of powers
of a and b, n being either whole or fractional, positive or negative,
commensurable or incommensurable.
The preceding case may be reduced to that of expanding (1-+.2)"
in a series of powers of x: for
b n “ b\"
a+b=a(1+-) (a+b) =a (+2)
Let r=, and (1+ x)" is the function to be expanded.
As this theorem is of particular importance, we shall give two
investigations of it: the first analytical, inquiring what is the series
which is equivalent to (1 +)"; the second synthetical, shewing that
the series so found is the one required.
If it be possible, let (1 +7)" be a series of whole powers of « of
the form
(1 + x)" = dy + a4% + a2? + 442° + &e.
in which a,, a,, &c. are functions of n and not of z.
ae a” — b*
Lemma. Whatever may be the value of, the limit of mary
to which it approaches as b is made more and more nearly equal
a c,18 na-—*.
; | -
[When a=), observe that this fraction takes the form a
page 162. |
First, let n be a whole number. Then, page 103,
a" — b”
a—b
= g"—14Q"—-2b +a" 3h? + «.. ab? + HR
as band a approach to equality, the limit of the second side is
a+ a" ata" a+ .-+» -aa"? +a"
208 ON THE BINOMIAL THEOREM.
or q?} + qr—l +q?-} + Pee A +qr- + qt-!
or Navas
[That there are n terms in the preceding is evident from this, that
there is a term for every power of 6 from 1 to »—1, both inclusive,
and one term independent of 5. |
Secondly, let 2 be a fraction E where p and g are whole numbers.
q
We have then
Sak l\p 7 ft 4?
a"—b" ai — 68 hid, (ar) ii )
a—b re G-—00) he 1\@ 1\9
(ar) — Gs")
1 1
== _ a? —bP
(Let a= b= b,) = ai —b
aP— bP
a,—b,
Ser ai—bs
a,— b,
now, as a approaches to 6, a, approaches to b, and as p and q are
whole numbers, the limits of the numerator and denominator of the
preceding are pa?—? and ga{—'; whence the limit of the fraction i
p—1 p—g P-1
a or tates or (a) or Pat = or na*-!
qa, q qg
Thirdly, let » be negative, and let the corresponding positive
quantity be p, so thatn==—p. Then
oy 1
ar— bh" __ 9 -b5? ae 1 bP —aP
Gp a Aa, Lo a? bP ab
= 1 a? — bP
7h a? bP € a—b
as a approaches towards b, the limit of the first factor is — ore or
a
—a-—*?, and that of the second (p being positive) is, as already
proved, pa?—'. Hence the limit of the preceding product is
—a~*? x pa?! or —pa—?—!, which, since n= —p, is na®—!.
We now resume the assumed series
(1 + x)" =a) + ax + dex? + asx? + &e.
ON THE BINOMIAL THEOREM, 209
Let y be a quantity which may be made as near to 2 as we
please ; the above series being supposed true for all values of x, we
may put y in the place of x, which gives
(1+ y)” =ay + ay + dey? + &e.
(l+2x) —(1+y)” =ay(a—y) + ao(x? — y’) + &e.
Divide both sides by «— y, or (1+ 7) — (1+).
(+ar)"—(i+y)" _ a? —y? P—y*
Ree) ot ay, 8 ey
which two sides being always equal, the limits to which they
‘approach, as # and y approach to equality (in which case 1+. and
+ &e.
1+ y approach to equality), are equal; or
n(1 + 2)"-! = a,+ 2agr + 3azx? + &e.
Multiply both sides by 1+ 2. |
n(1+ x)" =a, + 2aca + 3asx*® + 44423 + Ke.
+ aye + 2aga? + 3a52° + &e,
But
n(1 + 2v)"=Nd) + na,x + Nox” + Na3x> + &e.
therefore, page 188,
n—1 am—1 .
Q,=NM, 2a + a, = NA, or ag= a SN Ay
n—2 m—1 n—2
3d,+2d2=Nde or a3= “Tagline abst alts
n—3 na—1n—2n—3
4a,+93a,=na, or a, a, 72 3 5 ag
be
from which, by substitution and observing that a, is a factor of all the
terms, we have
(l+2)"=a)(1 +nx+n
m—1 n—1n—2
2 3
Beisel 3 +e.)
In which a, is not yet determined. To determine it, we must first see
whether the preceding series is ever convergent.
But previously, we must observe that what we have proved is
not, properly speaking, the truth of the above equation, but only that,
if (1+x)" can always be expanded in a series of whole powers of x,
the preceding series is the one. For we have assumed that (1+.)” is
a+ a,x +a,x* + &c., when, for any thing we know to the contrary,
it might be a series of fractional, or negative, or mixed powers which
T2
210 ON THE BINOMIAL THEOREM.
should have been chosen. Whenever, in our previous investigations,
we have made an impossible supposition, we have always been
warned of it at the end of the process by the appearance of some new
and unexplained anomaly. As we have not, before this one, made
any assumption of the form of an expansion other than we had either
reason or experience to justify, and as we have neither for the one we
have actually made, we know neither whether we are right, nor what
is the index of our being wrong. It may be that the appearance of
an always-divergent series would follow from a mistake, if there be
any. We therefore examine the preceding series.
The ratios of the several terms to those immediately preceding are
n—1 n—2 n—3
20k ric 4
NX x &e.
the general form of which is
(p+2) term n—p
(p+1) term = p+1
First, we observe, that if m be a positive whole number, the series is
finite: for the (n + 2)th term will contain n—vn or 0 as a factor, and
so will all the following terms. We therefore take the case where
n is fractional or negative. When p has passed n, the preceding ratio
will be always negative, shewing that the terms become alternately
positive and negative ; for that the ratio of two quantities is negative
indicates that they differ in sign. Neglect the sign of the preceding,
and make it positive. This we may do, as our object is to see
whether the series can be made convergent, which a series of terms
alternately positive and negative always can be, if the correspond-
ing series of positive terms can be made convergent. We have
x
then
| ema ri pr nx a x nL
p+l td Ags wi 14.
Pp
as we take higher and higher terms, the second term of the preceding
diminishes without limit, and the first has the limit z. Consequently,
if x be less than 1, the ratio above mentioned will, after a certain
number of terms, become less than unity, and will afterwards ap-
proximate continually to the limit 2. That is, the series obtained is
always convergent when z is less than 1.
This being the case, we may, page 189, use the result of making
ON THE BINOMIAL THEOREM. DA i
# == 0, which gives (1)"= a,. If bea whole number, a, =1; but
P Z x
if n be a fraction, as we have (1)?= a, =(1?)' =(1)", or a, may
be any gth root of 1; see page 113. If we choose the arithmetical
root, we find a, = 1; and this series is (if the doubtful assumption be
true) the arithmetical mth power of 1+ 2, when z is less than 1, or
the algebraical equivalent of (1 + x)” in all cases.
We shall now shew that the series is correct whenever n is a
whole number. Suppose it correct for any one whole number,
say m. Then (a, being 1),
(l+2)"= L-+ mar + ma? + m2 a + &e.
Multiply both sides by 1+ 2.
Q+ay"4=1l+mae+ m2 pm a + &e.
+x+m xt +m z+ &e.
oak —1 m—2 —1
=1+(m+1)x+ (m™ +m)a%+ (m™ "= +m™)a* + Be.
Burm +m=m(" +1) =m" = (m+)
m—1m—2 m—1 m—1 fm—2 = m m—1
ort merrpea ay (ae =m ("> +1)=(m41)5 3
whence
m m—1
Day} = 14+ (m41)r+(m+1)Fa%+(m+1)5 a+ be.
which, if we now write 2 for m-+1, or n—1 for m, becomes the same
series, or follows the same law as
(l+z2)'=1 $ne+n 0 +n.
Hence, if this expression be true for any one whole-value of n,
it is true for the next. But it is true when n =1; for
41 ny at ee
URe ayaa OE re aco aera Peers Pos &e.
therefore it is true when n = 2; but it is therefore true when n= 3,
and so on, ad infinitum.
If we consider (1-++ 27)" as a function of m, and call it gn, we
see that
212 ON THE BINOMIAL THEOREM:
(l+2)"x(1+2)"=(1+2)*™
or on xX om =9O(n+m)
but when x is a whole number (1+ 7)” is the series in question;
therefore, calling the above series gn, we have, when n is a whole
number,
onxX om = o(n+™m)
or
(14n2 an ga Te =. eae ais ke. )
x (ees 24m sa — ret Ke.)
=14(m+n)e+(m+ nts 2+(npn) FOE EN? + Be,
3
This may be verified to any extent we please, by actual multipli-
cation ; for the two first series multiplied together give
l4+(m+n)a+ (rn. +nmtm@)a°
n—1n—2 n— —1 m—1m =) 3
— — oe — ——_ &e
+(n. 5 3 n m+-nm 5 +m 5 3 a+
—1 n?—n +2nm+m—m
But n.— +nm-+m. —— = ee
2 2
n+m)?—(n +m Be
oS OHA)
m—1 n—2 —1 —1 m—jJ m—2
nN. tn m+ nm=“—— + m ae ee
__ W—3n?+2n+3n?m— 3nm + 3nm2— 3nm+m—3 m+ Im
a 2x3
= Stays tm nm) (n+ m)2teot A
and so on. We now lay down the following principle: When an
algebraical multiplication, or other operation, such as has hitherto
been defined, can be proved to produce a certain result in cases where
the letters stand for whole numbers, then the same result must be true
when the letters stand for fractions, or incommensurable numbers, and
also when they are negative. For we have never yet had occasion to
distinguish results into those which are true for whole numbers, and
those which are not true for whole numbers ; but all processes have
ON THE BINOMIAL THEOREM. 913
been, as stated in the introduction, true whether the letters are whole
numbers or fractions. There has been no such thing in any process
as a term of an equation, which exists when a letter stands for a
whole number, but does not exist when it stands fora fraction. If,
therefore, ¢m X gn will give ¢(m-+n) by the common process of
multiplication when m and n are whole numbers, that same process
will also give ¢(m-+n) when m and 7 are fractions, and also when
they are one or both negative : consequently, the series
L+natn— z+ &e.
has this property, that, considered as a function of n, and called gn,
it satisfies
QnX om = o(m+n)
in all cases. But in page 205, it has been proved that any solution
of the preceding equation must be ¢2 == c" where c = ¢(1), and ¢(1)
we find to be
t=
1+l2+1at+ &e. or l+az
therefore #7 is in all cases (1-++2)", which is the theorem in question.*
* Every proof which has ever been given of this theorem has been
contested; that is, no one has ever disputed the truth of the theorem
itself, but only the method of establishing it. And the general practice
is, for each proposer of a new proof to be very much astonished at the
want of logic of his predecessors. The proof given in the text is a
combination of two proofs, the first part, making use of limits, given
(according to Lacroix) in the Phil. Trans. for 1796; the second, the
well-known proof of Euler. The objection to the first part lies in the
assumption of a series of whole powers; to the second, in its being
synthetical, that is, not finding what (1+.)” is, but only proving that a
certain given series is the same as (1+)". But each part of this proof
answers the objection made to the other part ; in the first part analysis
is employed, but only so as to give strong grounds of conjecture that
1+nxr+ &c. is the required series; in the second part this conjectural
(not arbitrarily chosen) series is absolutely shewn to be that required.
The proof of Euler may be condensed into the following, of which the
several assertions are proved in the text.
1. If Ltnatn. 02+ &c. be called gn, then ¢(1) is 1+z, and
gnx@m is found to be ¢(n+m).
2. If gnxgm = ¢(n+m) then gn must be (1) }"
3. Therefore, 1+nzr+ &c. is (1+.2)"
In 1827, Messrs. Swinburne and Tylecote published their demon-
214 ON THE BINOMIAL THEOREM.
To use the preceding series, the readiest way is to form the
n—1 n—2
several factors n, :
2 3
for instance, let » = 3, or let it be V4 +.2, which is to be expanded.
, &c. previously to proceeding further 3
1 n—1 1 n—2 1 n—3 5 :
ee ee ee
a 7
Ar, «38 Q3. A 8
Vive = 1432+) (-D #4 CDC)
+9 CD CD#+
i
1
= Looe ee &e
If n= —1,
Log, ierett a Lord? a
n= —l, 9 = l, a 1 = —] &e.
1
(tay) or pt = 14+(-Ia+(-D(-Ne?
+(—1)(—1)(—]D) a3 + &e.
1 °
= ]—747°—# + &e.
as has been proved before.
stration of this theorem, the last original one with which I am acquainted.
Tt is much too laborious and difficult for a beginner, but is as unobjec-
tionable in point of logic as I conceive the one given in the text to be.
Their general objections to the theory of limits are not, I conceive, to
its logical soundness, but to its applicability in algebra; it being more
frequently than not a sort of convention that limits shall not be introduced
into algebra. But, until /10 is explained in arithmetic without limits,
I shall hold that limits are, and always have been, introduced into
algebra. Nay, until the symbol 0 is well explained in all its uses, with-
out limits, the same may be said. With regard to the objections made
by the above-named gentlemen to Euler’s proof, in their page 35, they
evidently hang upon the assumption of a finite number of terms of
the series, which is not Euler’s supposition, and which he therefore
“keeps out of sight,’ not “ dexterously shuffles out of the way,”
for it does not come in the way, any more than lines which meet come
in the way when we are expressly talking of parallel lines. After
quoting the last-mentioned phrase, it is but fair to those gentlemen to
state, that, judging by the general tenor of their work, it is not to be
presumed that they meant to accuse Euler of intentional deceit, which
the phrase “ dexterous shuffling”? generally means, and which, therefore,
was not the proper phrase to use.
ON THE BINOMIAL THEOREM. 215
Sino. 5,
op M1 4 m2 n—38_1 m4 _ 1 5
Meee 5 ak sg ss
Ma 1]+57+; S\x*+/ Slak?+]/ Sl\at+ Dlae 5\ x6 + &e
<2) x 2 x 2 x 2 x2
real x 1 x] ed:
X 2 3 x 2
x } X 3
x 0
= 14+5274+10277?4+1027° +524+2°+04+0+4 &e.
It is usual to prove the binomial theorem in the case of a
whole exponent, as follows: Multiply +a by r+, which gives
x?+ (a+ b)x+ ;in Ps
as many as there are combinations of 3 out of n quantities, or
n—1 n—2
2 3
equal to each other and to a, we have
n ; and so on. Hence, if a, a, &c. be all supposed
216 ON THE BINOMIAL THEOREM.
Pee "0. ae ae fm he A
eee AOL ah tna
containing 7 ertoe
In which, if we suppose + = 1, we have
—1
Cll) ea +na+n = a? + &e.
See Ar, 211, for the reason why the coefficients are the same,
whether we begin from the one end or the other of the series, as will
also appear in the following cases.
(l+2)? = 1422+ 2?
(ea =a oar aes
(l4+c2)t = 14424 6274+ 423+ 24
(l4+az)y = 14+52+1022+10273+ 524+ af
(lta) = 14+627+1527+20234+1524+62° +25
In order to find (1—.)", change the sign of x in the series; that
is, write —.z for x, leave x? unaltered ; write —z? for x3, and so on;
which gives
(l—z) = — ke.
We may write the general series in the following way when n is
a whole number, where Pn signifies the product of all whole numbers
between 1 and n, both inclusive.
” o x? ee any
(1 +2) = Pf 5. + P ea EP, ; a oe
which shews the similarity of the coefficients above alluded to.
We give the following as exercises :
1. When 7n is a whole number,
—1 n— -
a= ame
2
—1 I— 1h —
+ &e.
+ &e.
ON THE BINOMIAL THEOREM. 217
n—1 m—1 n—2 =
m—1 —.
2 l+n Sar ee :
+ Xe.
2. (a+b)"= ar nes &e.
3. Ifn bea whole number
(+a eshe2
(+a esha(e+)
(e+ haat hea(ee) +0
(c+ 2)" omy 4, pen(am24 4)
2n—1 1
2n—4
2 (« 2 =n-4) eee
2n(2Qn—1) .(2n—2) .... (n+1)
Nae © 2 IRE Sie Ria BR eg OL /2
2K
ending with
(e+ yn = gen “ont t(2n-+1) (22-1 +" a2 ih pe a ee
3 4
a lel ay ile = e+ &e.
5. The student may provide himself with examples and verifi-
cations in the following way: choose any exponent , whole or
fractional, positive or negative, and expand (1+ x)” and (1+ 2)"+1
by means of the theorem; then multiply the first series so found
by 1+-2, which should give the second series.
—1
6. a= 6" + n(a—b) bh" + n——(a— by? b"-2 + &e.
The following cases will afterwards require particular notice :
nx—1ln¢x—21
a 1 nxr—11
a —— a — onbtnadictnieeaiiieslaaien a
(1+ Sete Tee — 1: 2S, 5 ap tne 5 3 aa &
1
But ee — 7 nx — 4 x A oe he d
n o n? 2
Poult Sued
ne—1nx—221 n n
—— —- = —————— &e. & . i
NX 5 eas Lr 5 2 c. whence
218 ON THE BINOMIAL THEOREM.
2
1\2e 2 r— > 2—= \
(1+) =lte+e— "+a" 3 + &e.
In the preceding let 2 = 1, which gives
—- 1—= 1—=
(147) =141 :
But (page 105), (1 1 ry au i, ‘ a] :
or the first series is the «th power of the second, and
1
Lie = ee BE eh
5)
si gs
If n=0, A+2)" =1, or (1+27)" —1 =0, or the fraction
(1+a)"—1
n
O i
assumes the form i We now ask whether this frac-
tion has a limit when n is diminished without limit (page 162).
From the general theorem,
(A+ a)" —1 ew, See re Poe Urs
n—1 “=
w+ &e.
n
and when n is diminished without limit, the limit of the second side is
1G GG) + IGG)
x yt
or L——- +— — —+ Ke
2 o
That is,
As n diminishes without limit,
(1+a)"—1 f a x x
_—_*+—— approaches without —_— — —_—-— — &e.
A igi yor ae
limit to
If « = zg—1, then the limit of (m diminishing without
limit) is
(2—1)— 5(e—1)4+ s(@—1—
Of this series we know no property as yet, except that it is
convergent when # or g—1 is less than 1, or zg less than 2. We
shall proceed to examine the expression from which it is derived.
If in it we write 2” for z, we have
ON THE BINOMIAL THEOREM. 219
gmn gmn4
——— or ———
n mr
Let m be a fixed quantity and let » diminish without limit: then
mn also diminishes without limit. Now, if ¢n have the limit N
when 7 diminishes without limit, ¢(mn) must have the same limit.
The only difference is, that (say m= 6) for any very small value of n,
o(6n) will not have come so near to its limit as gz. For the nature
of the limit being, that, by taking m sufficiently small, we may make
gn within any given fraction (say k, which may be as small as we
please} of N, we may, by taking the sixth part of the requisite value
for n, make ¢(67) within the same degree of nearness to N. Hence
the limits of the two,
gn ‘ti smn — 1
and ~~ ===
nr mn
: et!
are the same. But if the first be called yz, the second is = wp (2) 5
and we have
ess! ayy 1
limit of .)2 = limit of rays
1
= — limit of td
him pz
or
1 1 1 B
2—1— 5(z—1)°+ &. =— ai phere La &e. )
a property of the series which will serve to verify succeeding
investigations.
We have not yet included in our results the case in which the
exponent is incommensurable, such as
Sz
(1 + 2)
but since we look upon V2 as the limit to which we approach nearer
and nearer in the series (obtained from arithmetical extraction in this
case),
1 14 1:41 #%1:414 1:°4142 &e.
we must regard (1 + 2) V2 as the limit to which we approach by
taking the successive expressions
os aie a
1 +2 (1 + a2) (1 + zy (1 + 1000 &e.
so that, whatever approximation k may be used for 2 or the limit
for which this symbol stands, then
220 ON THE BINOMIAL THEOREM.
hee hagas Kees &e.
is the corresponding approximation to (1+ 2) v2, When the series
is convergent, it is evident that if each term be found, say within its
thousandth part, the sum of the whole series is found within its
thousandth part. Suppose & and k +m to be two approximations to
V2, as in page 101, the first too small, the second too great, and
Suppose we compare the pth terms of the corresponding approxi-
mations to (1+2)¥2, or
‘erage eapiataP dee Gn) eee ey |
Dare pat
the ratio of which is
k+m k+m—t1 k+m—p+2
my Maer ey aT ee ee
Since m can be made as small as we please, it is evident that
each of these factors can be made as near to unity as we please, and,
consequently, their product can be brought as near to unity as we
please. That is, for any given number of terms, the terms of the two
approximations may be made as nearly equal as we please. Now,
suppose 2 less than 1, so that (page 210), both approximations are
convergent. Hence so many terms (say g terms) may be taken that
the limit of the sum of the remainder shall be as small as we please:
and as m may be taken so small that all the g terms of one approxi-
mation shall be within, say their millionth parts, of those of the other
approximation, it follows that we may reduce the two approximations
to the following form :
Fe remainder less
AE OHOH cesar 42 ef than the millionth |
| of the preceding
sum.
A remainder less
c(L+) 400.4 6)+ - beta) + | lane
sum.
where «@ ...... are severally less than one millionth. Let
a+b-+c+.... +2 (which we may call the approximation to the
first approximation) be called P; a#+68+ .... +2 is less than
one millionth of P, and the first approximation complete is
ON THE BINOMIAL THEOREM. 22)
PatxXxP (X_ is less than one millionth)
and the second
P+V)+YP(1+V) (Vand Y do. do.)
the difference of which is
PV+P(Y—X)+PYV
which is less than three millionths of P; because V, Y—X and Y V
are severally less than one millionth of P. But the limit (1+ 2) v2
must lie between these approximations, and therefore does not differ
from P by so much as the approximations differ from each other,
- that is, by three millionths of P.
It is not possible to shew, by the preceding method, the same
result in the case where x is greater than 1, or the series divergent:
but it must be remembered that in this case algebraical equality only
is asserted, not arithmetical equality: and all that is said is, that
{ets V2 2 4+ V32A1,0 + &e.
may be substituted without error for (1+ .2)*2; so that the proof of
this case falls within the general proof, as deduced from the principle
in page 212. What is done in the preceding process shews the
approximation to arithmetical equality, when the series is convergent.
The same arguments might be applied to any other case, with any
other degree of approximation. We now proceed to develope some
further consequences of the binomial theorem.
Doe EXPONENTIAL AND
CHAPTER XII.
EXPONENTIAL AND LOGARITHMIC SERIES.
Aw algebraic symbol may have different names, according to the
relation in which it stands to different symbols, or combinations of
symbols. Thus in ab, a, with respect to b, is called a coefficient ;
with respect to ab, it is called a factor. Similarly, in a®, b, with
respect to a, is called an exponent; with respect to a, it is called a
logarithm,* and a in reference to 6 is called the base of the logarithm.
Thus, in 3‘, 4 is the logarithm of 3¢ or 81, to the base 3; in
at, x is the logarithm of a to the base a. This we shall denote by
4 = log,81 and x = log, a”: the letters /og being an abbreviation of
logarithm, and the underwritten figure being the base.
ExampLes. ]0? = 1000 3 = log,.1000
If at= y z= logay
If pt = l—z q = log, (1—2z)
To construct a system of logarithms to a given base, say 10, we
must solve the series of equations
tee a LO? == 2 lO =3 107 = 4 &c.
and find the value of vw in each. This can, generally speaking, only
be done by approximation : that is, the logarithm is generally incom-
mensurable with the unit. By saying, then, that log,,2 =°30103, we
mean that
10:3 — 2 very nearly, or ee, 1028 = 2 nearly,
and that a fraction & can be found such that 10% shall be as near to 2
as we please, to which fraction *30103 is an approximation.
[In all the following theorems one base is supposed, namely, a. |
THeoreM I. Whatever the base may be, the logarithm of 1 is 0.
This is evidently another way of expressing a®=1, and may be
written log, 1 = 0.
* From asywy aeibuos, the number of the ratios, an idea derived from _
an old method of constructing logarithms, which cannot be here explained.
LOGARITHMIC SERIES. 2938
TueoreM II. The logarithm of the base itself is 1. This is
contained in a! =a, and is expressed thus: logg a = 1
TueoreM III. The logarithms of y and ; are of different signs,
but equal numerical value. For if y= a? or x = logay, we have
1 1 ; Ib
* =a or —r1 = logays that is, logay = — logy.
Turorem IV. If a be the base, und a number or fraction lie
between a™ and a®, the logarithm of that number or fraction lies
between m and x.
For if a?, the number, lie between a” and a”; then a, the
logarithm, lies between m and n (see page 89).
I 1
Base 10. Base on
Number _ Logarithm Number . Logarithm
has its has its
between between between between
1
1 and 10 0 and 1 1 and = Oo and 1
1 1
10 and 100 7 and. 2 5 and z A” ante 2
1 1
100 and 1000 2 and 3 P and a 2 and 3
&e. &e. &e. &e.
1 and 0 and —1 4 and 2 O and —1
1 1 P
Peas § 1 and —2 2 and 4 —1 and —2
10 100
ete and neo att) and ai} 4 and 8 —2 and —3
100 1000
&e. &e. &e. &e.
Turorem V. The logarithm of a product is equal to the sum of
the logarithms of the factors. Let a be the base, and p, q, and 7, the
logarithms of P, Q, and R. Then
1 P =a? aera! R = a’
PQR = a?t@*" or log(PQR) = pt+qt+7T = log P +logQ+logR
Turorem VI. The logarithm of a ratio, quotient, or fraction, is
the difference of the logarithms of the antecedent and consequent,
| dividend and divisor, or numerator and denominator.
924 EXPONENTIAL AND
For a —— i fh ee Bre) log & =p-q7= log P—log Q
TueoreM VIT. The logarithm of P™ is found by multiplying the
logarithm of P by m; that is, if P=ap, Pm = amp, or log P™ =
mp =m log P. -
TueoreM VIII. A negative number has no arithmetical logarithm :
nor is a system of logarithms with a negative base within the limits
of algebra as hitherto considered. This is rather a definition than a
theorem,* and it amounts to this: on account of certain anomalies,
of which the explanation cannot yet be understood by the student,
we are obliged to defer the consideration of the logarithms of
negative quantities, and all logarithms of positive quantities, except
only those which have arithmetical meaning. For the equation
a* = b has not been proved to have only one solution, though it has
only one arithmetical solution.
TueorEM IX. The logarithm of 0 is infinite ; by which we mean
that as y diminishes without limit, its logarithm (being always
negative on one side or other of 1), increases numerically without
a oe ee EYES ca ;
limit. To diminish (5) without limit, « must undergo numerical
increase without limit, and be positive: to diminish 2” without
limit, 2 must also undergo numerical increase without limit, being
negative. Hence (page 156) the meaning of the proposition stated ;
and it must be observed that the symbol o has either sign in
algebra, which may be explained as follows. If y= is then y and x
must have the same sign: if x diminish without limit, it approaches
a form (0) in which it has no sign, being the limiting boundary
between positive and negative quantity. Consequently y, which
increases without limit, approaches a similar boundary ; for since y
1s ; » eit § :
has the same sign as ve if there be any form of # in which either sign
may be supposed, the same may be supposed for y. But such a
comprehension of this point as can be derived from frequent instances
must be reserved until the student has seen more examples of the
application of algebra to geometry.
* Being usually considered as a theorem, we have stated it as such.
LOGARITHMIC SERIES. 925
The following examples will illustrate the preceding theorems.
ed bh ee loga+log1 = logz (log 1 = 0)
v= 2 1 od Reltabis (eS FEE
log,az = log,x+log,a = log, x +1
loga Vy = log «+logy? = log # + slog y
log a = log x+ = log y—log p—2 log 9
3\—1
log () =— {log z+3log y—log p—(—1) log y}
= —logxr—3 logy + log p—log 7
We shall now proceed to the series connected with logarithms.
In page 218, we have proved the following theorem for all values
of x and 2,
1 1 2 a } 2
2 jon S vn) ie 2 a, 18 see
1+1+
In which both series, being forms of (1+-)', will (page 210) be
Bi
convergent whenever 7; 38 less than 1, or m greater than 1. Now,
let n increase without limit, in which case the limit of
1 1 2
ee 4 i 3 + &e. isl+1+5t+5—3 + &
and that of
ae +, ine roe x2 x
eee a + &eislt+ery + oq t ke.
But 141-5 &e. has been calculated approximately in page
183, found to be 2°71828182 .......- , and called «. Therefore,
(page 183),
z= ] shay ere
Consequently, if « be the base, the number which has the logarithm x
2
as ifa+5 + &c. Hence a system of logarithms, having « for
the base, being arrived at in the course of investigation, has been
226 EXPONENTIAL AND
called the natural system of logarithms; it is also called the Naperian*
system of logarithms, and the hyperbolict system of logarithms.
And here let it be remembered, that in algebraical analysis, the
letters log, by themselves, imply the logarithm to the base «, any excep- -
tion being always specially mentioned.
The preceding equation being mE true, we have
= l+ka te + == re &e.
But e* is (e%)”; and if we take k to be the logarithm (base e) of a,
we have «* = a, and
(log or ney (log a) x3
(e*)? or a® = 1+ (log a)a+ ee
+ &e.
From which we find that “—~ will, if # be diminished without
limit, have the limit log a.
But (page 218) we have already found the series for this limit ;
whence
log a << (a—1)— 5 (a—1)? f. = (a—1}—
or, if a = 1+), we have
b? b3
log (1 +8) == Te + 3 — &c. te sae (1)
Substitute —b for b, and we have
b? Z
log (1—b) = —b—~ —* _&e. .......1 (2)
Subtract (2) from (1) [ tog (1+ b) — log (1i—b) = log sa
=) ae Nfl _ + &e.} ass (2)
List pewe 1+ : : 1
ea eas which gives 6 aE
log
let
* John Napier, commonly called Lord Napier, though not a peer,
or otherwise entitled to the appellation of lord than in the way in which
many landed proprietors are lords (of manors), was born in 1550, at his
lordship of Merchiston, near Edinburgh, and died in 1617. He published
the invention of logarithms (his method also leading him to the natural
system) in 1614.
t So called from a supposed analogy with the curve called the
hyperbola, but which analogy belongs equally to all systems of logarithms.
LOGARITHMIC SERIES. 907
ts
log = log —— = log(1+2)—logx
1 1 1 1
a Cs ee since Sad bal all
g(x +1) —loga = as +i +3 @rtiy 15 @epiy t Ke} ig
The last series gives
1 i et 1 1
z= 1log2?= e+su tem ts c.}
1 ab al 1
x = Qlog8 =log2+2{; +55 +5 met e.}
1 le, a! 1
x = 3log4 = log34+24- + 555 + saa t & c.
eet 1
x = 4logd = log 44244 sh strate = Soup + Be &e.}
Thus the logarithms of whole numbers may be successively cal-
culared with tolerable readiness, as the first example (which contains
the least convergent series), here given at length to ten places (that is
to eleven for the sake of accuracy), will shew.
. = ‘33333333333
> = — 01234567901
a = +00082304527
; = = -00006532105
; = — +00000564503
~ = = +00000051318
om = — -00000004825
=. os = 00000000465
= a = -00000000046
a = = — -00000000005
34657359028
2
log 2 = 69314718056 very nearly.
228 EXPONENTIAL AND
By this means the logarithm of 3 may be found from that of 2, .
that of 4 from that of 3, and so up to any given whole number. It
would be desirable,* as an exercise of arithmetic, that the student
should calculate them up to 10 inclusive; the results of which (to
eight places) would be as follows:
log 1 = 0:00000000 log 6 = 1:79175947
log 2 = 0°69314718 log 7 == 194591015
log 3 = 1:09861229 log 8 = 207944154
log 4 = 1°38629436 log 9 = 219722458
log 5 = 1:60943791 log 10 = 2°30258509
But the series need be employed only for prime numbers, and the
first tables of logarithms were thus constructed, as follows. Suppose
the logarithm of 59 to be required, or of 58+1. Now 58 is 2x29,
both factors being prime numbers; if, then, we have the logarithms
of 2 and 29, we have that of 58 from the equation
log 58 = log 2+ log 29
and that of 59 from
log 59 = log 584214 + 4 ay t Xe.
Beginning, then, with log 2, we have the following:
log 2 = a given series log 6 = log 3 + log 2
log 3 = log2 + a given series log 7 = log 6 + a given series
log 4 = 2 log 2 log 8 = 8 log 2
log 5 = log 4 + a given series log 9 = 2 log 3
log 10 = log 2 + log 5; and so on.
* Halley (Phil. Trans. 1695) thus expresses himself, after having
described the preceding method. ‘If the curiosity of any gentleman
that has leisure, would prompt him to undertake to do the logarithms of
all prime numbers under a hundred thousand to twenty-five or thirty
places of figures, I dare assure him that the facility of this method will
invite him thereto; nor can any thing more easy be desired.” Without
insisting upon any thing that would take up so much of a gentleman’s
leisure as the preceding, I should strongly recommend the student
always to work one example of every moderately convergent series
which occurs.
LOGARITHMIC SERIES. 229
In page 219, we proved independently that
1
e—1—s(2—1)*+ &e. = =(2"—1—F(2"—1)° + &e. )
2 m 2
which is, as we now see, the same thing as
a
logz = —logz™
and in it we also find further elucidation of the equation,
gm —1
limit of = z—l-— 5(2— 1)? &e. = loge.
The diminution of m without limit, or the supposing of m to be a
smaller and smaller fraction, implies the extraction of higher and
higher roots of . By extracting a sufficiently high root of z, we can
bring 2” as near to 1 as we please, or make z™—1 as small as we
please ; that is (page 187) 2”—1 may be made as nearly equal to
the sum of the whole series as we please. It was from this principle,
by continual extraction of the square root of z, that logarithms were
once calculated, by means of the formula
1
logz = ore 1) xX 140737488355328
very nearly,* the number named being 2°, equivalent to 47 extractions
of the square root.
The logarithm of any whole number being found, as in the last
page, that of a fraction can then be found by the subtraction of the
logarithm of the denominator from that of the numerator.
We also notice the following result: when z is a large number,
log(« +1) = logr+ — nearly (page 227).
The rest of this subject will be reserved for the next chapter,
on the practical use of logarithms in shortening arithmetical com-
putations. We now proceed with some uses of the preceding series.
Lemma. If f(x) be such a function of z, that f(v+y) can be
expanded in a series of the form
Ay tAyy + Acy?+ &e.
* Halley, in the memoir already cited. Each square root was ex-
tracted to 14 places.
x
230 . EXPONENTIAL AND:
where A,, A,, &c. are functions of x only, then
Sla+oVv—1) + f(a—bV=T) :
treated by common rules, will always represent a quantity, either
positive or negative, that is, all purely symbolical or impossible*
quantities disappear; while, on the other hand,
Sla+tV=1) —~f(@—bV =I)
will be of the form (possible quantity) x ~—1
Write the value of f(#-+ y) and afterwards that of f(r —y),
made by changing the sign of y;
S@t+y) = Avot Ary + Acy?+ Asy® + &e.
S(@—y) = Ay— Avy + Acy®— Asy?+ &e,
from which we find,
Sat+ytfa—y) = 2A, +2Acy?+ 2WAay*+ &e.
S(aty)—f(«—y) = 2Ayy+2Asy®+ 2Asy>+ Ke.
for @ write a, whence A,, A,, &c. become functions of a only ; for
y write b\/—1, that is, suppose
y= bV—=T yf = PVT
y= 2 x —1 = —2 yS = — 6
y= —BPxbV—-l=—-BVAT yl=—BV—1
yt = —BV 1x bV=1 yf = BF Xe.
= —b'x—1= bt
whence
f(atbV—1) + fla—bV =) = 2A,—2A,0274+2A,b4— ke
which is a possible quantity: and
flat bV—=1) -f(a-—bV=1) = 2A,bV—1—-2A,B3V —14
= {2A,6—2A5b°+ &.} V1
which is (a possible quantity) x V—1. -
* We shall now begin to make use of this common phrase: to the
student it must mean “ impossible till further explained.” See page 110.
LOGARITHMIC SERIES. pas
EXAMPLES sina + shales, alt
Wane Nea gb 21) Gee
1 1 hi: 2b 26 YT
- 5 a a PN ea a
(a+bV —1)"+(a—dV—1)"= Qa" —2n—al*b* + Ke.
| TAC ae ei
(Qna"b—2n™— = a8 + &e.)V 1
As the preceding theorem is true for all values of a, it is true
_ when a= 0; that is, for
f(bV—1) +f(—bV 1) ana f(bV —1)—f(—bV = 1)
Examptes. Let 7 be a positive whole number.
+26" when n is evenly even.*
Oars (DV =I =} ed eee
— 2b" when 7 is cddly even.t
O when 7 is even.
(bV —1)"— SE cea Waar aien a phy eoee
— 26"/—1 when vn is 3, 7,
115 cc:
If we apply the same process to ¢” V=l and e* ae we find
ee ae V i-=-v- ihapirimrrsp yen bo
i a/—1 —2rV7—1 2 4 6
é re _, er z ‘
2 | Boal pres OnE Be Aa aL be (A)
aV/7=1 —zvVv—1 5
eae ey hg ne Rc, Beat (25)
7G rey Sk OE ip Roa a
We have left the use of the symbol /—1, in this work, to be
justified by experience only (see page 111); we have now an oppor-
* Divisible by 4; 4, 8, 12, &c. are evenly even.
+ Divisible by 2, but not by 4; as 2, 6, 10, &c.
25494 EXPONENTIAL AND
tunity of examining the result of a long series of deduction, with a
view of ascertaining how far we shall produce consistent results.
\
The algebraical sign of equality is placed between the two sides of
the preceding equations; the question is, Will any relations we may
discover to exist between the two first sides also exist between the
two second sides ?
gv =i —27/—1
€ +e
Let SS TT ea
eval _ eval
and wear be called Px
we have then
2 Ve fev) oY oe pneev 3
ole ar 4
en shoe ME Val an e7eav 3
24
(x) bor A
ggv -l-aV-1
(pa)?+ (pay? = 2
Ze gy A
(pay — (pa)? = £__t# = 9 (22)
| 2% VHT 2eV/n1 1
pax Ne 4
of which three relations, namely,
(pxy+(zP = 1 (px)? — (px)? = g(2x)
p29LxX Vr = L222)
it is asked, are they true of the second sides of (A) and (B)?
The multiplication of a series by itself will be found to amount to
using the following rule: square each term, and multiply all that
follow by twice that term. Thus, the square of the series in (A) is
4 2 Fs oa + &e
TOOT a a eae ae
at w®
.— eee
22 5.3 A ee
+ &e.
LOGARITHMIC SERIES. 933
that of the series in (B) is
4 6
h ile ad Yi aR?
iz a4 kita So) Grohe
6
x
tacomaqiesm sc
— &c.
The first square increased by the second is
met 41° 4 s_f 1 ‘pachcae | Ve ¥
4 eS arr Sere ee oe Sn aye
gran 8 S15 6° 2.
eri 0 } =O} + &e.
The first diminished by the second is
r
| 1 Ags 1 1 1 1 \ 8
2 a Ne ae ce ee ‘ ee peered —_ Uv Ke.
2a t{satatsle Ve taaa tit ee ia
ee Ot) Ctl Ke.
2 weed ©. 9. 9) 4.5.6
The third relation may be similarly verified by multiplication.
The following results may also be proved in the same way both of
the first and second sides of equations (A) and (B)
o(a+y) = exoy—Vayy virty) = vrpy terry
e(a—y) = proytyayy va—y) = vrey—eary
Let <°”— be called p, aud let ~ be called x2, then (equations
A and B)
Lt 1
Si tal Liman D. = aye or Fea aa
— 7 = j
i p an pr +1
1+V 1
whence p= aged ate! and, page 226,
1—V —1 x2
log p* = 2 \vr1 ye +a(Volx2) + we.
= » eg | a —~ ; (x x) +5 (x x) — Ke}
But p?= paaee—1 or log p? = 22 VY —1, therefore
@ = x—5 (x2) +5 (x2) — &
x 2
234 EXPONENTIAL AND LOGARITHMIC SERIES.
The series A and B are always convergent, as may be proved by
the test in page 185; but the convergency may be made to begin at
any term, however distant, by making @ sufficiently great. Thus, if
a were 1000, the first series would not begin to converge before
the term
2 0,2 hes lege e ee ---- 263.264
but, notwithstanding the magnitude of the first terms of the series
(which it must be remembered, are alternately added and subtracted)
the actual value of ¢7 and vr can never exceed 1; for if.either were
arithmetically greater than 1 (positive or negative), the equation
(or)? +(x)? =1 could not be true.
In trigonometry, the properties of the preceding series are con-
nected with geometry in the following way. Leta circle be drawn ;
B
and from A let the point B set out until it has described an are equal
in length to # times the radius OA, going round the circle again if
necessary. Draw BD perpendicular to CA; then it will be proved
that BD is the fraction y2 of OA, and OD the fraction ¢ of OA.
Exercise 1. By means of the equation
1+2 ;
aa, 1 = &, prove the equation
14+
loge = « —=— 2 (a2 =) + 3 (4) —&e.
2, Prove the following, |
eV — ox+ Vl be
eV =1 = 97 Vol be
(gat V—1 ya)" = o(mz) + V=14 (mz)
ON THE USE OF LOGARITHMS. 235
CHAPTER XIII.
ON THE USE OF LOGARITHMS IN FACILITATING
COMPUTATIONS.
Tue natural system of logarithms, already explained, has this defect
as an instrument of calculation, that there is no method of finding
the logarithm of a fraction more simple than subtracting the loga-
rithm of the denominator from that of the numerator. For instance,
the logarithm of 3 can not be more shortly found than by taking
log3—log10 or 1'09861229 — 2:30258509 == — 1°20397280
We proceed to find another system in which there shall be some
more obvious connexion between the logarithm of 3 and those of
-3, 03, &e.
The fundamental definition of logaz gives
1
5 a Sie a= s"*
loggs logs x
We have then prego a a
log, 6b logs b log, z
But b=a e ard =a = Z
logac loge b logs x
Similarly x
loga ¢ loge c log, 5 logs x
a
which last result may be also thus verified :
logae loge c loge b logs x | log, sae ec log, b logs x
a = (a
log, ¢ loge b logs x
é —
loge 6 logs x logs x
c es
Now, there is but one arithmetical exponent which applied to «
will give «; for, if possible, let there be two, p and gq, and let
aP = 2, al = 2, whence a? =a%, and aP4=1; therefore p—g = 0,
or p=q, that is, p and qg do not differ, as was supposed. Hence,
since
loga x loga b logs x
“=a =a » we have
236 ON THE USE OF LOGARITHMS
log, « = log, b log, z = log,c log.b log, x &e.
these results may be remembered by means of the identical equations
a ac
—~m—--—- = —--—- = —* - — ke.
cb
b
remains true, if for each fraction be substituted the logarithm of its
numerator to the denominator as a base.
As an example, by means of
a
amet remember that log, a log, b = log,a fi |
1
or log, b = Togs a
We also have
loga r
loo z = —
OBE” log, 6
or; to convert a given system of logarithms having the base a, into
another which shall have the base b, divide every logarithm given by
the given logarithm of b.
Seeing that in practice it is convenient to reduce all fractions to
decimal fractions, the base chosen should be one in which the
logarithms of 10, 100, &c. are whole numbers, that is, it should be
10. For in that case we have
log10 = 1, log100 = 2, log1000 = 3, &c.
And if we call p the logarithm of any number, say 25, we have
log 2°5 = log25—log ~ 10 = p—1
log °25 = log25—log 100 = p—2
log -025 = log25—log1000 = p—3 &e.
log 250 = log25 +log 10 =p+l
log 2500 = log25+log 100 = pt &e.
So that, when the base is ten, any alteration of the place of the
decimal point in the number requires only the addition or sub-
traction of a whole number from the logarithm.
IN FACILITATING COMPUTATIONS. poe it
The system of logarithms to the base 10 is deduced from the
natural system by the following equation,
ia log, x a log, v “ :
logyox = log, 10 = 2°30258509 —! log. # x 4342944819
This system of logarithms is called the common, tabular, decimal,
or Brigg’s system, and *43429 .... is called its modulus, and gene-
rally 1 log, a, or logae is called the modulus of the system whose
base is a.
All the logarithms in the remainder of this chapter are common
logarithms.
The following are examples of the arrangements of some tables
of logarithms, for the purpose of explaining how to find the logarithms
of given numbers, or vice versd.
1. Lalande.*
Nomb.| Logarit. | D |Nomb.] Logarit. {| D
1080 | 3.03342 1110 | 3.04532
41 39
1 033 1111 | 3.04571
108 3 83 40 ff 39
1082 | 3.03423 1112 | 3.04610
40 40
1083 | 3.03463 1113 | 3.04650
40 39
1084 | 3.03503 1114 | 3.04689
&e. &e. &e.! &e. &e. &e.
2. Sherwin, Hutton, Babbage.t
Num. 0 LAR anc oe , 5” Gar 7 Sa Omens
fen Me eee
5150 | 7119072 |8157|824118325/84104 8494|8578|8663,8747\8831| 84
1 8915 [900019084/9168]9253 J 9337 |/9421]9506|9590|9674] 3 | 9
2 9759 |9843/9927|0011|0096 4 0180/0264|034910433|0517| 4| 34
7120601 |0686|0770/0854/0939 § 1023)1107|1191 1276|1360
1444 11528/1613)1697/1781 § 1865|1950|2034 2118)2202
&e. &e. &e. | &e. | &e. | &c. f &e. | &e. | &c. | &e.| &e.
wo
>
OCONTH OP WD
i
LS)
* Tables de Logarithmes, &c. par Jérome de ta Lanng, édition stéré-
otype, par Firmin Divor: Paris, chez Firmin Didot, &c. Rue Jacob,
No. 24, 1805 (tirage de 1831). This work is sufficient for most pur-
poses, but those who order it should remember to insist on having one of
the later tirages.
+ All these works are well known in this country. The first (an old
work) is frequently to be found with second-hand booksellers. The last
two can be obtained in the usual way from any bookseller.
238 ON THE USE OF LOGARITHMS
A logarithm usually consists of a whole number, followed by a
decimal fraction, both or either of which may be negative: but in
most tables nothing is put down but positive decimal fractions. We
proceed to shew how this arrangement includes all cases.
1. Take a negative logarithm, such as — 316804, which is
— 3—*16804, or —4+4+(1—16804), or —4 +-83196. © This is
usually written 4°83196, in which the negative sign over the four
means that that figure only is negative. [According to analogy, 13
would mean — 10 + 3, or —7; 136 would mean 106 — 30, or 76. |
Thus, every negative logarithm may be so converted as to have a
positive decimal part. |
2. When the number corresponding to the decimal part is known,
that corresponding to the whole can be immediately found by the
following table,
log a= or log 10-3 = —3 log 10. or logl0#'=1
log = or log10-? = —2 | log 100 or log 10? =2
log = or log 10-? = —1 /log 1000 or log 10° =3
&e.
Thus, the number to 30103 is 2 very nearly ; that is, *30103
= log 2: therefore, 1:30103 = log 10 + log2 = log 20, 1°301 03 = =
log = + log2 = log = == log'2, and so on.
If D simply stand for a decimal fraction less than unity, we have
(page 223) the following table:
The log. ofa No.}| Must lie And must there- | Instances of
lying between between [fore be of the form} such numbers
1 1
7000 and 100 —3 and —2 — Ss + D 0013, 0098
1 1 ,
aan and vi —2 and —1 —24+D 014, 0738
1
— and 1 —1 and 0 —1+D "103, "4296
1 and 10 O and 1 0+ D 2°56; 7°99
10 and 100 1 and 2 1+ D 11°03, 45°96
100 and 1000 2 and 3 2+ D 159, 159°108
&e. &e. &e. &e.
IN FACILITATING COMPUTATIONS. 239
Definition. The whole part of a common logarithm, whether
positive or negative, is called the characteristic of the logarithm.
The decimal part is called the mantissa.* We now give some
theorems which obviously follow from what precedes.
1. No alteration in the place of the decimal point (in which is
included the annexation of ciphers to a whole number) alters the
mantissa of the logarithm, but only the characteristic.
2. When the decimal point of the number zs preceded by signi-
ficant figures, the characteristic of the logarithm is positive, and a
unit less than the number of these figures. Thus, the logarithm of
12345:67 is 4+ mantissa ; that of 6°9 is O + mantissa.
3. When the decimal point of the number is not preceded by
significant figures, the characteristic of its logarithm is negative, and
a unit more than the number of ciphers which precedes the first
significant figure. Thus the logarithm of ‘00083 is —4 + mantissa,
that of -83 is —1-+ mantissa.
Lalande’s table first mentioned gives the characteristic, on the
supposition that the number mentioned is a whole number; thus, the
logarithm of 1081 (as given in the preceding specimen) is 3°03383.
But, to take the logarithm of 1-081 from this table, the mantissa
-03383 is all that must be taken, and the characteristic 0 applied.
Thus, the logarithm of 1°081 is 0°03383, and, from the rule laid
down, we have the following table:
log 1081000 = 603383 log 1:081 = 0:03383
log 108100 = 5:03383 log +1081 = 1:03383
log 10810 = 4:03383 log -01081 = 2°03383
log 1081 = 3:03383 log -001081 == 303383
log 108°1 = = 2°03383 log -0001081 = 403383
log 10°81 = 1:03383 log 00001081 = 503383
The second table gives the logarithms of numbers of five places
of figures. From the first table we might have found the logarithms
of 5153 and 5154, or (the difference being only in the characteristic)
of 51530 and 51540; from the second specimen we can find the
* This word is now seldom used, though there is not another single
word which means the same thing.
240 ON THE USE OF LOGARITHMS
logarithms of 51531, 51532, .... 51539, intermediate to the two
numbers just cited.
Now we must observe, that when a figure is changed in a
number, the first figure which changes in the logarithm will be nearer
to or further from the left hand, according to the figure changed in the
number. This amounts to saying that the smaller (in proportion to
the whole) the change of the number, the smaller the change in the
logarithm, and is shewn by the following theorem. Since (pages 227,
237) the common logarithm of 1+ a is (M =:43429 ....)
soe 1
2041 + 3 @rpiy kc.)
log (1+) = logx+2M (ee
the greater x is, the less the addition to log x by which log (a +1) is
formed. The following instances will illustrate this, in which the
figure undergoing change is marked with an accent. The columns
succeeding shew, 1st. By how much of itself the number is changed ;
2d. By how much of a unit the logarithm is changed (nearly),
Proportion to Absolute
the whole of the change
change in the in the
Number. Logarithm.
log 1’ = 0:0000000 ‘ 1
log 2’ = 0°3010300 3
log 10° = 1:0000000 i me?
log 11’ = 1:0413927 10 25
flog 100° = 20000000 ~ Phat eu
Ulog 101’ = -2-0043214 100 250
log 1000’ = 3:0000000 1 1
log 1001 == 3:0004341 1000 2500
log 10000’ = 4:0000000 1 1
log 10001’ = 4:0000434 10000 25000
Nee log 100000’ = 6:0000000 1 1
log 100001’ = 6:0000043 100000 250000
Hence, roughly speaking, the absolute change in the mantissa of
the logarithm is something less than one half of the relative change
in the number. Let the student try to ascertain this from the series
given above. Thus, if a number increase by its thousandth part,
IN FACILITATING COMPUTATIONS. 241
the increase in the logarithm is less than the absolute fraction
1
2000”
and so on. Hence we can ascertain with sufficient precision to how
many places of logarithmic figures it will be necessary to carry any
table. Let us suppose, for instance, our table is to give every
number of five places, from 10000 to 99999. At the end of the
table the relative increase of the number is about ———, the
100000
absolute increase of the logarithm is therefore about a hy or
250000
‘000004. Consequently six places of logarithmic figures are abso-
lutely necessary. With less than six places distinction would be
lost; for instance,
log 99846 = 5:9993307
log 99847 = 5°9993350
which only differ in the stxth place. In the first part of the table,
where the relative increase is little more than the absolute
1
10000”
increase of the logarithm is nearly :00004, or five places only would
be sufficient. But the tables must, for reasons of practical con-
venience, be of the same number of figures throughout, and, therefore,
must be at least of six places of figures.
In the second specimen, five places of figures in the number are
accompanied by seven places of figures in the logarithms. But, as
the three first places of the logarithm continue the same for some
time, even in the most changeable part of the table, they are placed
by themselves in the first column, at the point where a change takes
place: which saves much room, but is subject to this inconvenience,
that as a change in the third figure of the logarithm will seldom take
place exactly at the beginning ofa line, the new ¢hird figure cannot
be shewn till after it has really made its first appearance. The fol-
lowing instances, taken out of the second specimen, will shew both
the arrangement of the tables and this new difficulty, better than ny
verbal explanation.
242 ON THE USE OF LOGARITHMS
Number. Mant. of the log. Number. Mant. of the log.
51520 °711 9759 51526 "712 0264*
51521 *711 9843 51527 "712 03497
51522 “F711 9927 51528 212) 0435"
51523 ‘T12 0011” 51529 “712 0517"
51524 *712 0096* 51530 °712 0601
51525 *712 0180* 51531 °712 0686
In this list it will be observed that each logarithm differs from
the preceding either by ‘0000083, ‘0000084, or 0000085. In fact,
the whole difference between the logarithins of 51520 and 51520 +10
is ‘0000842, giving for each increase of a unit an average increase
of 0000084. We have then, at and near 51520, the following
equations :
log (51520 +1) = log 51520 +-0000084
log (51520 +2) = log 51520 +-0000084 x 2
or, if A be not greater than 10,
log (51520 + h) = log 51520 +-0000084 x h...... (A)
or, for a small part of the tables, the logarithms of numbers increasing:
by a unit increase in arithmetical progression very nearly. Now
(M being 43429 .... pages 226 and 237),
: Vee h A A\eee (25 )
Com. log (1+=)=M (4—$ ) ae cs = — &e.
L ar
ea NE a very nearly, when é is small (page 187)
or log (a +h) = loga + Me, very nearly,
an equation of the same form as (A); whence it follows that (A),
which is nearly true when / is 10, is even still more nearly true
when A is a fraction of a unit. Hence we have
eiiyery on =aiylealitcun a oe pe = -6000084
* In all these, the first three figures of the mantissa must be looked
for below. There are various devices in different tables for reminding
the reader of this, which we need not explain, as they are evident on
inspection,
IN FACILITATING COMPUTATIONS. 243
1og512505 = log 51520 + -0000084 x 5
log 51250-36 = log51520 + -0000084 x +36
The column marked Diff. (for difference) is meant to expedite
the multiplication which the last equation shews will become neces-
sary when the logarithm of six or seven places is sought. It consists
of the tenths of 84 to the nearest whole number: thus,
one tenth of 84 is 8:4 , nearest whole number 8
Portes ean) LOB, osc c eee seen bbe)
Tien aa DO' ay ooo wenn tee on Das
and so on. The hundredths of 84 may be got by striking off one
figure from the corresponding tenths (adding 1 where the figure
struck out is 5 or upwards), thus,
one hundredth of 84 is about °8, nearest whole number |
two hundredths ...... wcee 197, cece cece eee eees 9
ne eee es Le a 9°5 yet ee eee e ee neee 3"
and soon. Hence by inspection of the column of differences we can
immediately determine the tenths or hundredths of the difference in
question. And now let us determine the logarithm of 51:53946 from
the second specimen in page 237.
The mantissa is the same as that of the logarithm of 51539°46 and
log (51539 + :46) = log51539 ona tents x *46
= log 51539 +-0000084 (= + =.
but
4 6 4 6
0000084 (= re =~) = -(000001 (= x 844+ = x 84)
(from the table) = *0000001 (34 + 5)
But multiplying a whole number by ‘1, °01, ‘001, &c. is the
same as removing its unit's place to the first, second, third, &c. place
of decimals: from which we have
Mantissa of log 51539 from the table ........ = °7121360
Addition on account of the 4 which follows the 9 = 34
Ditto on account of the 6 which follows the 4 = 5
Sum °7121399
244 ON THE USE OF LOGARITHMS
This sum is the mantissa of the logarithm of 51539°46, which has
the same mantissa as that of 51°53946; therefore, taking the proper
characteristic for the last, we have
log51:53946 = 1:7121399
The following are other instances derived from the same rule, and
falling within the limits of the specimen.
log *5152748? log 5150008 ?
log *51527 1°7120349 log 5°1500 0°7118072
4 34 0 00
8 7 8 7
log 5152748 =: 1:7120390 log 5°150008 —- 07118079
log 5152768000 log :00005154899
log 5152700000 =: 97120349 ~— log -000051548 5 7122118
6 50 9 76
8 7 9 8
——————
log 5152768000 9°7120406 — log 00005154899 57122202
The inverse of this question is done as follows: Suppose it
required to find from the table the number corresponding to the
logarithm 1:7118366. Rejecting the characteristic we look in the
table for the mantissa which is nearest to 7118366 (but below it).
This we find to be 7118325, which is the logarithm of 51503; so
that the number required is (as to its significant figures) within a
unit of 51503. Let it be 51503 +4, then we know that / is to be se
taken* that
log (51503 + h) = 4°7118366
But (page 242),
log (51503 +h) = log 51503 + :0000084 x h very nearly.
= 4°7118325 + -0000084 x h
* We neglect the characteristic, or rather we make our logarithm
47118366. But an alteration of tHe characteristic is only an alteration
of the place of the decimal point.
IN FACILITATING COMPUTATIONS. 245
whence
y a 47118366 —4-7118325 _ 0000041 __ 4t
on “0000084 ~~ *0000084 ~ 84
Now, from the table of differences we see that
Z “A
34 is — of 84 nearly,
10
76 is —- of 84, or
10 _
ie 8
7 is Too of 84 nearly;
so that 34+7 or 41 is a $as of 84; that is, a 45 oi:
whence 51503 +h = 51503°48 and
4:7118366 isthelogof 51503°84
1:7118366 = ....., *5150384
But the readiest method of putting this into practice is by making
an inverse process to the method of finding a logarithm. We take an
instance from another part of the table.
What is the logarithm of 217483°6 ?
log 21748 (¥) 5°3374193
3 60
6 (1) 12
log 217483°6 is 5°3374265
What is the number whose logarithm is 5°3374265 ?
*3374265
Nearest log in table, belonging to Ne 21748 °3374193
Difference 72
Nearest N° in table of Diff. belonging
ine 3. : : F : : 60
120
Annex a cipher, because a figure was struck
off, and therefore a figure (we do not
know what) must be annexed. (See
subsequent remark). Opposite to this
we find 6. |
* Observe that we put in the right characteristic at once.
+ Here, as before, we look opposite to six, and find 120, from which
we strike off one figure.
ae
246 ON THE USE OF LOGARITHMS
Therefore the number to the logarithm required is (making six |
places before the decimal point, on account of the characteristic 5)
217483°6. :
The student will better understand, by forming a number of
logarithms and inverting each process: 1. Why a figure must be
annexed to the last difference, which comes after the table of differ-
ences has been used once. 2. Why that figure cannot be known.
3. Why 0 is most likely to be right. The above 12 might have been
the result of any tabular difference between 115 and 125, the mean
number of which is 120. The following are examples of the process,
without explanation :
What are the numbers to the logarithms 2°1183214 and
1:9648317 ?
°1183214 ‘90648317
Potow 1182978 92221 *9648298
236 . 19
tf Pipa 4 19
40 @)
1 ao
Ans: @O0lsisT71 Ans. 92°22140
The only unusual circumstance with which the student will now
meet is in the multiplication and division of such quantities as 2:9.
To multiply this by 5, carry to the negative term according to the
common algebraical rule of addition. Thus, five times 9 are 45, set
down 5, and carry 4; five times —2 are —10, and 4are —6. The
answer then is 6°5. Or
5(:9—2) = 4-5—10 = 4—104+°5 = 65
To divide 2:9 by 5, make the negative term divisible by 5, and
correct the expression by a corresponding addition. Thus, 2°9 is
5+3°9, and
ae =2+--=14+-8=1-78
The following are further instances :
: IN FACILITATING COMPUTATIONS. DAT
a
3°46 1417
8 10
6)21°68 5)6:170
MOIS 2: 2°834
The following is an example of the use of logarithms in multipli-
cation and division, &c. What is the result of -5729578 multiplied
by 20°62648, and the product divided by 7853982, after which the
tenth root of the ninth power is taken? or what is
9
{srese x 20°62648 |?
7853982 J
log *5729578 1'7581226
log 20°62648 1°3144251
Add 1:0725477
log 7853982 6 8950899
Subtract 6°1774578
9
10)53°5971202
6°7597120
57505 7597056
On
co
(oe)
Answer ‘000005750585
The student should furnish himself with examples for practice by
verifying such equations, for instance, as a(a+6) = a?+ab, where
a and b may be any given numbers. The logarithms of a and a+6
being added together, and a(a-+ b) thus found, a? and a6 should be
and if the whole be correct, the sum of the two
“=
separately found ;
last will be equal to the first.
~~ ee ee
248 ON THE USE OF LOGARITHMS.
The following equations may be thus used:
(a +b) (a—b) = &—B
(ab)" =a’ xb"
Nothing but practice will enable the student to work correctly
with logarithms, and most treatises on that subject contain detailed
examples of all the cases which arise in practice.
THE END.
'
Cv
LONDON:
PRINTED BY JAMES MOYES, CASTLE STREET,
LEICESTER SQUARE.
FE |
Page viii, line 9, The mistake alluded to is the saying that a multiplied
60,
. 108,
. 109,
... 109,
¥ 131,
ATES
... 144,
ADDENDA ET CORRIGENDA.
by nothing is a.
11, omit the word always.
16, for that 1 contains, read that is, 1 contains.
7, for shews, read shew.
4 from the end, for added, read added, &c.
8 from the end, ¢o on each share, add which he holds.
11, 12, and 14, for 1988 and 6988, read 1998 and 6998.
4, for as, read as often as.
. 20, for right, read left.
9, for (a+b), read (atc).
6, for C, read A.
13 from the end, for square root, read square.
Though the student omit what is said on the law of
continuity, he should not omit either this page, or
what immediately relates to it in the next.
4, for VJ 2, read NV 2°.
1
10, for a8, read et
7, for q, read r, in both places.
9 and 11, for q, read r.
1 v
7 from the end, for b + 55; read b+ a5:
er - e
ee
— ey
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