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Title: Elements of arithmetic

Author: Augustus De Morgan

Release date: August 1, 2022 [eBook #68662]
Most recently updated: October 18, 2024

Language: English

Original publication: United Kingdom: Walton and Maberly, 1858

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*** START OF THE PROJECT GUTENBERG EBOOK ELEMENTS OF
ARITHMETIC ***

ELEMENTS OF ARITHMETIC.

BY AUGUSTUS DE MORGAN,

OF TRINITY COLLEGE, CAMBRIDGE;
FELLOW OF THE ROYAL ASTRONOMICAL SOCIETY,
AND OF THE CAMBRIDGE PHILOSOPHICAL SOCIETY;
PROFESSOR OF MATHEMATICS IN UNIVERSITY COLLEGE, LONDON.

“Hominis studiosi est intelligere, quas utilitates proprie
afferat arithmetica his, qui solidam et perfectam doctrinam
in cæteris philosophiæ partibus explicant. Quod enim vulgo
dicunt, principium esse dimidium totius, id vel maxime in
philosophiæ partibus conspicitur.”—Melancthon.
“Ce n’est point par la routine qu’on e’instruit, c’est par
sa propre réflexion; et il est essentiel de contracter l’habitude
de se rendre raison de ce qu’on fait: cette habitude s’acquiert
plus facilement qu’on ne pense; et une fois acquise, elle ne
se perd plus.”—Condillac.

SEVENTEENTH THOUSAND.

LONDON:
WALTON AND MABERLY,
UPPER GOWER STREET, AND IVY LANE, PATERNOSTER ROW.

M.DCCC.LVIII.

LONDON:
PRINTED BY J. WERTHEIMER AND CO.,
CIRCUS-PLACE, FINSBURY-CIRCUS.

PREFACE.
The preceding editions of this work were published in 1830, 1832,
1835, and 1840. This fifth edition differs from the three preceding, as to the
body of the work, in nothing which need prevent the four, or any two of
them, from being used together in a class. But it is considerably augmented
by the addition of eleven new Appendixes,[1] relating to matters on which it
is most desirable that the advanced student should possess information. The
first Appendix, on Computation, and the sixth, on Decimal Money, should
be read and practised by every student with as much attention as any part of
the work. The mastery of the rules for instantaneous conversion of the usual
fractions of a pound sterling into decimal fractions, gives the possessor the
greater part of the advantage which he would derive from the introduction
of a decimal coinage.
At the time when this work was first published, the importance of
establishing arithmetic in the young mind upon reason and demonstration,
was not admitted by many. The case is now altered: schools exist in which
rational arithmetic is taught, and mere rules are made to do no more than
their proper duty. There is no necessity to advocate a change which is
actually in progress, as the works which are published every day
sufficiently shew. And my principal reason for alluding to the subject here,
is merely to warn those who want nothing but routine, that this is not the
book for their purpose.

A. De Morgan.
London, May 1, 1846.

TABLE OF CONTENTS.

BOOK I.
SECTION PAGE
I. Numeration 1
II. Addition and Subtraction 14
III. Multiplication 24
IV. Division 34
V. Fractions 51
VI. Decimal Fractions 65
VII. Square Root 89
VIII. Proportion 100
IX. Permutations and Combinations 118

BOOK II.
I. Weights and Measures, &c. 124
II. Rule of Three 144
III. Interest, &c. 150

APPENDIX.
I. On the mode of computing 161
II. On verification by casting out nines and elevens 166
III. On scales of notation 168
IV. On the definition of fractions 171
V. On characteristics 174
VI. On decimal money 176
VII. On the main principle of book-keeping 180
VIII. On the reduction of fractions to others of nearly equal value 190
IX. On some general properties of numbers 193
X. On combinations 201
XI. On Horner’s method of solving equations 210

XII. Rules for the application of arithmetic to geometry 217

ELEMENTS OF ARITHMETIC.

BOOK I.
PRINCIPLES OF ARITHMETIC.

SECTION I.
NUMERATION.

1. Imagine a multitude of objects of the same kind assembled together; for example,
a company of horsemen. One of the first things that must strike a spectator, although
unused to counting, is, that to each man there is a horse. Now, though men and horses
are things perfectly unlike, yet, because there is one of the first kind to every one of the
second, one man to every horse, a new notion will be formed in the mind of the
observer, which we express in words by saying that there is the same number of men as
of horses. A savage, who had no other way of counting, might remember this number
by taking a pebble for each man. Out of a method as rude as this has sprung our system
of calculation, by the steps which are pointed out in the following articles. Suppose that
there are two companies of horsemen, and a person wishes to know in which of them is
the greater number, and also to be able to recollect how many there are in each.
2. Suppose that while the first company passes by, he drops a pebble into a basket
for each man whom he sees. There is no connexion between the pebbles and the
horsemen but this, that for every horseman there is a pebble; that is, in common
language, the number of pebbles and of horsemen is the same. Suppose that while the
second company passes, he drops a pebble for each man into a second basket: he will
then have two baskets of pebbles, by which he will be able to convey to any other
person a notion of how many horsemen there were in each company. When he wishes
to know which company was the larger, or contained most horsemen, he will take a
pebble out of each basket, and put them aside. He will go on doing this as often as he
can, that is, until one of the baskets is emptied. Then, if he also find the other basket
empty, he says that both companies contained the same number of horsemen; if the
second basket still contain some pebbles, he can tell by them how many more were in
the second than in the first.
3. In this way a savage could keep an account of any numbers in which he was
interested. He could thus register his children, his cattle, or the number of summers and
winters which he had seen, by means of pebbles, or any other small objects which could
be got in large numbers. Something of this sort is the practice of savage nations at this
day, and it has in some places lasted even after the invention of better methods of
reckoning. At Rome, in the time of the republic, the prætor, one of the magistrates, used
to go every year in great pomp, and drive a nail into the door of the temple of Jupiter; a

way of remembering the number of years which the city had been built, which probably
took its rise before the introduction of writing.
4. In process of time, names would be given to those collections of pebbles which
are met with most frequently. But as long as small numbers only were required, the
most convenient way of reckoning them would be by means of the fingers. Any person
could make with his two hands the little calculations which would be necessary for his
purposes, and would name all the different collections of the fingers. He would thus get
words in his own language answering to one, two, three, four, five, six, seven, eight,
nine, and ten. As his wants increased, he would find it necessary to give names to larger
numbers; but here he would be stopped by the immense quantity of words which he
must have, in order to express all the numbers which he would be obliged to make use
of. He must, then, after giving a separate name to a few of the first numbers, manage to
express all other numbers by means of those names.
5. I now shew how this has been done in our own language. The English names of
numbers have been formed from the Saxon: and in the following table each number
after ten is written down in one column, while another shews its connexion with those
which have preceded it.
One eleven ten and one[2]
two twelve ten and two
three thirteen ten and three
four fourteen ten and four
five fifteen ten and five
six sixteen ten and six
seven seventeen ten and seven
eight eighteen ten and eight
nine nineteen ten and nine
ten twenty two tens

twenty-one two tens and one fifty five tens
twenty-two two tens and two sixty six tens
&c. &c. &c. &c. seventy seven tens
thirty three tens eighty eight tens
&c. &c. ninety nine tens
forty four tens a hundred ten tens
&c. &c.
a hundred and one ten tens and one
&c. &c.
a thousand ten hundreds
ten thousand

a hundred thousand
a million ten hundred thousand
or one thousand thousand
ten millions
a hundred millions
&c.

6. Words, written down in ordinary language, would very soon be too long for such
continual repetition as takes place in calculation. Short signs would then be substituted
for words; but it would be impossible to have a distinct sign for every number: so that
when some few signs had been chosen, it would be convenient to invent others for the
rest out of those already made. The signs which we use areas follow:
0 1 2 3 4 5 6 7 8 9
nought one two three four five six seven eight nine

I now proceed to explain the way in which these signs are made to represent other
numbers.
7. Suppose a man first to hold up one finger, then two, and so on, until he has held
up every finger, and suppose a number of men to do the same thing. It is plain that we
may thus distinguish one number from another, by causing two different sets of persons
to hold up each a certain number of fingers, and that we may do this in many different
ways. For example, the number fifteen might be indicated either by fifteen men each
holding up one finger, or by four men each holding up two fingers and a fifth holding
up seven, and so on. The question is, of all these contrivances for expressing the
number, which is the most convenient? In the choice which is made for this purpose
consists what is called the method of numeration.
8. I have used the foregoing explanation because it is very probable that our system
of numeration, and almost every other which is used in the world, sprung from the
practice of reckoning on the fingers, which children usually follow when first they
begin to count. The method which I have described is the rudest possible; but, by a little
alteration, a system may be formed which will enable us to express enormous numbers
with great ease.
9. Suppose that you are going to count some large number, for example, to measure
a number of yards of cloth. Opposite to yourself suppose a man to be placed, who keeps
his eye upon you, and holds up a finger for every yard which he sees you measure.
When ten yards have been measured he will have held up ten fingers, and will not be
able to count any further unless he begin again, holding up one finger at the eleventh
yard, two at the twelfth, and so on. But to know how many have been counted, you
must know, not only how many fingers he holds up, but also how many times he has
begun again. You may keep this in view by placing another man on the right of the

former, who directs his eye towards his companion, and holds up one finger the
moment he perceives him ready to begin again, that is, as soon as ten yards have been
measured. Each finger of the first man stands only for one yard, but each finger of the
second stands for as many as all the fingers of the first together, that is, for ten. In this
way a hundred may be counted, because the first may now reckon his ten fingers once
for each finger of the second man, that is, ten times in all, and ten tens is one hundred
(5).[3] Now place a third man at the right of the second, who shall hold up a finger
whenever he perceives the second ready to begin again. One finger of the third man
counts as many as all the ten fingers of the second, that is, counts one hundred. In this
way we may proceed until the third has all his fingers extended, which will signify that
ten hundred or one thousand have been counted (5). A fourth man would enable us to
count as far as ten thousand, a fifth as far as one hundred thousand, a sixth as far as a
million, and so on.
10. Each new person placed himself towards your left in the rank opposite to you.
Now rule columns as in the next page, and to the right of them all place in words the
number which you wish to represent; in the first column on the right, place the number
of fingers which the first man will be holding up when that number of yards has been
measured. In the next column, place the fingers which the second man will then be
holding up; and so on.
7th. 6th. 5th. 4th. 3rd. 2nd. 1st.
I. 5 7 fifty-seven
II. 1 0 4 one hundred and four.
III. 1 1 0 one hundred and ten.
IV. 2 3 4 8 two thousand three hundred
and forty-eight.
V. 1 5 9 0 6 fifteen thousand nine
hundred and six.
VI. 1 8 7 0 0 4 one hundred and eighty-seven
thousand and four.
VII. 3 6 9 7 2 8 5 three million, six hundred and
ninety-seven thousand,
two hundred and eighty-five.
11. In I. the number fifty-seven is expressed. This means (5) five tens and seven.
The first has therefore counted all his fingers five times, and has counted seven fingers
more. This is shewn by five fingers of the second man being held up, and seven of the
first. In II. the number one hundred and four is represented. This number is (5) ten tens
and four. The second person has therefore just reckoned all his fingers once, which is
denoted by the third person holding up one finger; but he has not yet begun again,

because he does not hold up a finger until the first has counted ten, of which ten only
four are completed. When all the last-mentioned ten have been counted, he then holds
up one finger, and the first being ready to begin again, has no fingers extended, and the
number obtained is eleven tens, or ten tens and one ten, or one hundred and ten. This is
the case in III. You will now find no difficulty with the other numbers in the table.
12. In all these numbers a figure in the first column stands for only as many yards as
are written under that figure in (6). A figure in the second column stands, not for as
many yards, but for as many tens of yards; a figure in the third column stands for as
many hundreds of yards; in the fourth column for as many thousands of yards; and so
on: that is, if we suppose a figure to move from any column to the one on its left, it
stands for ten times as many yards as before. Recollect this, and you may cease to draw
the lines between the columns, because each figure will be sufficiently well known by
the place in which it is; that is, by the number of figures which come upon the right
hand of it.
13. It is important to recollect that this way of writing numbers, which has become
so familiar as to seem the natural method, is not more natural than any other. For
example, we might agree to signify one ten by the figure of one with an accent, thus, 1′;
twenty or two tens by 2′; and so on: one hundred or ten tens by 1″; two hundred by 2″;
one thousand by 1‴; and so on: putting Roman figures for accents when they become
too many to write with convenience. The fourth number in the table would then be
written 2‴ 3′ 4′ 8, which might also be expressed by 8 4′ 3″ 2‴, 4′ 8 3″ 2‴; or the order
of the figures might be changed in any way, because their meaning depends upon the
accents which are attached to them, and not upon the place in which they stand. Hence,
a cipher would never be necessary; for 104 would be distinguished from 14 by writing
for the first 1″ 4, and for the second 1′ 4. The common method is preferred, not because
it is more exact than this, but because it is more simple.
14. The distinction between our method of numeration and that of the ancients, is in
the meaning of each figure depending partly upon the place in which it stands. Thus, in
44444 each four stands for four of something; but in the first column on the right it
signifies only four of the pebbles which are counted; in the second, it means four
collections of ten pebbles each; in the third, four of one hundred each; and so on.
15. The things measured in (11) were yards of cloth. In this case one yard of cloth is
called the unit. The first figure on the right is said to be in the units’ place, because it
only stands for so many units as are in the number that is written under it in (6). The
second figure is said to be in the tens’ place, because it stands for a number of tens of
units. The third, fourth, and fifth figures are in the places of the hundreds, thousands,
and tens of thousands, for a similar reason.
16. If the quantity measured had been acres of land, an acre of land would have
been called the unit, for the unit is one of the things which are measured. Quantities are
of two sorts; those which contain an exact number of units, as 47 yards, and those

which do not, as 47 yards and a half. Of these, for the present, we only consider the
first.
17. In most parts of arithmetic, all quantities must have the same unit. You cannot
say that 2 yards and 3 feet make 5 yards or 5 feet, because 2 and 3 make 5; yet you may
say that 2 yards and 3 yards make 5 yards, and that 2 feet and 3 feet make 5 feet. It
would be absurd to try to measure a quantity of one kind with a unit which is a quantity
of another kind; for example, to attempt to tell how many yards there are in a gallon, or
how many bushels of corn there are in a barrel of wine.
18. All things which are true of some numbers of one unit are true of the same
numbers of any other unit. Thus, 15 pebbles and 7 pebbles together make 22 pebbles;
15 acres and 7 acres together make 22 acres, and so on. From this we come to say that
15 and 7 make 22, meaning that 15 things of the same kind, and 7 more of the same
kind as the first, together make 22 of that kind, whether the kind mentioned be pebbles,
horsemen, acres of land, or any other. For these it is but necessary to say, once for all,
that 15 and 7 make 22. Therefore, in future, on this part of the subject I shall cease to
talk of any particular units, such as pebbles or acres, and speak of numbers only. A
number, considered without intending to allude to any particular things, is called an
abstract number: and it then merely signifies repetitions of a unit, or the number of
times a unit is repeated.
19. I will now repeat the principal things which have been mentioned in this
chapter.
I. Ten signs are used, one to stand for nothing, the rest for the first nine numbers.
They are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The first of these is called a cipher.
II. Higher numbers have not signs for themselves, but are signified by placing the
signs already mentioned by the side of each other, and agreeing that the first figure on
the right hand shall keep the value which it has when it stands alone; that the second on
the right hand shall mean ten times as many as it does when it stands alone; that the
third figure shall mean one hundred times as many as it does when it stands alone; the
fourth, one thousand times as many; and so on.
III. The right hand figure is said to be in the units’ place, the next to that in the tens’
place, the third in the hundreds’ place, and so on.
IV. When a number is itself an exact number of tens, hundreds, or thousands, &c.,
as many ciphers must be placed on the right of it as will bring the number into the place
which is intended for it. The following are examples:
Fifty, or five tens, 50: seven hundred, 700.
Five hundred and twenty-eight thousand, 528000.
If it were not for the ciphers, these numbers would be mistaken for 5, 7, and 528.

V. A cipher in the middle of a number becomes necessary when any one of the
denominations, units, tens, &c. is wanting. Thus, twenty thousand and six is 20006, two
hundred and six is 206. Ciphers might be placed at the beginning of a number, but they
would have no meaning. Thus 026 is the same as 26, since the cipher merely shews that
there are no hundreds, which is evident from the number itself.
20. If we take out of a number, as 16785, any of those figures which come together,
as 67, and ask, what does this sixty-seven mean? of what is it sixty-seven? the answer
is, sixty-seven of the same collections as the 7, when it was in the number; that is, 67
hundreds. For the 6 is 6 thousands, or 6 ten hundreds, or sixty hundreds; which, with
the 7, or 7 hundreds, is 67 hundreds: similarly, the 678 is 678 tens. This number may
then be expressed either as

1 ten thousand 6 thousands 7 hundreds 8 tens and 5;
or 16 thousands 78 tens and 5; or 1 ten thousand 678 tens and 5;
or 167 hundreds 8 tens and 5; or 1678 tens and 5, and so on.

21. EXERCISES.
I. Write down the signs for—four hundred and seventy-six; two thousand and
ninety-seven; sixty-four thousand three hundred and fifty; two millions seven hundred
and four; five hundred and seventy-eight millions of millions.
II. Write at full length 53, 1805, 1830, 66707, 180917324, 66713721, 90976390,
25000000.
III. What alteration takes place in a number made up entirely of nines, such as
99999, by adding one to it?
IV. Shew that a number which has five figures in it must be greater than one which
has four, though the first have none but small figures in it, and the second none but
large ones. For example, that 10111 is greater than 9879.
22. You now see that the convenience of our method of numeration arises from a
few simple signs being made to change their value as they change the column in which
they are placed. The same advantage arises from counting in a similar way all the
articles which are used in every-day life. For example, we count money by dividing it
into pounds, shillings, and pence, of which a shilling is 12 pence, and a pound 20
shillings, or 240 pence. We write a number of pounds, shillings, and pence in three
columns, generally placing points between the columns. Thus, 263 pence would not be
written as 263, but as £1. 1. 11, where £ shews that the 1 in the first column is a pound.
Here is a system of numeration in which a number in the second column on the right
means 12 times as much as the same number in the first; and one in the third column is
twenty times as great as the same in the second, or 240 times as great as the same in the
first. In each of the tables of measures which you will hereafter meet with, you will see
a separate system of numeration, but the methods of calculation for all will be the same.

23. In order to make the language of arithmetic shorter, some other signs are used.
They are as follow:
I. 15 + 38 means that 38 is to be added to 15, and is the same thing as 53. This is the
sum of 15 and 38, and is read fifteen plus thirty-eight (plus is the Latin for more).
II. 64-12 means that 12 is to be taken away from 64, and is the same thing as 52.
This is the difference of 64 and 12, and is read sixty-four minus twelve (minus is the
Latin for less).
III. 9 × 8 means that 8 is to be taken 9 times, and is the same thing as 72. This is the
product of 9 and 8, and is read nine into eight.
IV. 108/6 means that 108 is to be divided by 6, or that you must find out how many
sixes there are in 108; and is the same thing as 18. This is the quotient of 108 and 6; and
is read a hundred and eight by six.
V. When two numbers, or collections of numbers, with the foregoing signs, are the
same, the sign = is put between them. Thus, that 7 and 5 make 12, is written in this way,
7 + 5 = 12. This is called an equation, and is read, seven plus five equals twelve. It is
plain that we may construct as many equations as we please. Thus:
12
7 + 9 - 3 = 12 + 1; — - 1 + 3 × 2 = 11,
2
and so on.
24. It often becomes necessary to speak of something which is true not of any one
number only, but of all numbers. For example, take 10 and 7; their sum[4] is 17, their
difference is 3. If this sum and difference be added together, we get 20, which is twice
the greater of the two numbers first chosen. If from 17 we take 3, we get 14, which is
twice the less of the two numbers. The same thing will be found to hold good of any
two numbers, which gives this general proposition,—If the sum and difference of two
numbers be added together, the result is twice the greater of the two; if the difference be
taken from the sum, the result is twice the lesser of the two. If, then, we take any
numbers, and call them the first number and the second number, and let the first number
be the greater; we have
(1st No. + 2d No.) + (1st No. - 2d No.) = twice 1st No.
(1st No. + 2d No.) - (1st No. - 2d No.) = twice 2d No.
The brackets here enclose the things which must be first done, before the signs
which join the brackets are made use of. Thus, 8-(2 + 1) × (1 + 1) signifies that 2 + 1
must be taken 1 + 1 times, and the product must be subtracted from 8. In the same
manner, any result made from two or more numbers, which is true whatever numbers
are taken, may be represented by using first No., second No., &c., to stand for them,

and by the signs in (23). But this may be much shortened; for as first No., second No.,
&c., may mean any numbers, the letters a and b may be used instead of these words;
and it must now be recollected that a and b stand for two numbers, provided only that a
is greater than b. Let twice a be represented by 2a, and twice b by 2b. The equations
then become
(a + b) + (a - b) = 2a,
and (a + b) - (a - b) = 2b.
This may be explained still further, as follows:
25. Suppose a number of sealed packets, marked a, b, c, d, &c., on the outside, each
of which contains a distinct but unknown number of counters. As long as we do not
know how many counters each contains, we can make the letter which belongs to each
stand for its number, so as to talk of the number a, instead of the number in the packet
marked a. And because we do not know the numbers, it does not therefore follow that
we know nothing whatever about them; for there are some connexions which exist
between all numbers, which we call general properties of numbers. For example, take
any number, multiply it by itself, and subtract one from the result; and then subtract one
from the number itself. The first of these will always contain the second exactly as
many times as the original number increased by one. Take the number 6; this multiplied
by itself is 36, which diminished by one is 35; again, 6 diminished by 1 is 5; and 35
contains 5, 7 times, that is, 6 + 1 times. This will be found to be true of any number,
and, when proved, may be said to be true of the number contained in the packet marked
a, or of the number a. If we represent a multiplied by itself by aa,[5] we have, by (23)
aa - 1
——— = a + 1.
a-1
26. When, therefore, we wish to talk of a number without specifying any one in
particular, we use a letter to represent it. Thus: Suppose we wish to reason upon what
will follow from dividing a number into three parts, without considering what the
number is, or what are the parts into which it is divided. Let a stand for the number, and
b, c, and d, for the parts into which it is divided. Then, by our supposition,
a = b + c + d.
On this we can reason, and produce results which do not belong to any particular
number, but are true of all. Thus, if one part be taken away from the number, the other
two will remain, or
a - b = c + d.
If each part be doubled, the whole number will be doubled, or

2a = 2b + 2c + 2d.
If we diminish one of the parts, as d, by a number x, we diminish the whole number just
as much, or
a - x = b + c + (d - x).

27. EXERCISES.
What is a + 2b - c, where a = 12, b = 18, c = 7?—Answer, 41.

What is aa - bb
———
a-b
where a = 6 and b = 2?—Ans. 8.

What is the difference between (a + b)(c + d) and a + bc + d, for the following
values of a, b, c, and d?
a b c d Ans.
1 2 3 4 10
2 12 7 1 25
1 1 1 1 1

SECTION II.
ADDITION AND SUBTRACTION.

28. There is no process in arithmetic which does not consist entirely in the
increase or diminution of numbers. There is then nothing which might not be
done with collections of pebbles. Probably, at first, either these or the fingers
were used. Our word calculation is derived from the Latin word calculus,
which means a pebble. Shorter ways of counting have been invented, by which
many calculations, which would require long and tedious reckoning if pebbles
were used, are made at once with very little trouble. The four great methods
are, Addition, Subtraction, Multiplication, and Division; of which, the last two
are only ways of doing several of the first and second at once.
29. When one number is increased by others, the number which is as large
as all the numbers together is called their sum. The process of finding the sum
of two or more numbers is called Addition, and, as was said before, is denoted
by placing a cross (+) between the numbers which are to be added together.
Suppose it required to find the sum of 1834 and 2799. In order to add these
numbers, take them to pieces, dividing each into its units, tens, hundreds, and
thousands:
1834 is 1 thous. 8 hund. 3 tens and 4;
2799 is 2 thous. 7 hund. 9 tens and 9.
Each number is thus broken up into four parts. If to each part of the first
you add the part of the second which is under it, and then put together what
you get from these additions, you will have added 1834 and 2799. In the first
number are 4 units, and in the second 9: these will, when the numbers are
added together, contribute 13 units to the sum. Again, the 3 tens in the first and
the 9 tens in the second will contribute 12 tens to the sum. The 8 hundreds in
the first and the 7 hundreds in the second will add 15 hundreds to the sum; and
the thousand in the first with the 2 thousands in the second will contribute 3
thousands to the sum; therefore the sum required is
3 thousands, 15 hundreds, 12 tens, and 13 units.
To simplify this result, you must recollect that—
13 units are 1 ten and 3 units.

12 tens are 1 hund. and 2 tens.
15 hund. are 1 thous. and 5 hund.
3 thous. are 3 thous.
Now collect the numbers on the right hand side together, as was done
before, and this will give, as the sum of 1834 and 2799,
4 thousands, 6 hundreds, 3 tens, and 3 units,
which (19) is written 4633.
30. The former process, written with the signs of (23) is as follows:
1834 = 1 × 1000 + 8 × 100 + 3 × 10 + 4
2799 = 2 × 1000 + 7 × 100 + 9 × 10 + 9
Therefore,
1834 + 2799 = 3 × 1000 + 15 × 100 + 12 × 10 + 13
But
13 = 1 × 10 + 3
12 × 10 = 1 × 100 + 2 × 10
15 × 100 = 1 × 1000 + 5 × 100
3 × 1000 = 3 × 1000 Therefore,
1834 + 2799 = 4 × 1000 + 6 × 100 + 3 × 10 + 3
= 4633.
31. The same process is to be followed in all cases, but not at the same
length. In order to be able to go through it, you must know how to add together
the simple numbers. This can only be done by memory; and to help the
memory you should make the following table three or four times for yourself:
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9 10
2 3 4 5 6 7 8 9 10 11
3 4 5 6 7 8 9 10 11 12
4 5 6 7 8 9 10 11 12 13

5 6 7 8 9 10 11 12 13 14
6 7 8 9 10 11 12 13 14 15
7 8 9 10 11 12 13 14 15 16
8 9 10 11 12 13 14 15 16 17
9 10 11 12 13 14 15 16 17 18
The use of this table is as follows: Suppose you want to find the sum of 8
and 7. Look in the left-hand column for either of them, 8, for example; and
look in the top column for 7. On the same line as 8, and underneath 7, you find
15, their sum.
32. When this table has been thoroughly committed to memory, so that you
can tell at once the sum of any two numbers, neither of which exceeds 9, you
should exercise yourself in adding and subtracting two numbers, one of which
is greater than 9 and the other less. You should write down a great number of
such sentences as the following, which will exercise you at the same time in
addition, and in the use of the signs mentioned in (23).
12 + 6 = 18 22 + 6 = 28 19 + 8 = 27
54 + 9 = 63 56 + 7 = 63 22 + 8 = 30
100 - 9 = 91 27 - 8 = 19 44 - 6 = 38, &c.
33. When the last two articles have been thoroughly studied, you will be
able to find the sum of any numbers by the following process,[6] which is the
same as that in (29).
Rule I. Place the numbers under one another, units under units, tens under
tens, and so on.
II. Add together the units of all, and part the whole number thus obtained
into units and tens. Thus, if 85 be the number, part it into 8 tens and 5 units; if
136 be the number, part it into 13 tens and 6 units (20).
III. Write down the units of this number under the units of the rest, and
keep in memory the number of tens.
IV. Add together all the numbers in the column of tens, remembering to
take in (or carry, as it is called) the tens which you were told to recollect in III.,
and divide this number of tens into tens and hundreds. Thus, if 335 tens be the
number obtained, part this into 33 hundreds and 5 tens.

V. Place the number of tens under the tens, and remember the number of
hundreds.
VI. Proceed in this way through every column, and at the last column,
instead of separating the number you obtain into two parts, write it all down
before the rest.
Example.—What is
1805 + 36 + 19727 + 3 + 1474 + 2008

1805
36
19727
3
1474
2008
——-
25053

The addition of the units’ line, or 8 + 4 + 3 + 7 + 6 + 5, gives 33, that is, 3
tens and 3 units. Put 3 in the units’ place, and add together the line of tens,
taking in at the beginning the 3 tens which were created by the addition of the
units’ line. That is, find 3 + 0 + 7 + 2 + 3 + 0, which gives 15 for the number
of tens; that is, 1 hundred and 5 tens. Add the line of hundreds together, taking
care to add the 1 hundred which arose in the addition of the line of tens; that is,
find 1 + 0 + 4 + 7 + 8, which gives exactly 20 hundreds, or 2 thousands and no
hundreds. Put a cipher in the hundreds’ place (because, if you do not, the next
figure will be taken for hundreds instead of thousands), and add the figures in
the thousands’ line together, remembering the 2 thousands which arose from
the hundreds’ line; that is, find 2 + 2 + 1 + 9 + 1, which gives 15 thousands, or
1 ten thousand and 5 thousand. Write 5 under the line of thousands, and collect
the figures in the line of tens of thousands, remembering the ten thousand
which arose out of the thousands’ line; that is, find 1 + 1, or 2 ten thousands.
Write 2 under the ten thousands’ line, and the operation is completed.
34. As an exercise in addition, you may satisfy yourself that what I now
say of the following square is correct. The numbers in every row, whether

reckoned upright, or from right to left, or from corner to corner, when added
together give the number 24156.

2016 4212 1656 3852 1296 3492 936 3132 576 2772 216
252 2052 4248 1692 3888 1332 3528 972 3168 612 2412
2448 288 2088 4284 1728 3924 1368 3564 1008 2808 648
684 2484 324 2124 4320 1764 3960 1404 3204 1044 2844
2880 720 2520 360 2160 4356 1800 3600 1440 3240 1080
1116 2916 756 2556 396 2196 3996 1836 3636 1476 3276
3312 1152 2952 792 2592 36 2232 4032 1872 3672 1512
1548 3348 1188 2988 432 2628 72 2268 4068 1908 3708
3744 1584 3384 828 3024 468 2664 108 2304 4104 1944
1980 3780 1224 3420 864 3060 504 2700 144 2340 4140
4176 1620 3816 1260 3456 900 3096 540 2736 180 2376

35. If two numbers must be added together, it will not alter the sum if you
take away a part of one, provided you put on as much to the other. It is plain
that you will not alter the whole number of a collection of pebbles in two
baskets by taking any number out of one, and putting them into the other.
Thus, 15 + 7 is the same as 12 + 10, since 12 is 3 less than 15, and 10 is three
more than 7. This was the principle upon which the whole of the process in
(29) was conducted.
36. Let a and b stand for two numbers, as in (24). It is impossible to tell
what their sum will be until the numbers themselves are known. In the mean
while a + b stands for this sum. To say, in algebraical language, that the sum of
a and b is not altered by adding c to a, provided we take away c from b, we
have the following equation:
(a + c) + (b - c) = a + b;
which may be written without brackets, thus,
a + c + b - c = a + b.

For the meaning of these two equations will appear to be the same, on
consideration.
37. If a be taken twice, three times, &c., the results are represented in
algebra by 2a, 3a, 4a, &c. The sum of any two of this series may be expressed
in a shorter form than by writing the sign + between them; for though we do
not know what number a stands for, we know that, be it what it may, 2a + 2a =
4a, 3a + 2a = 5a, 4a + 9a = 13a; and generally, if a taken m times be added to
a taken n times, the result is a taken m + n times, or
ma + na = (m + n)a.
38. The use of the brackets must here be noticed. They mean, that the
expression contained inside them must be used exactly as a single letter would
be used in the same place. Thus, pa signifies that a is taken p times, and (m +
n)a, that a is taken m + n times. It is, therefore, a different thing from m + na,
which means that a, after being taken n times, is added to m. Thus (3 + 4) × 2
is 7 × 2 or 14; while 3 + 4 × 2 is 3 + 8, or 11.
39. When one number is taken away from another, the number which is left
is called the difference or remainder. The process of finding the difference is
called subtraction. The number which is to be taken away must be of course
the lesser of the two.
40. The process of subtraction depends upon these two principles.
I. The difference of two numbers is not altered by adding a number to the
first, if you add the same number to the second; or by subtracting a number
from the first, if you subtract the same number from the second. Conceive two
baskets with pebbles in them, in the first of which are 100 pebbles more than
in the second. If I put 50 more pebbles into each of them, there are still only
100 more in the first than in the second, and the same if I take 50 from each.
Therefore, in finding the difference of two numbers, if it should be convenient,
I may add any number I please to both of them, because, though I alter the
numbers themselves by so doing, I do not alter their difference.

II. Since 6 exceeds 4 by 2,
and 3 exceeds 2 by 1,
and 12 exceeds 5 by 7,

6, 3, and 12 together, or 21, exceed 4, 2, and 5 together, or 11, by 2, 1, and 7
together, or 10: the same thing may be said of any other numbers.

41. If a, b, and c be three numbers, of which a is greater than b (40), I.
leads to the following,
(a + c) - (b + c) = a - b.
Again, if c be less than a and b,
(a - c) - (b - c) = a - b.
The brackets cannot be here removed as in (36). That is, p- (q-r) is not the
same thing as p-q- r. For, in the first, the difference of q and r is subtracted
from p; but in the second, first q and then r are subtracted from p, which is the
same as subtracting as much as q and r together, or q + r. Therefore p-q-r is p-
(q + r). In order to shew how to remove the brackets from p -(q-r) without
altering the value of the result, let us take the simple instance 12-(8-5). If we
subtract 8 from 12, or form 12-8, we subtract too much; because it is not 8
which is to be taken away, but as much of 8 as is left after diminishing it by 5.
In forming 12-8 we have therefore subtracted 5 too much. This must be set
right by adding 5 to the result, which gives 12-8 + 5 for the value of 12-(8-5).
The same reasoning applies to every case, and we have therefore,
p - (q + r) = p - q - r.
p - (q - r) = p - q + r.
By the same kind of reasoning,
a - (b + c - d - e) = a - b - c + d + e.
2a + 3b - (a - 2b) = 2a + 3b - a + 2b = a + 5b.
4x + y - (17x - 9y) = 4x + y - 17x + 9y = 10y - 13x.
42. I want to find the difference of the numbers 57762 and 34631. Take
these to pieces as in (29) and
57762 is 5 ten-th. 7 th. 7 hund. 6 tens and 2 units.
34631 is 3 ten-th. 4 th. 6 hund. 3 tens and 1 unit.
Now 2 units exceed 1 unit by 1 unit.
6 tens 3 tens 3 tens.
7 hundreds 6 hundreds 1 hundred.
7 thousands 4 thousands 3 thousands.

5 ten-thousands 3 ten-thous. 2 ten-thous.
Therefore, by (40, Principle II.) all the first column together exceeds all the
second column by all the third column, that is, by
2 ten-th. 3 th. 1 hund. 3 tens and 1 unit,
which is 23131. Therefore the difference of 57762 and 34631 is 23131, or
57762-34631 = 23131.
43. Suppose I want to find the difference between 61274 and 39628. Write
them at length, and
61274 is 6 ten-th. 1 th. 2 hund. 7 tens and 4 units.
39628 is 3 ten-th. 9 th. 6 hund. 2 tens and 8 units.
If we attempt to do the same as in the last article, there is a difficulty
immediately, since 8, being greater than 4, cannot be taken from it. But from
(40) it appears that we shall not alter the difference of two numbers if we add
the same number to both of them. Add ten to the first number, that is, let there
be 14 units instead of four, and add ten also to the second number, but instead
of adding ten to the number of units, add one to the number of tens, which is
the same thing. The numbers will then stand thus,
6 ten-thous. 1 thous. 2 hund. 7 tens and 14 units.[7] 3 ten-thous. 9
thous. 6 hund. 3 tens and 8 units.
You now see that the units and tens in the lower can be subtracted from
those in the upper line, but that the hundreds cannot. To remedy this, add one
thousand or 10 hundred to both numbers, which will not alter their difference,
and remember to increase the hundreds in the upper line by 10, and the
thousands in the lower line by 1, which are the same things. And since the
thousands in the lower cannot be subtracted from the thousands in the upper
line, add 1 ten thousand or 10 thousand to both numbers, and increase the
thousands in the upper line by 10, and the ten thousands in the lower line by 1,
which are the same things; and at the close the numbers which we get will be,
6 ten-thous. 11 thous. 12 hund. 7 tens and 14 units.
4 ten-thous. 10 thous. 6 hund. 3 tens and 8 units.
These numbers are not, it is true, the same as those given at the beginning
of this article, but their difference is the same, by (40). With the last-mentioned

numbers proceed in the same way as in (42), which will give, as their
difference,
2 ten-thous. 1 thous. 6 hund. 4 tens, and 6 units, which is 21646.
44. From this we deduce the following rules for subtraction:
I. Write the number which is to be subtracted (which is, of
course, the lesser of the two, and is called the subtrahend)
under the other, so that its units shall fall under the units of the
other, and so on.
II. Subtract each figure of the lower line from the one
above it, if that can be done. Where that cannot be done, add
ten to the upper figure, and then subtract the lower figure; but
recollect in this case always to increase the next figure in the
lower line by 1, before you begin to subtract it from the upper
one.
45. If there should not be as many figures in the lower line as in the upper
one, proceed as if there were as many ciphers at the beginning of the lower
line as will make the number of figures equal. You do not alter a number by
placing ciphers at the beginning of it. For example, 00818 is the same number
as 818, for it means
0 ten-thous. 0 thous. 8 hunds. 1 ten and 8 units;
the first two signs are nothing, and the rest is
8 hundreds, 1 ten, and 8 units, or 818.
The second does not differ from the first, except in its being said that there
are no thousands and no tens of thousands in the number, which may be
known without their being mentioned at all. You may ask, perhaps, why this
does not apply to a cipher placed in the middle of a number, or at the right of
it, as, for example, in 28007 and 39700? But you must recollect, that if it were
not for the two ciphers in the first, the 8 would be taken for 8 tens, instead of 8
thousands; and if it were not for the ciphers in the second, the 7 would be
taken for 7 units, instead of 7 hundreds.

46. EXAMPLE.
What is the difference between 3708291640030174

and 30813649276188
Difference 3677477990753986

EXERCISES.
I. What is 18337 + 149263200 - 6472902?—Answer 142808635.
What is 1000 - 464 + 3279 - 646?—Ans. 3169.
II. Subtract
64 + 76 + 144 - 18 from 33 - 2 + 100037.—Ans. 99802.
III. What shorter rule might be made for subtraction when all the figures in
the upper line are ciphers except the first? for example, in finding
10000000 - 2731634.
IV. Find 18362 + 2469 and 18362-2469, add the second result to the first,
and then subtract 18362; subtract the second from the first, and then subtract
2469.—Answer 18362 and 2469.
V. There are four places on the same line in the order a, b, c, and d. From a
to d it is 1463 miles; from a to c it is 728 miles; and from b to d it is 1317
miles. How far is it from a to b, from b to c, and from c to d?—Answer. From
a to b 146, from b to c 582, and from c to d 735 miles.

VI. In the following table subtract b from a, and b from the remainder, and
so on until b can be no longer subtracted. Find how many times b can be
subtracted from a, and what is the last remainder.
No. of
A B Remainder.
times.
23604 9999 2 3606
209961 37173 5 24096
74712 6792 11 0
4802469 654321 7 222222
18849747 3141592 6 195
987654321 123456789 8 9

SECTION III.
MULTIPLICATION.

47. I have said that all questions in arithmetic require nothing but addition and
subtraction. I do not mean by this that no rule should ever be used except those
given in the last section, but that all other rules only shew shorter ways of finding
what might be found, if we pleased, by the methods there deduced. Even the last
two rules themselves are only short and convenient ways of doing what may be
done with a number of pebbles or counters.
48. I want to know the sum of five seventeens, or I ask the following question:
There are five heaps of pebbles, and seventeen pebbles in each heap; how many are
there in all? Write five seventeens in a column, and make the addition, which gives
85. In this case 85 is called the product of 5 and 17, and the process of finding the
product is called multiplication, which gives nothing more than the addition of a
number of the same quantities. Here 17 is called the multiplicand, and 5 is called
the multiplier.

17
17
17
17
17
85

49. If no question harder than this were ever proposed, there would be no
occasion for a shorter way than the one here followed. But if there were 1367 heaps
of pebbles, and 429 in each heap, the whole number is then 1367 times 429, or 429
multiplied by 1367. I should have to write 429 1367 times, and then to make an
addition of enormous length. To avoid this, a shorter rule is necessary, which I now
proceed to explain.
50. The student must first make himself acquainted with the products of all
numbers as far as 10 times 10 by means of the following table,[8] which must be
committed to memory.

1 2 3 4 5 6 7 8 9 10 11 12
2 4 6 8 10 12 14 16 18 20 22 24

3 6 9 12 15 18 21 24 27 30 33 36
4 8 12 16 20 24 28 32 36 40 44 48
5 10 15 20 25 30 35 40 45 50 55 60
6 12 18 24 30 36 42 48 54 60 66 72
7 14 21 28 35 42 49 56 63 70 77 84
8 16 24 32 40 48 56 64 72 80 88 96
9 18 27 36 45 54 63 72 81 90 99 108
10 20 30 40 50 60 70 80 90 100 110 120
11 22 33 44 55 66 77 88 99 110 121 132
12 24 36 48 60 72 84 96 108 120 132 144

If from this table you wish to know what is 7 times 6, look in the first upright
column on the left for either of them; 6 for example. Proceed to the right until you
come into the column marked 7 at the top. You there find 42, which is the product
of 6 and 7.
51. You may find, in this way, either 6 times 7, or 7 times 6, and for both you
find 42. That is, six sevens is the same number as seven sixes. This may be shewn
as follows: Place seven counters in a line, and repeat that line in all six times. The
number of counters in the whole is 6 times 7, or six sevens, if I reckon the rows
from the top to the bottom; but if I count the rows that stand side by side, I find
seven of them, and six in each row, the whole number of which is 7 times 6, or
seven sixes. And the whole number is 42, whichever way I count. The same method
may be applied to any other two numbers. If the signs of (23) were used, it would
be said that 7 × 6 = 6 × 7.

●●●●●●●
●●●●●●●
●●●●●●●
●●●●●●●
●●●●●●●
●●●●●●●

52. To take any quantity a number of times, it will be enough to take every one
of its parts the same number of times. Thus, a sack of corn will be increased fifty-
fold, if each bushel which it contains be replaced by 50 bushels. A country will be
doubled by doubling every acre of land, or every county, which it contains. Simple
as this may appear, it is necessary to state it, because it is one of the principles on
which the rule of multiplication depends.
53. In order to multiply by any number, you may multiply separately by any
parts into which you choose to divide that number, and add the results. For
example, 4 and 2 make 6. To multiply 7 by 6 first multiply 7 by 4, and then by 2,
and add the products. This will give 42, which is the product of 7 and 6. Again,
since 57 is made up of 32 and 25, 57 times 50 is made up of 32 times 50 and 25
times 50, and so on. If the signs were used, these would be written thus:
7 × 6 = 7 × 4 + 7 × 2.
50 × 57 = 50 × 32 + 50 × 25.
54. The principles in the last two articles may be expressed thus: If a be made
up of the parts x, y, and x, ma is made up of mx, my, and mz; or,
if a = x + y + z.
ma = mx + my + mz,
or, m(x + y + z) = mx + my + mz.
A similar result may be obtained if a, instead of being made up of x, y, and z, is
made by combined additions and subtractions, such as x + y-z, x- y + z, x-y-z, &c.
To take the first as an instance:
Let a = x + y - z,
then ma = mx + my - mz.
For, if a had been x + y, ma would have been mx + my. But since a is less than x
+ y by z, too much by z has been repeated every time that x + y has been repeated;
—that is, mz too much has been taken; consequently, ma is not mx + my, but mx +
my-mz. Similar reasoning may be applied to other cases, and the following results
may be obtained:
m(a + b + c - d) = ma + mb + mc - md.
a(a - b) = aa - ab.
b(a - b) = ba - bb.
3(2a - 4b) = 6a - 12b.

7a(7 + 2b) = 49a + 14ab.
(aa + a + 1)a = aaa + aa + a.
(3ab - 2c)4abc = 12aabbc - 8abcc.
55. There is another way in which two numbers may be multiplied together.
Since 8 is 4 times 2, 7 times 8 may be made by multiplying 7 and 4, and then
multiplying that product by 2. To shew this, place 7 counters in a line, and repeat
that line in all 8 times, as in figures I. and II.

I.
●●●●●●●
●●●●●●●
A ●●●●●●●
●●●●●●●

●●●●●●●
●●●●●●●
B ●●●●●●●
●●●●●●●

II.
●●●●●●●
●●●●●●●

●●●●●●●
●●●●●●●

●●●●●●●
●●●●●●●

●●●●●●●
●●●●●●●
The number of counters in all is 8 times 7, or 56. But (as in fig. I.) enclose each
four rows in oblong figures, such as a and b. The number in each oblong is 4 times
7, or 28, and there are two of those oblongs; so that in the whole the number of
counters is twice 28, or 28 x 2, or 7 first multiplied by 4, and that product
multiplied by 2. In figure II. it is shewn that 7 multiplied by 8 is also 7 first
multiplied by 2, and that product multiplied by 4. The same method may be applied
to other numbers. Thus, since 80 is 8 times 10, 256 times 80 is 256 multiplied by 8,
and that product multiplied by 10. If we use the signs, the foregoing assertions are
made thus:
7 × 8 = 7 × 4 × 2 = 7 × 2 × 4.
256 × 80 = 256 × 8 × 10 = 256 × 10 × 8.

EXERCISES.
Shew that 2 × 3 × 4 × 5 = 2 × 4 × 3 × 5 = 5 × 4 × 2 × 3, &c.
Shew that 18 × 100 = 18 × 57 + 18 × 43.

56. Articles (51) and (55) may be expressed in the following way, where by ab
we mean a taken b times; by abc, a taken b times, and the result taken c times.
ab = ba.
abc = acb = bca = bac, &c.
abc = a × (bc) = b × (ca) = c × (ab).
If we would say that the same results are produced by multiplying by b, c, and
d, one after the other, and by the product bcd at once, we write the following:
a × b × c × d = a × bcd.
The fact is, that if any numbers are to be multiplied together, the product of any
two or more may be formed, and substituted instead of those two or more; thus, the
product abcdef may be formed by multiplying

ab cde f
abf de c
abc def &c.

57. In order to multiply by 10, annex a cipher to the right hand of the
multiplicand. Thus, 10 times 2356 is 23560. To shew this, write 2356 at length
which is
2 thousands, 3 hundreds, 5 tens, and 6 units.
Take each of these parts ten times, which, by (52), is the same as multiplying the
whole number by 10, and it will then become
2 tens of thou. 3 tens of hun. 5 tens of tens, and 6 tens,
which is
2 ten-thou. 3 thous. 5 hun. and 6 tens.
This must be written 23560, because 6 is not to be 6 units, but 6 tens. Therefore
2356 × 10 = 23560.
In the same way you may shew, that in order to multiply by 100 you must affix
two ciphers to the right; to multiply by 1000 you must affix three ciphers, and so
on. The rule will be best caught from the following table:
13 × 10 = 130
13 × 100 = 1300
13 × 1000 = 13000
13 × 10000 = 130000
142 × 1000 = 142000
23700 × 10 = 237000
3040 × 1000 = 3040000
10000 × 100000 = 1000000000
58. I now shew how to multiply by one of the numbers, 2, 3, 4, 5, 6, 7, 8, or 9. I
do not include 1, because multiplying by 1, or taking the number once, is what is
meant by simply writing down the number. I want to multiply 1368 by 8. Write the
first number at full length, which is
1 thousand, 3 hundreds, 6 tens, and 8 units.
To multiply this by 8, multiply each of these parts by 8 (50) and (52), which
will give
8 thousands, 24 hundreds, 48 tens, and 64 units.

Now 64 units are written thus 64
48 tens 480
24 hundreds 2400
8 thousands 8000
Add these together, which gives 10944 as the product of 1368 and 8, or 1368 × 8 =
10944. By working a few examples in this way you will see for following rule.
59. I. Multiply the first figure of the multiplicand by the multiplier, write down
the units’ figure, and reserve the tens.
II. Do the same with the second figure of the multiplicand, and add to the
product the number of tens from the first; put down the units’ figure of this, and
reserve the tens.
III. Proceed in this way till you come to the last figure, and then write down the
whole number obtained from that figure.
IV. If there be a cipher in the multiplicand, treat it as if it were a number,
observing that 0 × 1 = 0, 0 × 2 = 0, &c.
60. In a similar way a number can be multiplied by a figure which is
accompanied by ciphers, as, for example, 8000. For 8000 is 8 × 1000, and therefore
(55) you must first multiply by 8 and then by 1000, which last operation (57) is
done by placing 3 ciphers on the right. Hence the rule in this case is, multiply by
the simple number, and place the number of ciphers which follow it at the right of
the product.

EXAMPLE.
Multiply 1679423800872
by 60000
100765428052320000

61. EXERCISES.
What is 1007360 × 7? Answer, 7051520.
123456789 × 9 + 10 and 123 × 9 + 4?—Ans. 1111111111 and 1111.
What is 136 × 3 + 129 × 4 + 147 × 8 + 27 × 3000?—Ans. 83100.
An army is made up of 33 regiments of infantry, each containing 800 men; 14
of cavalry, each containing 600 men; and 2 of artillery, each containing 300 men.

The enemy has 6 more regiments of infantry, each containing 100 more men; 3
more regiments of cavalry, each containing 100 men less; and 4 corps of artillery of
the same magnitude as those of the first: two regiments of cavalry and one of
infantry desert from the former to the latter. How many men has the second army
more than the first?—Answer, 13400.

62. Suppose it is required to multiply 23707 by 4567. Since 4567 is made up of
4000, 500, 60, and 7, by (53) we must multiply 23707 by each of these, and add the
products.

Now (58) 23707 × 7 is 165949
(60) 23707 × 60 is 1422420
23707 × 500 is 11853500
23707 × 4000 is 94828000
The sum of these is 108269869
which is the product required.
It will do as well if, instead of writing the ciphers at the end of each line, we
keep the other figures in their places without them. If we take away the ciphers, the
second line is one place to the left of the first, the third one place to the left of the
second, and so on. Write the multiplier and the multiplicand over these lines, and
the process will stand thus:

23707
4567
165949
142242
118535
94828
108269869

63. There is one more case to be noticed; that is, where there is a cipher in the
middle of the multiplier. The following example will shew that in this case nothing
more is necessary than to keep the first figure of each line in the column under the
figure of the multiplier from which that line arises. Suppose it required to multiply
365 by 101001. The multiplier is made up of 100000, 1000 and 1. Proceed as
before, and

365 × 1 is 365

(57) 365 × 1000 is 365000
365 × 100000 is 36500000
The sum of which is 36865365
and the whole process with the ciphers struck off is:

365
101001
365
365
365
36865365

64. The following is the rule in all cases:
I. Place the multiplier under the multiplicand, so that the units of one may be
under those of the other.
II. Multiply the whole multiplicand by each figure of the multiplier (59), and
place the unit of each line in the column under the figure of the multiplier from
which it came.
III. Add together the lines obtained by II. column by column.
65. When the multiplier or multiplicand, or both, have ciphers on the right hand,
multiply the two together without the ciphers, and then place on the right of the
product all the ciphers that are on the right both of the multiplier and multiplicand.
For example, what is 3200 × 3000? First, 3200 is 32 × 100, or one hundred times as
great as 32. Again, 32 × 13000 is 32 × 13, with three ciphers affixed, that is 416,
with three ciphers affixed, or 416000. But the product required must be 100 times
as great as this, or must have two ciphers affixed. It is therefore 41600000, having
as many ciphers as are in both multiplier and multiplicand.
66. When any number is multiplied by itself any number of times, the result is
called a power of that number. Thus:

6 is called the first power of 6
6×6 second power of 6
6×6×6 third power of 6
6×6×6×6 fourth power of 6
&c. &c.

The second and third powers are usually called the square and cube, which are
incorrect names, derived from certain connexions of the second and third power
with the square and cube in geometry. As exercises in multiplication, the following
powers are to be found.

Number
Square. Cube.
proposed.
972 944784 918330048
1008 1016064 1024192512
3142 9872164 31018339288
3163 10004569 31644451747
5555 30858025 171416328875
6789 46090521 312908547069
The fifth power of 36 is 60466176
fourth 50 6250000
fourth 108 136048896
fourth 277 5887339441
67. It is required to multiply a + b by c + d, that is, to take a + b as many times
as there are units in c + d. By (53) a + b must be taken c times, and d times, or the
product required is (a + b)c + (a + b)d. But (52) (a + b)c is ac + bc, and (a + b)d is
ad + bd; whence the product required is ac + bc + ad + bd; or,
(a + b)(c + d) = ac + bc + ad + bd.
By similar reasoning
(a - b)(c + d) is (a - b)c + (a - b)d; or,
(a - b)(c + d) = ac - bc + ad - bd.
To multiply a-b by c-d, first take a-b c times, which gives ac-bc. This is not
correct; for in taking it c times instead of c-d times, we have taken it d times too
many; or have made a result which is (a-b)d too great. The real result is therefore
ac-bc-(a -b)d. But (a-b)d is ad- bd, and therefore
(a - b)(c - d) = ac - bc - ad - bd
= ac - bc - ad + bd (41)
From these three examples may be collected the following rule for the
multiplication of algebraic quantities: Multiply each term of the multiplicand by

each term of the multiplier; when the two terms have both + or both-before them,
put + before their product; when one has + and the other-, put-before their product.
In using the first terms, which have no sign, apply the rule as if they had the sign +.
68. For example, (a + b)(a + b) gives aa + ab + ab + bb. But ab + ab is 2ab;
hence the square of a + b is aa + 2ab + bb. Again (a- b)(a-b) gives aa-ab-ab + bb.
But two subtractions of ab are equivalent to subtracting 2ab; hence the square of a-
b is aa-2ab + bb. Again, (a + b)(a-b) gives aa + ab-ab -bb. But the addition and
subtraction of ab makes no change; hence the product of a + b and a- b is aa-bb.
Again, the square of a + b + c + d or (a + b + c + d)(a + b + c + d) will be found
to be aa + 2ab + 2ac + 2ad + bb + 2bc + 2bd + cc + 2cd + dd; or the rule for
squaring such a quantity is: Square the first term, and multiply all that come after
by twice that term; do the same with the second, and so on to the end.

SECTION IV.
DIVISION.

69. Suppose I ask whether 156 can be divided into a number of parts each of
which is 13, or how many thirteens 156 contains; I propose a question, the solution
of which is called division. In this case, 156 is called the dividend, 13 the divisor,
and the number of parts required is the quotient; and when I find the quotient, I am
said to divide 156 by 13.
70. The simplest method of doing this is to subtract 13 from 156, and then to
subtract 13 from the remainder, and so on; or, in common language, to tell off 156
by thirteens. A similar process has already occurred in the exercises on subtraction,
Art. (46). Do this, and mark one for every subtraction that is made, to remind you
that each subtraction takes 13 once from 156, which operations will stand as
follows:

156
13 1
———
143
13 1
———
130
13 1
———
117
13 1

———
104
13 1
———
91
13 1
———
78
13 1
———
65
13 1
———
52
13 1
———
39
13 1
———
26
13 1
———
13
13 1
———
0

Begin by subtracting 13 from 156, which leaves 143. Subtract 13 from 143,
which leaves 130; and so on. At last 13 only remains, from which when 13 is
subtracted, there remains nothing. Upon counting the number of times which you
have subtracted 13, you find that this number is 12; or 156 contains twelve
thirteens, or contains 13 twelve times.
This method is the most simple possible, and might be done with pebbles. Of
these you would first count 156. You would then take 13 from the heap, and put
them into one heap by themselves. You would then take another 13 from the heap,
and place them in another heap by themselves; and so on until there were none left.
You would then count the number of heaps, which you would find to be 12.
71. Division is the opposite of multiplication. In multiplication you have a
number of heaps, with the same number of pebbles in each, and you want to know
how many pebbles there are in all. In division you know how many there are in all,

and how many there are to be in each heap, and you want to know how many heaps
there are.
72. In the last example a number was taken which contains an exact number of
thirteens. But this does not happen with every number. Take, for example, 159.
Follow the process of (70), and it will appear that after having subtracted 13 twelve
times, there remains 3, from which 13 cannot be subtracted. We may say then that
159 contains twelve thirteens and 3 over; or that 159, when divided by 13, gives a
quotient 12, and a remainder 3. If we use signs,
159 = 13 × 12 + 3.

EXERCISES.
146 = 24 × 6 + 2, or 146 contains six twenty-fours and 2 over.
146 = 6 × 24 + 2, or 146 contains twenty-four sixes and 2 over.
300 = 42 × 7 + 6, or 300 contains seven forty-twos and 6 over.
39624 = 7277 × 5 + 3239.
73. If a contain b q times with a remainder r, a must be greater than bq by r;
that is,
a = bq + r.
If there be no remainder, a = bq. Here a is the dividend, b the divisor, q the
quotient, and r the remainder. In order to say that a contains b q times, we write,
a/b = q, or a : b = q,
which in old books is often found written thus:
a ÷ b = q.
74. If I divide 156 into several parts, and find how often 13 is contained in each
of them, it is plain that 156 contains 13 as often as all its parts together. For
example, 156 is made up of 91, 39, and 26. Of these

91 contains 13 7 times,
39 contains 13 3 times,
26 contains 13 2 times;
therefore 91 + 39 + 26 contains 13 7 + 3 + 2 times, or 12 times.
Again, 156 is made up of 100, 50, and 6.

Now 100 contains 13 7 times and 9 over,
50 contains 13 3 times and 11 over,
6 contains 13 0 times[9] and 6 over.
Therefore 100 + 50 + 6 contains 13 7 + 3 + 0 times and 9 + 11 + 6 over; or 156
contains 13 10 times and 26 over. But 26 is itself 2 thirteens; therefore 156 contains
10 thirteens and 2 thirteens, or 12 thirteens.
75. The result of the last article is expressed by saying, that if
a = b + c + d, then
a b c d
= + +
m m m m
76. In the first example I did not take away 13 more than once at a time, in
order that the method might be as simple as possible. But if I know what is twice
13, 3 times 13, &c., I can take away as many thirteens at a time as I please, if I take
care to mark at each step how many I take away. For example, take away 13 ten
times at once from 156, that is, take away 130, and afterwards take away 13 twice,
or take away 26, and the process is as follows:

156
130 10 times 13.
26
26 2 times 13.
0

Therefore 156 contains 13 10 + 2, or 12 times.
Again, to divide 3096 by 18.

3096
1800 100 times 18.
1296
900 50 times 18.
396
360
36
36
0

Therefore 3096 contains 18 100 + 50 + 20 + 2, or 172 times.
77. You will now understand the following sentences, and be able to make
similar assertions of other numbers.
450 is 75 × 6; it therefore contains any number, as 5, 6 times as often as 75
contains it.

135 contains 3 more than 26 times; therefore,
Twice 135 ” 3 ” 52 or twice 26
10 times 135 ” 3 ” 260 or 10 times 26
50 times 135 ” 3 ” 1300 or 50 times 26
472 contains 18 more than 21 times; therefore,
4720 contains 18 more than 210 times,
47200 contains 18 more than 2100 times,
472000 contains 18 more than 21000 times,
32 contains 12 more than 2 times, and less than 3 times.
320 ” 12 ” 20 times, ” ” 30 times.
3200 ” 12 ” 200 times, ” ” 300 times.
32000 ” 12 ” 2000 times, ” ” 3000 times.
&c. &c. &c.
78. The foregoing articles contain the principles of division. The question now
is, to apply them in the shortest and most convenient way. Suppose it required to
divide 4068 by 18, or to find 4068/18 (23).
If we divide 4068 into any number of parts, we may, by the process followed in
(74), find how many times 18 is contained in each of these parts, and from thence
how many times it is contained in the whole. Now, what separation of 4068 into
parts will be most convenient? Observe that 4, the first figure of 4068, does not
contain 18; but that 40, the first and second figures together, does contain 18 more
than twice, but less than three times.[10] But 4068 (20) is made up of 40 hundreds,
and 68; of which, 40 hundreds (77) contains 18 more than 200 times, and less than
300 times. Therefore, 4068 also contains more than 200 times 18, since it must
contain 18 more times than 4000 does. It also contains 18 less than 300 times,
because 300 times 18 is 5400, a greater number than 4068. Subtract 18 200 times
from 4068; that is, subtract 3600, and there remains 468. Therefore, 4068 contains
18 200 times, and as many more times as 468 contains 18.

It remains, then, to find how many times 468 contains 18. Proceed exactly as
before. Observe that 46 contains 18 more than twice, and less than 3 times;
therefore, 460 contains it more than 20, and less than 30 times (77); as does also
468. Subtract 18 20 times from 468, that is, subtract 360; the remainder is 108.
Therefore, 468 contains 18 20 times, and as many more as 108 contains it. Now,
108 is found to contain 18 6 times exactly; therefore, 468 contains it 20 + 6 times,
and 4068 contains it 200 + 20 + 6 times, or 226 times. If we write down the process
that has been followed, without any explanation, putting the divisor, dividend, and
quotient, in a line separated by parentheses it will stand, as in example(A).
Let it be required to divide 36326599 by 1342 (B).
A. B.
)
18 4068 (200 + 20 + 6 )
1342 36326599 (20000 + 7000 + 60 + 9
3600 26840000
468 9486599
360 9394000
108 92599
108 80520
0 12079
12078
1
As in the previous example, 36326599 is separated into 36320000 and 6599; the
first four figures 3632 being separated from the rest, because it takes four figures
from the left of the dividend to make a number which is greater than the divisor.
Again, 36320000 is found to contain 1342 more than 20000, and less than 30000
times; and 1342 × 20000 is subtracted from the dividend, after which the remainder
is 9486599. The same operation is repeated again and again, and the result is found
to be, that there is a quotient 20000 + 7000 + 60 + 9, or 27069, and a remainder 1.
Before you proceed, you should now repeat the foregoing article at length in the
solution of the following questions. What are
10093874 66779922 2718218
, , ?
3207 114433 13352
the quotients of which are 3147, 583, 203; and the remainders 1445, 65483, 7762.
79. In the examples of the last article, observe, 1st, that it is useless to write
down the ciphers which are on the right of each subtrahend, provided that without

them you keep each of the other figures in its proper place: 2d, that it is useless to
put down the right hand figures of the dividend so long as they fall over ciphers,
because they do not begin to have any share in the making of the quotient until, by
continuing the process, they cease to have ciphers under them: 3d, that the quotient
is only a number written at length, instead of the usual way. For example, the first
quotient is 200 + 20 + 6, or 226; the second is 20000 + 7000 + 60 + 9, or 27069.
Strike out, therefore, all the ciphers and the numbers which come above them,
except those in the first line, and put the quotient in one line; and the two examples
of the last article will stand thus:

)
18 4068 (226 )
1342 36326599 (27069
36 2684
46 9486
36 9394
108 9259
108 8052
0 12079
12078
1
80. Hence the following rule is deduced:
I. Write the divisor and dividend in one line, and place parentheses on each side
of the dividend.
II. Take off from the left-hand of the dividend the least number of figures which
make a number greater than the divisor; find what number of times the divisor is
contained in these, and write this number as the first figure of the quotient.
III. Multiply the divisor by the last-mentioned figure, and subtract the product
from the number which was taken off at the left of the dividend.
IV. On the right of the remainder place the figure of the dividend which comes
next after those already separated in II.: if the remainder thus increased be greater
than the divisor, find how many times the divisor is contained in it; put this number
at the right of the first figure of the quotient, and repeat the process: if not, on the
right place the next figure of the dividend, and the next, and so on until it is greater;
but remember to place a cipher in the quotient for every figure of the dividend
which you are obliged to take, except the first.
V. Proceed in this way until all the figures of the dividend are exhausted.

In judging how often one large number is contained in another, a first and rough
guess may be made by striking off the same number of figures from both, and using
the results instead of the numbers themselves. Thus, 4,732 is contained in 14,379
about the same number of times that 4 is contained in 14, or about 3 times. The
reason is, that 4 being contained in 14 as often as 4000 is in 14000, and these last
only differing from the proposed numbers by lower denominations, viz. hundreds,
&c. we may expect that there will not be much difference between the number of
times which 14000 contains 4000, and that which 14379 contains 4732: and it
generally happens so. But if the second figure of the divisor be 5, or greater than 5,
it will be more accurate to increase the first figure of the divisor by 1, before trying
the method just explained. Nothing but practice can give facility in this sort of
guess-work.
81. This process may be made more simple when the divisor is not greater than
12, if you have sufficient knowledge of the multiplication table (50). For example, I
want to divide 132976 by 4. At full length the process stands thus:

)
4 132976 (33244
12
12
12
9
8
17
16
16
16
0

But you will recollect, without the necessity of writing it down, that 13 contains
4 three times with a remainder 1; this 1 you will place before 2, the next figure of
the dividend, and you know that 12 contains 4 3 times exactly, and so on. It will be
more convenient to write down the quotient thus:

)
4 132976
———
33244

While on this part of the subject, we may mention, that the shortest way to
multiply by 5 is to annex a cipher and divide by 2, which is equivalent to taking the

half of 10 times, or 5 times. To divide by 5, multiply by 2 and strike off the last
figure, which leaves the quotient; half the last figure is the remainder. To multiply
by 25, annex two ciphers and divide by 4. To divide by 25, multiply by 4 and strike
off the last two figures, which leaves the quotient; one fourth of the last two figures,
taken as one number, is the remainder. To multiply a number by 9, annex a cipher,
and subtract the number, which is equivalent to taking the number ten times, and
then subtracting it once. To multiply by 99, annex two ciphers and subtract the
number, &c.
In order that a number may be divisible by 2 without remainder, its units’ figure
must be an even number.[11] That it may be divisible by 4, its last two figures must
be divisible by 4. Take the example 1236: this is composed of 12 hundreds and 36,
the first part of which, being hundreds, is divisible by 4, and gives 12 twenty-fives;
it depends then upon 36, the last two figures, whether 1236 is divisible by 4 or not.
A number is divisible by 8 if the last three figures are divisible by 8; for every digit,
except the last three, is a number of thousands, and 1000 is divisible by 8; whether
therefore the whole shall be divisible by 8 or not depends on the last three figures:
thus, 127946 is not divisible by 8, since 946 is not so. A number is divisible by 3 or
9 only when the sum of its digits is divisible by 3 or 9. Take for example 1234; this
is
1 thousand, or 999 and 1
2 hundred, or twice 99 and 2
3 tens, or three times 9 and 3
and 4 or 4
Now 9, 99, 999, &c. are all obviously divisible by 9 and by 3, and so will be any
number made by the repetition of all or any of them any number of times. It
therefore depends on 1 + 2 + 3 + 4, or the sum of the digits, whether 1234 shall be
divisible by 9 or 3, or not. From the above we gather, that a number is divisible by 6
when it is even, and when the sum of its digits is divisible by 3. Lastly, a number is
divisible by 5 only when the last figure is 0 or 5.
82. Where the divisor is unity followed by ciphers, the rule becomes extremely
simple, as you will see by the following examples:

)
100 33429 (334
300
342
300
429

400
29

This is, then, the rule: Cut off as many figures from the right hand of the
dividend as there are ciphers. These figures will be the remainder, and the rest of
the dividend will be the quotient.

)
10 2717316
271731 and rem. 6.

Or we may prove these results thus: from (20), 2717316 is 271731 tens and 6;
of which the first contains 10 271731 times, and the second not at all; the quotient
is therefore 271731, and the remainder 6 (72). Again (20), 33429 is 334 hundreds
and 29; of which the first contains 100 334 times, and the second not at all; the
quotient is therefore 334, and the remainder 29.
83. The following examples will shew how the rule may be shortened when
there are ciphers in the divisor. With each example is placed another containing the
same process, all unnecessary figures being removed; and from the comparison of
the two, the rule at the end of this article is derived.

I. 1782000 ) 6424700000 (3605 )
1782 6424700 (3605
5346000 5346
10787000 10787
10692000 10692
9500000 9500
8910000 8910
590000 590000

)
II. 12300000 42176189300 (3428 123 ) 421761 (3428
36900000 369
52761893 527
49200000 492
35618930 356
24600000 246

110189300 1101
98400000 984
11789300 11789300
The rule, then, is: Strike out as many figures[12] from the right of the dividend as
there are ciphers at the right of the divisor. Strike out all the ciphers from the
divisor, and divide in the usual way; but at the end of the process place on the right
of the remainder all those figures which were struck out of the dividend.

84. EXERCISES.
Dividend. Divisor. Quotient. Remainder.
9694 47 206 12
175618 3136 56 2
23796484 130000 183 6484
14002564 1871 7484 0
310314420 7878 39390 0
3939040647 6889 571787 4
22876792454961 43046721 531441 0
Shew that
100 × 100 × 100 - 43 × 43 × 43
I. ———————————— = 100 × 100 + 100 × 43 + 43 × 43.
100 - 43

100 × 100 × 100 + 43 × 43 × 43
II. ———————————— = 100 × 100 - 100 × 43 + 43 × 43.
100 + 43

76 × 76 + 2 × 76 × 52 + 52 × 52
III. ———————————— = 76 + 52.
76 + 52

12 × 12 × 12 × 12 - 1
IV. 1 + 12 + 12 × 12 + 12 × 12 × 12 = ————————.
12 - 1

What is the nearest number to 1376429 which can be divided by 36300 without
remainder?—Answer, 1379400.
If 36 oxen can eat 216 acres of grass in one year, and if a sheep eat half as much
as an ox, how long will it take 49 oxen and 136 sheep together to eat 17550 acres?
—Answer, 25 years.
85. Take any two numbers, one of which divides the other without remainder;
for example, 32 and 4. Multiply both these numbers by any other number; for
example, 6. The products will be 192 and 24. Now, 192 contains 24 just as often as
32 contains 4. Suppose 6 baskets, each containing 32 pebbles, the whole number of
which will be 192. Take 4 from one basket, time after time, until that basket is
empty. It is plain that if, instead of taking 4 from that basket, I take 4 from each, the
whole 6 will be emptied together: that is, 6 times 32 contains 6 times 4 just as often
as 32 contains 4. The same reasoning applies to other numbers, and therefore we do
not alter the quotient if we multiply the dividend and divisor by the same number.
86. Again, suppose that 200 is to be divided by 50. Divide both the dividend and
divisor by the same number; for example, 5. Then, 200 is 5 times 40, and 50 is 5
times 10. But by (85), 40 divided by 10 gives the same quotient as 5 times 40
divided by 5 times 10, and therefore the quotient of two numbers is not altered by
dividing both the dividend and divisor by the same number.
87. From (55), if a number be multiplied successively by two others, it is
multiplied by their product. Thus, 27, first multiplied by 5, and the product
multiplied by 3, is the same as 27 multiplied by 5 times 3, or 15. Also, if a number
be divided by any number, and the quotient be divided by another, it is the same as
if the first number had been divided by the product of the other two. For example,
divide 60 by 4, which gives 15, and the quotient by 3, which gives 5. It is plain, that
if each of the four fifteens of which 60 is composed be divided into three equal
parts, there are twelve equal parts in all; or, a division by 4, and then by 3, is
equivalent to a division by 4 × 3, or 12.
88. The following rules will be better understood by stating them in an example.
If 32 be multiplied by 24 and divided by 6, the result is the same as if 32 had been
multiplied by the quotient of 24 divided by 6, that is, by 4; for the sixth part of 24
being 4, the sixth part of any number repeated 24 times is that number repeated 4
times; or, multiplying by 24 and dividing by 6 is equivalent to multiplying by 4.
89. Again, if 48 be multiplied by 4, and that product be divided by 24, it is the
same thing as if 48 were divided at once by the quotient of 24 divided by 4, that is,
by 6. For, every unit which is repeated 6 times in 48 is repeated 4 times as often, or
24 times, in 4 times 48, or the quotient of 48 and 6 is the same as the quotient of 48
× 4 and 6 × 4.

90. The results of the last five articles may be algebraically expressed thus:
ma a
= (85)
mb b
If n divide a and b without remainder,
a/n a
= (86)
b/n b
a/b a
= (87)
c bc
ab b
=a× (88)
c c
ac a
= (89)
b b/c
It must be recollected, however, that these have only been proved in the case
where all the divisions are without remainder.
91. When one number divides another without leaving any remainder, or is
contained an exact number of times in it, it is said to be a measure of that number,
or to measure it. Thus, 4 is a measure of 136, or measures 136; but it does not
measure 137. The reason for using the word measure is this: Suppose you have a
rod 4 feet long, with nothing marked upon it, with which you want to measure some
length; for example, the length of a street. If that street should happen to be 136 feet
in length, you will be able to measure it with the rod, because, since 136 contains 4
34 times, you will find that the street is exactly 34 times the length of the rod. But if
the street should happen to be 137 feet long, you cannot measure it with the rod; for
when you have measured 34 of the rods, you will find a remainder, whose length
you cannot tell without some shorter measure. Hence 4 is said to measure 136, but
not to measure 137. A measure, then, is a divisor which leaves no remainder.
92. When one number is a measure of two others, it is called a common measure
of the two. Thus, 15 is a common measure of 180 and 75. Two numbers may have
several common measures. For example, 360 and 168 have the common measures
2, 3, 4, 6, 24, and several others. Now, this question maybe asked: Of all the
common measures of 360 and 168, which is the greatest? The answer to this
question is derived from a rule of arithmetic, called the rule for finding the greatest
common measure, which we proceed to consider.

93. If one quantity measure two others, it measures their sum and difference.
Thus, 7 measures 21 and 56. It therefore measures 56 + 21 and 56-21, or 77 and 35.

This is only another way of saying what was said in (74).
94. If one number measure a second, it measures every number which the
second measures. Thus, 5 measures 15, and 15 measures 30, 45, 60, 75, &c.; all
which numbers are measured by 5. It is plain that if
15 contains 5 3 times,
30, or 15 + 15 contains 5 3 + 3 times, or 6 times,
45, or 15 + 15 + 15 contains 5 3 + 3 + 3 or 9 times;
and so on.
95. Every number which measures both the dividend and divisor measures the
remainder also. To shew this, divide 360 by 112. The quotient is 3, and the
remainder 24, that is (72) 360 is three times 112 and 24, or 360 = 112 × 3 + 24.
From this it follows, that 24 is the difference between 360 and 3 times 112, or 24 =
360-112 × 3. Take any number which measures both 360 and 112; for example, 4.
Then

4 measures 360,
4 measures 112, and therefore (94) measures 112 × 3,
or 112 + 112 + 112.

Therefore (93) it measures 360-112 × 3, which is the remainder 24. The same
reasoning may be applied to all other measures of 360 and 112; and the result is,
that every quantity which measures both the dividend and divisor also measures the
remainder. Hence, every common measure of a dividend and divisor is also a
common measure of the divisor and remainder.
96. Every common measure of the divisor and remainder is also a common
measure of the dividend and divisor. Take the same example, and recollect that 360
= 112 × 3 + 24. Take any common measure of the remainder 24 and the divisor 112;
for example, 8. Then

8 measures 24;
and 8 measures 112, and therefore (94) measures 112 × 3.

Therefore (93) 8 measures 112 × 3 + 24, or measures the dividend 360. Then
every common measure of the remainder and divisor is also a common measure of
the divisor and dividend, or there is no common measure of the remainder and
divisor which is not also a common measure of the divisor and dividend.

97. I. It is proved in (95) that the remainder and divisor have all the common
measures which are in the dividend and divisor.
II. It is proved in (96) that they have no others.
It therefore follows, that the greatest of the common measures of the first two is
the greatest of those of the second two, which shews how to find the greatest
common measure of any two numbers,[13] as follows:
98. Take the preceding example, and let it be required to find the g. c. m. of 360
and 112, and observe that
360 divided by 112 gives the remainder 24,
112 divided by 24 gives the remainder 16,
24 divided by 16 gives the remainder 8,
16 divided by 8 gives no remainder.
Now, since 8 divides 16 without remainder, and since it also divides itself
without remainder, 8 is the g. c. m. of 8 and 16, because it is impossible to divide 8
by any number greater than 8; so that, even if 16 had a greater measure than 8, it
could not be common to 16 and 8.

Therefore 8 is g. c. m. of 16 and 8,
(97) g. c. m. of 16 and 8 is g. c. m. of 24 and 16,
g. c. m. of 24 and 16 is g. c. m. of 112 and 24,
g. c. m. of 112 and 24 is g. c. m. of 360 and 112,
Therefore 8 is g. c. m. of 360 and 112.
The process carried on may be written down in either of the following ways:

)
112 360 (3
336

)
24 112 (4
96

)
16 24 (1
16

)
8 16 (2

16
0

112 360 3
96 336 4
16 24 1
16 16 2
0 8
The rule for finding the greatest common measure of two numbers is,
I. Divide the greater of the two by the less.
II. Make the remainder a divisor, and the divisor a dividend, and find another
remainder.
III. Proceed in this way until there is no remainder, and the last divisor is the
greatest common measure required.
99. You may perhaps ask how the rule is to shew when the two numbers have
no common measure. The fact is, that there are, strictly speaking, no such numbers,
because all numbers are measured by 1; that is, contain an exact number of units,
and therefore 1 is a common measure of every two numbers. If they have no other
common measure, the last divisor will be 1, as in the following example, where the
greatest common measure of 87 and 25 is found.

)
25 87 (3
75

)
12 25 (2
24

)
1 12 (12
12
0

EXERCISES.
Numbers. g. c. m.
6197 9521 1

Numbers. g. c. m.
58363 2602 1
5547 147008443 1849
6281 326041 571
28915 31495 5
1509 300309 3

What are 36 × 36 + 2 × 36 × 72 + 72 × 72
and 36 × 36 × 36 + 72 × 72 × 72;

and what is their greatest common measure?—Answer, 11664.

100. If two numbers be divisible by a third, and if the quotients be again
divisible by a fourth, that third is not the greatest common measure. For example,
360 and 504 are both divisible by 4. The quotients are 90 and 126. Now 90 and 126
are both divisible by 9, the quotients of which division are 10 and 14. By (87),
dividing a number by 4, and then dividing the quotient by 9, is the same thing as
dividing the number itself by 4 × 9, or by 36. Then, since 36 is a common measure
of 360 and 504, and is greater than 4, 4 is not the greatest common measure. Again,
since 10 and 14 are both divisible by 2, 36 is not the greatest common measure. It
therefore follows, that when two numbers are divided by their greatest common
measure, the quotients have no common measure except 1 (99). Otherwise, the
number which was called the greatest common measure in the last sentence is not
so in reality.
101. To find the greatest common measure of three numbers, find the g. c. m. of
the first and second, and of this and the third. For since all common divisors of the
first and second are contained in their g. c. m., and no others, whatever is common
to the first, second, and third, is common also to the third and the g. c. m. of the
first and second, and no others. Similarly, to find the g. c. m. of four numbers, find
the g. c. m. of the first, second, and third, and of that and the fourth.
102. When a first number contains a second, or is divisible by it without
remainder, the first is called a multiple of the second. The words multiple and
measure are thus connected: Since 4 is a measure of 24, 24 is a multiple of 4. The
number 96 is a multiple of 8, 12, 24, 48, and several others. It is therefore called a
common multiple of 8, 12, 24. 48, &c. The product of any two numbers is evidently
a common multiple of both. Thus, 36 × 8, or 288, is a common multiple of 36 and
8. But there are common multiples of 36 and 8 less than 288; and because it is

convenient, when a common multiple of two quantities is wanted, to use the least of
them, I now shew how to find the least common multiple of two numbers.
103. Take, for example, 36 and 8. Find their greatest common measure, which is
4, and observe that 36 is 9 × 4, and 8 is 2 × 4. The quotients of 36 and 8, when
divided by their greatest common measure, are therefore 9 and 2. Multiply these
quotients together, and multiply the product by the greatest common measure, 4,
which gives 9 × 2 × 4, or 72. This is a multiple of 8, or of 4 × 2 by (55); and also of
36 or of 4 × 9. It is also the least common multiple; but this cannot be proved to
you, because the demonstration cannot be thoroughly understood without more
practice in the use of letters to stand for numbers. But you may satisfy yourself that
it is the least in this case, and that the same process will give the least common
multiple in any other case which you may take. It is not even necessary that you
should know it is the least. Whenever a common multiple is to be used, any one
will do as well as the least. It is only to avoid large numbers that the least is used in
preference to any other.
When the greatest common measure is 1, the least common multiple of the two
numbers is their product.
The rule then is: To find the least common multiple of two numbers, find their
greatest common measure, and multiply one of the numbers by the quotient which
the other gives when divided by the greatest common measure. To find the least
common multiple of three numbers, find the least common multiple of the first two,
and find the least common multiple of that multiple and the third, and so on.

EXERCISES.
Least
Numbers proposed. common
multiple.
14, 21 42
16, 5, 24 240
1, 2, 3, 4, 5, 6, 7, 8, 9, 10 2520
6, 8, 11, 16, 20 2640
876, 864 63072
868, 854 52948
A convenient mode of finding the least common multiple of several numbers is
as follows, when the common measures are easily visible: Pick out a number of

common measures of two or more, which have themselves no divisors greater than
unity. Write them as divisors, and divide every number which will divide by one or
more of them. Bring down the quotients, and also the numbers which will not
divide by any of them. Repeat the process with the results, and so on until the
numbers brought down have no two of them any common measure except unity.
Then, for the least common multiple, multiply all the divisors by all the numbers
last brought down. For instance, let it be required to find the least common multiple
of all the numbers from 11 to 21.

)
2, 2, 3, 5, 7 11 12 13 14 15 16 17 18 19 20 21
11 1 13 1 1 4 17 3 19 1 1
There are now no common measures left in the row, and the least common
multiple required is the product of 2, 2, 3, 5, 7, 11, 13, 4, 17, 3, and 19; or
232792560.

SECTION V.
FRACTIONS.

104. Suppose it required to divide 49 yards into five equal parts, or, as it is called, to find the fifth part
of 49 yards. If we divide 45 by 5, the quotient is 9, and the remainder is 4; that is (72), 49 is made up of 5
times 9 and 4. Let the line a b represent 49 yards:

A————————————————————B
C ——————— I —
D ——————— K —
E ——————— L —
F ——————— M —
G ——————— N —

I K L M N
H | | | | | |
Take 5 lines, c, d, e, f, and g, each 9 yards in length, and the line h, 4 yards in length. Then, since 49 is
5 nines and 4, c, d, e, f, g, and h, are together equal to a b. Divide h, which is 4 yards, into five equal parts,
i, k, l, m, and n, and place one of these parts opposite to each of the lines, c, d, e, f, and g. It follows that
the ten lines, c, d, e, f, g, i, k, l, m, n, are together equal to a b, or 49 yards. Now d and k together are of
the same length as c and i together, and so are e and l, f and m, and g and n. Therefore, c and i together,
repeated 5 times, will be 49 yards; that is, c and i together make up the fifth part of 49 yards.
105. c is a certain number of yards, viz. 9; but i is a new sort of quantity, to which hitherto we have
never come. It is not an exact number of yards, for it arises from dividing 4 yards into 5 parts, and taking
one of those parts. It is the fifth part of 4 yards, and is called a fraction of a yard. It is written thus, ⁴/
₅(23), and is what we must add to 9 yards in order to make up the fifth part of 49 yards.
The same reasoning would apply to dividing 49 bushels of corn, or 49 acres of land, into 5 equal parts.
We should find for the fifth part of the first, 9 bushels and the fifth part of 4 bushels; and for the second, 9
acres and the fifth part of 4 acres.
We say, then, once for all, that the fifth part of 49 is 9 and ⁴/₅, or 9 + ⁴/₅; which is usually written (9⁴/₅),
or if we use signs, 49/5 = (9⁴/₅).

EXERCISES.
What is the seventeenth part of 1237?—Answer, (72-¹³/₁₇).

10032 663819 22773399
What are ———, ———, and ———— ?
1974 23710 2424

Answer, 162 23649 and 2343 .
(5 ——), (27 ———), (9394 ——)

1974 23710 2424
106. By the term fraction is understood a part of any number, or the sum of any of the equal parts into
which a number is divided. Thus, ⁴⁹/₅, ⁴/₅, ²⁰/₇, are fractions. The term fraction even includes whole
numbers:[14] for example, 17 is ¹⁷/₁, ³⁴/₂, ⁵¹/₃, &c.
The upper number is called the numerator, the lower number is called the denominator, and both of
these are called terms of the fraction. As long as the numerator is less than the denominator, the fraction is
less than a unit: thus, ⁶/₁₇ is less than a unit; for 6 divided into 6 parts gives 1 for each part, and must give
less when divided into 17 parts. Similarly, the fraction is equal to a unit when the numerator and
denominator are equal, and greater than a unit when the numerator is greater than the denominator.
107. By ⅔ is meant the third part of 2. This is the same as twice the third part of 1.
To prove this, let a b be two yards, and divide each of the yards a c and c b into three equal parts.

| | | | | | |
A D E C F G B
Then, because a e, e f, and f b, are all equal to one another, a e is the third part of 2. It is therefore ⅔.
But a e is twice a d, and a d is the third part of one yard, or ⅓; therefore ⅔ is twice ⅓; that is, in order to
get the length ⅔, it makes no difference whether we divide two yards at once into three parts, and take one
of them, or whether we divide one yard into three parts, and take two of them. By the same reasoning, ⅝
may be found either by dividing 5 into 8 parts, and taking one of them, or by dividing 1 into 8 parts, and
taking five of them. In future, of these two meanings I shall use that which is most convenient at the time,
as it is proved that they are the same thing. This principle is the same as the following: The third part of
any number may be obtained by adding together the thirds of all the units of which it consists. Thus, the
third part of 2, or of two units, is made by taking one-third out of each of the units, that is,

⅔ = ⅓ × 2.
This meaning appears ambiguous when the numerator is greater than the denominator: thus, ¹⁵/₇ would
mean that 1 is to be divided into 7 parts, and 15 of them are to be taken. We should here let as many units
be each divided into 7 parts as will give more than 15 of those parts, and take 15 of them.
108. The value of a fraction is not altered by multiplying the numerator and denominator by the same
quantity. Take the fraction ¾, multiply its numerator and denominator by 5, and it becomes ¹⁵/₂₀, which is
the same thing as ¾; that is, one-twentieth part of 15 yards is the same thing as one-fourth of 3 yards: or, if
our second meaning of the word fraction be used, you get the same length by dividing a yard into 20 parts
and taking 15 of them, as you get by dividing it into 4 parts and taking 3 of them. To prove this,

let a b represent a yard; divide it into 4 equal parts, a c, c d, d e, and e b, and divide each of these parts
into 5 equal parts. Then a e is ¾. But the second division cuts the line into 20 equal parts, of which a e
contains 15. It is therefore ¹⁵/₂₀. Therefore, ¹⁵/₂₀ and ¾ are the same thing.

Again, since ¾ is made from ¹⁵/₂₀ by dividing both the numerator and denominator by 5, the value of a
fraction is not altered by dividing both its numerator and denominator by the same quantity. This
principle, which is of so much importance in every part of arithmetic, is often used in common language,
as when we say that 14 out of 21 is 2 out of 3, &c.
109. Though the two fractions ¾ and ¹⁵/₂₀ are the same in value, and either of them may be used for
the other without error, yet the first is more convenient than the second, not only because you have a
clearer idea of the fourth of three yards than of the twentieth part of fifteen yards, but because the numbers
in the first being smaller, are more convenient for multiplication and division. It is therefore useful, when
a fraction is given, to find out whether its numerator and denominator have any common divisors or
common measures. In (98) was given a rule for finding the greatest common measure of any two numbers;
and it was shewn that when the two numbers are divided by their greatest common measure, the quotients
have no common measure except 1. Find the greatest common measure of the terms of the fraction, and
divide them by that number. The fraction is then said to be reduced to its lowest terms, and is in the state
in which the best notion can be formed of its magnitude.

EXERCISES.
With each fraction is written the same reduced to its lowest terms.

2794 22 × 127 22
= =
2921 23 × 127 23

2788 17 × 164 17
= =
4920 30 × 164 30

93280 764 × 122 764
= =
13786 113 × 122 113

888800 22 × 40400 22
= =
40359600 999 × 40400 999

95469 121 × 789 121
= =
359784 456 × 789 456
110. When the terms of the fraction given are already in factors,[15] any one factor in the numerator
may be divided by a number, provided some one factor in the denominator is divided by the same. This
follows from (88) and (108). In the following examples the figures altered by division are accented.

12 × 11 × 10 3′ × 11 × 10 1′ × 11 × 5′
= = = 55
2×3×4 2 × 3 × 1′ 1′ × 1′ × 1′

18 × 15 × 13 2′ × 3′ × 1′ 1′ × 1′ × 1′
= = = ¹/₁₆.
20 × 54 × 52 4′ × 6′ × 4′ 2′ × 2′ × 4′

27 × 28 3′ × 4′ 3′ × 2′
= = = ⁶/₅.
9 × 70 1′ × 10′ 1′ × 5′
111. As we can, by (108), multiply the numerator and denominator of a fraction by any number,
without altering its value, we can now readily reduce two fractions to two others, which shall have the
same value as the first two, and which shall have the same denominator. Take, for example, ⅔ and ⁴/₇;
multiply both terms of ⅔ by 7, and both terms of ⁴/₇ by 3. It then appears that

2×7
⅔ is or ¹⁴/₂₁
3×7

4×3
⁴/₇ is or ¹²/₂₁
7×3
Here are then two fractions ¹⁴/₂₁ and ¹²/₂₁, equal to ⅔ and ⁴/₇, and having the same denominator, 21; in
this case, ⅔ and ⁴/₇ are said to be reduced to a common denominator.
It is required to reduce ⅒, ⅚, and ⁷/₉ to a common denominator. Multiply both terms of the first by the
product of 6 and 9; of the second by the product of 10 and 9; and of the third by the product of 10 and 6.
Then it appears (108) that

1×6×9
⅒ is or ⁵⁴/₅₄₀.
10 × 6 × 9

5 × 10 × 9
⅚ is or ⁴⁵⁰/₅₄₀.
6 × 10 × 9

7 × 10 × 6
⁷/₉ is or ⁴²⁰/₅₄₀.
9 × 10 × 6
On looking at these last fractions, we see that all the numerators and the common denominator are
divisible by 6, and (108) this division will not alter their values. On dividing the numerators and
denominators of ⁵⁴/₅₄₀, ⁴⁵⁰/₅₄₀, and ⁴²⁰/₅₄₀ by 6, the resulting fractions are, ⁹/₉₀, ⁷⁵/₉₀, and ⁷⁰/₉₀. These are
fractions with a common denominator, and which are the same as ⅒, ⅚, and ⁷/₉; and therefore these are a
more simple answer to the question than the first fractions. Observe also that 540 is one common multiple
of 10, 6, and 9, namely, 10 × 6 × 9, but that 90 is the least common multiple of 10, 6, and 9 (103). The
following process, therefore, is better. To reduce the fractions ⅒, ⅚, and ⁷/₉, to others having the same
value and a common denominator, begin by finding the least common multiple of 10, 6, and 9, by the rule
in (103), which is 90. Observe that 10, 6, and 9 are contained in 90 9, 15, and 10 times. Multiply both
terms of the first by 9, of the second by 15, and of the third by 10, and the fractions thus produced are ⁹/₉₀,
⁷⁵/₉₀, and ⁷⁰/₉₀, the same as before.
If one of the numbers be a whole number, it may be reduced to a fraction having the common
denominator of the rest, by (106).

EXERCISES.

Fractions proposed reduced to a common denominator.

2 1 1 20 6 5
3 5 6 30 30 30

1 2 3 12 3 28 24 18 48 63
3 7 14 21 4 84 84 84 84 84

3 4 5 6 3000 400 50 6
10 100 1000 1000 1000 1000 1000

33 281 22341 106499
379 677 256583 256583
112. By reducing two fractions to a common denominator, we are able to compare them; that is, to tell
which is the greater and which the less of the two. For example, take ½ and ⁷/₁₅. These fractions reduced,
without alteration of their value, to a common denominator, are ¹⁵/₃₀ and ¹⁴/₃₁. Of these the first must be
the greater, because (107) it may be obtained by dividing 1 into 30 equal parts and taking 15 of them, but
the second is made by taking 14 of those parts.
It is evident that of two fractions which have the same denominator, the greater has the greater
numerator; and also that of two fractions which have the same numerator, the greater has the less
denominator. Thus, ⁸/₇ is greater than ⁸/⁹, since the first is a 7th, and the last only a 9th part of 8. Also, any
numerator may be made to belong to as small a fraction as we please, by sufficiently increasing the
denominator. Thus, ¹⁰/₁₀₀ is ¹/₁₀, ¹⁰/₁₀₀₀ is ¹/₁₀₀, and ¹⁰/₁₀₀₀₀₀₀ is ¹/₁₀₀₀₀₀₀ (108).
We can now also increase and diminish the first fraction by the second. For the first fraction is made
up of 15 of the 30 equal parts into which 1 is divided. The second fraction is 14 of those parts. The sum of
the two, therefore, must be 15 + 14, or 29 of those parts; that is, ½ + ⁷/₁₅ is ²⁹/₃₀. The difference of the two
must be 15-14, or 1 of those parts; that is, ½-⁷/₁₅ = ¹/₃₀.
113. From the last two articles the following rules are obtained:
I. To compare, to add, or to subtract fractions, first reduce them to a common denominator. When this
has been done, that is the greatest of the fractions which has the greatest numerator.
Their sum has the sum of the numerators for its numerator, and the common denominator for its
denominator.
Their difference has the difference of the numerators for its numerator, and the common denominator
for its denominator.

EXERCISES.
1 1 1 1 53
+ + - =
2 3 4 5 60

44 153 18329
- =
3 427 1282

8 3 4 1834
1 + 10 + 100 - 1000
=
1000

1 12 253
2- 7
+
13
=
91

1 8 94 3
+ + =
2 16 188 2

163 97 93066
- =
521 881 459001
114. Suppose it required to add a whole number to a fraction, for example, 6 to ⁴/₉. By (106) 6 is ⁵⁴/₉,
and ⁵⁴/₉ + ⁴/₉ is ⁵⁸/⁹; that is, 6 + ⁴/⁹, or as it is usually written, (6⁴/₉), is ⁵⁸/₉. The rule in this case is:
Multiply the whole number by the denominator of the fraction, and to the product add the numerator of
the fraction; the sum will be the numerator of the result, and the denominator of the fraction will be its
denominator. Thus, (3¼) = ¹³/₄, (22⁵/₉) = ²⁰³/₉, (74²/₅₅) = ⁴⁰⁷²/₅₅. This rule is the opposite of that in (105).
115. From the last rule it appears that
907 17230907
1723 10000 is 10000
,

225 667225
667 1000 is 1000
,

99 2300099
and 23 is ,
10000 10000
Hence, when a whole number is to be added to a fraction whose denominator is 1 followed by ciphers, the
number of which is not less than the number of figures in the numerator, the rule is: Write the whole
number first, and then the numerator of the fraction, with as many ciphers between them as the number of
ciphers in the denominator exceeds the number of figures in the numerator. This is the numerator of the
result, and the denominator of the fraction is its denominator. If the number of ciphers in the denominator
be equal to the number of figures in the numerator, write no ciphers between the whole number and the
numerator.

EXERCISES.
Reduce the following mixed quantities to fractions:
23707
1 100000 ,

6
2457 10 ,

2210
233 10000 .

116. Suppose it required to multiply ⅔ by 4. This by (48) is taking ⅔ four times; that is, finding ⅔ + ⅔
+ ⅔ + ⅔. This by (112) is ⁸/₃; so that to multiply a fraction by a whole number the rule is: Multiply the
numerator by the whole number, and let the denominator remain.
117. If the denominator of the fraction be divisible by the whole number, the rule may be stated thus:
Divide the denominator of the fraction by the whole number, and let the numerator remain. For example,
multiply ⁷/₃₆ by 6. This (116) is ⁴²/₃₆, which, since the numerator and denominator are now divisible by 6,
is (108) the same as ⁷/₆. It is plain that ⁷/₆ is made from ⁷/₃₆ in the manner stated in the rule.
118. Multiplication has been defined to be the taking as many of one number as there are units in
another. Thus, to multiply 12 by 7 is to take as many twelves as there are units in 7, or to take 12 as many
times as you must take 1 in order to make 7. Thus, what is done with 1 in order to make 7, is done with 12
to make 7 times 12. For example,

7 is 1 + 1 + 1 + 1 + 1 + 1 + 1
7 times 12 is 12 + 12 + 12 + 12 + 12 + 12 + 12.
When the same thing is done with two fractions, the result is still called their product, and the process
is still called multiplication. There is this difference, that whereas a whole number is made by adding 1 to
itself a number of times, a fraction is made by dividing 1 into a number of equal parts, and adding one of
these parts to itself a number of times. This being the meaning of the word multiplication, as applied to
fractions, what is ¾ multiplied by ⅞? Whatever is done with 1 in order to make ⅞ must now be done with
¾; but to make ⅞, 1 is divided into 8 parts, and 7 of them are taken. Therefore, to make ¾ × ⅞, ¾ must be
divided into 8 parts, and 7 of them must be taken. Now ¾ is, by (108), the same thing as ²⁴/₃₂. Since ²⁴/₃₂ is
made by dividing 1 into 32 parts, and taking 24 of them, or, which is the same thing, taking 3 of them 8
times, if ²⁴/₃₂ be divided into 8 equal parts, each of them is ³/₃₂; and if 7 of these parts be taken, the result
is ²¹/₃₂ (116): therefore ¾ multiplied by ⅞ is ²¹/₃₂; and the same reasoning may be applied to any other
fractions. But ²¹/₃₂ is made from ¾ and ⅞ by multiplying the two numerators together for the numerator,
and the two denominators for the denominator; which furnishes a rule for the multiplication of fractions.
119. If this product ²¹/₃₂ is to be multiplied by a third fraction, for example, by ⁵/₉, the result is, by the
same rule, ¹⁰⁵/₂₈₈; and so on. The general rule for multiplying any number of fractions together is
therefore:
Multiply all the numerators together for the numerator of the product, and all the denominators
together for its denominator.
120. Suppose it required to multiply together ¹⁵/₁₆ and ⁸/₁₀. The product may be written thus:

15 × 8 120
, and is, ,
16 × 10 160
which reduced to its lowest terms (109) is ¾. This result might have been obtained directly, by observing
that 15 and 10 are both measured by 5, and 8 and 16 are both measured by 8, and that the fraction may be
written thus:

3×5×8
2 × 8 × 2 × 5.

Divide both its numerator and denominator by 5 × 8 (108) and (87), and the result is at once ¾; therefore,
before proceeding to multiply any number of fractions together, if there be any numerator and any
denominator, whether belonging to the same fraction or not, which have a common measure, divide them
both by that common measure, and use the quotients instead of the dividends.
A whole number may be considered as a fraction whose denominator is 1; thus, 16 is ¹⁶/₁ (106); and
the same rule will apply when one or more of the quantities are whole numbers.

EXERCISES.
136 268 36448 18224
× = =
7470 919 6864930 3432465

1 2 3 4 1
× × × =
2 3 4 5 5

2 17 2
× =
17 45 45

2 13 241 6266
× × =
59 7 19 7874

13 601 7813
× =
461 11 5071
Fraction
Square. Cube.
proposed.
701 491401 344472101
158 24964 3944312

140 19600 2744000
141 19881 2803221

355 126025 44738875
113 12769 1442897
From 100 acres of ground, two-thirds of them are taken away; 50 acres are then added to the result,
and ⁵/₇ of the whole is taken; what number of acres does this produce?—Answer, (59¹¹/₂₁).
121. In dividing one whole number by another, for example, 108 by 9, this question is asked,—Can
we, by the addition of any number of nines, produce 108? and if so, how many nines will be sufficient for
that purpose?

Suppose we take two fractions, for example, ⅔ and ⅘ , and ask, Can we, by dividing ⅘ into some
number of equal parts, and adding a number of these parts together, produce ⅔? if so, into how many parts
must we divide ⅘ , and how many of them must we add together? The solution of this question is still
called the division of ⅔ by ⅘; and the fraction whose denominator is the number of parts into which ⅘ is
divided, and whose numerator is the number of them which is taken, is called the quotient. The solution of
this question is as follows: Reduce both these fractions to a common denominator (111), which does not
alter their value (108); they then become ¹⁰/₁₅ and ¹²/₁₅. The question now is, to divide ¹²/₁₅ into a number
of parts, and to produce ¹⁰/₁₅ by taking a number of these parts. Since ¹²/₁₅ is made by dividing 1 into 15
parts and taking 12 of them, if we divide ¹²/₁₅ into 12 equal parts, each of these parts is ¹/₁₅; if we take 10
of these parts, the result is ¹⁰/₁₅. Therefore, in order to produce ¹⁰/₁₅ or ⅔ (108), we must divide ¹²/₁₅ or ⅘
into 12 parts, and take 10 of them; that is, the quotient is ¹⁰/₁₂. If we call ⅔ the dividend, and ⅘ the divisor,
as before, the quotient in this case is derived from the following rule, which the same reasoning will shew
to apply to other cases:
The numerator of the quotient is the numerator of the dividend multiplied by the denominator of the
divisor. The denominator of the quotient is the denominator of the dividend multiplied by the numerator of
the divisor. This rule is the reverse of multiplication, as will be seen by comparing what is required in both
cases. In multiplying ⅘ by ¹⁰/₁₂, I ask, if out of ⅘ be taken 10 parts out of 12, how much of a unit is taken,
and the answer is ⁴⁰/⁶⁰, or ⅔. Again, in dividing ⅔ by ⅘, I ask what part of ⅘ is ⅔, the answer to which is
¹⁰/₁₂.
122. By taking the following instance, we shall see that this rule can be sometimes simplified. Divide
¹⁶/₃₃ by ²⁸/₁₅. Observe that 16 is 4 × 4, and 28 is 4 × 7; 33 is 3 × 11, and 15 is 3 × 5; therefore the two
fractions are

4×4 4×7
and ,
3 × 11 3×5
and their quotient, according to the rule, is

4×4×3×5
3 × 11 × 4 × 7,

in which 4 × 3 is found both in the numerator and denominator. The fraction is therefore (108) the same as

4×5 20
or
11 × 7 77
The rule of the last article, therefore, admits of this modification: If the two numerators or the two
denominators have a common measure, divide by that common measure, and use the quotients instead of
the dividends.
123. In dividing a fraction by a whole number, for example, ⅔ by 15, consider 15 as the fraction ¹⁵/₁.
The rule gives ²/⁴⁵ as the quotient. Therefore, to divide a fraction by a whole number, multiply the
denominator by that whole number.

EXERCISES.

Dividend. Divisor. Quotient.
41 63 41
33 11 189

467 907 47167
151 101 136957

7813 601 13
5071 11 461

What are ¹/₅ × ¹/₅ × ¹/₅ - ²/₁₇× ²/₁₇ × ²/₁₇
,
¹/₅ - ²/₁₇

and ⁸/₁₁ × ⁸/₁₁ - ³/₁₁ × ³/₁₁
?
⁸/₁₁ - ³/₁₁

Answer, 559
and 1.
7225
A can reap a field in 12 days, B in 6, and C in 4 days; in what time can they all do it together?[16]—
Answer, 2 days.
In what time would a cistern be filled by cocks which would separately fill it in 12, 11, 10, and 9
hours?—Answer, (2⁴⁵⁴/₇₆₃) hours.
124. The principal results of this section may be exhibited algebraically as follows; let a, b, c, &c.
stand for any whole numbers. Then
a 1
(107) = ×a
b a

a ma
(108) =
b ma
a c ad bc
(111) and are the same as and
b d bd bd
a b a+b
(112) + =
c c c
a b a-b
- =
c c c

a c ad + bc
(113) + =
b d bd
a - c = ad - bc

b d bd

a c ac
(118) × =
b d bd
a c a/b ad
(121) divᵈ. by or =
b d c/d bc
125. These results are true even when the letters themselves represent fractions. For example, take the
fraction

a/b
c/d

whose numerator and denominator are fractional, and multiply its numerator and denominator by the
fraction
e ae/bf
, which gives
f ce/df

aedf
which (121) is
bfce
which, dividing the numerator and denominator by ef (108), is

ad
bc

But the original fraction itself is

ad
bc

hence
a/b a/b × e/f
=
c/d c/d × e/f
which corresponds to the second formula[17] in (124). In a similar manner it may be shewn, that the other
formulæ of the same article are true when the letters there used either represent fractions, or are removed
and fractions introduced in their place. All formulæ established throughout this work are equally true
when fractions are substituted for whole numbers. For example (54), (m + n)a = ma + na. Let m, n, and a
be respectively the fractions
p r b
, , and
q s c
Then m + n is
p r ps + qr
+ , or
q s qs

and (m + n)a is
ps + qr b (ps + qr)b
× , or
qs c qsc

psb + qrb
or .
qsc
psb qrb pb rb
But this (112) is + , which is + ,
qsc qsc qc sc

psb pb qrb rb
since = , and = (103).
qsc qc qsc sc

pb p b rb r b
But = × , and ,= × .
qc q c sc s c
Therefore (m + n)a, or

( pq + rs ) bc = pq × bc + rs × bc .
In a similar manner the same may be proved of any other formula.
The following examples may be useful:
a c e g
× + ×
b d f h acfh + bdeg
———————— = —————
a e c g
× + × aedh + bcfg
b f d h

1 b
=
1 ab + 1
a+—
b

1 1 bc + 1
———————— = —————— = ——————
1 c abc + a + c
a + ———— a + ———
1 bc + 1
b+ —
c
Thus,
1 = 1 = 57

———————— —————— ———
1 8 350
6 + ———— 6 + ——
1 57
7+ —
8
The rules that have been proved to hold good for all numbers may be applied when the numbers are
represented by letters.

SECTION VI.
DECIMAL FRACTIONS.

126. We have seen (112) (121) the necessity of reducing fractions to a common denominator, in order to
compare their magnitudes. We have seen also how much more readily operations are performed upon fractions
which have the same, than upon those which have different, denominators. On this account it has long been
customary, in all those parts of mathematics where fractions are often required, to use none but such as either
have, or can be easily reduced to others having, the same denominators. Now, of all numbers, those which can
be most easily managed are such as 10, 100, 1000, &c., where 1 is followed by ciphers. These are called decimal
numbers; and a fraction whose denominator is any one of them, is called a decimal fraction, or more
commonly, a decimal.
127. A whole number may be reduced to a decimal fraction, or one decimal fraction to another, with the
greatest ease. For example,
940 9400 94000
94 is , or , or (106);
10 100 1000

3 30 300 3000
is , or , or (108).
30 100 1000 10000
The placing of a cipher on the right hand of any number is the same thing as multiplying that number by 10
(57), and this may be done as often as we please in the numerator of a fraction, provided it be done as often in
the denominator (108).
128. The next question is, How can we reduce a fraction which is not decimal to another which is, without
altering its value? Take, for example, the fraction ⁷/₁₆, multiply both the numerator and denominator
successively by 10, 100, 1000, &c., which will give a series of fractions, each of which is equal to ⁷/₁₆ (108), viz.
⁷⁰/₁₆₀, ⁷⁰⁰/₁₆₀₀, ⁷⁰⁰⁰/₁₆₀₀₀, ⁷⁰⁰⁰⁰/₁₆₀₀₀₀, &c. The denominator of each of these fractions can be divided without
remainder by 16, the quotients of which divisions form the series of decimal numbers 10, 100, 1000, 10000, &c.
If, therefore, one of the numerators be divisible by 16, the fraction to which that numerator belongs has a
numerator and denominator both divisible by 16. When that division has been made, which (108) does not alter
the value of the fraction, we shall have a fraction whose denominator is one of the series 10, 100, 1000, &c., and
which is equal in value to ⁷/₁₆. The question is then reduced to finding the first of the numbers 70, 700, 7000,
70000, &c., which can be divided by 16 without remainder.
Divide these numbers, one after the other, by 16, as follows:

16) 70(4 16) 700(43 16) 7000(437 16) 70000(4375
64 64 64 64
6 60 60 60
48 48 48
12 120 120
112 112
8 80
80
0
It appears, then, that 70000 is the first of the numerators which is divisible by 16. But it is not necessary to
write down each of these divisions, since it is plain that the last contains all which came before. It will do, then,
to proceed at once as if the number of ciphers were without end, to stop when the remainder is nothing, and then
count the number of ciphers which have been used. In this case, since 70000 is 16 × 4375,

70000 16 × 4375 4375
, which is , or ,
160000 16 × 10000 10000
gives the fraction required.
Therefore, to reduce a fraction to a decimal fraction, annex ciphers to the numerator, and divide by the
denominator until there is no remainder. The quotient will be the numerator of the required fraction, and the
denominator will be unity, followed by as many ciphers as were used in obtaining the quotient.

EXERCISES.
Reduce to decimal fractions

½, ¼, ²/₂₅, ¹/₅₀, ³⁹²⁷/₁₂₅₀, and ⁴⁵³/₆₂₅.
Answer, ⁵/₁₀, ²⁵/₁₀₀, ⁸/₁₀₀, ²/₁₀₀, ³¹⁴¹⁶/₁₀₀₀₀, and ⁷²⁴⁸/₁₀₀₀₀.
129. It will happen in most cases that the annexing of ciphers to the numerator will never make it divisible
by the denominator without remainder. For example, try to reduce ¹/₇ to a decimal fraction.

7)1000000000000000000, &c.
142857142857142857, &c.

The quotient here is a continual repetition of the figures 1, 4, 2, 8, 5, 7, in the same order; therefore ¹/₇
cannot be reduced to a decimal fraction. But, nevertheless, if we take as a numerator any number of figures from
the quotient 142857142857, &c., and as a denominator 1 followed by as many ciphers as were used in making
that part of the quotient, we shall get a fraction which differs very little from ¹/₇, and which will differ still less
from it if we put more figures in the numerator and more ciphers in the denominator.
Thus,
1 is less 1 3 which is not 1
by
10 than 7 70 so much as 10
14 1 2 1
100 7 700 100
142 1 6 1
1000 7 7000 1000
1428 1 4 1
10000 7 70000 10000
14285 1 5 1
100000 7 700000 100000
142857 1 1 1
1000000 7 7000000 1000000
&c. &c. &c. &c.
In the first column is a series of decimal fractions, which come nearer and nearer to ¹/₇, as the third column
shews. Therefore, though we cannot find a decimal fraction which is exactly ¹/₇, we can find one which differs
from it as little as we please.
This may also be illustrated thus: It is required to reduce ¹/₇ to a decimal fraction without the error of say a
millionth of a unit; multiply the numerator and denominator of ¹/₇ by a million, and then divide both by 7; we
have then
1 1000000 1428571¹/₇
= =
7 7000000 1000000

If we reject the fraction ¹/₇ in the numerator, what we reject is really the 7th part of the millionth part of a
unit; or less than the millionth part of a unit. Therefore ¹⁴²⁸⁵⁷/₁₀₀₀₀₀₀ is the fraction required.

EXERCISES.
Make similar tables 3 17 1
, , and .
with these fractions 91 143 247

The recurring 3
is 329670,329670, &c.
quotient of 91
17
118881,118881, &c.
143
1
404858299595141700,4048582 &c.
247
130. The reason for the recurrence of the figures of the quotient in the same order is as follows: If 1000, &c.
be divided by the number 247, the remainder at each step of the division is less than 247, being either 0, or one
of the first 246 numbers. If, then, the remainder never become nothing, by carrying the division far enough, one
remainder will occur a second time. If possible, let the first 246 remainders be all different, that is, let them be 1,
2, 3, &c., up to 246, variously distributed. As the 247th remainder cannot be so great as 247, it must be one of
these which have preceded. From the step where the remainder becomes the same as a former remainder, it is
evident that former figures of the quotient must be repeated in the same order.
131. You will here naturally ask, What is the use of decimal fractions, if the greater number of fractions
cannot be reduced at all to decimals? The answer is this: The addition, subtraction, multiplication, and division
of decimal fractions are much easier than those of common fractions; and though we cannot reduce all common
fractions to decimals, yet we can find decimal fractions so near to each of them, that the error arising from using
the decimal instead of the common fraction will not be perceptible. For example, if we suppose an inch to be
divided into ten million of equal parts, one of those parts by itself will not be visible to the eye. Therefore, in
finding a length, an error of a ten-millionth part of an inch is of no consequence, even where the finest
measurement is necessary. Now, by carrying on the table in (129), we shall see that
1428571 1 1
does not differ from by ;
10000000 7 10000000
and if these fractions represented parts of an inch, the first might be used for the second, since the difference is
not perceptible. In applying arithmetic to practice, nothing can be measured so accurately as to be represented in
numbers without any error whatever, whether it be length, weight, or any other species of magnitude. It is
therefore unnecessary to use any other than decimal fractions, since, by means of them, any quantity may be
represented with as much correctness as by any other method.

EXERCISES.
Find decimal fractions which do not differ from the following fractions by ¹/₁₀₀₀₀₀₀₀₀.

⅓ ³³³³³³³³/₁₀₀₀₀₀₀₀₀.
Answer,
⁴/₇ ⁵⁷¹⁴²⁸⁵⁷/₁₀₀₀₀₀₀₀₀.
¹¹³/₃₅₅ ³¹⁸³⁰⁹⁸⁵/₁₀₀₀₀₀₀₀₀.
³⁵⁵/₁₁₃ ³¹⁴¹⁵⁹²⁹²/₁₀₀₀₀₀₀₀₀.
132. Every decimal may be immediately reduced to a quantity consisting either of a whole number and more
simple decimals, or of more simple decimals alone, having one figure only in each of the numerators. Take, for

example,
147326 147326 326
. By (115) is 147 ;
1000 1000 1000
and since 326 is made up of 300, and 20, and 6; by (112) ³²⁶/₁₀₀₀₀ = ³⁰⁰/₁₀₀₀ + ²⁰/₁₀₀₀ + ⁶/₁₀₀₀. But (108) ³⁰⁰/₁₀₀₀ is ³/
₁₀, and ²⁰/₁₀₀₀ is ²/₁₀₀. Therefore, ¹¹⁴⁷³²6/₁₀₀₀ is made up of 147 + ³/₁₀ + ²/₁₀₀ + 6/₁₀₀₀. Now, take any number, for
example, 147326, and form a number of fractions having for their numerators this number, and for their
denominators 1, 10, 100, 1000, 10000, &c., and reduce these fractions into numbers and more simple decimals,
in the foregoing manner, which will give the table below.

DECOMPOSITION OF A DECIMAL FRACTION.
147326
= 147326
1
147326 6
= 14732 +
10 10
147326 2 6
= 1473 + +
100 10 100
147326 3 2 6
= 147 + + +
1000 10 100 1000
147326 7 3 2 6
= 14 + + + +
10000 10 100 1000 10000
147326 4 7 3 2 6
= 1+ + + + +
100000 10 100 1000 10000 100000
147326 1 4 7 3 2 6
= + + + + +
1000000 10 100 1000 10000 100000 1000000
147326 1 4 7 3 2 6
= + + + + +
10000000 100 1000 10000 100000 1000000 10000000
N.B. The student should write this table himself, and then proceed to make
similar tables from the following exercises.

EXERCISES.
Reduce the following fractions into a series of numbers and more simple fractions:
31415926 31415926
, , &c.
10 100
2700031 2700031
, , &c.
10 100
2073000 2073000
, , &c.
10 100
3331303 3331303
, , &c.
1000 10000
133. If, in this table, and others made in the same manner, you look at those fractions which contain a whole
number, you will see that they may be made thus: Mark off, from the right hand of the numerator, as many
figures as there are ciphers in the denominator by a point, or any other convenient mark.
This will give 14732·6 when the fraction is 147326

10
147326
1473·26
100
147326
147·326
1000
&c. &c.
The figures on the left of the point by themselves make the whole number which the fraction contains. Of
those on its right, the first is the numerator of the fraction whose denominator is 10, the second of that whose
denominator is 100, and so on. We now come to those fractions which do not contain a whole number.
134. The first of these is ¹⁴⁷³²⁶/₁₀₀₀₀₀₀ which the number of ciphers in the denominator is the same as the
number of figures in the numerator. If we still follow the same rule, and mark off all the figures, by placing the
point before them all, thus, ·147326, the observation in (133) still holds good; for, on looking at ¹⁴⁷³²⁶/₁₀₀₀₀₀₀ in
the table, we find it is
1 4 7 3 2 6
+ + + + +
10 100 1000 10000 100000 1000000
The next fraction is ¹⁴⁷³²⁶/₁₀₀₀₀₀₀₀, which we find by the table to be
1 4 7 3 2 6
+ + + + +
100 1000 10000 100000 1000000 10000000
In this, 1 is not divided by 10, but by 100; if, therefore, we put a point before the whole, the rule is not true,
for the first figure on the left of the point has the denominator which, according to the rule, the second ought to
have, the second that which the third ought to have, and so on. In order to keep the same rule for this case, we
must contrive to make 1 the second figure on the right of the point instead of the first. This may be done by
placing a cipher between it and the point, thus, ·0147326. Here the rule holds good, for by that rule this fraction
is
0 1 4 7 3 2 6
+ + + + + +
10 100 1000 10000 100000 1000000 10000000
which is the same as the preceding line, since ⁰/₁₀ is 0, and need not be reckoned.
Similarly, when there are two ciphers more in the denominator than there are figures in the numerator, the
rule will be true if we place two ciphers between the point and the numerator. The rule, therefore, stated fully, is
this:
To reduce a decimal fraction to a whole number and more simple decimals, or to more simple decimals alone
if it do not contain a whole number, mark off by a point as many figures from the numerator as there are ciphers
in the denominator. If the numerator have not places enough for this, write as many ciphers before it as it wants
places, and put the point before these ciphers. Then, if there be any figures before the point, they make the whole
number which the fraction contains. The first figure after the point with the denominator 10, the second with the
denominator 100, and so on, are the fractions of which the first fraction is composed.
135. Decimal fractions are not usually written at full length. It is more convenient to write the numerator
only, and to cut off from the numerator as many figures as there are ciphers in the denominator, when that is
possible, by a point. When there are more ciphers in the denominator than figures in the numerator, as many
ciphers are placed before the numerator as will supply the deficiency, and the point is placed before the ciphers.
Thus, ·7 will be used in future to denote ⁷/₁₀, ·07 for ⁷/₁₀₀, and so on. The following tables will give the whole of
this notation at one view, and will shew its connexion with the decimal notation explained in the first section.
You will observe that the numbers on the right of the units’ place stand for units divided by 10, 100, 1000, &c.
while those on the left are units multiplied by 10, 100, 1000, &c.
The student is recommended always to write the decimal point in a line with the top of the figures or in the
middle, as is done here, and never at the bottom. The reason is, that it is usual in the higher branches of

mathematics to use a point placed between two numbers or letters which are multiplied together; thus, 15. 16, a.
b, (a + b). (c + d) stand for the products of those numbers or letters.
1234 4 4
I. 123·4 stands for or 123 or 123 +
10 10 10
1234 34 3 4
12·34 or 12 or 12 + +
100 100 10 100
1234 234 2 3 4
1·234 or 1 or 1 + + +
1000 1000 10 100 1000
1234 1 2 3 4
·1234 or + + +
10000 10 100 1000 10000
1234 1 2 3 4
·01234 or + + +
100000 100 1000 10000 100000
1234 1 2 3 4
·001234 or + + +
1000000 1000 10000 100000 1000000

1003 1 3
II. ·01003 is or +
100000 100 100000
1003 1 3
·1003 is or +
10000 10 10000
1003 3
10·03 is or 10 +
100 100
1003 3
100·3 is or 100 +
10 10

1 2 8 3
III. ·1238 = + + +
10 100 1000 10000
= ·1 + ·02 + ·008 + ·0003
= ·1 + ·0283 = ·12 + ·0083
= ·128 + ·0003 = ·108 + ·0203
= ·1003 + ·028 = ·1203 + ·008

1 is 1000 inches
IV.
2 is 200
3 is 30
4 is 4
In 1234·56789 5 is ⁵/₁₀ of an inch
inches the 6 is ⁶/₁₀₀
7 is ⁷/₁₀₀₀
8 is ⁸/₁₀₀₀₀
9 is ⁹/₁₀₀₀₀₀

136. The ciphers on the right hand of the decimal point serve the same purpose as the ciphers in (10). They
are not counted as any thing themselves, but serve to shew the place in which the accompanying numbers stand.
They might be dispensed with by writing the numbers in ruled columns, as in the first section. They are
distinguished from the numbers which accompany them by calling the latter significant figures. Thus, ·0003747
is a decimal of seven places with four significant figures, ·346 is a decimal of three places with three significant
figures, &c.
137. The value of a decimal is not altered by putting any number of ciphers on its right. Take, for example,
·3 and ·300. The first (135) is ³/₁₀, and the second ³⁰⁰/₁₀₀₀, which is made from the first by multiplying both its
numerator and denominator by 100, and (108) is the same quantity.
138. To reduce two decimals to a common denominator, put as many ciphers on the right of that which has
the smaller number of places as will make the number of places in both fractions the same. Take, for example,
·54 and 4·3297. The first is ⁵⁴/₁₀₀, and the second ⁴³²⁹⁷/₁₀₀₀₀. Multiply the numerator and denominator of the
first by 100 (108), which reduces it to ⁵⁴⁰⁰/₁₀₀₀₀, which has the same denominator as ⁴³²⁹⁷/₁₀₀₀₀. But ⁵⁴⁰⁰/₁₀₀₀₀
is ·5400 (135). In whole numbers, the decimal point should be placed at the end: thus, 129 should be written
129·. It is, however, usual to omit the point; but you must recollect that 129 and 129·000 are of the same value,
since the first is 129 and the second ¹²⁹⁰⁰⁰/₁₀₀₀.
139. The rules which were given in the last chapter for addition, subtraction, multiplication, and division,
apply to all fractions, and therefore to decimal fractions among the rest. But the way of writing decimal
fractions, which is explained in this chapter, makes the application of these rules more simple. We proceed to the
different cases.
Suppose it required to add 42·634, 45·2806, 2·001, and 54. By (112) these must be reduced to a common
denominator, which is done (138) by writing them as follows: 42·6340, 45·2806, 2·0010, and 54·0000. These
are decimal fractions, whose numerators are 426340, 452806, 20010, and 540000, and whose common
denominator is 10000. By (112) their sum is
426340 + 452806 + 20010 + 540000 1439156
, which is
10000 10000
or 143·9156. The simplest way of doing this is as follows: write the decimals down under one another, so that
the decimal points may fall under one another, thus:

42·634
45·2806
2·001
54
143·9156

Add the different columns together as in common addition, and place the decimal point under the other decimal
points.

EXERCISES.
What are 1527 + 64·732094 + 2·0013 + ·00001974;
2276·3 + ·107 + ·9 + 26·3172 + 56732·001;
and 1·11 + 7·7 + ·0039 + ·00142 + ·8838?

Answer, 1593·73341374, 59035·6252, 9·69912.
140. Suppose it required to subtract 91·07324 from 137·321. These fractions when reduced to a common
denominator are 91·07324 and 137·32100 (138). Their difference is therefore
13732100 - 9107324 4624776
, which is
100000 100000

or 46·24776. This may be most simply done as follows: write the less number under the greater, so that its
decimal point may fall under that of the greater, thus:

137·321
91·07324
46·24776

Subtract the lower from the upper line, and wherever there is a figure in one line and not in the other, proceed as
if there were a cipher in the vacant place.

EXERCISES.
What is 12362 - 274·22107 + ·5;
9976·2073942 - ·00143976728;
and 1·2 + ·03 + ·004 - ·0005?

Answer, 12088·27893, 9976·20595443272; and 1·2335.
141. The multiplication of a decimal by 10, 100, 1000, &c., is performed by merely moving the decimal
point to the right. Suppose, for example, 13·2079 is to be multiplied by 100. The decimal is ¹³²⁰⁷⁹/₁₀₀₀₀, which
multiplied by 100 is (117) ¹³²⁰⁷⁹/₁₀₀, or 1320·79. Again, 1·309 × 100000 is ¹³⁰⁹/₁₀₀₀ × 100000, or (116)
¹³⁰⁹⁰⁰⁰⁰⁰/₁₀₀₀ or 130900. From these and other instances we get the following rule: To multiply a decimal
fraction by a decimal number (126), move the decimal point as many places to the right as there are ciphers in
the decimal number. When this cannot be done, annex ciphers to the right of the decimal (137) until it can.
142. Suppose it required to multiply 17·036 by 4·27. The first of these decimals is ¹⁷⁰³⁶/₁₀₀₀, and the second
⁴²⁷/₁₀₀. By (118) the product of these fractions has for its numerator the product of 17036 and 427, and for its
denominator the product of 1000 and 100; therefore this product is ⁷²⁷⁴³⁷²/₁₀₀₀₀₀, or 72·74372. This may be
done more shortly by multiplying the two numbers 17036 and 427, and cutting off by the decimal point as many
places as there are decimal places both in 17·036 and 4·27, because the product of two decimal numbers will
contain as many ciphers as there are ciphers in both.
143. This question now arises: What if there should not be as many figures in the product as there are
decimal places in the multiplier and multiplicand together? To see what must be done in this case, multiply ·172
by ·101, or ¹⁷²/₁₀₀₀ by ¹⁰¹/₁₀₀₀. The product of these two is ¹⁷³⁷²/₁₀₀₀₀₀₀, or ·017372 (135). Therefore, when the
number of places in the product is not sufficient to allow the rule of the last article to be followed, as many
ciphers must be placed at the beginning as will make up the deficiency.

ADDITIONAL EXAMPLES.
·001 × ·01 is ·00001
56 × ·0001 is ·0056.

EXERCISES.
Shew that
3·002 × 3·002 = 3 × 3 + 2 × 3 × ·002 + ·002 × ·002
11·5609 × 5·3191 = 8·44 × 8·44 - 3·1209 × 3·1209
8·217 × 10·001 = 8 × 10 + 8 × ·001 + 10 × ·217 + ·001 × ·217.

Fraction. Square. Cube.
82·92 6875·7264 570135·233088
·0173 ·00029929 ·000005177717

Fraction. Square. Cube.
1·43 2·0449 2·924207
·009 ·000081 ·000000729

15·625 × 64 = 1000
1·5625 × ·64 = 1
·015625 × ·0064 = ·0001
·15625 × ·64 = ·1
1562·5 × ·064 = 100
15625000 × ·064 = 1000000
144. The division of a decimal by a decimal number, such as 10, 100, 1000, &c., is performed by moving the
decimal point as many places to the left as there are ciphers in the decimal number. If there are not places
enough in the dividend to allow of this, annex ciphers to the beginning of it until there are. For example, divide
1734·229 by 1000: the decimal fraction is ¹⁷³⁴²²⁹/₁₀₀₀, which divided by 1000 (123) is ¹⁷³⁴²²⁹/₁₀₀₀₀₀₀, or
1·734229. If, in the same way, 1·2106 be divided by 10000, the result is ·00012106.
145. Before proceeding to shorten the rule for the division of one decimal fraction by another, it will be
necessary to resume what was said in (128) upon the reduction of any fraction to a decimal fraction. It was there
shewn that ⁷/₁₆ is the same fraction as ⁴³⁷⁵/₁₀₀₀₀ or ·4375. As another example, convert ³/₁₂₈ into a decimal
fraction. Follow the same process as in (128), thus:

128)300000000000(234375
256
440
384
560
512
480
384
960
896
640
640
0

Since 7 ciphers are used, it appears that 30000000 is the first of the series 30, 300, &c., which is divisible by
128; and therefore ³/₁₂₈ or, which is the same thing (108), ³⁰⁰⁰⁰⁰⁰⁰/₁₂₈₀₀₀₀₀₀₀ is equal to ²³⁴³⁷⁵/₁₀₀₀₀₀₀₀ or
·0234375 (135).
From these examples the rule for reducing a fraction to a decimal is: Annex ciphers to the numerator; divide
by the denominator, and annex a cipher to each remainder after the figures of the numerator are all used,
proceeding exactly as if the numerator had an unlimited number of ciphers annexed to it, and was to be divided
by the denominator. Continue this process until there is no remainder, and observe how many ciphers have been
used. Place the decimal point in the quotient so as to cut off as many figures as you have used ciphers; and if
there be not figures enough for this, annex ciphers to the beginning until there are places enough.
146. From what was shewn in (129), it appears that it is not every fraction which can be reduced to a decimal
fraction. It was there shewn, however, that there is no fraction to which we may not find a decimal fraction as
near as we please. Thus, ¹/₁₀, ¹⁴/₁₀₀, ¹⁴²/₁₀₀₀, ¹⁴²⁸/₁₀₀₀₀, ¹⁴²⁸⁵/₁₀₀₀₀₀, &c., or ·1, ·14, ·142, ·1428, ·14285, were
shewn to be fractions which approach nearer and nearer to ¹/₇. To find either of these fractions, the rule is the
same as that in the last article, with this exception, that, I. instead of stopping when there is no remainder, which
never happens, stop at any part of the process, and make as many decimal places in the quotient as are equal in
number to the number of ciphers which have been used, annexing ciphers to the beginning when this cannot be

done, as before. II. Instead of obtaining a fraction which is exactly equal to the fraction from which we set out,
we get a fraction which is very near to it, and may get one still nearer, by using more of the quotient. Thus,
·1428 is very near to ¹/₇, but not so near as ·142857; nor is this last, in its turn, so near as ·142857142857, &c.
147. If there should be ciphers in the numerator of a fraction, these must not be reckoned with the number of
ciphers which are necessary in order to follow the rule for changing it into a decimal fraction. Take, for example,
¹⁰⁰/₁₂₅; annex ciphers to the numerator, and divide by the denominator. It appears that 1000 is divisible by 125,
and that the quotient is 8. One cipher only has been annexed to the numerator, and therefore 100 divided by 125
is ·8. Had the fraction been ¹/₁₂₅, since 1000 divided by 125 gives 8, and three ciphers would have been annexed
to the numerator, the fraction would have been ·008.
148. Suppose that the given fraction has ciphers at the right of its denominator; for example, ³¹/₂₅₀₀. Then
annexing a cipher to the numerator is the same thing as taking one away from the denominator; for, (108) ³¹⁰/₂₅₀₀
is the same thing as ³¹/₂₅₀, and ³¹⁰/₂₅₀ as ³¹/₂₅. The rule, therefore, is in this case: Take away the ciphers from the
denominator.

EXERCISES.
Reduce the following fractions to decimal fractions:
1 36 297 1
, , , and .
800 1250 64 128

Answer, ·00125, ·0288, 4·640625,
and·0078125.
Find decimals of 6 places very near to the following fractions:
27 156 22 194 2637 1 1 3
, , , , , , , and .
49 33 37000 13 9907 2908 466 277

Answer,·551020, 4·727272, ·000594, 14·923076, ·266175,
·000343, ·002145, and ·010830.
149. From (121) it appears, that if two fractions have the same denominator, the first may be divided by the
second by dividing the numerator of the first by the numerator of the second. Suppose it required to divide
17·762 by 6·25. These fractions (138), when reduced to a common denominator, are 17·762 and 6·250, or ¹⁷⁷⁶²/
₁₀₀₀ and ⁶²⁵⁰/₁₀₀₀. Their quotient is therefore ¹⁷⁷⁶²/₆₂₅₀, which must now be reduced to a decimal fraction by the
last rule. The process at full length is as follows: Leave out the cipher in the denominator, and annex ciphers to
the numerator, or, which will do as well, to the remainders, when it becomes necessary, and divide as in (145).

625)17762(284192
1250
5262
5000
2620
2500
1200
625
5750
5625
1250
1250
0

Here four ciphers have been annexed to the numerator, and one has been taken from the denominator. Make
five decimal places in the quotient, which then becomes 2·84192, and this is the quotient of 17·762 divided by
6·25.
150. The rule for division of one decimal by another is as follows: Equalise the number of decimal places in
the dividend and divisor, by annexing ciphers to that which has fewest places. Then, further, annex as many
ciphers to the dividend[18] as it is required to have decimal places, throw away the decimal point, and operate as
in common division. Make the required number of decimal places in the quotient.
Thus, to divide 6·7173 by ·014 to three decimal places, I first write 6·7173 and ·0140, with four places in
each. Having to provide for three decimal places, I should annex three ciphers to 6·7173; but, observing that the
divisor ·0140 has one cipher, I strike that one out and annex two ciphers to 6·7173. Throwing away the decimal
points, then divide 6717300 by 014 or 14 in the usual way, which gives the quotient 479807 and the remainder
2. Hence 479·807 is the answer.
The common rule is: Let the quotient contain as many decimal places as there are decimal places in the
dividend more than in the divisor. But this rule becomes inoperative except when there are more decimals in the
dividend than in the divisor, and a number of ciphers must be annexed to the former. The rule in the text
amounts to the same thing, and provides for an assigned number of decimal places. But the student is
recommended to make himself familiar with the rule of the characteristic given in the Appendix, and also to
accustom himself to reason out the place of the decimal point. Thus, it should be visible, that 26·119 ÷ 7·2436
has one figure before the decimal point, and that 26·119 ÷ 724·36 has one cipher after it, preceding all
significant figures.
Or the following rule may be used: Expunge the decimal point of the divisor, and move that of the dividend
as many places to the right as there were places in the divisor, using ciphers if necessary. Then proceed as in
common division, making one decimal place in the quotient for every decimal place of the final dividend which
is used. Thus 17·314 divided by 61·2 is 173·14 divided by 612, and the decimal point must precede the first
figure of the quotient. But 17·314 divided by 6617·5 is 173·14 by 66175; and since three decimal places of
173·14000 ... must be used before a quotient figure can be found, that quotient figure is the third decimal place,
or the quotient is ·002.....

EXAMPLES.
3·1 ·00062
= 1240, = ·00096875
·0025 ·64

EXERCISES.
Shew that
15·006 × 15·006 - ·004 × ·004
= 15·002,
15·01
and that
·01 × ·01 × ·01 + 2·9 × 2·9 × 2·9
= 2·9 × 2·9 - 2·9 × ·01 + ·01 × ·01
2·91
What are
1 1 365
, , and ,
3·14159 2·7182818 ·18349
as far as 6 places of decimals?—Answer, ·318310, ·367879, and 1989·209221.
Calculate 10 terms of each of the following series, as far as 5 places of decimals.

1+ 1 + 1 + 1 + 1 + &c. = ·71824.

2 2×3 2×3×4 2×3×4×5
1 1 1 1
1 + 2 + 3 + 4 + 5 + &c. = 2·92895.
80 81 82 83 84
+ + + + + &c. = 9·88286.
81 82 83 84 85
151. We now enter upon methods by which unnecessary trouble is saved in the computation of decimal
quantities. And first, suppose a number of miles has been measured, and found to be 17·846217 miles. If you
were asked how many miles there are in this distance, and a rough answer were required which should give
miles only, and not parts of miles, you would probably say 17. But this, though the number of whole miles
contained in the distance, is not the nearest number of miles; for, since the distance is more than 17 miles and 8
tenths, and therefore more than 17 miles and a half, it is nearer the truth to say, it is 18 miles. This, though too
great, is not so much too great as the other was too little, and the error is not so great as half a mile. Again, if the
same were required within a tenth of a mile, the correct answer is 17·8; for though this is too little by ·046217,
yet it is not so much too little as 17·9 is too great; and the error is less than half a tenth, or ¹/₂₀. Again, the same
distance, within a hundredth of a mile, is more correctly 17·85 than 17·84, since the last is too little by ·006217,
which is greater than the half of ·01; and therefore 17·84 + ·01 is nearer the truth than 17·84. Hence this general
rule: When a certain number of the decimals given is sufficiently accurate for the purpose, strike off the rest
from the right hand, observing, if the first figure struck off be equal to or greater than 5, to increase the last
remaining figure by 1.
The following are examples of a decimal abbreviated by one place at a time.
3·14159, 3·1416, 3·142, 3·14, 3·1, 3·0
2·7182818, 2·718282, 2·71828, 2·7183, 2·718, 2·72, 2·7, 3·0
1·9919, 1·992, 1·99, 2·00, 2·0
152. In multiplication and division it is useless to retain more places of decimals in the result than were
certainly correct in the multiplier, &c., which gave that result. Suppose, for example, that 9·98 and 8·96 are
distances in inches which have been measured correctly to two places of decimals, that is, within half a
hundredth of an inch each way. The real value of that which we call 9·98 may be any where between 9·975 and
9·985, and that of 8·96 may be any where between 8·955 and 8·965. The product, therefore, of the numbers
which represent the correct distances will lie between 9·975 × 8·955 and 9·985 × 8·965, that is, taking three
decimal places in the products, between 89·326 and 89·516. The product of the actual numbers given is
89·4208. It appears, then, that in this case no more than the whole number 89 can be depended upon in the
product, or, at most, the first place of decimals. The reason is, that the error made in measuring 8·96, though
only in the third place of decimals, is in the multiplication increased at least 9·975, or nearly 10 times; and
therefore affects the second place. The following simple rule will enable us to judge how far a product is to be
depended upon. Let a be the multiplier, and b the multiplicand; if these be true only to the first decimal place,
the product is within (a + b)/20[19] of the truth; if to two decimal places, within (a + b)/200; if to three, within (a
+ b)/2000; and so on. Thus, in the above example, we have 9·98 and 8·96, which are true to two decimal places:
their sum divided by 200 is ·0947, and their product is 89·4208, which is therefore within ·0947 of the truth. If,
in fact, we increase and diminish 89·4208 by ·0947, we get 89·5155 and 89·3261, which are very nearly the
limits found within which the product must lie. We see, then, that we cannot in this case depend upon the first
place of decimals, as (151) an error of ·05 cannot exist if this place be correct; and here is a possible error of ·09
and upwards. It is hardly necessary to say, that if the numbers given be exact, their product is exact also, and that
this article applies where the numbers given are correct only to a certain number of decimal places. The rule is:
Take half the sum of the multiplier and multiplicand, remove the decimal point as many places to the left as
there are correct places of decimals in either the multiplier or multiplicand; the result is the quantity within
which the product can be depended upon. In division, the rule is: Proceed as in the last rule, putting the dividend
and divisor in place of the multiplier and multiplicand, and divide by the square of the divisor; the quotient will
be the quantity within which the division of the first dividend and divisor may be depended upon. Thus, if
17·324 be divided by 53·809, both being correct to the third place, their half sum will be 35·566, which, by the
last rule, is made ·035566, and is to be divided by the square of 53·809, or, which will do as well for our

purpose, the square of 50, or 2500. The result is something less than ·00002, so that the quotient of 17·324 and
53·809 can be depended on to four places of decimals.
153. It is required to multiply two decimal fractions together, so as to retain in the product only a given
number of decimal places, and dispense with the trouble of finding the rest. First, it is evident that we may write
the figures of any multiplier in a contrary order (for example, 4321 instead of 1234), provided that in the
operation we move each line one place to the right instead of to the left, as in the following example:
2221 2221
1234 4321
8884 2221
6663 4442
4442 6663
2221 8884
2740714 2740714
Suppose now we wish to multiply 348·8414 by 51·30742, reserving only four decimal places in the product.
If we reverse the multiplier, and proceed in the manner just pointed out, we have the following:
3488414
2470315
17442070
3488414
1046524 2
24418 898
1395 3656
69 76828
17898·1522 23188
Cut off, by a vertical line, the first four places of decimals, and the columns which produced them. It is plain
that in forming our abbreviated rule, we have to consider only, I. all that is on the left of the vertical line; II. all
that is carried from the first column on the right of the line. On looking at the first column to the left of the line,
we see 4, 4, 8, 5, 9, of which the first 4 comes from 4 × 1′,[20] the second 4 from 1 × 3′, the 8 from 8 × 7′, the 5
from 8 × 4′, and the 9 from 4 × 2′. If, then, we arrange the multiplicand and the reversed multiplier thus,

3488414
2470315

each figure of the multiplier is placed under the first figure of the multiplicand which is used with it in forming
the first four places of decimals. And here observe, that the units’ figure in the multiplier 51·30742, viz. 1,
comes under 4, the fourth decimal place in the multiplicand. If there had been no carrying from the right of the
vertical line, the rule would have been: Reverse the multiplier, and place it under the multiplicand, so that the
figure which was the units’ figure in the multiplier may stand under the last place of decimals in the multiplicand
which is to be preserved; place ciphers over those figures of the multiplier which have none of the multiplicand
above them, if there be any: proceed to multiply in the usual way, but begin each figure of the multiplier with the
figure of the multiplicand which comes above it, taking no account of those on the right: place the first figures of
all the lines under one another. To correct this rule, so as to allow for what is carried from the right of the
vertical line, observe that this consists of two parts, 1st, what is carried directly in the formation of the different
lines, and 2dly, what is carried from the addition of the first column on the right. The first of these may be taken
into account by beginning each figure of the multiplier with the one which comes on its right in the
multiplicand, and carrying the tens to the next figure as usual, but without writing down the units. But both may
be allowed for at once, with sufficient correctness, on the principle of (151), by carrying 1 from 5 up to 15, 2
from 15 up to 25, &c.; that is, by carrying the nearest ten. Thus, for 37, 4 would be carried, 37 being nearer to 40

than to 30. This will not always give the last place quite correctly, but the error may be avoided by setting out so
as to keep one more place of decimals in the product than is absolutely required to be correct. The rule, then, is
as follows:
154. To multiply two decimals together, retaining only n decimal places.
I. Reverse the multiplier, strike out the decimal points, and place the multiplier under the multiplicand, so
that what was its units’ figure shall fall under the nᵗʰ decimal place of the multiplicand, placing ciphers, if
necessary, so that every place of the multiplier shall have a figure or cipher above it.
II. Proceed to multiply as usual, beginning each figure of the multiplier with the one which is in the place to
its right in the multiplicand: do not set down this first figure, but carry its nearest ten to the next, and proceed.
III. Place the first figures of all the lines under one another; add as usual; and mark off n places from the
right for decimals.
It is required to multiply 136·4072 by 1·30609, retaining 7 decimal places.

1364072000
906031
1364072000
409221600
8184432
122766
178·1600798

In the following examples the first two lines are the multiplicand and multiplier; and the number of decimals
to be retained will be seen from the results.
·4471618 33·166248 3·4641016
3·7719214 1·4142136 1732·508
37719214 033166248 346410160
8161744 63124141 8052371
15087686 3316625 346410160
1508768 1326650 242487112
264034 33166 10392305
3772 13266 692820
2263 663 173205
38 33 2771
30 10 6001·58373
1·6866591 2
46·90415
Exercises may be got from article (143).

155. With regard to division, take any two numbers, for example, 16·80437921 and 3·142, and divide the
first by the second, as far as any required number of decimal places, for example, five. This gives the following:
3·142)16·80437921(5·34830
15·710
1·0943
9426
15177
(A) 12568

2609 2609 9
2514 2513 6
95 96 32
94 94 26
1 2 061
Now cut off by a vertical line, as in (153), all the figures which come on the right of the first figure 2, in the
last remainder 2061. As in multiplication, we may obtain all that is on the left of the vertical line by an
abbreviated method, as represented at (A). After what has been said on multiplication, it is useless to go further
into the detail; the following rule will be sufficient: To divide one decimal by another, retaining only n places:
Proceed one step in the ordinary division, and determine, by (150), in what place is the quotient so obtained;
proceed in the ordinary way, until the number of figures remaining to be found in the quotient is less than the
number of figures in the divisor: if this should be already the case, proceed no further in the ordinary way.
Instead of annexing a figure or cipher to the remainder, cut off a figure from the divisor, and proceed one step
with this curtailed divisor as usual, remembering, however, in multiplying this divisor, to carry the nearest ten,
as in (154), from the figure which was struck off; repeat this, striking off another figure of the divisor, and so on,
until no figures are left. Since we know from the beginning in what place the first figure of the quotient is, and
also how many decimals are required, we can tell from the beginning how many figures there will be in the
whole quotient. If the divisor contain more figures than the quotient, it will be unnecessary to use them: and they
may be rejected, the rest being corrected as in (151): if there be ciphers at the beginning of the divisor, if it be,
for example,
·3178
·003178, since this is ,
100
divide by ·3178 in the usual way, and afterwards multiply the quotient by 100, or remove the decimal point two
places to the right. If, therefore, six decimals be required, eight places must be taken in dividing by ·3178, for an
obvious reason. In finding the last figure of the quotient, the nearest should be taken, as in the second of the
subjoined examples.
Places required, 2 8
Divisor, ·41432 3·1415927
Dividend, 673·1489 2·71828180
41432 2·51327416
258828 20500764
248592 18849556
10237 [21] 1651208
8286 1570796
1951 80412
1657 62832
294 17580
290 15708
4 1872
4 1571
0 301
283
18
19
Quotient, 1624·71 ·86525596

Examples may be obtained from (143) and (150).

SECTION VII.
ON THE EXTRACTION OF
THE SQUARE ROOT.

156. We have already remarked (66), that a number multiplied by itself
produces what is called the square of that number. Thus, 169, or 13 × 13, is
the square of 13. Conversely, 13 is called the square root of 169, and 5 is
the square root of 25; and any number is the square root of another, which
when multiplied by itself will produce that other. The square root is
signified by the sign
√ or √ ; thus, √25 means the square root of 25, or 5; √16 + 9
means the square root of 16 + 9, and is 5, and must not be confounded with
√16 + √9, which is 4 + 3, or 7.
157. The following equations are evident from the definition:
√a × √a = a
√aa = a
√ab × √ab = ab
(√a × √b) × (√a × √b) = √a × √a × √b × √b = ab
whence
√a × √b = √ab
158. It does not follow that a number has a square root because it has a
square; thus, though 5 can be multiplied by itself, there is no number which
multiplied by itself will produce 5. It is proved in algebra, that no
fraction[22] multiplied by itself can produce a whole number, which may be
found true in any number of instances; therefore 5 has neither a whole nor a
fractional square root; that is, it has no square root at all. Nevertheless, there
are methods of finding fractions whose squares shall be as near to 5 as we
please, though not exactly equal to it. One of these methods gives ¹⁵¹²⁷/₆₇₆₅,
whose square, viz.
15127× 15127 or 228826129 ,

6765 6765 45765225
differs from 5 by only ⁴/₄₅₇₆₅₂₂₅, which is less than ·0000001: hence we are
enabled to use √5 in arithmetical and algebraical reasoning: but when we
come to the practice of any problem, we must substitute for √5 one of the
fractions whose square is nearly 5, and on the degree of accuracy we want,
depends what fraction is to be used. For some purposes, ¹²³/₅₅ may be
sufficient, as its square only differs from 5 by ⁴/₃₀₂₅; for others, the fraction
first given might be necessary, or one whose square is even nearer to 5. We
proceed to shew how to find the square root of a number, when it has one,
and from thence how to find fractions whose squares shall be as near as we
please to the number, when it has not. We premise, what is sufficiently
evident, that of two numbers, the greater has the greater square; and that if
one number lie between two others, its square lies between the squares of
those others.
159. Let x be a number consisting of any number of parts, for example,
four, viz. a, b, c, and d; that is, let
x=a+b+c+d
The square of this number, found as in (68), will be

aa + 2a(b + c + d)
+ bb + 2b(c + d)
+ cc + 2cd
+ dd

The rule there found for squaring a number consisting of parts was:
Square each part, and multiply all that come after by twice that part, the
sum of all the results so obtained will be the square of the whole number. In
the expression above obtained, instead of multiplying 2a by each of the
succeeding parts, b, c, and d, and adding the results, we multiplied 2a by the
sum of all the succeeding parts, which (52) is the same thing; and as the
parts, however disposed, make up the number, we may reverse their order,
putting the last first, &c.; and the rule for squaring will be: Square each
part, and multiply all that come before by twice that part. Hence a reverse
rule for extracting the square root presents itself with more than usual

simplicity. It is: To extract the square root of a number N, choose a number
A, and see if N will bear the subtraction of the square of A; if so, take the
remainder, choose a second number B, and see if the remainder will bear
the subtraction of the square of B, and twice B multiplied by the preceding
part A: if it will, there is a second remainder. Choose a third number C, and
see if the second remainder will bear the subtraction of the square of C, and
twice C multiplied by A + B: go on in this way either until there is no
remainder, or else until the remainder will not bear the subtraction arising
from any new part, even though that part were the least number, which is 1.
In the first case, the square root is the sum of A, B, C, &c.; in the second,
there is no square root.
160. For example, I wish to know if 2025 has a square root. I choose 20
as the first part, and find that 400, the square of 20, subtracted from 2025,
gives 1625, the first remainder. I again choose 20, whose square, together
with twice itself, multiplied by the preceding part, is 20 × 20 + 2 × 20 × 20,
or 1200; which subtracted from 1625, the first remainder, gives 425, the
second remainder. I choose 7 for the third part, which appears to be too
great, since 7 × 7, increased by 2 × 7 multiplied by the sum of the preceding
parts 20 + 20, gives 609, which is more than 425. I therefore choose 5,
which closes the process, since 5 × 5, together with 2 × 5 multiplied by 20 +
20, gives exactly 425. The square root of 2025 is therefore 20 + 20 + 5, or
45, which will be found, by trial, to be correct; since 45 × 45 = 2025. Again,
I ask if 13340 has, or has not, a square root. Let 100 be the first part, whose
square is 10000, and the first remainder is 3340. Let 10 be the second part.
Here 10 × 10 + 2 × 10 × 100 is 2100, and the second remainder, or 3340-
2100, is 1240. Let 5 be the third part; then 5 × 5 + 2 × 5 × (100 + 10) is
1125, which, subtracted from 1240, leaves 115. There is, then, no square
root; for a single additional unit will give a subtraction of 1 × 1 + 2 × 1 ×
(100 + 10 + 5), or 231, which is greater than 115. But if the number
proposed had been less by 115, each of the remainders would have been 115
less, and the last remainder would have been nothing. Therefore 13340-115,
or 13225, has the square root 100 + 10 + 5, or 115; and the answer is, that
13340 has no square root, and that 13225 is the next number below it which
has one, namely, 115.
161. It only remains to put the rule in such a shape as will guide us to
those parts which it is most convenient to choose. It is evident (57) that any

number which terminates with ciphers, as 4000, has double the number of
ciphers in its square. Thus, 4000 × 4000 = 16000000; therefore, any square
number,[23] as 49, with an even number of ciphers annexed, as 490000, is a
square number. The root[24] of 490000 is 700. This being premised, take any
number, for example, 76176; setting out from the right hand towards the
left, cut off two figures; then two more, and so on, until one or two figures
only are left: thus, 7,61,76. This number is greater than 7,00,00, of which
the first figure is not a square number, the nearest square below it being 4.
Hence, 4,00,00 is the nearest square number below 7,00,00, which has four
ciphers, and its square root is 200. Let this be the first part chosen: its
square subtracted from 76176 leaves 36176, the first remainder; and it is
evident that we have obtained the highest number of the highest
denomination which is to be found in the square root of 76176; for 300 is
too great, its square, 9,00,00, being greater than 76176: and any
denomination higher than hundreds has a square still greater. It remains,
then, to choose a second part, as in the examples of (160), with the
remainder 36176. This part cannot be as great as 100, by what has just been
said; its highest denomination is therefore a number of tens. Let N stand for
a number of tens, which is one of the simple numbers 1, 2, 3, &c.; that is,
let the new part be 10N, whose square is 10N × 10N, or 100NN, and whose
double multiplied by the former part is 20N × 200, or 4000N; the two
together are 4000N + 100NN. Now, N must be so taken that this may not be
greater than 36176: still more 4000N must not be greater than 36176. We
may therefore try, for N, the number of times which 36176 contains 4000,
or that which 36 contains 4. The remark in (80) applies here. Let us try 9
tens or 90. Then, 2 × 90 × 200 + 90 × 90, or 44100, is to be subtracted,
which is too great, since the whole remainder is 36176. We then try 8 tens
or 80, which gives 2 × 80 × 200 + 80 × 80, or 38400, which is likewise too
great. On trying 7 tens, or 70, we find 2 × 70 × 200 + 70 × 70, or 32900,
which subtracted from 36176 gives 3276, the second remainder. The rest of
the square root can only be units. As before, let N be this number of units.
Then, the sum of the preceding parts being 200 + 70, or 270, the number to
be subtracted is 270 × 2N + NN, or 540N + NN. Hence, as before, 540N
must be less than 3276, or N must not be greater than the number of times
which 3276 contains 540, or (80) which 327 contains 54. We therefore try if
6 will do, which gives 2 × 6 × 270 + 6 × 6, or 3276, to be subtracted. This
being exactly the second remainder, the third remainder is nothing, and the

process is finished. The square root required is therefore 200 + 70 + 6, or
276.
The process of forming the numbers to be subtracted may be shortened
thus. Let A be the sum of the parts already found, and N a new part: there
must then be subtracted 2AN + NN, or (54) 2A + N multiplied by N. The
rule, therefore, for forming it is: Double the sum of all the preceding parts,
add the new part, and multiply the result by the new part.
162. The process of the last article is as follows:
7,61,76(200 7,61,76(276
4 00 00 70 4
400 3,61,76 6 47)361
70 ) 3 29 00 329
400 32 76 546)3276
140 ) 32 76 3276
6 0 0
In the first of these, the numbers are written at length, as we found
them; in the second, as in (79), unnecessary ciphers are struck off, and the
periods 61, 76, are not brought down, until, by the continuance of the
process, they cease to have ciphers under them. The following is another
example, to which the reasoning of the last article may be applied.
34,86,78,44,01 (50000 34,86,78,44,01(59049
25 00 00 00 00 9000 25
100000 9 86 78 44 01 40 109)986
9000 ) 9 81 00 00 00 9 981
100000 5 78 44 01 11804)57844
18000 ) 4 72 16 00 47216
40 —————— ————
100000 5 78 44 01
18000 ) 4 72 16 00
40 ——————

100000 1 06 28 01 118089)1062801
18000 1 06 28 01 1062801
80 ) —————— —————
9 0 0
163. The rule is as follows: To extract the square root of a number;—
I. Beginning from the right hand, cut off periods of two
figures each, until not more than two are left.
II. Find the root of the nearest square number next below
the number in the first period. This root is the first figure of
the required root; subtract its square from the first period,
which gives the first remainder.
III. Annex the second period to the right of the
remainder, which gives the first dividend.
IV. Double the first figure of the root; see how often this
is contained in the number made by cutting one figure from
the right of the first dividend, attending to IX., if necessary;
use the quotient as the second figure of the root; annex it to
the right of the double of the first figure, and call this the
first divisor.
V. Multiply the first divisor by the second figure of the
root; if the product be greater than the first dividend, use a
lower number for the second figure of the root, and for the
last figure of the divisor, until the multiplication just
mentioned gives the product less than the first dividend;
subtract this from the first dividend, which gives the second
remainder.
VI. Annex the third period to the second remainder,
which gives the second dividend.
VII. Double the first two figures of the root;[25] see how
often the result is contained in the number made by cutting
one figure from the right of the second dividend; use the
quotient as the third figure of the root; annex it to the right

of the double of the first two figures, and call this the second
divisor.
VIII. Get a new remainder, as in V., and repeat the
process until all the periods are exhausted; if there be then
no remainder, the square root is found; if there be a
remainder, the proposed number has no square root, and the
number found as its square root is the square root of the
proposed number diminished by the remainder.
IX. When it happens that the double of the figures of the
root is not contained at all in all the dividend except the last
figure, or when, being contained once, 1 is found to give
more than the dividend, put a cipher in the square root and in
the divisor, and bring down the next period; should the same
thing still happen, put another cipher in the root and divisor,
and bring down another period; and so on.

EXERCISES.
Numbers proposed. Square roots.
73441 271
2992900 1730
6414247921 80089
903687890625 950625
42420747482776576 205962976
13422659310152401 115856201
164. Since the square of a fraction is obtained by squaring the
numerator and the denominator, the square root of a fraction is found by
taking the square root of both. Thus, the square root of ²⁵/₆₄ is ⅝, since 5 × 5
is 25, and 8 × 8 is 64. If the numerator or denominator, or both, be not
square numbers, it does not therefore follow that the fraction has no square
root; for it may happen that multiplication or division by the same number
may convert both the numerator and denominator into square numbers
(108). Thus, ²⁷/₄₈, which appears at first to have no square root, has one in
reality, since it is the same as ⁹/₁₆, whose square root is ¾.

165. We now proceed from (158), where it was stated that any number
or fraction being given, a second may be found, whose square is as near to
the first as we please. Thus, though we cannot solve the problem, “Find a
fraction whose square is 2,” we can solve the following, “Find a fraction
whose square shall not differ from 2 by so much as ·00000001.” Instead of
this last, a still smaller fraction may be substituted; in fact, any one however
small: and in this process we are said to approximate to the square root of 2.
This can be done to any extent, as follows: Suppose we wish to find the
square root of 2 within ¹/₅₇ of the truth; by which I mean, to find a fraction
a/b whose square is less than 2, but such that the square of a/b + ¹/₅₇ is
greater than 2. Multiply the numerator and denominator of ²/₁ by the square
of 57, or 3249, which gives ⁶⁴⁹⁸/₃₂₄₉. On attempting to extract the square
root of the numerator, I find (163) that there is a remainder 98, and that the
square number next below 6498 is 6400, whose root is 80. Hence, the
square of 80 is less than 6498, while that of 81 is greater. The square root of
the denominator is of course 57. Hence, the square of ⁸⁰/⁵⁷ is less than ⁶⁴⁹⁸/
₃₂₄₉, or 2, while that of ⁸¹/₅₇ is greater, and these two fractions only differ
by ¹/₅₇; which was required to be done.
166. In practice, it is usual to find the square root true to a certain
number of places of decimals. Thus, 1·4142 is the square root of 2 true to
four places of decimals, since the square of 1·4142, or 1·99996164, is less
than 2, while an increase of only 1 in the fourth decimal place, giving
1·4143, gives the square 2·00024449, which is greater than 2. To take a
more general case: Suppose it required to find the square root of 1·637 true
to four places of decimals. The fraction is ¹⁶³⁷/₁₀₀₀, whose square root is to
be found within ·0001, or ¹/₁₀₀₀₀. Annex ciphers to the numerator and
denominator, until the denominator becomes the square of ¹/₁₀₀₀₀, which
gives ¹⁶³⁷⁰⁰⁰⁰⁰/₁₀₀₀₀₀₀₀₀, extract the square root of the numerator, as in
(163), which shews that the square number nearest to it is 163700000-
13564, whose root is 12794. Hence, ¹²⁷⁹⁴/₁₀₀₀₀, or 1·2794, gives a square
less than 1·637, while 1·2795 gives a square greater. In fact, these two
squares are 1·63686436 and 1·63712025.
167. The rule, then, for extracting the square root of a number or
decimal to any number of places is: Annex ciphers until there are twice as

many places following the units’ place as there are to be decimal places in
the root; extract the nearest square root of this number, and mark off the
given number of decimals. Or, more simply: Divide the number into
periods, so that the units’ figure shall be the last of a period; proceed in the
usual way; and if, when decimals follow the units’ place, there is one figure
on the right, in a period by itself, annex a cipher in bringing down that
period, and afterwards let each new period consist of two ciphers. Place the
decimal point after that figure in forming which the period containing the
units was used.
168. For example, what is the square root of (1⅜) to five places of
decimals? This is (145) 1·375, and the process is the first example over leaf.
The second example is the extraction of the root of ·081 to seven places, the
first period being 08, from which the cipher is omitted as useless.

1,37,5(1·17260
1
21) 37
21
227) 1650
1589
2342) 6100
4684
23446) 141600
140676
23452) 92400

8,1(·2846049
4
48)410
384
564) 2600
2256
5686) 34400
34116
569204) 2840000
2276816

569208) 56318400

·000002413672221(·001553599
1
25) 141
125
305) 1636
1525
3103) 11172
9309
31065) 186322
155325
310709) 3099710
2796381
30332900

169. When more than half the decimals required have been found, the
others may be simply found by dividing the dividend by the divisor, as in
(155). The extraction of the square root of 12 to ten places, which will be
found in the next page, is an example. It must, however, be observed in this
process, as in all others where decimals are obtained by approximation, that
the last place cannot always be depended upon: on which account it is
advisable to carry the process so far, that one or even two more decimals
shall be obtained than are absolutely required to be correct.

A
12(3·46410161513
9
64) 300
256
686) 4400
4116
6924) 28400
27696
69281) 70400
69281

6928201) 11190000
6928201
69282026) 4261799|00
4156921|56
692820321) 104877|4400
69282|0321
6928203225) 35595|407900
34641|016125
69282032301) 954|39177500
692|82032301
692820323023) 261|5714519900
207|8460969069
53|7253550831

B
692820323026)537253550831(77545870549
484974226118
52279324713
48497422611;
3781902102
3464101615
317800487
277128129
40672358
34641016
6031342
5542562
488780
484974
3806
3464
342
277
65
62
3

If from any remainder we cut off the ciphers, and all figures which
would come under or on the right of these ciphers, by a vertical line, we
find on the left of that line a contracted division, such as those in (155).
Thus, after having found the root as far as 3·464101, we have the remainder
4261799, and the divisor 6928202. The figures on the left of the line are
nothing more than the contracted division of this remainder by the divisor,
with this difference, however, that we have to begin by striking a figure off
the divisor, instead of using the whole divisor once, and then striking off the
first figure. By this alone we might have doubled our number of decimal
places, and got the additional figures 615137, the last 7 being obtained by
carrying the contracted division one step further with the remainder 53. We
have, then, this rule: When half the number of decimal places have been
obtained, instead of annexing two ciphers to the remainder, strike off a
figure from what would be the divisor if the process were continued at
length, and divide the remainder by this contracted divisor, as in (155).
As an example, let us double the number of decimal places already
obtained, which are contained in 3·46410161513. The remainder is
537253550831, the divisor 692820323026, and the process is as in (B).
Hence the square root of 12 is,
3·4641016151377545870549;
which is true to the last figure, and a little too great; but the substitution of 8
instead of 9 on the right hand would make it too small.

EXERCISES.
Numbers. Square roots.
·001728 ·0415692194
64·34 8·02122185
8074 89·8554394
10 3·16227766
1·57 1·2529964086141667788495

SECTION VIII.
ON THE PROPORTION OF NUMBERS.

170. When two numbers are named in any problem, it is usually
necessary, in some way or other, to compare the two; that is, by considering
the two together, to establish some connexion between them, which may be
useful in future operations. The first method which suggests itself, and the
most simple, is to observe which is the greater, and by how much it differs
from the other. The connexion thus established between two numbers may
also hold good of two other numbers; for example, 8 differs from 19 by 11,
and 100 differs from 111 by the same number. In this point of view, 8 stands
to 19 in the same situation in which 100 stands to 111, the first of both
couples differing in the same degree from the second. The four numbers
thus noticed, viz.:
8, 19, 100, 111,
are said to be in arithmetical[26] proportion. When four numbers are thus
placed, the first and last are called the extremes, and the second and third
the means. It is obvious that 111 + 8 = 100 + 19, that is, the sum of the
extremes is equal to the sum of the means. And this is not accidental,
arising from the particular numbers we have taken, but must be the case in
every arithmetical proportion; for in 111 + 8, by (35), any diminution of 111
will not affect the sum, provided a corresponding increase be given to 8;
and, by the definition just given, one mean is as much less than 111 as the
other is greater than 8.
171. A set or series of numbers is said to be in continued arithmetical
proportion, or in arithmetical progression, when the difference between
every two succeeding terms of the series is the same. This is the case in the
following series:

1, 2, 3, 4, 5, &c.
3, 6, 9, 12, 15, &c.
1½, 2, 2½, 3, 3½, &c.

The difference between two succeeding terms is called the common
difference. In the three series just given, the common differences are, 1, 3,
and ½.
172. If a certain number of terms of any arithmetical series be taken, the
sum of the first and last terms is the same as that of any other two terms,
provided one is as distant from the beginning of the series as the other is
from the end. For example, let there be 7 terms, and let them be,
a b c d e f g.
Then, since, by the nature of the series, b is as much above a as f is below g
(170), a + g = b + f. Again, since c is as much above b as e is below f (170),
b + f = c + e. But a + g = b + f; therefore a + g = c + e, and so on. Again,
twice the middle term, or the term equally distant from the beginning and
the end (which exists only when the number of terms is odd), is equal to the
sum of the first and last terms; for since c is as much below d as e is above
it, we have c + e = d + d = 2d. But c + e = a + g; therefore, a + g = 2d. This
will give a short rule for finding the sum of any number of terms of an
arithmetical series. Let there be 7, viz. those just given. Since a + g, b + f,
and c + e, are the same, their sum is three times (a + g), which with d, the
middle term, or half a + g, is three times and a half (a + g), or the sum of
the first and last terms multiplied by (3½), or ⁷/₂, or half the number of
terms. If there had been an even number of terms, for example, six, viz. a,
b, c, d, e, and f, we know now that a + f, b + e, and c + d, are the same,
whence the sum is three times (a + f), or the sum of the first and last terms
multiplied by half the number of terms, as before. The rule, then, is: To sum
any number of terms of an arithmetical progression, multiply the sum of the
first and last terms by half the number of terms. For example, what are 99
terms of the series 1, 2, 3, &c.? The 99th term is 99, and the sum is
99 100 × 99
(99 + 1) , or , or 4950.
2 2
The sum of 50 terms of the series
1 2 4 5 1 50 50
,
3 3
, 1, ,
3 3
, 2, &c. is (3 + 3 ) 2 ,

or 17 × 25, or 425.
173. The first term being given, and also the common difference and
number of terms, the last term may be found by adding to the first term the
common difference multiplied by one less than the number of terms. For it
is evident that the second term differs from the first by the common
difference, the third term by twice, the fourth term by three times the
common difference; and so on. Or, the passage from the first to the nth term
is made by n-1 steps, at each of which the common difference is added.

EXERCISES.
Given. To find.
Series. No. of terms. Last term. Sum.
4, 6½, 9, &c. 33 84 1452
1, 3, 5, &c. 28 55 784
2, 20, 38, &c. 100,000 1799984 89999300000
174. The sum being given, the number of terms, and the first term, we
can thence find the common difference. Suppose, for example, the first term
of a series to be one, the number of terms 100, and the sum 10,000. Since
10,000 was made by multiplying the sum of the first and last terms by ¹⁰⁰/₂,
if we divide by this, we shall recover the sum of the first and last terms.
Now, ¹⁰,⁰⁰⁰/₁ divided by ¹⁰⁰/₂ is (122) 200, and the first term being 1, the
last term is 199. We have then to pass from 1 to 199, or through 198, by 99
equal steps. Each step is, therefore, ¹⁹⁸/⁹⁹, or 2, which is the common
difference; or the series is 1, 3, 5, &c., up to 199.
Given. To find.
Sum. No. of terms. First term. Last term. Common diff.
1809025 1345 1 2689 2
29 14
44 10 3
5 45
7075600 1330 4 10636 8
175. We now return to (170), in which we compared two numbers
together by their difference. This, however, is not the method of comparison

which we employ in common life, as any single familiar instance will shew.
For example, we say of A, who has 10 thousand pounds, that he is much
richer than B, who has only 3 thousand; but we do not say that C, who has
107 thousand pounds, is much richer than D, who has 100 thousand, though
the difference of fortune is the same in both cases, viz. 7 thousand pounds.
In comparing numbers we take into our reckoning not only the differences,
but the numbers themselves. Thus, if B and D both received 7 thousand
pounds, B would receive 233 pounds and a third for every 100 pounds
which he had before, while D for every 100 pounds would receive only 7
pounds. And though, in the view taken in (170), 3 is as near to 10 as 100 is
to 107, yet, in the light in which we now regard them, 3 is not so near to 10
as 100 is to 107, for 3 differs from 10 by more than twice itself, while 100
does not differ from 107 by so much as one-fifth of itself. This is expressed
in mathematical language by saying, that the ratio or proportion of 10 to 3
is greater than the ratio or proportion of 107 to 100. We proceed to define
these terms more accurately.
176. When we use the term part of a number or fraction in the
remainder of this section, we mean, one of the various sets of equal parts
into which it may be divided, either the half, the third, the fourth, &c.: the
term multiple has been already explained (102). By the term multiple-part
of a number we mean, the abbreviation of the words multiple of a part.
Thus, 1, 2, 3, 4, and 6, are parts of 12; ½ is also a part of 12, being
contained in it 24 times; 12, 24, 36, &c., are multiples of 12; and 8, 9, ⁵/₂,
&c. are multiple parts of 12, being multiples of some of its parts. And when
multiple parts generally are spoken of, the parts themselves are supposed to
be included, on the same principle that 12 is counted among the multiples
of 12, the multiplier being 1. The multiples themselves are also included in
this term; for 24 is also 48 halves, and is therefore among the multiple parts
of 12. Each part is also in various ways a multiple-part; for one-fourth is
two-eighths, and three-twelfths, &c.
177. Every number or fraction is a multiple-part of every other number
or fraction. If, for example, we ask what part 12 is of 7, we see that on
dividing 7 into 7 parts, and repeating one of these parts 12 times, we obtain
12; or, on dividing 7 into 14 parts, each of which is one-half, and repeating
one of these parts 24 times, we obtain 24 halves, or 12. Hence, 12 is ¹²/₇, or
²⁴/₁₄, or ³⁶/₂₁ of 7; and so on. Generally, when a and b are two whole

numbers, a/b expresses the multiple-part which a is of b, and b/a that which
b is of a. Again, suppose it required to determine what multiple-part (2⅐) is
of (3⅕), or ¹⁵/₇ of ¹⁶/₅. These fractions, reduced to a common denominator,
are ⁷⁵/₃₅ and ¹¹²/₃₅, of which the second, divided into 112 parts, gives ¹/₃₅,
which repeated 75 times gives ⁷⁵/₃₅, the first. Hence, the multiple-part
which the first is of the second is ⁷⁵/₁₁₂, which being obtained by the rule
given in (121), shews that a/b, or a divided by b, according to the notion of
division there given, expresses the multiple-part which a is of b in every
case.
178. When the first of four numbers is the same multiple-part of the
second which the third is of the fourth, the four are said to be
geometrically[27] proportional, or simply proportional. This is a word in
common use; and it remains to shew that our mathematical definition of it,
just given, is, in fact, the common notion attached to it. For example,
suppose a picture is copied on a smaller scale, so that a line of two inches
long in the original is represented by a line of one inch and a half in the
copy; we say that the copy is not correct unless all the parts of the original
are reduced in the same proportion, namely, that of 2 to (1½). Since, on
dividing two inches into 4 parts, and taking 3 of them, we get (1½), the
same must be done with all the lines in the original, that is, the length of
any line in the copy must be three parts out of four of its length in the
original. Again, interest being at 5 per cent, that is, £5 being given for the
use of £100, a similar proportion of every other sum would be given; the
interest of £70, for example, would be just such a part of £70 as £5 is of
£100.
Since, then, the part which a is of b is expressed by the fraction a/b, or
any other fraction which is equivalent to it, and that which c is of d by c/d,
it follows, that when a, b, c, and d, are proportional, a/b = c/d. This
equation will be the foundation of all our reasoning on proportional
quantities; and in considering proportionals, it is necessary to observe not
only the quantities themselves, but also the order in which they come. Thus,
a, b, c, and d, being proportionals, that is, a being the same multiple-part of
b which c is of d, it does not follow that a, d, b, and c are proportionals, that
is, that a is the same multiple-part of d which b is of c. It is plain that a is
greater than, equal to, or less than b, according as c is greater than, equal to,
or less than d.

179. Four numbers, a, b, c, and d, being proportional in the order
written, a and d are called the extremes, and b and c the means, of the
proportion. For convenience, we will call the two extremes, or the two
means, similar terms, and an extreme and a mean, dissimilar terms. Thus, a
and d are similar, and so are b and c; while a and b, a and c, d and b, d and
c, are dissimilar. It is customary to express the proportion by placing dots
between the numbers, thus:
a:b∷c:d
180. Equal numbers will still remain equal when they have been
increased, diminished, multiplied, or divided, by equal quantities. This
amounts to saying that if

a = b and p = q,
a + p = b + q,
a - p = b - q,
ap = bq,

a b
and — = —.
p q

It is also evident, that a + p-p, a -p + p, ap/p, and a/p × p, are all
equal to a.
181. The product of the extremes is equal to the product of the means.
Let a/b = c/d, and multiply these equal numbers by the product bd. Then,
a abd
× bd = (116) = ad,
b b
c cbd
and × bd = = cb:
d d
hence (180), ad = bc.
Thus, 6, 8, 21, and 28, are proportional, since

6 3 3 × 7 21
= = = (180);
8 4 4 × 7 28
and it appears that 6 × 28 = 8 × 21, since both products are 168.
182. If the product of two numbers be equal to the product of two
others, these numbers are proportional in any order whatever, provided the
numbers in the same product are so placed as to be similar terms; that is, if
ab = pq, we have the following proportions:—

a:p∷q:b
a:q∷p:b
b:p∷q:a
b:q∷p:a
p:a∷b:q
p:b∷a:q
q:a∷b:p
q:b∷a:p
To prove any one of these, divide both ab and pq by the product of its
second and fourth terms; for example, to shew the truth of a: q ∷ p: b,
divide both ab and pq by bq. Then,
ab a pq p
= , and = ; hence (180),
bq q bq b

a p
= , or a : q ∷ p : b .
q b
The pupil should not fail to prove every one of the eight cases, and to verify
them by some simple examples, such as 1 × 6 = 2 × 3, which gives 1: 2 ∷ 3:
6, 3: 1 ∷ 6: 2, &c.

183. Hence, if four numbers be proportional, they are also proportional
in any other order, provided it be such that similar terms still remain similar.
For since, when
a c
= ,
b d
it follows (181) that ad = bc, all the proportions which follow from ad = bc,
by the last article, follow also from
a c
= .
b d
184. From (114) it follows that
a b+a
1+ b = b
,
a
and if be less than 1,
b
a b-a
1- b = b ,
a
while if be greater than 1,
b
a a-b
-1 = .
b b
a+b a-b
Also (122), if be divided by
b b
a+b
the result is
a-b
Hence, a, b, c, and d, being proportionals, we may obtain other proportions,
thus:
a c
Let =
b d

a c
Then (114) 1+ b = 1+ d
a+b c+d
or =
b d
or a + b: b ∷ c + d: d
That is, the sum of the first and second is to the second as the sum of the
third and fourth is to the fourth. For brevity, we shall not state in words any
more of these proportions, since the pupil will easily supply what is
wanting.
Resuming the proportion a: b ∷ c: d
a c
or =
b d
a c a
1 - b = 1 - d , if b be less than 1,
b-a d-c
or =
b d
that is, b-a: b ∷ d-c: d
or, a-b: b ∷ c-d: d,
a
if be greater than 1.
b

a+b c+d
Again, since =
b d
a-b c-d a
and
b
=
d ( b being greater than 1 )
dividing the first by the second we have
a+b = c+d

a-b c-d
or a + b : a - b ∷ c + d : c - d
and also a + b : b - a ∷ c + d : d - c,
a
if be less than 1.
b

185. Many other proportions might be obtained in the same manner. We
will, however, content ourselves with writing down a few which can be
obtained by combining the preceding articles.

a+b : a ∷ c+d : c
a : a-b ∷ c : c-d
a + c : a - c ∷ b + d : b - d.
In these and all others it must be observed, that when such expressions as a-
b and c-d occur, it is supposed that a is greater than b, and c greater than d.
186. If four numbers be proportional, and any two dissimilar terms be
both multiplied, or both divided by the same quantity, the results are
proportional. Thus, if a: b ∷ c: d, and m and n be any two numbers, we have
also the following:

ma : b ∷ mc : d
a : mb ∷ c : md
a c
n
: mb ∷ n : md

ma : nb ∷ mc : nd
a b c d
m
: m
∷ m
: m

a b c d
m
: m
∷ n
: n
and various others. To prove any one of these, recollect that nothing more is
necessary to make four numbers proportional except that the product of the
extremes should be equal to that of the means. Take the third of those just
given; the product of its extremes is
a mbc
× md, or ,
n n
while that of the means is
c mad
mb × , or .
n n
But since a : b ∷ c : d, by (181) ad = bc,
mad mbc
whence, by (180), mad = mbc, and = .
n n
a c
Hence , , and md, are proportionals.
n n
187. If the terms of one proportion be multiplied by the terms of a
second, the products are proportional; that is, if a: b ∷ c: d, and p: q ∷ r: s, it
follows that ap: bq ∷ cr: ds. For, since ad = bc, and ps = qr, by (180) adps =
bcqr, or ap × ds = bq × cr, whence (182) ap: bq ∷ cr: ds.
188. If four numbers be proportional, any similar powers of these
numbers are also proportional; that is, if
a : b ∷ c : d
Then aa : bb ∷ cc : dd
aaa : bbb ∷ ccc : ddd
&c. &c.
For, if we write the proportion twice, thus,

a : b ∷ c : d
a : b ∷ c : d
by (187) aa : bb ∷ cc : dd
But a : b ∷ c : d
Whence (187) aaa : bbb ∷ ccc : ddd and so on.
189. An expression is said to be homogeneous with respect to any two
or more letters, for instance, a, b, and c, when every term of it contains the
same number of letters, counting a, b, and c only. Thus, maab + nabc + rccc
is homogeneous with respect to a, b, and c; and of the third degree, since in
each term there is either a, b, and c, or one of these repeated alone, or with
another, so as to make three in all. Thus, 8aaabc, 12abccc, maaaaa,
naabbc, are all homogeneous, and of the fifth degree, with respect to a, b,
and c only; and any expression made by adding or subtracting these from
one another, will be homogeneous and of the fifth degree. Again ma + mnb
is homogeneous with respect to a and b, and of the first degree; but it is not
homogeneous with respect to m and n, though it is so with respect to a and
n. This being premised, we proceed to a theorem,[28] which will contain all
the results of (184), (185), and (188).
190. If any four numbers be proportional, and if from the first two, a
and b, any two homogeneous expressions of the same degree be formed;
and if from the last two, two other expressions be formed, in precisely the
same manner, the four results will be proportional. For example, if a: b ∷ c:
d, and if 2aaa + 3aab and bbb + abb be chosen, which are both
homogeneous with respect to a and b, and both of the third degree; and if
the corresponding expressions 2ccc + 3ccd and ddd + cdd be formed, which
are made from c and d precisely in the same manner as the two former ones
from a and b, then will
2aaa + 3aab : bbb + abb ∷ 2ccc + 3ccd : ddd + cdd
a
To prove this, let be called x.
b
Then, since a = x, and a = c ,

b b d
c
it follows that = x.
d
But since a divided by b gives x, x multiplied by b will give a, or a = bx.
For a similar reason, c = dx. Put bx and dx instead of a and c in the four
expressions just given, recollecting that when quantities are multiplied
together, the result is the same in whatever order the multiplications are
made; that, for example, bxbxbx is the same as bbbxxx.
Hence, 2aaa + 3aab = 2bxbxbx + 3bxbxb
= 2bbbxxx + 3bbbxx
which is bbb multiplied by 2xxx + 3xx

or bbb (2xxx + 3xx)[29]
Similarly, 2ccc + 3ccd = ddd (2xxx + 3xx)
Also, bbb + abb = bbb + bxbb
= bbb multiplied by 1 + x
or bbb(1 + x)
Similarly, ddd + cdd = ddd (1 + x)
Now, bbb : bbb ∷ ddd : ddd
Whence (186), bbb(2xxx + 3xx): bbb(1 + x) ∷ ddd(2xxx + 3xx):
ddd(1 + x), which, when instead of these expressions their equals
just found are substituted, becomes 2aaa + 3aab: bbb + abb ∷ 2ccc
+ 3ccd: ddd + cdd.
The same reasoning may be applied to any other case, and the pupil
may in this way prove the following theorems:
If
a:b∷c:d

2a + 3b : b ∷ 2c + 3d : d
aa + bb : aa - bb ∷ cc + dd : cc - dd
mab : 2aa + bb ∷ mcd : 2cc + dd

191. If the two means of a proportion be the same, that is, if a : b ∷ b: c,
the three numbers, a, b, and c, are said to be in continued proportion, or in
geometrical progression. The same terms are applied to a series of numbers,
of which any three that follow one another are in continued proportion, such
as

1 2 4 8 16 32 64 &c.

2 2 2 2 2 2
2 &c.
3 9 27 81 243 729
Which are in continued proportion, since
2 2 2
1:2∷2:4 2 : ∷ :
3 3 9
2 2 2 2
2:4∷4:8 : ∷ :
3 9 9 27
&c. &c.
192. Let a, b, c, d, e be in continued proportion; we have then
a b
a : b ∷ b : c or = or ac = bb
b c
b c
b:c∷c:d = bd = cc
c d
c d
c:d∷d:e = ce = dd
d e
Each term is formed from the preceding, by multiplying it by the same
number. Thus,

b c
b= × a (180); c = × b;
a b
a b b c
and since = , =
b c a b
b
or c = × b.
a
d
Again, d = ×c,
c
d c b
but = , which is = ;
c b a
b
therefore, d = × c, and so on
a
b
If, then, (which is called the common ratio of the series)
a
be denoted by r, we have
b = ar c = br = arr d = cr = arrr
and so on; whence the series

a b c d &c.
is a ar arr arrr &c.
Hence a : c ∷ a : arr
(186) ∷ aa : aarr
∷ aa : bb
because, b being ar, bb is arar or aarr. Again,

a : d ∷ a : arrr
(186) ∷ aaa : aaarrr
∷ aaa : bbb
a : e ∷ aaaa : bbbb , and so on;

that is, the first bears to the nᵗʰ term from the first the same proportion as the
nᵗʰ power of the first to the nᵗʰ power of the second.
193. A short rule may be found for adding together any number of terms
of a continued proportion. Let it be first required to add together the terms
1, r, rr, &c. where r is greater than unity. It is evident that we do not alter
any expression by adding or subtracting any numbers, provided we
afterwards subtract or add the same. For example,
p=p-q+q-r+r-s+s
Let us take four terms of the series, 1, r, rr, &c. or,
1 + r + rr + rrr
It is plain that
rrrr - 1 = rrrr - rrr + rrr - rr + rr - r + r - 1
Now (54), rr-r = r(r-1), rrr -rr = rr(r-1), rrrr-rrr = rrr(r-1), and
the above equation becomes rrrr -1 = rrr(r-1) + rr (r-1) + r (r-1) + r-1;
which is (54) rrr + rr + r + 1 taken r-1 times. Hence, rrrr-1 divided by r-1
will give 1 + r + rr + rrr, the sum of the terms required. In this way may
be proved the following series of equations:
rr - 1
1+r =
r-1
rrr - 1
1 + r + rr =
r-1
rrrr - 1
1 + r + rr + rrr =
r-1
rrrrr - 1
1 + r + rr + rrr + rrrr =
r-1
If r be less than unity, in order to find 1 + r + rr + rrr, observe that
1 - rrrr = 1 - r + r - rr + rr - rrr + rrr - rrrr
= 1 - r + r(1 - r) + rr(1 - r) + rrr(1 - r);

whence, by similar reasoning, 1 + r + rr + rrr is found by dividing 1-rrrr by
1-r; and equations similar to these just given may be found, which are,
1 - rr
1+r =
1-r
1 - rrr
1 + r + rr =
1-r
1 - rrrr
1 + r + rr + rrr =
1-r
1 - rrrrr
1 + r + rr + rrr + rrrr =
1-r
The rule is: To find the sum of n terms of the series, 1 + r + rr + &c.,
divide the difference between 1 and the (n + 1)ᵗʰ term by the difference
between 1 and r.
194. This may be applied to finding the sum of any number of terms of
a continued proportion. Let a, b, c, &c. be the terms of which it is required
to sum four, that is, to find a + b + c + d, or (192) a + ar + arr + arrr, or
(54) a(1 + r + rr + rrr), which (193) is
rrrr - 1 1 - rrrr
× a, or × a,
r-1 1-r
according as r is greater or less than unity. The first fraction is
arrrr - a e-a
, or (192) .
r-1 r-1
Similarly, the second is
a-e
.
1-r
The rule, therefore, is: To sum n terms of a continued proportion, divide
the difference of the (n + 1)ᵗʰ and first terms by the difference between unity
and the common measure. For example, the sum of 10 terms of the series 1
+ 3 + 9 + 27 + &c. is required. The eleventh term is 59049, and ⁽⁵⁹⁰⁴⁹ ⁻ ¹⁾/

₍₃₋₁₎ is 29524. Again, the sum of 18 terms of the series 2 + 1 + ½ + ½ + &c.
of which the nineteenth term is ¹/₁₃₁₀₇₂, is
1
———
2
- 131072 131070
————— ——— .
1-½ = 3 131072

EXAMPLES.
9 terms of 1 + 4 + 16 + &c. are 87381
6 12 847422675
10 ...... 3 + 7 + 49 + &c. ... 201768035
1 1 1 1048575
20 ...... + + + &c. ...
2 4 8 1048576
195. The powers of a number or fraction greater than unity increase; for
since 2½ is greater than 1, 2½ × 2½ is 2½ taken more than once, that is, is
greater than 2½, and so on. This increase goes on without limit; that is,
there is no quantity so great but that some power of 2½ is greater. To prove
this, observe that every power of 2½ is made by multiplying the preceding
power by 2½, or by 1 + 1½, that is, by adding to the former power that
power itself and its half. There will, therefore, be more added to the 10th
power to form the 11th, than was added to the 9th power to form the 10th.
But it is evident that if any given quantity, however small, be continually
added to 2½, the result will come in time to exceed any other quantity that
was also given, however great; much more, then, will it do so if the quantity
added to 2½ be increased at each step, which is the case when the
successive powers of 2½ are formed. It is evident, also, that the powers of 1
never increase, being always 1; thus, 1 × 1 = 1, &c. Also, if a be greater
than m times b, the square of a is greater than mm times the square of b.
Thus, if a = 2b + c, where a is greater than 2b, the square of a, or aa, which
is (68) 4bb + 4bc + cc is greater than 4bb, and so on.

196. The powers of a fraction less than unity continually decrease; thus,
the square of ⅖, or ⅖ × ⅖, is less than ⅖, being only two-fifths of it. This
decrease continues without limit; that is, there is no quantity so small but
that some power of ⅖ is less. For if
5
= x,
2
2 1
= ,
5 x
1 1
and the powers of ⅖ are , ,
xx xxx
and so on. Since x is greater than 1 (195), some power of x may be found
which shall be greater than a given quantity. Let this be called m; then 1/m
is the corresponding power of ⅖; and a fraction whose denominator can be
made as great as we please, can itself be made as small as we please (112).
197. We have, then, in the series
1 r rr rrr rrrr &c.
I. A series of increasing terms, if r be greater than 1. II. Of terms having
the same value, if r be equal to 1. III. A series of decreasing terms, if r be
less than 1. In the first two cases, the sum
1 + r + rr + rrr + &c.
may evidently be made as great as we please, by sufficiently increasing the
number of terms. But in the third this may or may not be the case; for
though something is added at each step, yet, as that augmentation
diminishes at every step, we may not certainly say that we can, by any
number of such augmentations, make the result as great as we please. To
shew the contrary in a simple instance, consider the series,
1 + ½ + ¼ + ⅛ + ¹/₁₆ + &c.
Carry this series to what extent we may, it will always be necessary to add
the last term in order to make as much as 2. Thus,

(1 + ½ + ¼) + ¼ = 1 + ½ + ½ = 1 + 1 = 2
(1 + ½ + ¼ + ⅛) + ⅛ = 2.
(1 + ½ + ¼ + ⅛ + ¹/₁₆) + ¹/₁₆ = 2, &c.
But in the series, every term is only the half of the preceding; consequently
no number of terms, however great, can be made as great as 2 by adding
one more. The sum, therefore, of 1, ½, ¼, ⅛ &c. continually approaches to
2, diminishing its distance from 2 at every step, but never reaching it.
Hence, 2 is celled the limit of 1 + ½ + ¼ + &c. We are not, therefore, to
conclude that every series of decreasing terms has a limit. The contrary may
be shewn in the very simple series, 1 + ½ + ⅓ + ¼ + &c. which may be
written thus:
1 + ½ + (⅓ + ¼) + (⅕ + ... up to ⅛) + (⅑ + ... up to ¹/₁₆)
+ (¹/₁₇ + ... up to ¹/₃₂) + &c.
We have thus divided all the series, except the first two terms, into lots,
each containing half as many terms as there are units in the denominator of
its last term. Thus, the fourth lot contains 16 or ³²/₂2 terms. Each of these
lots may be shewn to be greater than ½. Take the third, for example,
consisting of ⅑, ¹/₁₀, ¹/₁₁, ¹/₁₂, ¹/₁₃, ¹/₁₄, ¹/₁₅, and ¹/₁₆. All except ¹/₁₆, the last,
are greater than ¹/₁₆; consequently, by substituting ¹/₁₆ for each of them, the
amount of the whole lot would be lessened; and as it would then become ⁸/
₁₆, or ½, the lot itself is greater than ½. Now, if to 1 + ½, ½ be continually
added, the result will in time exceed any given number. Still more will this
be the case if, instead of ½, the several lots written above be added one after
the other. But it is thus that the series 1 + ½ + ⅓, &c. is composed, which
proves what was said, that this series has no limit.
198. The series 1 + r + rr + rrr + &c. always has a limit when r is less
than 1. To prove this, let the term succeeding that at which we stop be a,
whence (194) the sum is
1-a 1 a
, or (112) - .
1-r 1-r 1-r
The terms decrease without limit (196), whence we may take a term so far
distant from the beginning, that a, and therefore

1-a
,
1-r
shall be as small as we please. But it is evident that in this case

1 a
-
1-r 1-r
though always less than

1
1-r
may be brought as near to

1
1-r
as we please; that is, the series 1 + r + rr + &c. continually approaches to
the limit

1
.
1-r
Thus 1 + ½ + ¼ + ⅛ + &c. where r = ½, continually approaches to

1
or 2,
1-½
as was shewn in the last article.

EXERCISES.
2 2
The limit of 2 +
3
+
9
+ &c.

or
2 ( 1 + 1 + 1 + &c. ) is 3

3 9
9 81
... 1 +
10
+
100
+ &c. ... 10
15 45
... 5 +
7
+
49
+ &c. ... 8¾
199. When the fraction a/b is not equal to c/d, but greater, a is said to
have to b a greater ratio than c has to d; and when a/b is less than c/d, a is
said to have to b a less ratio than c has to d. We propose the following
questions as exercises, since they follow very simply from this definition.
I. If a be greater than b, and c less than or equal to d, a will have a
greater ratio to b than c has to d.
II. If a be less than b, and c greater than or equal to d, a has a less ratio
to b than c has to d.
III. If a be to b as c is to d, and if a have a greater ratio to b than c has to
x, d is less than x; and if a have a less ratio to b than c to x, d is greater than
x.
IV. a has to b a greater ratio than ax to bx + y, and a less ratio than ax to
bx- y.
200. If a have to b a greater ratio than c has to d, a + c has to b + d a
less ratio than a has to b, but a greater ratio than c has to d; or, in other
words, if a/b be the greater of the two fractions a/b and c/d,

a+c
b+d
will be greater than c/d, but less than a/b. To shew this, observe that (mx +
ny)/(m + n) must lie between x and y, if x and y be unequal: for if x be the
less of the two, it is certainly greater than

mx + nx
m+n
or than x; and if y be the greater of the two, it is certainly less than

my + ny
m+n
or than y. It therefore lies between x and y. Now let a/b be x, and let c/d be
y: then a = bx, c = dy. Now

bx + dy
b+d
is something between x and y, as was just proved; therefore

a+c
b+d
is something between a/b and c/d. Again, since a/b and c/d are respectively
equal to ap/bp and cq/dq, and since, as has just been proved,

ap + cq
bp + dq
lies between the two last, it also lies between the two first; that is, if p and q
be any numbers or fractions whatsoever,

ap + cq
bp + dq
lies between a/b and c/d.
201. By the last article we may often form some notion of the value of
an expression too complicated to be easily calculated. Thus,

1+x 1 x 1
lies between and , or 1 and ;
1 + xx 1 xx x
ax + by ax by
lies between and ,
axx + bbyy axx bbyy

that is, between 1/x and 1/by. And it has been shewn that (a + b)/2 lies
between a and b, the denominator being considered as 1 + 1.
202. It may also be proved that a fraction such as

a+b+c+d
p+q+r+s
always lies among
a b c d
, , , and ,
p q r s
that is, is less than the greatest of them, and greater than the least. Let these
fractions be arranged in order of magnitude; that is, let a/p be greater than
b/q, b/q be greater than c/r, and c/r greater than d/s. Then by (200)
is and
less greater
than than
a+b a b c
and
p+q p q r
a+b+c a+b a c d
and and
p+q+r p+q p r s
a+b+c+d a+b+c a d
and
p+q+r+s p+q+r p s
whence the proposition is evident.
203. It is usual to signify “a is greater than b” by a > b and “a is less
than b” by a < b; the opening of V being turned towards the greater
quantity. The pupil is recommended to make himself familiar with these
signs.

SECTION IX.
ON PERMUTATIONS AND
COMBINATIONS.

204. If a number of counters, distinguished by different letters, be
placed on the table, and any number of them, say four, be taken away, the
question is, to determine in how many different ways this can be done. Each
way of doing it gives what is called a combination of four, but which might
with more propriety be called a selection of four. Two combinations or
selections are called different, which differ in any way whatever; thus, abcd
and abce are different, d being in one and e in the other, the remaining parts
being the same. Let there be six counters, a, b, c, d, e, and f; the
combinations of three which can be made out of them are twenty in number,
as follow:

abc ace bcd bef
abd acf bce cde
abe ade bcf cdf
abf adf bde cef
acd aef bdf def
The combinations of four are fifteen in number, namely,

abcd abde acde adef bcef
abce abdf acdf bcde bdcf
abcf abef acef bcdf cdef
and so on.
205. Each of these combinations may be written in several different
orders; thus, abcd may be disposed in any of the following ways:

abcd acbd acdb abdc adbc adcb
bacd cabd cadb badc dabc dacb

bcad cbad cdab bdac dbac dcab
bcda cbda cdba bdca dbca dcba
of which no two are entirely in the same order. Each of these is said to be a
distinct permutation of abcd. Considered as a combination, they are all the
same, as each contains a, b, c, and d.
206. We now proceed to find how many permutations, each containing
one given number, can be made from the counters in another given number,
six, for example. If we knew how to find all the permutations containing
four counters, we might make those which contain five thus: Take any one
which contains four, for example, abcf in which d and e are omitted; write d
and e successively at the end, which gives abcfd, abcfe, and repeat the same
process with every other permutation of four; thus, dabc gives dabce and
dabcf. No permutation of five can escape us if we proceed in this manner,
provided only we know those of four; for any given permutation of five, as
dbfea, will arise in the course of the process from dbfe, which, according to
our rule, furnishes dbfea. Neither will any permutation be repeated twice,
for dbfea, if the rule be followed, can only arise from the permutation dbfe.
If we begin in this way to find the permutations of two out of the six,
a b c d e f
each of these gives five; thus,
a gives ab ac ad ae af
b ... ba bc bd be bf
and the whole number is 6 × 5, or 30.
Again,
ab gives abc abd abe abf
ac ... acb acd ace acf
and here are 30, or 6 × 5 permutations of 2, each of which gives 4
permutations of 3; the whole number of the last is therefore 6 × 5 × 4, or
120.
Again,

abc gives abcd abce abcf
abd ... abdc abde abdf
and here are 120, or 6 × 5 × 4, permutations of three, each of which gives 3
permutations of four; the whole number of the last is therefore 6 × 5 × 4 ×
3, or 360.
In the same way, the number of permutations of 5 is 6 × 5 × 4 × 3 × 2,
and the number of permutations of six, or the number of different ways in
which the whole six can be arranged, is 6 × 5 × 4 × 3 × 2 × 1. The last two
results are the same, which must be; for since a permutation of five only
omits one, it can only furnish one permutation of six. If instead of six we
choose any other number, x, the number of permutations of two will be x(x-
1), that of three will be x(x-1)(x-2), that of four x(x -1)(x-2)(x-3), the rule
being: Multiply the whole number of counters by the next less number, and
the result by the next less, and so on, until as many numbers have been
multiplied together as there are to be counters in each permutation: the
product will be the whole number of permutations of the sort required.
Thus, out of 12 counters, permutations of four may be made to the number
of 12 × 11 × 10 × 9, or 11880.

EXERCISES.
207. In how many different ways can eight persons be arranged on eight
seats?

Answer, 40320.
In how many ways can eight persons be seated at a round table, so that
all shall not have the same neighbours in any two arrangements?[30]

Answer, 5040.
If the hundredth part of a farthing be given for every different
arrangement which can be made of fifteen persons, to how much will the
whole amount?

Answer, £13621608.

Out of seventeen consonants and five vowels, how many words can be
made, having two consonants and one vowel in each?

Answer, 4080.
208. If two or more of the counters have the same letter upon them, the
number of distinct permutations is less than that given by the last rule. Let
there be a, a, a, b, c, d, and, for a moment, let us distinguish between the
three as thus, a, a′, a″. Then, abca′a″d, and a″bcaa′d are reckoned as
distinct permutations in the rule, whereas they would not have been so, had
it not been for the accents. To compute the number of distinct permutations,
let us make one with b, c, and d, leaving places for the as, thus, ( ) bc ( ) ( )
d. If the as had been distinguished as a, a′, a″, we might have made 3 × 2 ×
1 distinct permutations, by filling up the vacant places in the above, all
which six are the same when the as are not distinguished. Hence, to deduce
the number of permutations of a, a, a, b, c, d, from that of aa′a″bcd, we
must divide the latter by 3 × 2 × 1, or 6, which gives

6×5×4×3×2×1
3×2×1

or 120. Similarly, the number of permutations of aaaabbbcc is
9×8×7×6×5×4×3×2×1
.
4×3×2×1×3×2×1×2×1

EXERCISE.
How many variations can be made of the order of the letters in the word
antitrinitarian?

Answer, 126126000.
209. From the number of permutations we can easily deduce the number
of combinations. But, in order to form these combinations independently,
we will shew a method similar to that in (206). If we know the
combinations of two which can be made out of a, b, c, d, e, we can find the
combinations of three, by writing successively at the end of each
combination of two, the letters which come after the last contained in it.

Thus, ab gives abc, abd, abe; ad gives ade only. No combination of three
can escape us if we proceed in this manner, provided only we know the
combinations of two; for any given combination of three, as acd, will arise
in the course of the process from ac, which, according to our rule, furnishes
acd. Neither will any combination be repeated twice, for acd, if the rule be
followed, can only arise from ac, since neither ad nor cd furnishes it. If we
begin in this way to find the combinations of the five,

a b c d e
a gives ab ac ad ae
b ···· bc bd be
c ···· cd ce
d ···· de
Of these, ab gives abc abd abe
ac ···· acd ace
ad ···· ade
bc ···· bcd bce
bd ···· bde
cd ···· cde
ae be ce and de give none.
Of these, abc gives abcd abce
abd ···· abde
acd ···· acde
bcd ···· bcde
Those which contain e give none, as before.
Of the last, abcd gives abcde, and the others none, which is evidently true,
since only one selection of five can be made out of five things.
210. The rule for calculating the number of combinations is derived
directly from that for the number of permutations. Take 7 counters; then,

since the number of permutations of two is 7 × 6, and since two
permutations, ba and ab, are in any combination ab, the number of
combinations is half that of the permutations, or (7 × 6)/2. Since the number
of permutations of three is 7 × 6 × 5, and as each combination abc has 3 × 2
× 1 permutations, the number of combinations of three is
7×6×5
.
1×2×3
Also, since any combination of four, abcd, contains 4 × 3 × 2 × 1
permutations, the number of combinations of four is
7×6×5×4
,
1×2×3×4
and so on. The rule is: To find the number of combinations, each containing
n counters, divide the corresponding number of permutations by the product
of 1, 2, 3, &c. up to n. If x be the whole number, the number of
combinations of two is
x(x - 1)
;
1×2
that of three is
x(x - 1)(x - 2)
;
1×2×3
that of four is
x(x - 1)(x - 2)(x - 3)
;
1×2×3×4
211. The rule may in half the cases be simplified, as follows. Out of ten
counters, for every distinct selection of seven which is taken, a distinct
combination of 3 is left. Hence, the number of combinations of seven is as
many as that of three. We may, therefore, find the combinations of three
instead of those of seven; and we must moreover expect, and may even
assert, that the two formulæ for finding these two numbers of combinations

are the same in result, though different in form. And so it proves; for the
number of combinations of seven out of ten is
10 × 9 × 8 × 7 × 6 × 5 × 4
,
1×2×3×4×5×6×7
in which the product 7 × 6 × 5 × 4 occurs in both terms, and therefore may
be removed from both (108), leaving
10 × 9 × 8
,
1×2×3
which is the number of combinations of three out of ten. The same may be
shewn in other cases.

EXERCISES.
How many combinations of four can be made out of twelve things?

Answer, 495.
What number of 6 8 28
combinations 4 11 330
out of Answer,
can be made of 26 28 378
6 15 5005
How many combinations can be made of 13 out of 52; or how many
different hands may a person hold at the game of whist?

Answer, 635013559600.

BOOK II.
COMMERCIAL ARITHMETIC.

SECTION I.
WEIGHTS, MEASURES, &C.

212. In making the calculations which are necessary in commercial affairs, no more
processes are required than those which have been explained in the preceding book.
But there is still one thing wanted—not to insure the accuracy of our calculations, but
to enable us to compare and judge of their results. We have hitherto made use of a
single unit (15), and have treated of other quantities which are made up of a number of
units, in Sections II., III., and IV., and of those which contain parts of that unit in
Sections V. and VI. Thus, if we are talking of distances, and take a mile as the unit, any
other length may be represented,[31] either by a certain number of miles, or a certain
number of parts of a mile, and (1 meaning one mile) may be expressed either by a
whole number or a fraction. But we can easily see that in many cases inconveniences
would arise. Suppose, for example, I say, that the length of one room is ¹/₁₈₀ of a mile,
and of another ¹/₁₇₄ of a mile, what idea can we form as to how much the second is
longer than the first? It is necessary to have some smaller measure; and if we divide a
mile into 1760 equal parts, and call each of these parts a yard, we shall find that the
length of the first room is 9 yards and ⁷/₉ of a yard, and that of the second 10 yards and
¹⁰/₈₇ of a yard. From this we form a much better notion of these different lengths, but
still not a very perfect one, on account of the fractions ⁷/₉ and ¹⁰/₈₇. To get a clearer idea
of these, suppose the yard to be divided into three equal parts, and each of these parts
to be called a foot; then ⁷/₉ of a yard contains 2⅓ feet, and ¹⁰/₈₇ of a yard contains ³⁰/₈₇
of a foot, or a little more than ⅓ of a foot. Therefore the length of the first room is now
9 yards, 2 feet, and ⅓ of a foot; that of the second is 10 yards and a little more than ⅓
of a foot. We see, then, the convenience of having large measures for large quantities,
and smaller measures for small ones; but this is done for convenience only, for it is
possible to perform calculations upon any sort of quantity, with one measure alone, as
certainly as with more than one; and not only possible, but more convenient, as far as
the mere calculation is concerned.
The measures which are used in this country are not those which would have been
chosen had they been made all at one time, and by a people well acquainted with
arithmetic and natural philosophy. We proceed to shew how the results of the latter
science are made useful in our system of measures. Whether the circumstances
introduced are sufficiently well known to render the following methods exact enough

for the recovery of astronomical standards, may be matter of opinion; but no doubt can
be entertained of their being amply correct for commercial purposes.
It is evidently desirable that weights and measures should always continue the
same, and that posterity should be able to replace any one of them when the original
measure is lost. It is true that a yard, which is now exact, is kept by the public
authorities; but if this were burnt by accident,[32] how are those who shall live 500
years hence to know what was the length which their ancestors called a yard? To
ensure them this knowledge, the measure must be derived from something which
cannot be altered by man, either from design or accident. We find such a quantity in the
time of the daily revolution of the earth, and also in the length of the year, both of
which, as is shewn in astronomy, will remain the same, at least for an enormous
number of centuries, unless some great and totally unknown change take place in the
solar system. So long as astronomy is cultivated, it is impossible to suppose that either
of these will be lost, and it is known that the latter is 365·24224 mean solar days, or
about 365¼ of the average interval which elapses between noon and noon, that is,
between the times when the sun is highest in the heavens. Our year is made to consist
of 365 days, and the odd quarter is allowed for by adding one day to every fourth year,
which gives what we call leap-year. This is the same as adding ¼ of a day to each year,
and is rather too much, since the excess of the year above 365 days is not ·25 but
·24224 of a day. The difference is ·00776 of a day, which is the quantity by which our
average year is too long. This amounts to a day in about 128 years, or to about 3 days
in 4 centuries. The error is corrected by allowing only one out of four of the years
which close the centuries to be leap-years. Thus, a.d. 1800 and 1900 are not leap-years,
but 2000 is so.
213. The day is therefore the first measure obtained, and is divided into 24 parts or
hours, each of which is divided into 60 parts or minutes, and each of these again into
60 parts or seconds. One second, marked thus, 1″,[33] is therefore the 86400ᵗʰ part of a
day, and the following is the

MEASURE OF TIME.[34]
60 seconds are 1 minute 1 m.
60 minutes ” 1 hour 1 h.
24 hours ” 1 day 1 d.
7 days ” 1 week 1 wk.
365 days ” 1 year 1 yr.
214. The second having been obtained, a pendulum can be constructed which shall,
when put in motion, perform one vibration in exactly one second, in the latitude of
Greenwich.[35] If we were inventing measures, it would be convenient to call the length
of this pendulum a yard, and make it the standard of all our measures of length. But as

there is a yard already established, it will do equally well to tell the length of the
pendulum in yards. It was found by commissioners appointed for the purpose, that this
pendulum in London was 39·1393 inches, or about one yard, three inches, and ⁵/₃₆ of
an inch. The following is the division of the yard.

MEASURES OF LENGTH.
The lowest measure is a barleycorn.[36]
3 barleycorns are 1 inch 1 in.
12 inches 1 foot 1 ft.
3 feet 1 yard 1 yd.
5½ yards 1 pole 1 po.
40 poles or 220 yards 1 furlong 1 fur.
8 furlongs or 1760 yards 1 mile 1 mi.
Also
6 feet 1 fathom 1 fth.
69⅓ miles 1 degree 1 deg. or 1°.
A geographical mile is ¹/₆₀th of a degree, and three such miles are one nautical
league.
In the measurement of cloth or linen the following are also used:
2¼ inches are 1 nail 1 nl.
4 nails 1 quarter (of a yard) 1 qr.
3 quarters 1 Flemish ell 1 Fl. e.
5 quarters 1 English ell 1 E. e.
6 quarters 1 French ell 1 Fr. e.

215. MEASURES OF SURFACE, OR SUPERFICIES.
All surfaces are measured by square inches, square feet, &c.; the square inch being
a square whose side is an inch in length, and so on. The following measures may be
deduced from the last, as will afterwards appear.
144 square inches are 1 square foot 1 sq. ft.
9 square feet 1 square yard 1 sq. yd.
30¼ square yards 1 square pole 1 sq. p.
40 square poles 1 rood 1 rd.
4 roods 1 acre 1 ac.

Thus, the acre contains 4840 square yards, which is ten times a square of 22 yards
in length and breadth. This 22 yards is the length which land-surveyors’ chains are
made to have, and the chain is divided into 100 links, each ·22 of a yard or 7·92 inches.
An acre is then 10 square chains. It may also be noticed that a square whose side is 69⁴/
₇ yards is nearly an acre, not exceeding it by ⅕ of a square foot.

216. MEASURES OF SOLIDITY OR CAPACITY.[37]
Cubes are solids having the figure of dice. A cubic inch is a cube each of whose
sides is an inch, and so on.
1728 cubic inches are 1 cubic foot 1 c. ft.
27 cubic feet 1 cubic yard 1 c. yd.
This measure is not much used, except in purely mathematical questions. In the
measurements of different commodities various measures were used, which are now
reduced, by act of parliament, to one. This is commonly called the imperial measure,
and is as follows:

MEASURE OF LIQUIDS AND
OF ALL DRY GOODS.
4 gills are 1 pint 1 pt.
2 pints 1 quart 1 qt.
4 quarts 1 gallon 1 gall.
2 gallons 1 peck[38] 1 pk.
4 pecks 1 bushel 1 bu.
8 bushels 1 quarter 1 qr.
5 quarters 1 load 1 ld.
The gallon in this measure is about 277·274 cubic inches; that is, very nearly 277¼
cubic inches.[39]
217. The smallest weight in use is the grain, which is thus determined. A vessel
whose interior is a cubic inch, when filled with water,[40] has its weight increased by
252·458 grains. Of the grains so determined, 7000 are a pound averdupois, and 5760 a
pound troy. The first pound is always used, except in weighing precious metals and
stones, and also medicines. It is divided as follows:

AVERDUPOIS WEIGHT.
27¹¹/₃₂ grains are 1 dram 1 dr.

6 drams, or drachms 1 ounce[41] 1 oz.
16 ounces 1 pound 1 lb.
28 pounds 1 quarter 1 qr.
4 quarters 1 hundred-weight 1 cwt.
20 hundred-weight 1 ton 1 ton.
The pound averdupois contains 7000 grains. A cubic foot of water weighs
62·3210606 pounds averdupois, or 997·1369691 ounces.
For the precious metals and for medicines, the pound troy, containing 5760 grains,
is used, but is differently divided in the two cases. The measures are as follow:

TROY WEIGHT.
24 grains are 1 pennyweight 1 dwt.
20 pennyweights 1 ounce 1 oz.
12 ounces 1 pound 1 lb.
The pound troy contains 5760 grains. A cubic foot of water weighs 75·7374 pounds
troy, or 908·8488 ounces.

APOTHECARIES’ WEIGHT.
20 grains are 1 scruple ℈
3 scruples 1 dram ʒ
8 drams 1 ounce ℥
12 ounces 1 pound lb
218. The standard coins of copper, silver, and gold, are,—the penny, which is 10⅔
drams of copper; the shilling, which weighs 3 pennyweights 15 grains, of which 3 parts
out of 40 are alloy, and the rest pure silver; and the sovereign, weighing 5
pennyweights and 3¼ grains, of which 1 part out of 12 is copper, and the rest pure
gold.

MEASURES OF MONEY.
The lowest coin is a farthing, which is marked thus, ¼, being one fourth of a penny.

2 farthings are 1 halfpenny ½d.
2 halfpence 1 penny 1d.
12 pence 1 shilling 1s.

20 shillings 1 pound[42] or sovereign £1
21 shillings 1 guinea.[43]
219. When any quantity is made up of several others, expressed in different units,
such as £1. 14. 6, or 2cwt. 1qr. 3lbs., it is called a compound quantity. From these
tables it is evident that any compound quantity of any substance can be measured in
several different ways. For example, the sum of money which we call five pounds four
shillings is also 104 shillings, or 1248 pence, or 4992 farthings. It is easy to reduce any
quantity from one of these measurements to another; and the following examples will
be sufficient to shew how to apply the same process, usually called Reduction, to all
sorts of quantities.
I. How many farthings are there in £18. 12. 6¾?[44]
Since there are 20 shillings in a pound, there are, in £18, 18 × 20, or 360 shillings;
therefore, £18. 12 is 360 + 12, or 372 shillings. Since there are 12 pence in a shilling, in
372 shillings there are 372 × 12, or 4464 pence; and, therefore, in £18. 12. 6 there are
4464 + 6, or 4470 pence.
Since there are 4 farthings in a penny, in 4470 pence there are 4470 × 4, or 17880
farthings; and, therefore, in £18. 12. 6¾ there are 17880 + 3, or 17883 farthings. The
whole of this process may be written as follows:

£18 . 12 . 6¾
20
360 + 12 = 372
12
4464 + 6 = 4470
4
17880 + 3 = 17883

II. In 17883 farthings, how many pounds, shillings, pence, and farthings are there?
Since 17883, divided by 4, gives the quotient 4470, and the remainder 3, 17883
farthings are 4470 pence and 3 farthings (218).
Since 4470, divided by 12, gives the quotient 372, and the remainder 6, 4470 pence
is 372 shillings and 6 pence.
Since 372, divided by 20, gives the quotient 18, and the remainder 12, 372 shillings
is 18 pounds and 12 shillings.
Therefore, 17883 farthings is 4470¾d., which is 372s. 6¾d., which is £18. 12. 6¾.
The process may be written as follows:

4)17883
——
12)4470 ... 3
——
20)372 ... 6
£18 . 12 . 6¾

EXERCISES.
A has £100. 4. 11½, and B has 64392 farthings. If A receive 1492 farthings, and B
£1. 2. 3½, which will then have the most, and by how much?—Answer, A will have
£33. 12. 3 more than B.
In the following table the quantities written opposite to each other are the same:
each line furnishes two exercises.

£15 . 18 . 9½ 15302 farthings.
115ˡᵇˢ 1ᵒᶻ 8ᵈᵚᵗ 663072 grains.
3ˡᵇˢ 14ᵒᶻ 9ᵈʳ 1001 drams.
3ᵐ 149 yds 2ᶠᵗ 9 in 195477 inches.
19ᵇᵘ 2ᵖᵏˢ 1 gall 2 qᵗˢ 1260 pints.
16 ʰ 23ᵐ 47ˢ 59027 seconds.
220. The same may be done where the number first expressed is fractional. For
example, how many shillings and pence are there in ⁴/₁₅ of a pound? Now, ⁴/₁₅ of a
pound is ⁴/₁₅ of 20 shillings; ⁴/₁₅ of 20 is
4 × 20 4×4 16
, or (110), or ,
15 3 3
or (105) 5⅓ of a shilling. Again, ⅓ of a shilling is ⅓ of 12 pence, or 4 pence.
Therefore, £⁴/₁₅ = 5s. 4d.
Also, ·23 of a day is ·23 × 24 in hours, or 5ʰ·52; and ·52 of an hour is ·52 × 60 in
minutes, or 3ᵐ·2; and ·2 of a minute is ·2 × 60 in seconds, or 12ˢ; whence ·23 of a day
is 5ʰ 31ᵐ 12ˢ.
Again, suppose it required to find what part of a pound 6s. 8d. is. Since 6s. 8d. is 80
pence, and since the whole pound contains 20 × 12 or 240 pence, 6s. 8d. is made by
dividing the pound into 240 parts, and taking 80 of them. It is therefore £⁸⁰/₂₄₀ (107),
but ⁸⁰/₂₄₀ = ⅓ (108); therefore, 6s. 8d. = £⅓.

EXERCISES.
⅖ of a day is 9ʰ 36ᵐ
·12841 of a day 3ʰ 4ᵐ 54ᔆ·624[45]
·257 of a cwt. 28ˡᵇˢ 12ᵒᶻ 8ᵈʳ·704
£·14936 2ˢ 11ᵈ 3ᶠ·3856
221, 222. I have thought it best to refer the mode of converting shillings, pence, and
farthings into decimals of a pound to the Appendix (See Appendix On Decimal
Money). I should strongly recommend the reader to make himself perfectly familiar
with the modes given in that Appendix. To prevent the subsequent sections from being
altered in their numbering, I have numbered this paragraph as above.
223. The rule of addition[46] of two compound quantities of the same sort will be
evident from the following example. Suppose it required to add £192. 14. 2½ to £64.
13. 11¾. The sum of these two is the whole of that which arises from adding their
several parts. Now
¾d. + ½d. = ⁵/₄d. = £0 . 0 . 1¼ (219)
11d. + 2d. = 13d. = 0.1.1
13s. + 14s. = 27s. = 1.7.0
£64 + £192 = = 256 . 0 . 0
The sum of all of which is £257. 8 . 2¼
This may be done at once, and written as follows:

£192.14. 2½
64.13.11¾
£257. 8. 2¼

Begin by adding together the farthings, and reduce the result to pence and farthings.
Set down the last only, carry the first to the line of pence, and add the pence in both
lines to it. Reduce the sum to shillings and pence; set down the last only, and carry the
first to the line of shillings, and so on. The same method must be followed when the
quantities are of any other sort; and if the tables be kept in memory, the process will be
easy.
224. Subtraction is performed on the same principle as in (40), namely, that the
difference of two quantities is not altered by adding the same quantity to both. Suppose
it required to subtract £19 . 13. 10¾ from £24. 5. 7½. Write these quantities under one
another thus:

£24. 5. 7½
19. 13. 10¾

Since ¾ cannot be taken from ½ or ²/₄, add 1d. to both quantities, which will not
alter their difference; or, which is the same thing, add 4 farthings to the first, and 1d. to
the second. The pence and farthings in the two lines then stand thus: 7⁶/₄d. and 11¾d.
Now subtract ¾ from ⁶/₄, and the difference is ¾ which must be written under the
farthings. Again, since 11d. cannot be subtracted from 7d., add 1s. to both quantities by
adding 12d. to the first, and 1s. to the second. The pence in the first line are then 19,
and in the second 11, and the difference is 8, which write under the pence. Since the
shillings in the lower line were increased by 1, there are now 14s. in the lower, and 5s.
in the upper one. Add 20s. to the upper and £1 to the lower line, and the subtraction of
the shillings in the second from those in the first leaves 11s. Again, there are now £20
in the lower, and £24 in the upper line, the difference of which is £4; therefore the
whole difference of the two sums is £4. 11. 8¾. If we write down the two sums with all
the additions which have been made, the process will stand thus:

£24 . 25 . 19⁶/₄
20 . 14 . 11¾
Difference £4 . 11 . 8¾

225. The same method may be applied to any of the quantities in the tables. The
following is another example:
From 7 cwt. 2 qrs. 21 lbs. 14 oz. Subtract 2 cwt. 3 qrs. 27 lbs. 12 oz.
After alterations have been made similar to those in the last article, the question
becomes:
From 7 cwt. 6 qrs. 49 lbs. 14 oz.
Subtract 3 cwt. 4 qrs. 27 lbs. 12 oz.
The difference is 4 cwt. 2 qrs. 22 lbs. 2 oz.
In this example, and almost every other, the process may be a little shortened in the
following way. Here we do not subtract 27 lbs. from 21 lbs., which is impossible, but
we increase 21 lbs. by 1 qr. or 28 lbs. and then subtract 27 lbs. from the sum. It would
be shorter, and lead to the same result, first to subtract 27 lbs. from 1 qr. or 28 lbs. and
add the difference to 21 lbs.

226. EXERCISES.
A man has the following sums to receive: £193. 14. 11¼, £22. 0. 6¾, £6473. 0. 0,
and £49. 14. 4½; and the following debts to pay: £200 . 19. 6¼, £305. 16. 11, £22, and
£19. 6. 0½. How much will remain after paying the debts?

Answer, £6190. 7. 4¾.
There are four towns, in the order A, B, C, and D. If a man can go from A to B in 5ʰ
20ᵐ 33ˢ, from B to C in 6ʰ 49ᵐ 2ˢ and from A to D in 19ʰ 0ᵐ 17ˢ, how long will he be in
going from B to D, and from C to D?

Answer, 13ʰ 39ᵐ 44ˢ, and 6ʰ 50ᵐ 42ˢ.
227. In order to perform the process of Multiplication, it must be recollected that,
as in (52), if a quantity be divided into several parts, and each of these parts be
multiplied by a number, and the products be added, the result is the same as would
arise from multiplying the whole quantity by that number.
It is required to multiply £7. 13. 6¼ by 13. The first quantity is made up of 7
pounds, 13 shillings, 6 pence, and 1 farthing. And
1 farth. × 13 is 13 farth. or £0 . 0 . 3¼ (219)
6 pence × 13 is 78 pence, or 0. 6.6
13 shill. × 13 is 169 shill. or 8. 9.0
7 pounds × 13 is 91 pounds, or 91 . 0 . 0
The sum of all these is £99 . 15 . 9¼
which is therefore £7. 13. 6¼ × 13.
This process is usually written as follows:

£ 7 . 13 . 6¼
13
£99 . 15 . 9¼

228. Division is performed upon the same principle as in (74), viz. that if a quantity
be divided into any number of parts, and each part be divided by any number, the
different quotients added together will make up the quotient of the whole quantity
divided by that number. Suppose it required to divide £99. 15. 9¼ by 13. Since 99
divided by 13 gives the quotient 7, and the remainder 8, the quantity is made up of £13
× 7, or £91, and £8. 15. 9¼. The quotient of the first, 13 being the divisor, is £7: it
remains to find that of the second. Since £8 is 160s., £8. 15. 9¼ is 175s. 9¼d., and 175
divided by 13 gives the quotient 13, and the remainder 6; that is, 175s. 9¼d. is made up
of 169s. and 6s. 9¼d., the quotient of the first of which is 13s., and it remains to find
that of the second. Since 6s. is 72d., 6s. 9¼d. is 81¼d., and 81 divided by 13 gives the
quotient 6 and remainder 3; that is, 81¼d. is 78d. and 3¼d., of the first of which the
quotient is 6d. Again, since 3d. is ¹²/₄, or 12 farthings, 3¼d. is 13 farthings, the quotient
of which is 1 farthing, or ¼, without remainder. We have then divided £99. 15. 9¼ into
four parts, each of which is divisible by 13, viz. £91, 169s., 78d., and 13 farthings; so

that the thirteenth part of this quantity is £7. 13. 6¼. The whole process may be written
down as follows; and the same sort of process may be applied to the exercises which
follow:

£ s. d. £ s. d.
13)99 15 9¼(7 13 6¼
91
8
20
160 + 15 = 175
13
45
39
6
12
72 + 9 = 81
78
3
4
12 + 1 = 13
13
0

Here, each of the numbers 99, 175, 81, and 13, is divided by 13 in the usual way,
though the divisor is only written before the first of them.

EXERCISES.
2 cwt. 1 qr. 21 lbs. 7 oz. × 53 = 129 cwt. 1 qr. 16 lbs. 3 oz.
2ᵈ 4ʰ 3ᵐ 27ˢ × 109 = 236ᵈ 10ʰ 16ᵐ 3ˢ
£27 . 10 . 8 × 569 = £15666 . 9 . 4
£7 . 4 . 8 × 123 = £889 . 14
£166 × ₈/₃₃ = £40 . 4 . 10⁶/₃₃
£187 . 6 . 7 × ³/₁₀₀ = £5 . 12 . 4¾ ²/₂₅
4s. 6½d. × 1121 = £254 . 11 . 2½
4s. 4d. × 4260 = 6s. 6d. × 2840

229. Suppose it required to find how many times 1s. 4¼d. is contained in £3. 19.
10¾. The way to do this is to find the number of farthings in each. By 219, in the first
there are 65, and in the second 3835 farthings. Now, 3835 contains 65 59 times; and
therefore the second quantity is 59 times as great as the first. In the case, however, of
pounds, shillings, and pence, it would be best to use decimals of a pound, which will
give a sufficiently exact answer. Thus 1s. 4¼d. is £·067, and £3. 19. 10¾ is £3·994,
and 3·994 divided by ·067 is 3994 by 67, or 59⁴¹/₆₇. This is an extreme case, for the
smaller the divisor, the greater the effect of an error in a given place of decimals.

EXERCISES.
How many times does 6 cwt. 2 qrs. contain 1 qr. 14 lbs. 1 oz.? and 1ᵈ 2ʰ 0ᵐ 47ˢ
contain 3ᵐ 46ˢ?

Answer, 17·30758 and 414·367257.
If 2 cwt. 3 qrs. 1 lb. cost £150. 13. 10, how much does 1 lb. cost?

Answer, 9s. 9d. ¹³/₃₀₉.
A grocer mixes 2 cwt. 15 lbs. of sugar at 11d. per pound with 14 cwt. 3 lbs. at 5d.
per pound. At how much per pound must he sell the mixture so as not to lose by mixing
them?

Answer, 5d. ¾ ¹⁵³/₉₀₅.
230. There is a convenient method of multiplication called Practice. Suppose I ask,
How much do 153 tons cost if each ton cost £2. 15. 7½? It is plain that if this sum be
multiplied by 153, the product is the price of the whole. But this is also evident, that, if
I buy 153 tons at £2. 15. 7½ each ton, payment may be made by first putting down £2
for each ton, then 10s. for each, then 5s., then 6d., and then 1½d. These sums together
make up £2. 15. 7½, and the reason for this separation of £2. 15 . 7½ into different
parts will be soon apparent. The process may be carried on as follows:
1. 153 tons, at £2 each ton, will cost

£306 0 0
2. Since 10s. is £½, 153 tons, at 10s. each, will cost £15³/₂,
which is

76 10 0
3. Since 5s. is ½ of 10s., 153 tons, at 5s., will cost half as
much as the same number at 10s. each, that is, ½ of £76 .
10, which is

38 5 0
4. Since 6d. is ⅒ of 5s., 153 tons, at 6d. each, will cost ⅒ of
what the same number costs at 5s. each, that is, ⅒ of £38 .
5, which is

3 16 6
5. Since 1½ or 3 halfpence is ¼ of 6d. or 12 halfpence, 153
tons, at 1½d. each, will cost ¼ of what the same number
costs at 6d. each, that is, ¼ of £3 . 16 . 6, which is

0 19 1½

The sum of all these quantities is 425 10 7½
which is, therefore, £2 . 15 . 7½ × 153.
The whole process may be written down as follows:
or what
153 tons
would
cost at
£153 0 0 £1 per ton.
£2 is 2 × £1 306 0 0 2 0 0
10s. is ½ of £1 76 10 0 0 10 0
10s. is ½ of £1 76 10 0 0 10 0
5s. is ½ of 10s. 38 5 0 0 5 0
6d. is ⅒ of 5s. 3 16 6 0 0 6
1½d. is ¼ of 6d. 0 19 1½ 0 0 1½
Sum £425 10 7½ 0 0 1½

ANOTHER EXAMPLE.
What do 1735 lbs. cost at 9s. 10¾d. per lb.? The price 9s. 10¾d. is made up of 5s.,
4s., 10d., ½d., and ¼d.; of which 5s. is ¼ of £1, 4s. is ⅕ of £1, 10d. is ⅙ of 5s., ½d. is
¹/₂₀ of 10d., and ¼d. is ½ of ½d. Follow the same method as in the last example, which
gives the following:
or what
1735 tons

would
cost at
£1735 0 0 £1 per ton.
5s. is ¼ of £1 433 15 0 0 5 0
4s. is ⅕ of £1 347 0 0 0 4 0
10d. is ⅙ of 5s. 72 5 10 0 0 10
½d. is ¹/₂₀ of 10d. 3 12 3½ 0 0 0½
¼d. is ½ of ½d. 1 16 1¾ 0 0 0¼
by addition ... £858 9 3¼ £0 9 10¾
In all cases, the price must first be divided into a number of parts, each of which is
a simple fraction[47] of some one which goes before. No rule can be given for doing
this, but practice will enable the student immediately to find out the best method for
each case. When that is done, he must find how much the whole quantity would cost if
each of these parts were the price, and then add the results together.

EXERCISES.
What is the cost of
243 cwt. at £14 . 18 . 8¼ per cwt.?—Answer, £3629 . 1 . 0¾.
169 bushels at £2 . 1 . 3¼ per bushel?—Answer, £348 . 14 . 9¼.
273 qrs. at 19s. 2d. per quarter?—Answer, £261 . 12. 6.
2627 sacks at 7s. 8½d. per sack?—Answer, £1012 . 9 . 9½.

231. Throughout this section it must be observed, that the rules can be applied to
cases where the quantities given are expressed in common or decimal fractions, instead
of the measures in the tables. The following are examples:
What is the price of 272·3479 cwt. at £2. 1. 3½ per cwt.?

Answer, £562·2849, or
£562. 5. 8¼. 66½lbs. at 1s. 4½d. per lb. cost £4. 11. 5¼.
How many pounds, shillings, and pence, will 279·301 acres let for if each acre lets
for £3·1076?—Answer, £867·9558, or £867. 19. 1¼.
What does ¼ of ³/₁₃ of 17 bush. cost at ⅙ of ⅔ of £17. 14 per bushel?

Answer, £2·3146, or £2. 6. 3½.

What is the cost of 19lbs. 8oz. 12dwt. 8gr. at £4. 4. 6 per ounce?—Answer, £999.
14. 1¼ ⅙.
232. It is often required to find to how much a certain sum per day will amount in a
year. This may be shortly done, since it happens that the number of days in a year is
240 + 120 + 5; so that a penny per day is a pound, half a pound, and 5 pence per year.
Hence the following rule: To find how much any sum per day amounts to in a year,
turn it into pence and fractions of a penny; to this add the half of itself, and let the
pence be pounds, and each farthing five shillings; then add five times the daily sum,
and the total is the yearly amount. For example, what does 12s. 3¾d. amount to in a
year? This is 147¾d., and its half is 73⅞d., which added to 147¾d. gives 221⅝d.,
which turned into pounds is £221. 12. 6. Also, 12s. 3¾d. × 5 is £3. 1. 6¾, which added
to the former sum gives £224. 14. 0¾ for the yearly amount. In the same way the
yearly amount of 2s. 3½d. is £41. 16. 5½; that of 6¾d. is £10. 5. 3¾; and that of 11d. is
£16. 14. 7.
233. An inverse rule may be formed, sufficiently correct for every purpose, in the
following way: If the year consisted of 360 days, or ³/₂ of 240, the subtraction of one-
third from any sum per year would give the proportion which belongs to 240 days; and
every pound so obtained would be one penny per day. But as the year is not 360, but
365 days, if we divide each day’s share into 365 parts, and take 5 away, the whole of
the subtracted sum, or 360 × 5 such parts, will give 360 parts for each of the 5 days
which we neglected at first. But 360 such parts are left behind for each of the 360 first
days; therefore, this additional process divides the whole annual amount equally among
the 365 days. Now, 5 parts out of 365 is one out of 73, or the 73d part of the first result
must be subtracted from it to produce the true result. Unless the daily sum be very
large, the 72d part will do equally well, which, as 72 farthings are 18 pence, is
equivalent to subtracting at the rate of one farthing for 18d., or ½d. for 3s., or 10d. for
£3. The rule, then, is as follows: To find how much per day will produce a given sum
per year, turn the shillings, &c. in the given sum into decimals of a pound (221);
subtract one-third; consider the result as pence; and diminish it by one farthing for
every eighteen pence, or ten pence for every £3. For example, how much per day will
give £224. 14. 0¾ per year? This is 224·703, and its third is 74·901, which subtracted
from 224·703, gives 149·802, which, if they be pence, amounts to 12s. 5·802d., in
which 1s. 6d. is contained 8 times. Subtract 8 farthings, or 2d., and we have 12s.
3·802d., which differs from the truth only about ¹/₂₀ of a farthing. In the same way,
£100 per year is 5s. 5¾d. per day.
234. The following connexion between the measures of length and the measures of
surface is the foundation of the application of arithmetic to geometry.

Suppose an oblong figure, a, b, c, d, as here drawn (which is called a rectangle in
geometry), with the side a b 6 inches, and the side a c 4 inches. Divide a b and c d
(which are equal) each into 6 inches by the points a, b, c, l, m, &c.; and a c and b d
(which are also equal) into 4 inches by the points f, g, h, x, y, and z. Join a and l, b and
m, &c., and f and x, &c. Then, the figure a b c d is divided into a number of squares;
for a square is a rectangle whose sides are equal, and therefore a a f e is square, since a
a is of the same length as a f, both being 1 inch. There are also four rows of these
squares, with six squares in each row; that is, there are 6 × 4, or 24 squares altogether.
Each of these squares has its sides 1 inch in length, and is what was called in (215) a
square inch. By the same reasoning, if one side had contained 6 yards, and the other 4
yards, the surface would have contained 6 × 4 square yards; and so on.
235. Let us now suppose that the sides of a b c d, instead of being a whole number
of inches, contain some inches and a fraction. For example, let a b be 3½ inches, or
(114) ⁷/₂ of an inch, and let a c contain 2½ inches, or ⁹/₄ of an inch. Draw a e twice as
long as a b, and a f four times as long as a c, and complete the rectangle a e f g. The
rest of the figure needs no description. Then, since a e is twice a b, or twice ⁷/₂ inches,
it is 7 inches. And since a f is four times a c, or four times ⁹/₄ inches, it is 9 inches.
Therefore, the whole rectangle a e f g contains, by (234), 7 × 9 or 63 square inches.
But the rectangle a e f g contains 8 rectangles, all of the same figure as a b c d; and
therefore a b c d is one-eighth part of a e f g, and contains ⁶³/₈ square inches. But ⁶³/₈ is
made by multiplying ⁹/₄ and ⁷/₂ together (118). From this and the last article it appears,
that, whether the sides of a rectangle be a whole or a fractional number of inches, the

number of square inches in its surface is the
product of the numbers of inches in its sides.
The square itself is a rectangle whose sides are
all equal, and therefore the number of square
inches which a square contains is found by
multiplying the number of inches in its side by
itself. For example, a square whose side is 13
inches in length contains 13 × 13 or 169 square
inches.

236. EXERCISES.
What is the content, in square feet and
inches, of a room whose sides are 42 ft. 5 inch.
and 31 ft. 9 inch.? and supposing the piece
from which its carpet is taken to be three
quarters of a yard in breadth, what length of it
must be cut off?—Answer, The content is 1346
square feet 105 square inches, and the length of
carpet required is 598 feet 6⁵/₉ inches.
The sides of a rectangular field are 253
yards and a quarter of a mile; how many acres
does it contain?—Answer, 23.
What is the difference between 18 square
miles, and a square of 18 miles long, or 18
miles square?—Answer, 306 square miles.
237. It is by this rule that the measure in (215) is deduced from that in (214); for it
is evident that twelve inches being a foot, the square foot is 12 × 12 or 144 square
inches, and so on. In a similar way it may be shewn that the content in cubic inches of a
cube, or parallelepiped,[48] may be found by multiplying together the number of inches
in those three sides which meet in a point. Thus, a cube of 6 inches contains 6 × 6 × 6,
or 216 cubic inches; a chest whose sides are 6, 8, and 5 feet, contains 6 × 8 × 5, or 240
cubic feet. By this rule the measure in (216) was deduced from that in (214).

SECTION II.
RULE OF THREE.

238. Suppose it required to find what 156 yards will cost, if 22 yards cost 17s. 4d.
This quantity, reduced to pence, is 208d.; and if 22 yards cost 208d., each yard costs
²⁰⁸/₂₂d. But 156 yards cost 156 times the price of one yard, and therefore cost

208 208 × 156
× 156 pence, or pence (117).
22 22
Again, if 25½ French francs be 20 shillings sterling, how many francs are in £20. 15?
Since 25½ francs are 20 shillings, twice the number of francs must be twice the number
of shillings; that is, 51 francs are 40 shillings, and one shilling is the fortieth part of 51
francs, or ⁵¹/₄₀ francs. But £20 15s. contain 415 shillings (219); and since 1 shilling is
⁵¹/₄₀ francs, 415 shillings is
51 × 415
⁵¹/₄₀ × 415 francs, or (117) francs.
40
239. Such questions as the last two belong to the most extensive rule in
Commercial Arithmetic, which is called the Rule of Three, because in it three
quantities are given, and a fourth is required to be found. From both the preceding
examples the following rule may be deduced, which the same reasoning will shew to
apply to all similar cases.
It must be observed, that in these questions there are two quantities which are of the
same sort, and a third of another sort, of which last the answer must be. Thus, in the
first question there are 22 and 156 yards and 208 pence, and the thing required to be
found is a number of pence. In the second question there are 20 and 415 shillings and
25½ francs, and what is to be found is a number of francs. Write the three quantities in
a line, putting that one last which is the only one of its kind, and that one first which is
connected with the last in the question.[49] Put the third quantity in the middle. In the
first question the quantities will be placed thus:
22 yds. 156 yds. 17s. 4d.
In the second question they will be placed thus:
20s. £20 15s. 25½ francs.
Reduce the first and second quantities, if necessary, to quantities of the same
denomination. Thus, in the second question, £20 15s. must be reduced to shillings
(219). The third quantity may also be reduced to any other denomination, if
convenient; or the first and third may be multiplied by any quantity we please, as was
done in the second question; and, on looking at the answer in (238), and at (108), it will
be seen that no change is made by that multiplication. Multiply the second and third
quantities together, and divide by the first. The result is a quantity of the same sort as
the third in the line, and is the answer required. Thus, to the first question the answer is
(238)
208 × 156 17s. 4d. × 156
pence, or, which is the same thing, .
22 22

240. The whole process in the first question is as follows:[50]

yds. yds. s. d.
22 : 156 ∷ 17 . 4
12
208 pence.
156
1248
1040
208
22)32448(1474¾d. and ¹⁴/₂₂, or ⁷/₁₁ of a farthing,
22 or (219) £6 . 2 . 10¾-⁷/₁₁.
104
88
164
154
108
88
20
(228) 4
80
66
14

The question might have been solved without reducing 17s. 4d. to pence, thus:

yds. yds. s. d.
22 : 156 ∷ 17 . 4
156 (227)
22) £135 . 4 . 0(£6 . 2 . 10¾-⁷/₁₁ (228)
132
3 × 20 + 4 = 64
44
20 × 12 = 240
220
20 × 4 = 80
66

14

The student must learn by practice which is the most convenient method for any
particular case, as no rule can be given.
241. It may happen that the three given quantities are all of one denomination;
nevertheless it will be found that two of them are of one, and the third of another sort.
For example: What must an income of £400 pay towards an income-tax of 4s. 6d. in
the pound? Here the three given quantities are, £400, 4s. 6d., and £1, which are all of
the same species, viz. money. Nevertheless, the first and third are income; the second is
a tax, and the answer is also a tax; and therefore, by (152), the quantities must be
placed thus:
£1 : £400 ∷ 4s. 6d.
242. The following exercises either depend directly upon this rule, or can be shewn
to do so by a little consideration. There are many questions of the sort, which will
require some exercise of ingenuity before the method of applying the rule can be
found.

EXERCISES.
If 15 cwt. 2 qrs. cost £198. 15. 4, what does 1 qr. 22 lbs. cost?

Answer, £5 . 14 . 5 ¾ ¹⁸⁵/₂₁₇.
If a horse go 14 m. 3 fur. 27 yds. in 3ʰ 26ᵐ 12ˢ, how long will he be in going 23
miles?

Answer, 5ʰ 29ᵐ 34ˢ(²⁴⁶²/₂₅₃₂₇).
Two persons, A and B, are bankrupts, and owe exactly the same sum; A can pay
15s. 4½d. in the pound, and B only 7s. (6¾)d. At the same time A has in his possession
£1304. 17 more than B; what do the debts of each amount to?

Answer, £3340 . 8 . 3 ¾ ⁹/₂₅.
For every (12½) acres which one country contains, a second contains (56¼). The
second country contains 17,300 square miles. How much does the first contain? Again,
for every 3 people in the first, there are 5 in the second; and there are in the first 27
people on every 20 acres. How many are there in each country?—Answer, The number
of square miles in the first is 3844⁴/₉, and its population 3,321,600; and the population
of the second is 5,536,000.
If (42½) yds. of cloth, 18 in. wide, cost £59. 14. 2, how much will (118¼) yds. cost,
if the width be 1 yd.?

Answer, £332. 5. (2⁴/₁₇).
If £9. 3. 6 last six weeks, how long will £100 last?

Answer, (65¹⁴⁵/₃₆₇) weeks.
How much sugar, worth (9¾d). a pound, must be given for 2 cwt. of tea, worth 10d.
an ounce?

Answer, 32 cwt. 3 qrs. 7 lbs. ³⁵/₃₉.
243. Suppose the following question asked: How long will it take 15 men to do that
which 45 men can finish in 10 days? It is evident that one man would take 45 × 10, or
450 days, to do the same thing, and that 15 men would do it in one-fifteenth part of the
time which it employs one man, that is, in (450 ÷ 15) or 30 days. By this and similar
reasoning the following questions can be solved.

EXERCISES.
If 15 oxen eat an acre of grass in 12 days, how long will it take 26 oxen to eat 14
acres?

Answer, (96¹²/₁₃) days.
If 22 masons build a wall 5 feet high in 6 days, how long will it take 43 masons to
build 10 feet?

Answer, (6⁶/₄₃) days.
244. The questions in the preceding article form part of a more general class of
questions, whose solution is called the Double Rule of Three, but which might, with
more correctness, be called the Rule of Five, since five quantities are given, and a sixth
is to be found. The following is an example: If 5 men can make 30 yards of cloth in 3
days, how long will it take 4 men to make 68 yards? The first thing to be done is to find
out, from the first part of the question, the time it will take one man to make one yard.
Now, since one man, in 3 days, will do the fifth part of what 5 men can do, he will in 3
days make ³⁰/₅, or 6 yards. He will, therefore, make one yard in ³/₆6 or (3 × 5)/30 of a
day. From this we are to find how long it will take 4 men to make 68 yards. Since one
man makes a yard in
3×5 3×5
of a day, he will make 68 yards in × 68 days,
30 30
3 × 5 × 68
or (116) in days; and 4 men will do this in one-fourth;
30
of the time, that is (123), in 3 × 5 × 68 days, or in 8½ days.

30 × 4
Again, suppose the question to be: If 5 men can make 30 yards in 3 days, how
much can 6 men do in 12 days? Here we must first find the quantity one man can do in
one day, which appears, on reasoning similar to that in the last example, to be 30/(3 ×
5) yards. Hence, 6 men, in one day, will make
6 × 30 12 × 6 × 30
yards, and in 12 days will make or 144 yards.
5×3 5×3
From these examples the following rule may be drawn. Write the given quantities
in two lines, keeping quantities of the same sort under one another, and those which are
connected with each other, in the same line. In the two examples above given, the
quantities must be written thus:

SECOND EXAMPLE.

Draw a curve through the middle of each line, and the extremities of the other.
There will be three quantities on one curve and two on the other. Divide the product of
the three by the product of the two, and the quotient is the answer to the question.
If necessary, the quantities in each line must be reduced to more simple
denominations (219), as was done in the common Rule of Three (238).

EXERCISES.
If 6 horses can, in 2 days, plough 17 acres, how many acres will 93 horses plough
in 4½ days?

Answer, 592⅞.
If 20 men, in 3¼ days, can dig 7 rectangular fields, the sides of each of which are
40 and 50 yards, how long will 37 men be in digging 53 fields, the sides of each of
which are 90 and 125½ yards?
2451
Answer, 75 20720 days.

If the carriage of 60 cwt. through 20 miles cost £14 10s., what weight ought to be
carried 30 miles for £5. 8. 9?

Answer, 15 cwt.
If £100 gain £5 in a year, how much will £850 gain in 3 years and 8 months?

Answer, £155. 16. 8.

SECTION III.
INTEREST, ETC.

245. In the questions contained in this Section, almost the only process which will
be employed is the taking a fractional part of a sum of money, which has been done
before in several cases. Suppose it required to take 7 parts out of 40 from £16, that is,
to divide £16 into 40 equal parts, and take 7 of them. Each of these parts is
16 16 16 × 7
£ 40 and 7 of them make 40 × 7, or 40
pounds (116).

The process may be written as below:

£16
7
40)112(£2 . 16s.
80
32
20
640
40

240
240
0

Suppose it required to take 13 parts out of a hundred from £56. 13. 7½.

56 . 13 . 7½
13
100) 736 . 17 . 1½ ( £7 . 7 . 4 ¼ ¹/₄₁
700
36 × 20 + 17 = 737
700
37 × 12 + 1 = 445
400
45 × 4 × 2 = 182
100
82

Let it be required to take 2½ parts out of a hundred from £3 12s. The result, by the
same rule is
£3 12s. × 2½ 5
, or 123 £3 12s. × ;
100 200
so that taking 2½ out of a hundred is the same as taking 5 parts out of 200.

EXERCISES.
Take 7⅓ parts out of 53 from £1 10s.
129
Answer, 4s. 1 d.
159
Take 5 parts out of 100 from £107 13s. 4¾d.

Answer, £5. 7. 8 and ³/₂₀ of a farthing.
£56 3s. 2d. is equally divided among 32 persons. How much does the share of 23 of
them exceed that of the rest?

Answer, £24. 11. 4½ ½.
246. It is usual, in mercantile business, to mention the fraction which one sum is of
another, by saying how many parts out of a hundred must be taken from the second in

order to make the first. Thus, instead of saying that £16 12s. is the half of £33 4s., it is
said that the first is 50 per cent of the second. Thus, £5 is 2½ per cent of £200; because,
if £200 be divided into 100 parts, 2½ of those parts are £5. Also, £13 is 150 per cent of
£8. 13. 4, since the first is the second and half the second. Suppose it asked, How much
per cent is 23 parts out of 56 of any sum? The question amounts to this: If he who has
£56 gets £100 for them, how much will he who has 23 receive? This, by 238, is 23 ×
¹⁰⁰/₅₆ or ²³⁰⁰/₅₆ or 41¹/₁₄. Hence, 23 out of 56 is 41¹/₁₄ per cent.
Similarly 16 parts out of 18 is 16 × ¹⁰⁰/₁₈, or 88⁸/₉ per cent, and 2 parts out of 5 is 2
× ¹⁰⁰/₅, or 40 per cent.
From which the method of reducing other fractions to the rate per cent is evident.
Suppose it asked, How much per cent is £6. 12. 2 of £12. 3? Since the first contains
1586d., and the second 2916d., the first is 1586 out of 2916 parts of the second; that is,
by the last rule, it is ¹⁵⁸⁶⁰⁰/₂₉₁₆, or 54¹¹³⁶/₂₉₁₆, or £54. 7. 9½ per cent, very nearly. The
more expeditious way of doing this is to reduce the shillings, &c. to decimals of a
pound. Three decimal places will give the rate per cent to the nearest shilling, which is
near enough for all practical purposes. For instance, in the last example, which is to
find how much £6·608 is of £12·15, 6·608 × 100 is 660·8, which divided by 12·15
gives £54·38, or £54. 7. Greater correctness may be had, if necessary, as in the
Appendix.

EXERCISES.
How much per cent is 198¼ out of 233 parts?—Ans. £85. 1. 8¾.
Goods which are bought for £193. 12, are sold for £216. 13. 4; how much per cent
has been gained by them?

Answer, A little less than £11. 18. 6.
A sells goods for B to the amount of £230. 12, and is allowed a commission[51] of 3
per cent; what does that amount to?

Answer, £6 . 18. 4¼ ⁷/₂₅.
A stockbroker buys £1700 stock, brokerage being at £⅛ per cent; what does he
receive?—Answer, £2. 2. 6.
A ship whose value is £15,423 is insured at 19⅔ per cent; what does the insurance
amount to?—Answer, £3033. 3. 9½ ²/₅.
247. In reckoning how much a bankrupt is able to pay his creditors, as also to how
much a tax or rate amounts, it is usual to find how many shillings in the pound is paid.
Thus, if a person who owes £100 can only pay £50, he is said to pay 10s. in the pound.

The rule is easily derived from the same reasoning as in 246. For example, £50 out of
£82 is
50 50 × 20
£ out of £1, or shillings,
82 82
or 12s. 2½ ¹⁵/₄₁ in the pound.
248. Interest is money paid for the use of other money, and is always a per-centage
upon the sum lent. It may be paid either yearly, half-yearly, or quarterly; but when it is
said that £100 is lent at 4 per cent, it must be understood to mean 4 per cent per annum;
that is, that 4 pounds are paid every year for the use of £100.
The sum lent is called the principal, and the interest upon it is of two kinds. If the
borrower pay the interest as soon as, from the agreement, it becomes due, it is evident
that he has to pay the same sum every year; and that the whole of the interest which he
has to pay in any number of years is one year’s interest multiplied by the number of
years. But if he do not pay the interest at once, but keeps it in his hands until he returns
the principal, he will then have more of his creditor’s money in his hands every year,
and if it were so agreed will have to pay interest upon each year’s interest for the time
during which he keeps it after it becomes due. In the first case, the interest is called
simple, and in the second compound. The interest and principal together are called the
amount.
249. What is the simple interest of £1049. 16. 6 for 6 years and one-third, at 4½ per
cent? This interest must be 6⅓ times the interest of the same sum for one year, which
(245) is found by multiplying the sum by 4½, and dividing by 100. The process is as
follows:
(230) (a) £1049 .16 . 6
a × 4 4199 . 6 . 0
a×½ 524 .18 . 3
(82) 100) 47,24 . 4 . 3(£47 . 4 . 10¹¹/₁₀₀
20
(228) 4,84 [52]
12
10,11 [53]

(b) £47 . 4 . 10¹¹/₁₀₀ Int. for one yr.
b×6 283 . 9 . 0⁶⁶/₁₀₀

b×⅓ 15 . 14 . 11³⁷/₁₀₀
£299 . 4 . 0³/₁₀₀ Int. for 6⅓ yrs.

EXERCISES.
What is the interest of £105. 6. 2 for 19 years and 7 weeks at 3 per cent?

Answer, £60. 9, very nearly.
What is the difference between the interest of £50. 19 for 7 years at 3 per cent, and
for 8 years at 2½ per cent?

Answer, 10s. (2½)d.
What is the interest of £157. 17. 6 for one year at 5 per cent?

Answer, £7. 17. 10½.
Shew that the interest of any sum for 9 years at 4 per cent is the same as that of the
same sum for 4 years at 9 per cent.
250. In order to find the interest of any sum at compound interest, it is necessary to
find the amount of the principal and interest at the end of every year; because in this
case (248) it is the amount of both principal and interest at the end of the first year,
upon which interest accumulates during the second year. Suppose, for example, it is
required to find the interest, for 3 years, on £100, at 5 per cent, compound interest. The
following is the process:

£100 First principal.
5 First year’s interest.
105 Amount at the end of the first year.
(249) 5 . 5 Interest for the second year on £105.
110 . 5 Amount at the end of two years.
5 . 10 . 3 Interest due for the third year.
115 . 15 . 3 Amount at the end of three years.
100 . 0 . 0 First principal.
15 . 15 . 3 Interest gained in the three years.
When the number of years is great, and the sum considerable, this process is very
troublesome; on which account tables[54] are constructed to shew the amount of one
pound, for different numbers of years, at different rates of interest. To make use of
these tables in the present example, look into the column headed “5 per cent;” and
opposite to the number 3, in the column headed “Number of years,” is found 1·157625;

meaning that £1 will become £1·157625 in 3 years. Now, £100 must become 100 times
as great; and 1·157625 × 100 is 115·7625 (141); but (221) £·7625 is 15s. 3d.; therefore
the whole amount of £100 is £115. 15. 3, as before.
251. Suppose that a sum of money has lain at simple interest 4 years, at 5 per cent,
and has, with its interest, amounted to £350; it is required to find what the sum was at
first. Whatever the sum was, if we suppose it divided into 100 parts, 5 of those parts
were added every year for 4 years, as interest; that is, 20 of those parts have been added
to the first sum to make £350. If, therefore, £350 be divided into 120 parts, 100 of
those parts are the principal which we want to find, and 20 parts are interest upon it;
that is, the principal is £(350 × 100)/150, or £291. 13. 4.
252. Suppose that A was engaged to pay B £350 at the end of four years from this
time, and that it is agreed between them that the debt shall be paid immediately;
suppose, also, that money can be employed at 5 per cent, simple interest; it is plain that
A ought not to pay the whole sum, £350, because, if he did, he would lose 4 years’
interest of the money, and B would gain it. It is fair, therefore, that he should only pay
to B as much as will, with interest, amount in four years to £350, that is (251), £291.
13. 4. Therefore, £58. 6. 8 must be struck off the debt in consideration of its being paid
before the time. This is called Discount;[55] and £291. 13. 4 is called the present value
of £350 due four years hence, discount being at 5 per cent. The rule for finding the
present value of a sum of money (251) is: Multiply the sum by 100, and divide the
product by 100 increased by the product of the rate per cent and number of years. If the
time that the debt has yet to run be expressed in years and months, or months only, the
months must be reduced to the equivalent fraction of a year.

EXERCISES.
What is the discount on a bill of £138. 14. 4, due 2 years hence, discount being at
4½ per cent?

Answer, £11. 9. 1.
What is the present value of £1031. 17, due 6 months hence, interest being at 3 per
cent?

Answer, £1016. 12.
253. If we multiply by a + b, or by a-b, when we should multiply by a, the result is
wrong by the fraction
b b
+ b, or ,
a a-b
of itself: being too great in the first case, and too small in the second. Again, if we
divide by a + b, where we should have divided by a, the result is too small by the

fraction b/a of itself; while, if we divide by a-b instead of a, the result is too great by
the same fraction of itself. Thus, if we divide by 20 instead of 17, the result is ³/₁₇ of
itself too small; and if we divide by 360 instead of 365, the result is too great by ⁵/₃₆₅,
or ¹/₇₃ of itself.
If, then, we wish to find the interest of a sum of money for a portion of a year, and
have not the assistance of tables, it will be found convenient to suppose the year to
contain only 360 days, in which case its 73d part (the 72d part will generally do) must
be subtracted from the result, to make the alteration of 360 into 365. The number 360
has so large a number of divisors, that the rule of Practice (230) may always be readily
applied. Thus, it is required to find the portion which belongs to 274 days, the yearly
interest being £18. 9. 10, or 18·491.
274 18·491
180 is ½ of 360 9·246
94
90 is ½ of 180 4·623
4 is ¹/₉₀ of 360 ·205
9)14·074
8)1·564
·196
13·878 = £13 . 17 . 7 Answer.
But if the nearest farthing be wanted, the best way is to take 2-tenths of the number
of days as a multiplier, and 73 as a divisor; since m ÷ 365 is 2m ÷ 730, or (²/₁₀)m ÷ 73.
Thus, in the preceding instance, we multiply by 54·8 and divide by 73; and 54·8 ×
18·491 = 1013·3068, which divided by 73 gives 13·881, very nearly agreeing with the
former, and giving £13. 17. 7½, which is certainly within a farthing of the truth.
254. Suppose it required to divide £100 among three persons in such a way that
their shares may be as 6, 5, and 9; that is, so that for every £6 which the first has, the
second may have £5, and the third £9. It is plain that if we divide the £100 into 6 + 5 +
9, or 20 parts, the first must have 6 of those parts, the second 5, and the third 9.
Therefore (245) their shares are respectively,
100 × 6 100 × 5 100 × 9
£ ,£ and £
20 20 20
or £30, £25, and £45.

EXERCISES.

Divide £394. 12 among four persons, so that their shares may be as 1, 6, 7, and 18.
—Answer, £12. 6. 7½; £73. 19. 9; £86. 6. 4½; £221. 19. 3.
Divide £20 among 6 persons, so that the share of each may be as much as those of
all who come before put together.—Answer, The first two have 12s. 6d.; the third £1. 5;
the fourth £2. 10; the fifth £5; and the sixth £10.
255. When two or more persons employ their money together, and gain or lose a
certain sum, it is evidently not fair that the gain or loss should be equally divided
among them all, unless each contributed the same sum. Suppose, for example, A
contributes twice as much as B, and they gain £15, A ought to gain twice as much as B;
that is, if the whole gain be divided into 3 parts, A ought to have two of them and B
one, or A should gain £10 and B £5. Suppose that A, B, and C engage in an adventure,
in which A embarks £250, B £130, and C £45. They gain £1000. How much of it ought
each to have? Each one ought to gain as much for £1 as the others. Now, since there are
250 + 130 + 45, or 425 pounds embarked, which gain £1000, for each pound there is a
gain of £¹⁰⁰⁰/₄₂₄. Therefore A should gain 1000 × ²⁵⁰/₄₂₅ pounds, B should gain 1000 ×
¹³⁰/₄₂₅ pounds, and C 1000 × ⁴⁵/₄₂₅ pounds. On these principles, by the process in
(245), the following questions may be answered.
A ship is to be insured, in which A has ventured £1928, and B £4963. The expense
of insurance is £474. 10. 2. How much ought each to pay of it?
Answer, A must pay £132. 15. (2½).
A loss of £149 is to be made good by three persons, A, B, and C. Had there been a
gain, A would have gained 4 times as much as B, and C as much as A and B together.
How much of the loss must each bear?
Answer, A pays £59. 12, B £14. 18, and C £74. 10.
256. It may happen that several individuals employ several sums of money together
for different times. In such a case, unless there be a special agreement to the contrary, it
is right that the more time a sum is employed, the more profit should be made upon it.
If, for example, A and B employ the same sum for the same purpose, but A’s money is
employed twice as long as B’s, A ought to gain twice as much as B. The principle is,
that one pound employed for one month, or one year, ought to give the same return to
each. Suppose, for example, that A employs £3 for 6 months, B £4 for 7 months, and C
£12 for 2 months, and the gain is £100; how much ought each to have of it? Now, since
A employs £3 for six months, he must gain 6 times as much as if he employed it one
month only; that is, as much as if he employed £6 × 3, or £18, for one month; also, B
gains as much as if he had employed £4 × 7 for one month; and C as if he had
employed £12 × 2 for one month. If, then, we divide £100 into 6 × 3 + 4 × 7 + 12 × 2,
or 70 parts, A must have 6 × 3, or 18, B must have 4 × 7, or 28, and C 12 × 2, or 24 of
those parts. The shares of the three are, therefore,

6 × 3 × 100 4 × 7 × 100 12 × 2 × 100
£ ,£ , and £
6 × 3 + 4 × 7 + 12 × 6 × 3 + 4 × 7 + 12 × 2 6 × 3 + 4 × 7 + 12 × 2

EXERCISES.
A, B, and C embark in an undertaking; A placing £3. 6 for 2 years, B £100 for 1
year, and C £12 for 1½ years. They gain £4276. 7 How much must each receive of the
gain?

Answer, A £226. 10. 4; B £3432. 1. 3; C £617. 15. 5.
A, B, and C rent a house together for 2 years, at £150 per annum. A remains in it
the whole time, B 16 months, and C 4½ months, during the occupancy of B. How
much must each pay of the rent?[56]

Answer, A should pay £190. 12. 6; B £90. 12. 6; C £18. 15.
257. These are the principal rules employed in the application of arithmetic to
commerce. There are others, which, as no one who understands the principles here laid
down can fail to see, are virtually contained in those which have been given. Such is
what is commonly called the Rule of Exchange, for such questions as the following: If
20 shillings be worth 25½ francs, in France, what is £160 worth? This may evidently
be done by the Rule of Three. The rules here given are those which are most useful in
common life; and the student who understands them need not fear that any ordinary
question will be above his reach. But no student must imagine that from this or any
other book of arithmetic he will learn precisely the modes of operation which are best
adapted to the wants of the particular kind of business in which his future life may be
passed. There is no such thing as a set of rules which are at once most convenient for a
butcher and a banker’s clerk, a grocer and an actuary, a farmer and a bill-broker; but a
person with a good knowledge of the principles laid down in this work, will be able to
examine and meet his own future wants, or, at worst, to catch with readiness the
manner in which those who have gone before him have done so for themselves.

APPENDIX TO
THE FIFTH EDITION OF
DE MORGAN’S ELEMENTS OF
ARITHMETIC.

I. ON THE MODE OF COMPUTING.

The rules in the preceding work are given in the usual form, and the
examples are worked in the usual manner. But if the student really wish to
become a ready computer, he should strictly follow the methods laid down
in this Appendix; and he may depend upon it that he will thereby save
himself trouble in the end, as well as acquire habits of quick and accurate
calculation.
I. In numeration learn to connect each primary decimal number, 10, 100,
1000, &c. not with the place in which the unit falls, but with the number of
ciphers following. Call ten a one-cipher number, a hundred a two-cipher
number, a million a six-cipher number, and so on. If five figures be cut off
from a number, those that are left are hundred-thousands; for 100,000 is a
five-cipher number. Learn to connect tens, hundreds, thousands, tens of
thousands, hundreds of thousands, millions, &c. with 1, 2, 3, 4, 5, 6, &c. in
the mind. What is a seventeen-cipher number? For every 6 in seventeen say
million, for the remaining 5 say hundred-thousand: the answer is a hundred
thousand millions of millions. If twelve places be cut off from the right of a
number, what does the remaining number stand for?—Answer, As many
millions of millions as there are units in it when standing by itself.
II. After learning to count forwards and backwards with rapidity, as in 1,
2, 3, 4, &c. or 30, 29, 28, 27, &c., learn to count forwards or backwards by
twos, threes, &c. up to nines at least, beginning from any number. Thus,
beginning from four and proceeding by sevens, we have 4, 11, 18, 25, 32,
&c., along which series you must learn to go as easily as along the series 1,
2, 3, 4, &c.; that is, as quick as you can pronounce the words. The act of
addition must be made in the mind without assistance: you must not permit
yourself to say, 4 and 7 are 11, 11 and 7 are 18, &c.; but only 4, 11, 18, &c.
And it would be desirable, though not so necessary, that you should go back
as readily as forward; by sevens for instance, from sixty, as in 60, 53, 46,
39, &c.
III. Seeing a number and another both of one figure, learn to catch
instantly the number you must add to the smaller to get the greater. Seeing 3
and 8, learn by practice to think of 5 without the necessity of saying 3 from
8 and there remains 5. And if the second number be the less, as 8 and 3,

learn also by practice how to pass up from 8 to the next number which ends
with 3 (or 13), and to catch the necessary augmentation, five, without the
necessity of formally undertaking in words to subtract 8 from 13. Take rows
of numbers, such as
4260501864
and practise this rule upon every figure and the next, not permitting yourself
in this simple case ever to name the higher one. Thus, say 4 and 8 (4 first, 2
second, 4 from the next number that ends with 2, or 12, leaves 8), 2 and 4, 6
and 4, 0 and 5, 5 and 5, 0 and 1, 1 and 7, 8 and 8, 6 and 8.
IV. Study the same exercise as the last one with two figures and one.
Thus, seeing 27 and 6, pass from 27 up to the next number that ends with 6
(or 36), catch the 9 through which you have to pass, and allow yourself to
repeat as much as “27 and 9 are 36.” Thus, the row of figures 17729638109
will give the following practice: 17 and 0 are 17; 77 and 5 are 82; 72 and 7
are 79; 29 and 7 are 36; 96 and 7 are 103; 63 and 5 are 68; 38 and 3 are 41;
81 and 9 are 90; 10 and 9 are 19.
V. In a number of two figures, practise writing down the units at the
moment that you are keeping the attention fixed upon the tens. In the
preceding exercise, for instance, write down the results, repeating the tens
with emphasis at the instant of writing down the units.
VI. Learn the multiplication table so well as to name the product the
instant the factors are seen; that is, until 8 and 7, or 7 and 8, suggest 56 at
once, without the necessity of saying “7 times 8 are 56.” Thus looking
along a row of numbers, as 39706548, learn to name the products of every
successive pair of digits as fast as you can repeat them, namely, 27, 63, 0, 0,
30, 20, 32.
VII. Having thoroughly mastered the last exercise, learn further, on
seeing three numbers, to augment the product of the first and second by the
third without any repetition of words. Practise until 3, 8, 4, for instance,
suggest 3 times 8 and 4, or 28, without the necessity of saying “3 times 8
are 24, and 4 is 28.” Thus, 179236408 will suggest the following practice,
16, 65, 21, 12, 22, 24, 8.
VIII. Now, carry the last still further, as follows: Seeing four figures, as
2, 7, 6, 9, catch up the product of the first and second, increased by the

third, as in the last, without a helping word; name the result, and add the
next figure, name the whole result, laying emphasis upon the tens. Thus, 2,
7, 6, 9, must immediately suggest “20 and 9 are 29.” The row of figures
773698974 will give the instances 52 and 6 are 58; 27 and 9 are 36; 27 and
8 are 35; 62 and 9 are 71; 81 and 7 are 88; 79 and 4 are 83.
IX. Having four numbers, as 2, 4, 7, 9, vary the last exercise as follows:
Catch the product of the first and second, increased by the third; but instead
of adding the fourth, go up to the next number that ends with the fourth, as
in exercise IV. Thus, 2, 4, 7, 9, are to suggest “15 and 4 are 19.” And the
row of figures 1723968929 will afford the instances 9 and 4 are 13; 17 and
2 are 19; 15 and 1 are 16; 33 and 5 are 38; 62 and 7 are 69; 57 and 5 are 62;
74 and 5 are 79.
X. Learn to find rapidly the number of times a digit is contained in
given units and tens, with the remainder. Thus, seeing 8 and 53, arrive at
and repeat “6 and 5 over.” Common short division is the best practice. Thus,
in dividing 236410792 by 7,

7)236410792
33772970, remainder 2.

All that is repeated should be 3 and 2; 3 and 5; 7 and 5; 7 and 2; 2 and 6; 9
and 4; 7 and 0; 0 and 2.
In performing the several rules, proceed as follows:
Addition. Not one word more than repeating the numbers written in the
following process: the accented figure is the one to be written down; the
doubly accented figure is carried (and don’t say “carry 3,” but do it).

47963 6, 15, 17, 23, 31, 3″ 4′;
1598 11, 12, 21, 22, 31, 3″7′;
26316 9, 17, 24, 27, 32, 4″1′;
54792 10, 14, 20, 21, 2″8′; 7, 9, 1′3′.
819
6686

138174
In verifying additions, instead of the usual way of omitting one line,
adding without it, and then adding the line omitted, verify each column by
adding it both upwards and downwards.
Subtraction. The following process is enough. The carriages, being
always of one, need not be mentioned.

From 79436258190
Take 58645962738
20790295452

8 and 2′, 4 and 5′, 7 and 4′, 3 and 5′, 6 and 9′, 10 and 2′, 6
and 0′, 4 and 9′, 7 and 7′, 9 and 0′, 5 and 2′. It is useless to
stop and say, 8 and 2 make 10; for as soon as the 2 is
obtained, there is no occasion to remember what it came
from.
Multiplication. The following, put into words, is all that need be
repeated in the multiplying part; the addition is then done as usual. The
unaccented figures are carried.

670383
9876
4022298 18′, 49′, 22′, 2′, 42′, 4′0′,
4692681 21′, 58′, 26′, 2′, 49′, 4′6′,
5363064 24′, 66′, 30′, 3′, 56′, 5′3′,
6033447 27′, 74′, 34′, 3′, 63′, 6′0′.
6620702508
Verify each line of the multiplication and the final result by casting out
the nines. (Appendix II. p. 166.)
It would be almost as easy, for a person who has well practised the 8th
exercise, to add each line to the one before in the process, thus:

670383
9876
4022298 8; 21 and 9 are 30′; 59 and 2
50949108 are 61′; 27 and 2 are 29; 2
587255508 and 2 are 4′; 49 and 0 are 49′;
6620702508 46 and 4 are 5′0′.
On the right is all the process of forming the second line, which
completes the multiplication by 76, as the third line completes that by 876,
and the fourth line that by 9876.
Division. Make each multiplication and the following subtraction in one
step, by help of the process in the 9th exercise, as follows:

27693)441972809662(15959730
165042
265778
165410
269459
202226
83756
6772

The number of words by which 26577 is obtained from 165402 (the
multiplier being 5) is as follows: 15 and 7′ are 2″2; 47 and 7′ are 5″4; 35
and 5′ are 4″0; 39 and 6′ are 4″5; 14 and 2′ are 16.
The processes for extracting the square root, and for the solution of
equations (Appendix XI.), should be abbreviated in the same manner as the
division.[57]

APPENDIX II.
ON VERIFICATION BY CASTING OUT
NINES AND ELEVENS.

The process of casting out the nines, as it is called, is one which the
young computer should learn and practise, as a check upon his
computations. It is not a complete check, since if one figure were made too
small, and another as much too great, it would not detect this double error;
but as it is very unlikely that such a double error should take place, the
check furnishes a strong presumption of accuracy.
The proposition upon which this method depends is the following: If a,
b, c, d be four numbers, such that
a = bc + d,
and if m be any other number whatsoever, and if a, b, c, d, severally divided
by m, give the remainders p, q, r, s, then
p and qr + s
give the same remainder when divided by m (and perhaps are themselves
equal).
For instance, 334 = 17 × 19 + 11;
divide these four numbers by 7, the remainders are 5, 3, 5, and 4. And 5 and
5 × 3 + 4, or 5 and 19, both leave the remainder 5 when divided by 7.
Any number, therefore, being used as a divisor, may be made a check
upon the correctness of an operation. To provide a check which may be
most fit for use, we must take a divisor the remainder to which is most
easily found. The most convenient divisors are 3, 9, and 11, of which 9 is
far the most useful.
As to the numbers 3 and 9, the remainder is always the same as that of
the sum of the digits. For instance, required the remainder of 246120377
divided by 9. The sum of the digits is 2 + 4 + 6 + 1 + 2 + 0 + 3 + 7 + 7, or
32, which gives the remainder 5. But the easiest way of proceeding is by
throwing out nines as fast as they arise in the sum. Thus, repeat 2, 6 (2 + 4),

12 (6 + 6), say 3 (throwing out 9), 4, 6, 9 (throw this away), 7, 14, (or
throwing out the 9) 5. This is the remainder required, as would appear by
dividing 246120377 by 9. A proof may be given thus: It is obvious that each
of the numbers, 1, 10, 100, 1000, &c. divided by 9, leaves a remainder 1,
since they are 1, 9 + 1, 99 + 1, &c. Consequently, 2, 20, 200, &c. leave the
remainder 2; 3, 30, 300, the remainder 3; and so on. If, then, we divide, say
1764 by 9 in parcels, 1000 will be one more than an exact number of nines,
700 will be seven more, and 60 will be six more. So, then, from 1, 7, 6, 4,
put together, and the nines taken out, comes the only remainder which can
come from 1764.
To apply this process to a multiplication: It is asserted, in page 32, that
10004569 × 3163 = 31644451747.
In casting out the nines from the first, all that is necessary to repeat is,
one, five, ten, one, seven; in the second, three, four, ten, one, four; in the
third, three, four, ten, one, five, nine, four, nine, eight, twelve, three, ten,
one. The remainders then are, 7, 4, 1. Now, 7 × 4 is 28, which, casting out
the nines, gives 1, the same as the product.
Again, in page 43, it is asserted that
23796484 = 130000 × 183 + 6484.
Cast out the nines from 13000, 183, 6484, and we have 4, 3, and 4. Now, 4
× 3 + 4, with the nines cast out, gives 7; and so does 23796484.
To avoid having to remember the result of one side of the equation, or to
write it down, in order to confront it with the result of the other side,
proceed as follows: Having got the remainder of the more complicated side,
into which two or more numbers enter, subtract it from 9, and carry the
remainder into the simple side, in which there is only one number. Then the
remainder of that side ought to be 0. Thus, having got 7 from the left-hand
of the preceding, take 2, the rest of 9, forget 7, and carry in 2 as a beginning
to the left-hand side, giving 2, 4, 7, 14, 5, 11, 2, 6, 14, 5, 9, 0.
Practice will enable the student to cast out nines with great rapidity.
This process of casting out the nines does not detect any errors in which
the remainder to 9 happens to be correct. If a process be tedious, and some
additional check be desirable, the method of casting out elevens may be

followed after that of casting out the nines. Observe that 10 + 1, 100-1,
1000 + 1, 10000-1, &c. are all divisible by eleven. From this the following
rule for the remainder of division by 11 may be deduced, and readily used
by those who know the algebraical process of subtraction. For those who
have not got so far, it may be doubted whether the rule can be made easier
than the actual division by 11.
Subtract the first figure from the second, the result from the third, the
result from the fourth, and so on. The final result, or the rest of 11 if the
figure be negative, is the remainder required. Thus, to divide 1642915 by
11, and find the remainder, we have 1 from 6, 5; 5 from 4, -1; -1 from 2, 3;
3 from 9, 6; 6 from 1, -5; -5 from 5, 10; and 10 is the remainder. But 164
gives-1, and 10 is the remainder; 164291 gives-5, and 6 is the remainder.
With very little practice these remainders may be read as rapidly as the
number itself. Thus, for 127619833424 need only be repeated, 1, 6, 0, 1, 8,
0, 3, 0, 4, -2, 6, and 6 is the remainder.
When a question has been tried both by nines and elevens, there can be
no error unless it be one which makes the result wrong by a number of
times 99 exactly.

APPENDIX III.
ON SCALES OF NOTATION.

We are so well accustomed to 10, 100, &c., as standing for ten, ten tens,
&c., that we are not apt to remember that there is no reason why 10 might
not stand for five, 100 for five fives, &c., or for twelve, twelve twelves, &c.
Because we invent different columns of numbers, and let units in the
different columns stand for collections of the units in the preceding
columns, we are not therefore bound to allow of no collections except in
tens.
If 10 stood for 2, that is, if every column had its unit double of the unit
in the column on the right, what we now represent by 1, 2, 3, 4, 5, 6, &c.,
would be represented by 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010,
1011, 1100, &c. This is the binary scale. If we take the ternary scale, in
which 10 stands for 3, we have 1, 2, 10, 11, 12, 20, 21, 22, 100, 101, 102,
110, &c. In the quinary scale, in which 10 is five, 234 stands for 2 twenty-
fives, 3 fives, and 4, or sixty-nine. If we take the duodenary scale, in which
10 is twelve, we must invent new symbols for ten and eleven, because 10
and 11 now stand for twelve and thirteen; use the letters t and e. Then 176
means 1 twelve-twelves, 7 twelves, and 6, or two hundred and thirty-four;
and 1te means two hundred and seventy-five.
The number which 10 stands for is called the radix of the scale of
notation. To change a number from one scale into another, divide the
number, written as in the first scale, by the number which is to be the radix
of the new scale; repeat this division again and again, and the remainders
are the digits required. For example, what, in the quinary scale, is that
number which, in the decimal scale, is 17036?

5)17036
5)3407 Remʳ. 1
5)681 2
5)136 1
5)27 1
5)5 2

5)1 0
0 1

Answer 1021121
Quinary. Decimal.
Verification, 1000000 means 15625
20000 1250
1000 125
100 25
20 10
1 1
1021121 17036
The reason of this rule is easy. Our process of division is nothing but
telling off 17036 into 3407 fives and 1 over; we then find 3407 fives to be
681 fives of fives and 2 fives over. Next we form 681 fives of fives into 136
fives of fives of fives and 1 five of fives over; and so on.
It is a useful exercise to multiply and divide numbers represented in
other scales of notation than the common or decimal one. The rules are in
all respects the same for all systems, the number carried being always the
radix of the system. Thus, in the quinary system we carry fives instead of
tens. I now give an example of multiplication and division:

Quinary. Decimal.
42143 means 2798
1234 194
324232 11192
232034 25182
134341 2798
42143
114332222 542812

Duodecimal. Decimal.
4t9)76t4e08(16687 705)22610744(32071
4t9 1460
2814 5074
2546 1394
28te 689
2546
3650
3320
3308
2t33
495
Another way of turning a number from one scale into another is as
follows: Multiply the first digit by the old radix in the new scale, and add
the next digit; multiply the result again by the old radix in the new scale,
and take in the next digit, and so on to the end, always using the radix of the
scale you want to leave, and the notation of the scale you want to end in.
Thus, suppose it required to turn 16687 (duodecimal) into the decimal
scale, and 16432 (septenary) into the quaternary scale:

16687 16432
Duodecimals into Decimals. Septenaries into Quaternaries.
1 × 12 + 6 = 18 1 × 7 + 6 = 31
× 12 + 6 ×7+4
222 1133
× 12 + 8 ×7+3
2672 22130
× 12 + 7 ×7+2

Answer 32071 1021012
Owing to our division of a foot into 12 equal parts, the duodecimal scale
often becomes very convenient. Let the square foot be also divided into 12
parts, each part is 12 square inches, and the 12th of the 12th is one square
inch. Suppose, now, that the two sides of an oblong piece of ground are 176
feet 9 inches 7-12ths of an inch, and 65 feet 11 inches 5-12ths of an inch.
Using the duodecimal scale, and duodecimal fractions, these numbers are
128·97 and 55·e5. Their product, the number of square feet required, is thus
found:

128·97
55·e5
617ee
116095
617ee
617ee
68e8144e

Answer, 68e8·144e (duod.) square feet, or 11660 square
feet 16 square inches ⁴/₁₂ and ¹¹/₁₄₄ of a square inch.
It would, however, be exact enough to allow 2-hundredths of a foot for
every quarter of an inch, an additional hundredth for every 3 inches,[58] and
1-hundredth more if there be a 12th or 2-12ths above the quarter of an inch.
Thus, 9⁷/₁₂ inches should be ·76 + ·03 + ·01, or ·80, and 11⁵/₁₂ would be
·95; and the preceding might then be found decimally as 176·8 × 65·95 as
11659·96 square feet, near enough for every practical purpose.

APPENDIX IV.
ON THE DEFINITION OF FRACTIONS.

The definition of a fraction given in the text shews that ⁷/₉, for instance,
is the ninth part of seven, which is shewn to be the same thing as seven-
ninths of a unit. But there are various modes of speech under which a
fraction may be signified, all of which are more or less in use.

1. In ⁷/₉ we have the 9th part of 7.
2. 7-9ths of a unit.
3. The fraction which 7 is of 9.
4. The times and parts of a time (in this case part of a
time only) which 7 contains 9.
5. The multiplier which turns nines into sevens.
6. The ratio of 7 to 9, or the proportion of 7 to 9.
7. The multiplier which alters a number in the ratio of 9
to 7.
8. The 4th proportional to 9, 1, and 7.
The first two views are in the text. The third is deduced thus: If we
divide 9 into 9 equal parts, each is 1, and 7 of the parts are 7; consequently
the fraction which 7 is of 9 is ⁷/₉. The fourth view follows immediately: For
a time is only a word used to express one of the repetitions which take place
in multiplication, and we allow ourselves, by an easy extension of language,
to speak of a portion of a number as being that number taken a part of a
time. The fifth view is nothing more than a change of words: A number
reduced to ⁷/₉ of its amount has every 9 converted into a 7, and any fraction
of a 9 which may remain over into the corresponding fraction of 7. This is
completely proved when we prove the equation ⁷/₉ of a = 7 times a/9. The
sixth, seventh, and eighth views are illustrated in the chapter on proportion.
When the student comes to algebra, he will find that, in all the
applications of that science, fractions such as a/b most frequently require
that a and b should be themselves supposed to be fractions. It is, therefore,

of importance that he should learn to accommodate his views of a fraction
to this more complicated case.

2½
Suppose we take .
4³/₅
We shall find that we have, in this case, a better idea of the views from and
after the third inclusive, than of the first and second, which are certainly the
most simple ways of conceiving ⁷/₉. We have no notion of the (4³/₅)th part
of 2½,
1 3
nor of 22(45)
of a unit; indeed, we coin a new species of adjective when we talk of the
(4³/₅)th part of anything. But we can readily imagine that 2½ is some
fraction of 4³/₅; that the first is some part of a time the second; that there
must be some multiplier which turns every 4³/₅ in a number into 2½; and so
on. Let us now see whether we can invent a distinct mode of applying the
first and second views to such a compound fraction as the above.
We can easily imagine a fourth part of a length, and a fifth part,
meaning the lines of which 4 and 5 make up the length in question; and
there is also in existence a length of which four lengths and two-fifths of a
length make up the original length in question. For instance, we might say
that 6, 6, 2 is a division of 14 into 2⅓ equal parts—2 equal parts, 6, 6, and a
third of a part, 2. So we might agree to say, that the (2⅓)th, or (2⅓)rd, or
(2⅓)st (the reader may coin the adjective as he pleases) part of 14 is 6. If
we divide the line a b into eleven equal parts in c, d, e, &c., we must then
say that a c is the 11th part,

a d the (5½)th, a e the (3⅔)th, a f the (2¾)th, a g the (2⅕)th, a h the
(1⅚)th, a i the (1⁴/₇)th, a k the (1⅜)th, a l the (1²/₉)th, a m the (1⅒)th, and
a b itself the 1st part of a b. The reader may refuse the language if he likes

(though it is not so much in defiance of etymology as talking of multiplying
by ½); but when a b is called 1, he must either call a f 1/(2¾), or make one
definition of one class of fractions and another of another. Whatever
abbreviations they may choose, all persons will agree that a/b is a direction
to find such a fraction as, repeated b times, will give 1, and then to take that
fraction a times.
So, to get 2½/4⅗, the simplest way is to divide the whole unit into 46
parts; 10 of these parts, repeated 4⅗ times, give the whole. The

4⅗th is then ¹⁰/₄₆, and 2½ such parts is ²⁵/₄₆, or a c. The student should
try several examples of this mode of interpreting complex fractions.
But what are we to say when the denominator itself is less than unity, as
in 3¼/⅖? Are we to have a (⅖)th part of a unit? and what is it? Had there
been a 5 in the denominator, we should have taken the part of which 5 will
make a unit. As there is ⅖ in the denominator, we must take the part of
which ⅖ will be a unit. That part is larger than a unit; it is 2½ units; 2½ is
that of which ⅖ is 1. The above fraction then directs us to repeat 2½ units
3¼ times. By extending our word ‘multiplication’ to the taking of a part of a
time, all multiplications are also divisions, and all divisions multiplications,
and all the terms connected with either are subject to be applied to the
results of the other.
If 2⅓ yards cost 3½ shillings, how much does one yard cost? In such a
case as this, the student looks at a more simple question. If 5 yards cost 10
shillings, he sees that each yard costs ¹⁰/₅, or 2 shillings, and, concluding
that the same process will give the true result when the data are fractional,
he forms 3½/2⅓, reduces it by rules to ³/₂ or 1½, and concludes that 1 yard
costs 18 pence. The answer happens to be correct; but he is not to suppose

that this rule of copying for fractions whatever is seen to be true of integers
is one which requires no demonstration. In the above question we want
money which, repeated 2⅓ times, shall give 3½ shillings. If we divide the
shilling into 14 equal parts, 6 of these parts repeated 2⅓ times give the
shilling. To get 3½ times as much by the same repetition, we must take 3½
of these 6 parts at each step, or 21 parts. Hence, ²¹/₁₄, or 1½, is the number
of shillings in the price.

APPENDIX V.
ON CHARACTERISTICS.

When the student comes to use logarithms, he will find what follows
very useful. In the mean while, I give it merely as furnishing a rapid rule for
finding the place of a decimal point in the quotient before the division is
commenced.
When a bar is written over a number, thus, 7 let the number be called
negative, and let it be thus used: Let it be augmented by additions of its own
species, and diminished by subtractions; thus, 7 and 2 give 9 , and let 7
with 2 subtracted give 5 . But let the addition of a number without the bar
diminish the negative number, and the subtraction increase it. Thus, 7 and
4 are 3 , 7 and 12 make 5, 7 with 8 subtracted is 15 . In fact, consider 1,
2, 3, &c., as if they were gains, and 1 , 2 , 3 , as if they were losses: let the
addition of a gain or the removal of a loss be equivalent things, and also the
removal of a gain and the addition of a loss. Thus, when we say that 4
diminished by 11 gives 7, we say that a loss of 4 incurred at the moment
when a loss of 11 is removed, is, on the whole, equivalent to a gain of 7;
and saying that 4 diminished by 2 is 6 , we say that a loss of 4,
accompanied by the removal of a gain of 2, is altogether a loss of 6.
By the characteristic of a number understand as follows: When there
are places before the decimal point, it is one less than the number of such
places. Thus, 3·214, 1·0083, 8 (which is 8·00 ...) 9·999, all have 0 for their
characteristics. But 17·32, 48, 93·116, all have 1; 126·03 and 126 have 2;
11937264·666 has 7. But when there are no places before the decimal point,
look at the first decimal place which is significant, and make the
characteristic negative accordingly. Thus, ·612, ·121, ·9004, in all of which
significance begins in the first decimal place, have the characteristic 1 ; but
·018 and ·099 have 2 ; ·00017 has 4 ; ·000000001 has 9 .
To find the characteristic of a quotient, subtract the characteristic of the
divisor from that of the dividend, carrying one before subtraction if the first
significant figures of the divisor are greater than those of the dividend. For
instance, in dividing 146·08 by ·00279. The characteristics are 2 and 3 ;
and 2 with 3 removed would be 5. But on looking, we see that the first

significant figures of the divisor, 27, taken by themselves, and without
reference to their local value, mean a larger number than 14, the first two
figures of the dividend. Consequently, to 3 we carry 1 before subtracting,
and it then becomes 2 , which, taken from 2, gives 4. And this 4 is the
characteristic of the quotient, so that the quotient has 5 places before the
decimal point. Or, if abcdef be the first figures of the quotient, the decimal
point must be thus placed, abcde·f. But if it had been to divide ·00279 by
146·08, no carriage would have been required; and 3 diminished by 2 is
5 ; that is, the first significant figure of the quotient is in the 5th place. The
quotient, then, has ·0000 before any significant figure. A few applications
of this rule will make it easy to do it in the head, and thus to assign the
meaning of the first figure of the quotient even before it is found.

APPENDIX VI.
ON DECIMAL MONEY.

Of all the simplifications of commercial arithmetic, none is comparable
to that of expressing shillings, pence, and farthings as decimals of a pound.
The rules are thereby put almost upon as good a footing as if the country
possessed the advantage of a real decimal coinage.
Any fraction of a pound sterling may be decimalised by rules which can
be made to give the result at once.
Two shillings is £·100
One shilling is £·050
Sixpence is £·025
One farthing is £·001 04⅙
Thus, every pair of shillings is a unit in the first decimal place; an odd
shilling is a 50 in the second and third places; a farthing is so nearly the
thousandth part of a pound, that to say one farthing is ·001, two farthings is
·002, &c., is so near the truth that it makes no error in the first three
decimals till we arrive at sixpence, and then 24 farthings is exactly ·025 or
25 thousandths. But 25 farthings is ·026, 26 farthings is ·027, &c. Hence
the rule for the first three places is
One in the first for every pair of shillings; 50 in the second and third for
the odd shilling, if any; and 1 for every farthing additional, with 1 extra for
sixpence.
Thus, 0s. 3½d. = £·014
0s. 7¾d. = £·032
1s. 2½d. = £·060
1s. 11¼d. = £·096
2s. 6d. = £·125
2s. 9½d. = £·139
3s. 2¾d. = £·161

13s. 10¾d. = £·694
In the fourth and fifth places, and those which follow, it is obvious that we
have no produce from any farthings except those above sixpence. For at
every sixpence, ·00004⅙ is converted into ·001, and this has been already
accounted for. Consequently, to fill up the fourth and fifth places,
Take 4 for every farthing[59] above the last sixpence, and an additional 1
for every six farthings, or three halfpence.
The remaining places arise altogether from ·00000⅙ for every farthing
above the last three halfpence; for at every three halfpence complete,
·00000 ⅙ is converted into ·00001, and has been already accounted for.
Consequently, to fill up all the places after the fifth,
Let the number of farthings above the last three halfpence be a
numerator, 6 a denominator, and annex the figures of the corresponding
decimal fraction.
It may be easily remembered that

The figures of ¹/₆ are 166666...
” ²/₆ ... 333333...
” ³/₆ ... 5
” ⁴/₆ ... 666666...
” ⁵/₆ ... 833333...

0s. 3½d. = ·014 58 3333...
0s. 7¾d. = ·032 29 1666...
1s. 2½d. = ·060 41 6666...
1s. 11¼d. = ·096 87 83333...

2s. 6d. = ·125 00 0000...
2s. 9½d. = ·139 58 3333...
3s. 2¾d. = ·161 45 83333...
13s. 10¾d. = 694 79 1666...
The following examples will shew the use of this rule, if the student will
also work them in the common way.
To turn pounds, &c., into farthings: Multiply the pounds by 960, or by
1000-40, or by 1000(1-⁴/₁₀₀); that is, from 1000 times the pounds subtract 4
per cent of itself. Thus, required the number of farthings in £1663. 11. 9¾.
1663·590625 × 1000 = 1663590·625
4 per cent of this, 66543·625
No. of farthings required, 1597047
What is 47½ per cent of £166. 13. 10 and ·6148 of £2971. 16. 9?

166·691
40 p. c. 66·6764
5 p. c. 8·3346
2½ p. c. 4·1673
79·1783
£79.3.6¾

2971·837
·6 1783·1022
·01 29·7184
·004 11·8873
·0008 2·3775
1827·0854

£1827.1.8½
The inverse rule for turning the decimal of a pound into shillings, pence,
and farthings, is obviously as follows:
A pair of shillings for every unit in the first place; an odd shilling for 50
(if there be 50) in the second and third places; and a farthing for every
thousandth left, after abating 1 if the number of thousandths so left exceed
24.
The direct rule (with three places) gives too little, the inverse rule too
much, except at the end of a sixpence, when both are accurate. Thus, £·183
is rather less than 3s. 8d., and 6s. 4¾d. is rather greater than £319; or when
the two do not exactly agree, the common money is the greatest. But £·125
and £·35 are exactly 2s. 6d. and 7s.
Required the price of 17 cwt. 81 lb. 13½ oz. at £3.11.9¾ per cwt. true to
the hundredth of a farthing.
3·590625
17
61·040625
lb. 56 ½ 1·795313
16 ⅐ ·512946
7⅛ ·224414
2⅛ ·064118
oz. 8¼ ·016029
4½ ·008015
1¼ ·002004
½½ ·001002
£63·664466
£63.13.3½
Three men, A, B, C, severally invest £191.12.7¾, £61.14.8, and
£122.1.9½ in an adventure which yields £511.12.6½. How ought the
proceeds to be divided among them?

A, 191·63229
B, 61·73333
C, 122·08958 Produce of £1.
375·45520)511·62708(1·362686
136·17188
23·53532
1·00801
25710
3183
180

1·362686 1·362686 1·362686
92·236191 33·33716 85·980221
1·362686 8·17612 1·362686
1·226417 13627 272537
13627 9538 27254
8176 409 1090
409 41 122
27 4 7
3 8·41231 1
1 1·663697
2·611346

261·1346 ... A’s share £261.2.8¼
84·1231 ... B’s ... 84.2.5¾
166·3697 ... C’s ... 166.7.4¾
511·6274 £511.2.6¾
If ever the fraction of a farthing be wanted, remember that the coinage-
result is larger than the decimal of a pound, when we use only three places.
From 1000 times the decimal take 4 per cent, and we get the exact number

of farthings, and we need only look at the decimal then left to set the
preceding right. Thus, in
134·6 123·1 369·7
5·38 4·92 14·79
·22 ·18 ·91
we see that (if we use four decimals only) the pence of the above results are
nearly 8d. ·22 of a farthing, 5½d. ·18, and 4½d. ·91.
A man can pay £2376. 4. 4½, his debts being £3293. 11. 0¾. How much
per cent can he pay, and how much in the pound?

3293·553)2376·2180(·7214756
70·7309
4·8598
1·5662
2488
183
18

Answer, £72. 2.11½ per cent.
0.14. 5¼ per pound.

APPENDIX VII.
ON THE MAIN PRINCIPLE
OF BOOK-KEEPING.

A brief notice of the principle on which accounts are kept (when they are properly kept) may
perhaps be useful to students who are learning book-keeping, as the treatises on that subject frequently
give too little in the way of explanation.
Any person who is engaged in business must desire to know accurately, whenever an investigation
of the state of his affairs is made.
1, What he had at the commencement of the account, or immediately after the last investigation was
made; 2, What he has gained and lost in the interval in all the several branches of his business; 3, What
he is now worth. From the first two of these things he obviously knows the third. In the interval
between two investigations, he may at any one time desire to know how any one account stands.
An account is a recital of all that has happened, in reference to any class of dealings, since the last
investigation. It can only consist of receipts and expenditures, and so it is said to have two sides, a
debtor and a creditor side.
All accounts are kept in money. If goods be bought, they are estimated by the money paid for them.
If a debtor give a bill of exchange, being a promise to pay a certain sum at a certain time, it is put down
as worth that sum of money. All the tools, furniture, horses, &c. used in the business are rated at their
value in money. All the actual coin, bank-notes, &c., which are in or come in, being the only money in
the books which really is money, is called cash.
The accounts are kept as if every different sort of account belonged to a separate person, and had an
interest of its own, which every transaction either promotes or injures. If the student find that it helps
him, he may imagine a clerk to every account: one to take charge of, and regulate, the actual cash;
another for the bills which the house is to receive when due; another for those which it is to pay when
due; another for the cloth (if the concern deal in cloth); another for the sugar (if it deal in sugar); one for
every person who has an account with the house; one for the profits and losses; and so on.
All these clerks (or accounts) belonging to one merchant, must account to him in the end—must
either produce all they have taken in charge, or relieve themselves by shewing to whom it went. For all
that they have received, for every responsibility they have undertaken to the concern itself, they are
bound, or are debtors; for everything which has passed out of charge, or about which they are relieved
from answering to the concern, they are unbound, or are creditors. These words must be taken in a very
wide sense by any one to whom book-keeping is not to be a mystery. Thus, whenever any account
assumes responsibility to any parties out of the concern, it must be creditor in the books, and debtor
whenever it discharges any other parties of their responsibility. But whenever an account removes
responsibility from any other account in the same books it is debtor, and creditor whenever it imposes
the same.
To whom are all these parties, or accounts, bound, and from whom are they released? Undoubtedly
the merchant himself, or, more properly, the balance-clerk, presently mentioned. But it is customary to
say that the accounts are debtors to each other, and creditors by each other. Thus, cash debtor to bills
receivable, means that the cash account (or the clerk who keeps it) is bound to answer for a sum which
was paid on a bill of exchange due to the house. At full length it would be: “Mr. C (who keeps the cash-
box) has received, and is answerable for, this sum which has been paid in by Mr. A, when he paid his
bill of exchange.” On the other hand, the corresponding entry in the account of bills receivable runs—
bills receivable, creditor by cash. At full length: “Mr. B (who keeps the bills receivable) is freed from
all responsibility for Mr. A’s bill, which he once held, by handing over to Mr. C, the cash-clerk, the

money with which Mr. A took it up.” Bills receivable creditor by cash is intelligible, but cash debtor to
bills receivable is a misnomer. The cash account is debtor to the merchant by the sum received for the
bill, and it should be cash debtor by bill receivable. The fiction of debts, not one of which is ever paid to
the party to whom it is said to be owing, though of no consequence in practice, is a stumbling-block to
the learner; but he must keep the phrase, and remember its true meaning.
The account which is made debtor, or bound, is said to be debited; that which is made creditor, or
released, is said to be credited. All who receive must be debited; all who give must be credited.
No cancel is ever made. If cash received be afterwards repaid, the sum paid is not struck off the
receipts (or debtor-side of the cash account), but a discharge, or credit, is written on the expenditure (or
credit) side.
The book in which the accounts are kept is called a ledger. It has double columns, or else the
debtor-side is on one page, and the creditor side on the opposite, of each account. The debtor-side is
always the left. Other books are used, but they are only to help in keeping the ledger correct. Thus there
may be a waste-book, in which all transactions are entered as they occur, in common language; a
journal, in which the transactions described in the waste-book are entered at stated periods, in the
language of the ledger. The items entered in the journal have references to the pages of the ledger to
which they are carried, and the items in the ledger have also references to the pages of the journal from
which they come; and by this mode of reference it is easy to make a great deal of abbreviation in the
ledger. Thus, when it happens, in making up the journal to a certain date, that several different sums
were paid or received at or near the same time, the totals may be entered in the ledger, and the cash
account may be made debtor to, or creditor by, sundry accounts, or sundries; the sundry accounts being
severally credited or debited for their shares of the whole. The only book that need be explained is the
ledger. All the other books, and the manner in which they are kept, important as they may be, have
nothing to do with the main principle of the method. Let us, then, suppose that all the items are entered
at once in the ledger as they arise. It has appeared that every item is entered twice. If A pay on account
of B, there is an entry, “A, creditor by B;” and another, “B, debtor to A.” This is what is called double-
entry; and the consequence of it is, that the sum of all the debtor items in the whole book is equal to the
sum of all the creditor items. For what is the first set but the second with the items in a different order?
If it were convenient, one entry of each sum might be made a double-entry. The multiplication table is
called a table of double-entry, because 42, for instance, though it occurs only once, appears in two
different aspects, namely, as 6 times 7 and as 7 times 6. Suppose, for example, that there are five
accounts, A, B, C, D, E, and that each account has one transaction of its own with every other account;
and let the debits be in the columns, the credits in the rows, as follows:
Debtor A B C D E
A, Creditor 23 19 32 4
B, Creditor 17 6 11 25
C, Creditor 9 41 10 2
D, Creditor 14 28 16 3
E, Creditor 15 4 60 1

Here the 16 is supposed to appear in D’s account as D creditor by C, and in C’s account as C debtor
to D. And to say that the sum of debtor items is the same as that of creditor items, is merely to say that
the preceding numbers give the same sum, whether the rows or the columns be first added up.
If it be desired to close the ledger when it stands as above, the following is the way the accounts
will stand: the lines in italics will presently be explained.

A, Debtor. A, Creditor. B, Debtor. B, Creditor.
To B 17 By B 23 To A 23 By A 17
To C 9 By C 19 To C 41 By C 6
To D 14 By D 32 To D 28 By D 11
To E 15 By E 4 To E 4 By E 25
To Balance 23 By Balance 37
78 78 96 96
C, Debtor. C, Creditor. D, Debtor. D, Creditor.
To A 19 By A 9 To A 32 By A 14
To B 6 By B 41 To B 11 By B 28
To D 16 By D 10 To C 10 By C 16
To E 60 By E 2 To E 1 By E 3
By Balance 39 To Balance 7
101 101 61 61
E, Debtor. E, Creditor. Balance, Debtor. Balance, Cred.
To A 4 By A 15 To B 37 By A 23
To B 25 By B 4 To C 39 By D 7
To C 2 By C 60 By E 46
To D 3 By D 1 76 76
To Balance 46
80 80

In all the part of the above which is printed in Roman letters we see nothing but the preceding table
repeated. But when all the accounts have been completed, and no more entries are left to be made, there
remains the last process, which is termed balancing the ledger. To get an idea of this, suppose a new
clerk, who goes round all the accounts, collecting debts and credits, and taking them all upon himself,
that he alone may be entitled to claim the debts and to be responsible for the assets of the concern. To
this new clerk, whom I will call the balance-clerk, every account gives up what it has, whether the same
be debt or credit. The cash-clerk gives up all the cash; the clerks of the two kinds of bills give up all
their documents, whether bills receivable or entries of bills payable (remember that any entry against
which there is money set down in the books counts as money when given up, that is, as money due or
money owing); the clerks of the several accounts of goods give up all their unsold remainders at cost
prices; the clerks of the several personal accounts give up vouchers for the sums owing to or from the
several parties; and so on. But where more has been paid out than received, the balance-clerk adjusts
these accounts by giving instead of receiving; in fact, he so acts as to make the debtor and creditor sides
of the accounts he visits equal in amount. For instance, the A account is indebted to the concern 55,
while payments or discharges to the amount of 78 have been made by it. The balance-clerk accordingly
hands over 23 to that account, for which it becomes debtor, while the balance enters itself as creditor to
the same amount. But in the B account there is 96 of receipt, and only 59 of payment or discharge. The
balance-clerk then receives 37 from this account, which is therefore credited by balance, while the
balance acknowledges as much of debt. The balance account must, of course, exactly balance itself, if
the accounts be all right; for of all the equal and opposite entries of which the ledger consists, so far as

they do not balance one another, one goes into one side of the balance account, and the other into the
other. Thus the balance account becomes a test of the accuracy of one part of the work: if its two sides
do not give the same sums, either there have been entries which have not had their corresponding
balancing entries correctly made, or else there has been error in the additions.
But since the balance account must thus always give the same sum on both sides, and since balance
debtor implies what is favourable to the concern, and balance creditor what is unfavourable, does it not
appear as if this system could only be applied to cases in which there is neither loss nor gain? This
brings us to the two accounts in which are entered all that the concern began with, and all that it gains
or loses—the stock account, and the profit-and-loss account. In order to make all that there was to
begin with a matter of double entry, the opening of the ledger supposes the merchant himself to put his
several clerks in charge of their several departments. In the stock account, stock, which here stands for
the owner of the books, is made creditor by all the property, and debtor by all the liabilities; while the
several accounts are made debtors for all they take from the stock, and creditors by all the
responsibilities they undertake. Suppose, for instance, there are £500 in cash at the commencement of
the ledger. There will then appear that the merchant has handed over to the cash-box £500, and in the
stock account will appear, “Stock creditor by cash, £500;” while in the cash account will appear, “Cash
debtor to stock, £500.” Suppose that at the beginning there is a debt outstanding of £50 to Smith and
Co., then there will appear in the stock account, “Stock debtor to Smith and Co. £50,” and in Smith and
Co.’s account will appear “Smith and Co. creditors by stock, £50.” Thus there is double entry for all
that the concern begins with by this contrivance of the stock account.
The account to which everything is placed for which an actual equivalent is not seen in the books is
the profit-and-loss account. This profit-and-loss account, or the clerk who keeps it, is made answerable
for every loss, and the supposed cause of every gain. This account, then, becomes debtor for every loss,
and creditor by every gain. If goods be damaged to the amount of £20 by accident, and a loss to that
amount occur in their sale, say they cost £80 and sell for £60 cash, it is clear that there is an entry “Cash
debtor to goods £60,” and “Goods creditor by cash £60.” Now, there is an entry of £80 somewhere to
the debit of the goods for cash laid out, or bills given, for the whole of the goods. It would affect the
accuracy of the accounts to take no notice of this; for when the balance-clerk comes to adjust this
account, he would find he receives £20 less than he might have reckoned upon, without any explanation
of the reason; and there would be a failure of the principle of double-entry. Since it is convenient that
the balance account of the goods should merely represent the stock in hand at the close, the account of
goods therefore lays the responsibility of £20 upon the profit-and-loss account, or there is the entry
“Goods creditor by profit-and-loss, £20,” and also “Profit-and-loss debtor to goods, £20.” Again, in all
payments which are not to bring in a specific return, such as house and trade expenses, wages, &c. these
several accounts are supposed to adjust matters with the profit-and-loss account before the balance
begins. Thus, suppose the outgoings from the mere premises occupied exceed anything those premises
yield by £200, or the debits of the house account exceed its credits by £200, the account should be
balanced by transferring the responsibility to the profit-and-loss account, under the entries “House
expenses creditor by profit-and-loss, £200”, “Profit-and-loss debtor to house expenses, £200.” In this
way the profit-and-loss account steps in from time to time before the balance account commences its
operations, in order that that same balance account may consist of nothing but the necessary matters of
account for the next year’s ledger.
This transference of accounts, or transfusion of one account into another, requires attentive
consideration. The receiving account becomes creditor for the credits, and debtor for the debits, of the
transmitting account. The rule, therefore, is: Make the transmitting account balance itself, and, on
whichever side it is necessary to enter a balancing sum, make the account debtor or creditor, as the case
may be, to the receiving account, and the latter creditor or debtor to the former. Thus, suppose account
A is to be transferred to account B, and the latter is to arrange with the balance account. If the two stand
as in Roman letters, the processes in Italic letters will occur before the final close.

A, Debtor. A, Creditor. B, Debtor. B, Creditor.
To sundries £100 By sundries £500 To sundries £600 By sundries £400
To B 400 To Balance 200 By A 400
£500 £500 £800 £800

And the entry in the balance account will be, “Creditor by B, £200,” shewing that, on these two
accounts, the credits exceed the debits by £200.
Still, before the balance account is made up, it is desirable that the profit-and-loss account should be
transferred to the stock account; for the profit and loss of this year is of no moment as a part of next
year’s ledger, except in so far as it affects the stock at the commencement of the latter. Let this be done,
and the balance account may then be made in the form required.
The stock account and the profit-and-loss account, the latter being the only direct channel of
alteration for the former, differ in a peculiar manner[60] from the other preliminary accounts, and the
balance account is a species of umpire. They represent the merchant: their interests are his interests; he
is solvent upon the excess of their credits over their debits, insolvent upon the excess of their debits
over their credits. It is exactly the reverse in all the other accounts. If a malicious person were to get at
the ledger, and put on a cipher to the pounds in various items, with a view of making the concern appear
worse than it really is, he would make his alterations on the debtor sides of the stock and profit-and-loss
accounts, and on the creditor sides of all the others. Accordingly, in the balance account, the net stock,
after the incorporation of the profit-and-loss account, appears on the creditor side (if not, it should be
called amount of insolvency, not stock), and the debts of the concern appear on the same side. But on
the debit side of the balance account appear all the assets of the concern (for which the balance-clerk is
debtor to the clerks from whom he has taken them).
The young student must endeavour to get the enlarged view of the words debtor and creditor which
is requisite, and must then learn by practice (for nothing else will give it) facility in allotting the actual
entries in the waste-book to the proper sides of the proper accounts. I do not here pretend to give more
than such a view of the subject as may assist him in studying a treatise on book-keeping, which he will
probably find to contain little more than examples.

APPENDIX VIII.
ON THE REDUCTION OF FRACTIONS TO
OTHERS OF NEARLY EQUAL VALUE.

There is a useful method of finding fractions which shall be nearly equal to
a given fraction, and with which the computer ought to be acquainted. Proceed
as in the rule for finding the greatest common measure of the numerator and
denominator, and bring all the quotients into a line. Then write down,
1 2nd Quot.
1st Quot. 1st Quot. × 2d Quot. + 1
Then take the third quotient, multiply the numerator and denominator of the
second by it, and add to the products the preceding numerator and
denominator. Form a third fraction with the results for a numerator and
denominator. Then take the fourth quotient, and proceed with the third and
second fractions in the same way; and so on till the quotients are exhausted.
For example, let the fraction be ⁹¹³¹/₁₃₁₂₈.

9131)13128(1, 2
1137 3997(3, 1
551 586(1, 15
201 35(1, 2
26 9(1, 8
8 1

This is the process for finding the greatest common measure of 9131 and
13128 in its most compact form, and the quotients and fractions are:

1 2 3 1 1 15 1 2 1 8
1 2 7 9 16 249 265 779 1044 9131
1 3 10 13 23 358 381 1120 1501 13128
It will be seen that we have thus a set of fractions ending with the original
fraction itself, and formed by the above rule, as follows:

1 1
1st Fraction = =
1st Quot. 1
2d Quot. 2
2d Fraction = =
1st Quot. × 2d Quot. + 1 3
2d Numʳ. × 3d Quot. + 1st Numʳ. 2×3+1 7
3d Fraction = = =
2d Denʳ. × 3d Quot. + 1st Denʳ. 3×3+1 10
3d Numʳ. × 4th Quot. + 2d Numʳ. 7×1+2 9
4th Fraction = = = ;
3d Denʳ. × 4th Quot. + 2d Denʳ. 10 × 1 + 3 13
and so on. But we have done something more than merely reascend to the
original fraction by means of the quotients. The set of fractions, ¹/₁, ²/₃, ⁷/₁₀, ⁹/
₁₃, &c. are continually approaching in value to the original fraction, the first
being too great, the second too small, the third too great, and so on alternately,
but each one being nearer to the given fraction than any of those before it.
Thus, ¹/₁ is too great, and ²/₃ is too small; but ²/₃ is not so much too small as ¹/₁
is too great. And again, ⁷/₁₀, though too great, is not so much too great as ²/₃ is
too small.
Moreover, the difference of any of the fractions from the original fraction is
never greater than a fraction having unity for its numerator and the product of
the denominator and the next denominator for its denominator. Thus, ¹/₁ does
not err by so much as ¹/₃, nor ²/₃ by so much as ¹/₃₀, nor ⁷/₁₀ by so much as ¹/₁₃₀,
nor ⁹/₁₃ by so much as ¹/₂₉₉, &c.
Lastly, no fraction of a less numerator and denominator can come so near
to the given fraction as any one of the fractions in the list. Thus, no fraction
with a less numerator than 249, and a less denominator than 358, can come so
near to

9131 249
as .
13128 358
The reader may take any example for himself, and the test of the accuracy
of the process is the ultimate return to the fraction begun with. Another test is
as follows: The numerator of the difference of any two consecutive
approximating fractions ought to be unity. Thus, in our instance, we have ¹⁶/₂₃

and ²⁴⁹/₃₅₈, which, with a common denominator, 23 × 358, have 5728 and
5727 for their numerators.
As another example, let us examine this question: The length of the year is
365·24224 days, which is called in common life 365¼ days. Take the fraction
²⁴²²⁴/₁₀₀₀₀₀, and proceed as in the rule.

24224)100000(4, 7, 1, 4, 9, 2
2496 3104
64 608
0 32

1 7 8 39 359 757
4 29 33 161 1482 3125
and ⁷⁵⁷/₃₁₂₅ is ·24224 in its lowest terms. Hence, it appears that the excess of
the year over 365 days amounts to about 1 day in 4 years, which is not wrong
by so much as 1 day in 116 years; more accurately, to 7 days in 29 years, which
is not wrong by so much as 1 day in 957 years; more accurately still, to 8 days
in 33 years, which is not wrong by so much as 1 day in 5313 years; and so on.
This method may be applied to finding fractions nearly equal to the square
roots of integers, in the following manner:

√43 = 6 + ...

6 1 5 4 5 5 4 5 1 6 6 1 5 4, &c.
1 7 6 3 9 2 9 3 6 7 1 7 6 3, &c.
6 1 1 3 1 5 1 3 1 1 1 2 1 1 3, &c.
Set down the number whose square root is wanted, say 43. This square root
is 6 and a fraction. Set down the integer 6 in the first and third row, and 1 in the
second row always. Form the successive rows each from the one before, in the
following manner:

One row The next row has b′, a′, c′, formed
being in this order, thus,
a a′ = excess of b′c′, already formed, over a.
b b′ = quotient of 43 - a² divided by b.
c c′ = integer in the quotient of 6 + a divided by b′.

Thus the second row is formed from the first, as under:
6 1 = excess of 7 × 1 (both just found) over 6.
1 7 = 43 - 6 × 6 divided by 1.
6 1 = integer of 6 + 6 divided by 7 (just found).

The third row is formed from the second, thus:
1 5 = excess of 1 × 6 over 1.
7 6 = 43 - 1 × 1 divided by 7.
1 1 = integer of 6 + 1 divided by 6;
and so on. In process of time the second column, 1, 7, 1, occurs again, after
which the several columns are repeated in the same order. As a final process,
take the set in the lowest line (excluding the first, 6), namely, 1, 1, 3, 1, 5, 1, 3,
&c. and use them by the rule given at the beginning of this article, as follows:

1 1 3 1 5 1 3 1 1, &c.
1 1 4 5 29 34 131 165 296
1 2 7 9 52 61 235 296 531

Hence, 6¹⁶⁵/₂₉₆ is very near the square root of 43, not erring by so much as

1
.
296 × 531
If we try it, we shall find (⁶¹⁶⁵/₂₉₆) to be ¹⁹⁴¹/₂₉₆, the square of which is
³⁷⁶⁷⁴⁸¹/₈₇₆₁₆, or 43⁷/₈₇₆₁₆.
This rule is of use when it is frequently wanted to use one square root, and
therefore desirable to ascertain whether any easy approximation exists by
means of a common fraction. For example, √2 is often used.

√2 = 1 + ...
11 1
11 1
12 2 2 2 2 2
1 2 5 12 29 70
&c.
2 5 12 29 70 169
Here it appears that

29 1
1 70 does not err by 70 × 169 ;

99 100 - 1
consequently, or is,
70 70
considering the ease of the operation, a fair approximation. In fact, ⁹⁹/₇₀ is
1·4142857 ... the truth being 1·4142135 ...
The following is an additional example:

√ 19 = 4 + ...
42 3 3 2 4 4 2
13 5 2 5 3 1 3
42 1 3 1 2 8 2 1 3 1 2, &c.
1 1 4 5 14
, &c.
2 3 11 14 39

APPENDIX IX.
ON SOME GENERAL PROPERTIES OF NUMBERS.

Prop. 1. If a fraction be reduced to its lowest terms, so called,[61] that is,
if neither, numerator nor denominator be divisible by any integer greater
than unity, then no fraction of a smaller numerator and denominator can
have the same value.
Let a/b be a fraction in which a and b have no common measure greater
than unity: and, if possible, let c/d be a fraction of the same value, c being
less than a, and d less than b. Now, since
a c a b
= we have = ;
b d c d
let m be the integer quotient of these last fractions (which must exist, since
a > c, b > d), and let e and f be the remainders. Then
a mc + e c mc
or = =
b md + f d md
Hence,
e mc
and must be equal, for if not,
f md
mc + e mc e
would lie between and ,
md + f md f
instead of being equal to the former. Hence,
a e
= ;
b f
so that if a fraction whose numerator and denominator have no common
measure greater than unity, be equal to a fraction of lower numerator and
denominator, it is equal to another in which the numerator and denominator
are still lower. If we proceed with

a e
= in a similar manner, we find
b f
a g
= where g < e, h < f,
b h
and so on. Now, if there be any process which perpetually diminishes the
terms of a fraction by one or more units at every step, it must at last bring
either the numerator or denominator, or both, to 0. Let
a v
=
b w
be one of the steps, and let a = kv + x, b = kw + y; so that
kv + x v
=
kw + y w
Now, if x = 0 but not y, this is absurd, for it gives
kv kv
= .
kw + y kw
A similar absurdity follows if y be 0, but not x; and if both x and y be =
0, then a = kv, b = kw, or a and b have a common measure, k. Now k must
be greater than 1, for v and w are less than c and d, which by hypothesis are
less than a and b. Consequently a and b have a common measure k greater
than 1, which by hypothesis they have not. If, then, a and b be integers not
divisible by any integer greater than 1, the fraction a/b is really in its lowest
terms. Also a and b are said to be prime to one another.
Prop. 2. If the product ab be divisible by c, and if c be prime to b, it
must divide a. Let
ab b d
= d, then = .
c c c
Now b/c is in its lowest terms; therefore, by the last proposition, d and a
must have a common measure. Let the greatest common measure be k, and
let a = kl, d = km. Then

b km m m
= = , and
c kl l l
is also in its lowest terms; but so is b/c; therefore we must have m = b, l = c,
for otherwise a fraction in its lowest terms would be equal to another of
lower terms. Therefore a = kc, or a is divisible by c. And from this it
follows, that if a number be prime to two others, it is prime to their product.
Let a be prime to b and c, then no measure of a can measure either b or c,
and no such measure can measure the product bc; for any measure of bc
which is prime to one must measure the other.
Prop. 3. If a be prime to b, it is prime to all the powers of b. Every
measure[62] of a is prime to b, and therefore does not divide b. Hence, by
the last, no measure of a divides b²; hence, a is prime to b², and so is every
measure of it; therefore, no measure of a divides bb², consequently a is
prime to b³, and so on.
Hence, if a be prime to b, a cannot divide without remainder any power
of b. This is the reason why no fraction can be made into a decimal unless
its denominator be measured by no prime[63] numbers except 2 and 5. For if
a c
= ,
b 10ⁿ
which last is the general form of a decimal fraction, let
a 10ⁿa
be in its lowest terms; then ,
b b
is an integer, whence (Prop. 2) b must divide 10ⁿ, and so must all the
divisors of b. If, then, among the divisors of b there be any prime numbers
except 2 and 5, we have a prime number (which is of course a number
prime to 10) not dividing 10, but dividing one of its powers, which is
absurd.
Prop. 4. If b be prime to a, all the multiples of b, as b, 2b, ... up to (a-
1)b must leave different remainders when divided by a. For if, m being
greater than n, and both less than a, we have mb and nb giving the same
remainder, it follows that mb-nb, or (m-n)b, is divisible by a; whence
(Prop. 2), a divides m-n, a number less than itself, which is absurd.

If a number be divided into its prime factors, or reduced to a product of
prime numbers only (as in 360 = 2 × 2 × 2 × 3 × 3 × 5), and if a, b, c, &c.
be the prime factors, and α, β, γ, &c. the number of times they severally
enter, so that the number is aα × bᵝ × cᵞ × &c., then this can be done in
only one way: For any prime number v, not included in the above list, is
prime to a, and therefore to aα, to b and therefore to bᵝ and therefore to aα
× bᵝ Proceeding in this way, we prove that v is prime to the complete
product above, or to the given number itself.

The number of divisors which the preceding number aαbᵝcᵞ ... can
have, 0 and itself included, is (α + 1)(β+ 1)(γ + 1).... For aα as the
divisors 1, a, a² ... aα and no others, α + 1 in all. Similarly, bᵝ has β+ 1
divisors, and so on. Now as all the divisors are made by multiplying
together one out of each set, their number (page 202) is (α + 1)(β + 1)(γ+
1)....
If a number, n, be divisible by certain prime numbers, say 3, 5, 7, 11,
then the third part of all the numbers up to n is divisible by 3, the fifth part
by 5, and so on. But more than this: when the multiples of 3 are omitted,
exactly the fifth part of those which remain are divisible by 5; for the fifth
part of the whole are divisible by 5, and the fifth part of those which are
removed are divisible by 5, therefore the fifth part of those which are left
are divisible by 5. Again, because the seventh part of the whole are divisible
by 7, and the seventh part of those which are divisible by 3, or by 5, or by
15, it follows that when all those which are multiples of 3 or 5, or both, are
removed, the seventh part of those which remain are divisible by 7; and so
on. Hence, the number of numbers not exceeding n, which are not divisible
by 3, 5, 7, or 11, is ¹⁰/₁₁ of ⁶/₇ of ⁴/₅ of ²/₃ of n. Proceeding in this way, we
find that the number of numbers which are prime to n, that is, which are not
divisible by any one of its prime factors, a, b, c, ... is
a -1 b - 1 c - 1
n ...
a b c

or aα-1bβ-1cγ-1 ... (a - 1)(b - 1)(c - 1)....
Thus, 360 being 2³3²5, its number of divisors is 4 × 3 × 2, or 24, and there
are 2³3.1.2.4 or 96 numbers less than 360 which are prime to it.
Prop. 5. If a be prime to b, then the terms of the series, a, a², a³, ...
severally divided by b, must all leave different remainders, until 1 occurs as
a remainder, after which the cycle of remainders will be again repeated.
Let a + b give the remainder r (not unity); then a² ÷ b gives the same
remainder as ra + b, which (Prop. 4) cannot be r: let it be s. Then aˢ ÷ b
gives the same remainder as sa ÷ b, which (Prop. 4) cannot be either r or s,
unless s be 1: let it be t. Then aᵗ ÷ b gives the same remainder as ta ÷ b; if t
be not 1, this cannot be either r, s, or t: let it be u. So we go on getting
different remainders, until 1 occurs as a remainder; after which, at the next
step, the remainder of a ÷ b is repeated. Now, 1 must come at last; for
division by b cannot give any remainders but 0, 1, 2, ... b- 1; and 0 never
arrives (Prop. 3), so that as soon as b-2 different remainders have occurred,
no one of which is unity, the next, which must be different from all that
precede, must be 1. If not before, then at aᵇ⁻¹ we must have a remainder 1;
after which the cycle will obviously be repeated.
Thus, 7, 7², 7³, 7⁴, &c. will, when divided by 5, be found to give the
remainders 2, 4, 3, 1, &c.
Prop. 6. The difference of two mth powers is always divisible without
remainder by the difference of the roots; or aᵐ -bᵐ is divisible by a-b; for
aᵐ - bᵐ = aᵐ - aᵐ⁻¹b + aᵐ⁻¹b - bᵐ
= aᵐ⁻¹(a - b) + b(aᵐ⁻¹ - bᵐ⁻¹)
From which, if aᵐ⁻¹-bᵐ⁻¹ is divisible by a - b, so is aᵐ-bᵐ. But a - b is
divisible by a - b; so therefore is a²- b²; so therefore is a³-b³; and so on.
Therefore, if a and b, divided by c, leave the same remainder, a² and b²,
a³ and b³, &c. severally divided by c, leave the same remainders; for this
means that a - b is divisible by c. But aᵐ - bᵐ is divisible by a - b, and
therefore by every measure of a-b, or by c; but aᵐ - bᵐ cannot be divisible
by c, unless aᵐ and bᵐ, severally divided by c, give the same remainder.

Prop. 7. If b be a prime number, and a be not divisible by b, then aᵇ and
(a-1)ᵇ + 1 leave the same remainder when divided by b. This proposition
cannot be proved here, as it requires a little more of algebra than the reader
of this work possesses.[64]
Prop. 8. In the last case, aᵇ⁻¹ divided by b leaves a remainder 1. From
the last, aᵇ-a leaves the same remainder as (a-1)ᵇ + 1-a or (a-1)ᵇ- (a-1);
that is, the remainder of aᵇ - a is not altered if a be reduced by a unit. By
the same rule, it may be reduced another unit, and so on, still without any
alteration of the remainder. At last it becomes 1ᵇ-1, or 0, the remainder of
which is 0. Accordingly, aᵇ - a, which is a(aᵇ⁻¹- 1), is divisible by b; and
since b is prime to a, it must (Prop. 2) divide aᵇ⁻¹-1; that is, aᵇ⁻¹, divided
by b, leaves a remainder 1, if b be a prime number and a be not divisible by
b.
From the above it appears (Prop. 5 and 7), that if a be prime to b, the
set 1, a, a², a³, &c. successively divided by b, give a set of remainders
beginning with 1, and in which 1 occurs again at aᵇ⁻¹, if not before, and at
aᵇ⁻¹ certainly (whether before or not), if b be a prime number. From the
point at which 1 occurs, the cycle of remainders recommences, and 1 is
always the beginning of a cycle. If, then, aᵐ be the first power which gives
1 for remainder, m must either be b-1, or a measure of it, when b is a prime
number.
But if we divide the terms of the series m, ma, ma², ma³, &c. by b, m
being less than b, we have cycles of remainders beginning with m. If 1, r, s,
t, &c. be the first set of remainders, then the second set is the set of
remainders arising from m, mr, ms, mt, &c. If 1 never occur in the first set
before aᵇ⁻¹ (except at the beginning), then all the numbers under b-1
inclusive are found among the set 1, r, s, t, &c.; and if m be prime to b
(Prop. 4), all the same numbers are found, in a different order, among the
remainders of m, mr, &c. But should it happen that the set 1, r, s, t, &c. is
not complete, then m, mr, ms, &c. may give a different set of remainders.
All these last theorems are constantly verified in the process for
reducing a fraction to a decimal fraction. If m be prime to b, or the fraction
m/b in its lowest terms, the process involves the successive division of m, m

× 10, m × 10², &c. by b. This process can never come to an end unless some
power of 10, say 10ⁿ, is divisible by b; which cannot be, if b contain any
prime factors except 2 and 5. In every other case the quotient repeats itself,
the repeating part sometimes commencing from the first figure, sometimes
from a later figure. Thus, ¹/₇ yields ·142857142857, &c., but ¹/₁₄ gives
·07(142857)(142857), &c., and ¹/₂₈ gives ·03(571428)(571428), &c.
In m/b, the quotient always repeats from the very beginning whenever b
is a prime number and m is less than b; and the number of figures in the
repeating part is then always b-1, or a measure of it. That it must be so,
appears from the above propositions.
Before proceeding farther, we write down the repeating part of a
quotient, with the remainders which are left after the several figures are
formed. Let the fraction be ¹/₁₇, we have
0₁₀5₁₅8₁₄8₄2₆3₉5₅2₁₆9₇4₂1₃1₁₃7₁₁6₈4₁₂7₁
This may be read thus: 10 by 17, quotient 0, remainder 10; 10² by 17,
quotient 05, remainder 15; 10³ by 17, quotient 058, remainder 14; and so
on. It thus appears that 10¹⁶ by 17 leaves a remainder 1, which is according
to the theorem.
If we multiply 0588, &c. by any number under 17, the same cycle is
obtained with a different beginning. Thus, if we multiply by 13, we have
7647058823529411
beginning with what comes after remainder 13 in the first number. If we
multiply by 7, we have 4117, &c. The reason is obvious: ¹/₁₇ × 13, or ¹³/₁₇,
when turned into a decimal fraction, starts with the divisor 130, and we
proceed just as we do in forming ¹/₁₇, when within four figures of the close
of the cycle.
It will also be seen, that in the last half of the cycle the quotient figures
are complements to 9 of those in the first half, and that the remainders are
complements to 17. Thus, in 0₁₀5₁₅8₁₄8₄, &c. and 9₇4₂1₃1₁₃, &c. we see 0 +
9 = 9, 5 + 4 = 9, 8 + 1 = 9, &c., and 10 + 7 = 17, 15 + 2 = 17, 14 + 3 = 17,
&c. We may shew the necessity of this as follows: If the remainder 1 never
occur till we come to use aᵇ⁻¹, then, b being prime, b-1 is even; let it be 2k.

Accordingly, a²ᵏ-1 is divisible by b; but this is the product of aᵏ-1 and aᵏ +
1, one of which must be divisible by b. It cannot be aᵏ - 1, for then a power
of a preceding the (b - 1)th would leave remainder 1, which is not the case
in our instance: it must then be aᵏ + 1, so that aᵏ divided by b leaves a
remainder b-1; and the kth step concludes the first half of the process.
Accordingly, in our instance, we see, b being 17 and a being 10, that
remainder 16 occurs at the 8th step of the process. At the next step, the
remainder is that yielded by 10(b-1), or 9b + b - 10, which gives the
remainder b-10. But the first remainder of all was 10, and 10 + (b - 10) = b.
If ever this complemental character occur in any step, it must continue,
which we shew as follows: Let r be a remainder, and b - r a subsequent
remainder, the sum being b. At the next step after the first remainder, we
divide 10r by b, and, at the next step after the second remainder, we divide
10b - 10r by b. Now, since the sum of 10r and 10b - 10r is divisible by b,
the two remainders from these new steps must be such as added together
will give b, and so on; and the quotients added together must give 9, for the
sum of the remainders 10r and 10b - 10r yields a quotient 10, of which the
two remainders give 1.
If ¹/₅₉ and ¹/₆₁ be taken, the repeating parts will be found to contain 58
and 60 figures. Of these we write down only the first halves, as the reader
may supply the rest by the complemental property just given.
01694915254237288135593220338, &c.
016393442622950819672131147540, &c.
Here, then, are two numbers, the first of which multiplied by any number
under 59, and the second by any number under 61, can have the products
formed by carrying certain of the figures from one end to the other.
But, b being still prime, it may happen that remainder 1 may occur
before b - 1 figures are obtained; in which case, as shewn, the number of
figures must be a measure of b - 1. For example, take ¹/₄₁. The repeating
quotient, written as above, has only 5 figures, and 5 measures 41 - 1.
0₁₀2₁₈4₁₆3₃₇9₁
Now, this period, it will be found, has its figures merely transposed, if we
multiply by 10, 18, 16, or 37. But if we multiply by any other number under

41, we convert this period into the period of another fraction whose
denominator is 41. The following are 8 periods which may be found.
0₁₀2₁₈4₁₆3₃₇9₁ 1₉2₈1₃₉9₂₁5₅
0₂₀4₃₆8₃₂7₃₃8₂ 1₁₉4₂₆6₁₄3₁₇4₆
0₃₀7₁₃3₇1₂₉7₃ 2₂₈6₃₄8₁₂2₃₈9₁₁
0₄₀9₈₁7₂₃5₂₅6₄ 3₂₇6₂₄5₃₅8₂₂5₁₅
To find m/41, look out for m among the remainders, and take the period
in which it is, beginning after the remainder. Thus, ³⁴/₄₁ is ·8292682926,
&c., and ¹⁵/₄₁ is ·3658536585, &c. These periods are complemental, four
and four, as 02439 and 97560, 07317 and 92682, &c. And if the first
number, 02439, be multiplied by any number under 41, look for that
number among the remainders, and the product is found in the period of
that remainder by beginning after the remainder. Thus, 02439 multiplied by
23 gives 56097, and by 6 gives 14634.
The reader may try to decipher for himself how it is that, with no more
figures than the following, we can extend the result of our division. The
fraction of which the period is to be found is ¹/₈₇.

87)100(01149425
130
430
820 01149425 × 25
370 28735625 × 25
220 718390625 × 25
460 17959765625 × 25
25 448994140625
0114942528735625
718390625
1795976 5625
448994
0114942528735632183908045977|011494
|

APPENDIX X.
ON COMBINATIONS.

There are some things connected with combinations which I place in an
appendix, because I intend to demonstrate them more briefly than the
matters in the text.
Suppose a number of boxes, say 4, in each of which there are counters,
say 5, 7, 3, and 11 severally. In how many ways can one counter be taken
out of each box, the order of going to the boxes not being regarded. Answer,
in 5 × 7 × 3 × 11 ways. For out of the first box we may draw a counter in 5
different ways, and to each such drawing we may annex a drawing from the
second in 7 different ways—giving 5 × 7 ways of making a drawing from
the first two. To each of these we may annex a drawing from the third box
in 3 ways—giving 5 × 7 × 3 drawings from the first three; and so on. The
following statements may now be easily demonstrated, and similar ones
made as to other cases.
If the order of going to the boxes make a difference, and if a, b, c, d be
the numbers of counters in the several boxes, there are 4 × 2 × 3 × 1 × a × b
× c × d distinct ways. If we want to draw, say 2 out of the first box, 3 out of
the second, 1 out of the third, and 3 out of the fourth, and if the order of the
boxes be not considered, the number of ways is
a -1 b-1 b-2 d-1 d-2
a 2
×
2 3
×c×d
2 3
If the order of going to the boxes be considered, we must multiply the
preceding by 4 × 3 × 2 × 1. If the order of the drawings out of the boxes
makes a difference, but not the order of the boxes, then the number of ways
is
a(a-1)b(b-1)(b-2)cd(d-1)(d-2)
The nth power of a, or aⁿ, represents the number of ways in which a
counters differently marked can be distributed in n boxes, order of placing
them in each box not being considered. Suppose we want to distribute 4
differently-marked counters among 7 boxes. The first counter may go into
either box, which gives 7 ways; the second counter may go into either; and

any of the first 7 allotments may be combined with any one of the second 7,
giving 7 × 7 distinct ways; the third counter varies each of these in 7
different ways, giving 7 × 7 × 7 in all; and so on. But if the counters be
undistinguishable, the problem is a very different thing.
Required the number of ways in which a number can be compounded of
other numbers, different orders counting as different ways. Thus, 1 + 3 + 1
and 1 + 1 + 3 are to be considered as distinct ways of making 5. It will be
obvious, on a little examination, that each number can be composed in
exactly twice as many ways as the preceding number. Take 8 for instance. If
every possible way of making 7 be written down, 8 may be made either by
increasing the last component by a unit, or by annexing a unit at the end.
Thus, 1 + 3 + 2 + 1 may yield 1 + 3 + 2 + 2, or 1 + 3 + 2 + 1 + 1: and all the
ways of making 8 will thus be obtained; for any way of making 8, say a + b
+ c + d, must proceed from the following mode of making 7, a + b + c + (d
- 1). Now, (d - 1) is either 0—that is, d is unity and is struck out—or (d - 1)
remains, a number 1 less than d. Hence it follows that the number of ways
of making n is 2ⁿ⁻¹. For there is obviously 1 way of making 1, 2 of making
2; then there must be, by our rule, 2² ways of making 3, 2³ ways of making
4; and so on.
1+1+1+1
1+1+1 1+1+2
1+1 1+2+1
1+2 1+3
1
2+1 2+1+1
2 2+2
3 3+1
4
This table exhibits the ways of making 1, 2, 3, and 4. Hence it follows
(which I leave the reader to investigate) that there are twice as many ways
of forming a + b as there are of forming a and then annexing to it a
formation of b; four times as many ways of forming a + b + c as there are of
annexing to a formation of a formations of b and of c; and so on. Also, in

summing numbers which make up a + b, there are ways in which a is a rest,
and ways in which it is not, and as many of one as of the other.
Required the number of ways in which a number can be compounded of
odd numbers, different orders counting as different ways. If a be the
number of ways in which n can be so made, and b the number of ways in
which n + 1 can be made, then a + b must be the number of ways in which
n + 2 can be made; for every way of making 12 out of odd numbers is either
a way of making 10 with the last number increased by 2, or a way of
making 11 with a 1 annexed. Thus, 1 + 5 + 3 + 3 gives 12, formed from 1 +
5 + 3 + 1 giving 10. But 1 + 9 + 1 + 1 is formed from 1 + 9 + 1 giving 11.
Consequently, the number of ways of forming 12 is the sum of the number
of ways of forming 10 and of forming 11. Now, 1 can only be formed in 1
way, and 2 can only be formed in 1 way; hence 3 can only be formed in 1 +
1 or 2 ways, 4 in only 1 + 2 or 3 ways. If we take the series 1, 1, 2, 3, 5, 8,
13, 21, 34, 55, 89, &c. in which each number is the sum of the two
preceding, then the nth number of this set is the number of ways (orders
counting) in which n can be formed of odd numbers. Thus, 10 can be
formed in 55 ways, 11 in 89 ways, &c.
Shew that the number of ways in which mk can be made of numbers
divisible by m (orders counting) is 2ᵏ⁻¹.
In the two series, 1 1 1 2 3 4 6 9 13 19 28, &c.
0 1 0 1 1 1 2 2 3 4 5, &c.,
the first has each new term after the third equal to the sum of the last and
last but two; the second has each new term after the third equal to the sum
of the last but one and last but two. Shew that the nth number in the first is
the number of ways in which n can be made up of numbers which, divided
by 3, leave a remainder 1; and that the nth number in the second is the
number of ways in which n can be made up of numbers which, divided by
3, leave a remainder 2.
It is very easy to shew in how many ways a number can be made up of a
given number of numbers, if different orders count as different ways.
Suppose, for instance, we would know in how many ways 12 can be thus
made of 7 numbers. If we write down 12 units, there are 11 intervals
between unit and unit. There is no way of making 12 out of 7 numbers
which does not answer to distributing 6 partition-marks in the intervals, 1 in

each of 6, and collecting all the units which are not separated by partition-
marks. Thus, 1 + 1 + 3 + 2 + 1 + 2 + 2, which is one way of making 12 out
of 7 numbers, answers to

1 1 111 11 1 11 11

in which the partition-marks come in the 1st, 2d, 5th, 7th, 8th, and 10th of
the 11 intervals. Consequently, to ask in how many ways 12 can be made of
7 numbers, is to ask in how many ways 6 partition-marks can be placed in
11 intervals; or, how many combinations or selections can be made of 6 out
of 11. The answer is,
11 × 10 × 9 × 8 × 7 × 6
, or 462.
1×2×3×4×5×6
Let us denote by mₙ the number of ways in which m things can be taken
out of n things, so that mₙ is the abbreviation for
n-1 n-2 n-m+1
n× × ... as far as
2 3 m
Then mₙ also represents the number of ways in which m + 1 numbers can
be put together to make n + 1. What we proved above is, that 6₁₁ is the
number of ways in which we can put together 7 numbers to make 12. There
will now be no difficulty in proving the following:
2ⁿ = 1 + 1ₙ + 2ₙ + 3ₙ ... + nₙ
In the preceding question, 0 did not enter into the list of numbers used.
Thus, 3 + 1 + 0 + 0 was not considered as one of the ways of putting
together four numbers to make 5. But let us now ask, what is the number of
ways of putting together 7 numbers to make 12, allowing 0 to be in the list
of numbers. There can be no more (nor fewer) ways of doing this than of
putting 7 numbers together, among which 0 is not included, to make 19.
Take every way of making 12 (0 included), and put on 1 to each number,
and we get a way of making 19 (0 not included). Take any way of making
19 (0 not included), and strike off 1 from each number, and we have one of

the ways of making 12 (0 included). Accordingly, 6₁₈ is the number of ways
of putting together 7 numbers (0 being allowed) to make 12. And (m-
1)ₙ₊ₘ₋₁ is the number of ways of putting together m numbers to make n, 0
being included.
This last amounts to the solution of the following: In how many ways
can n counters (undistinguishable from each other) be distributed into m
boxes? And the following will now be easily proved: The number of ways
of distributing c undistinguishable counters into b boxes is (b - 1)b + c - 1,
if any box or boxes may be left empty. But if there must be 1 at least in each
box, the number of ways is (b - 1)c - 1; if there must be 2 at least in each
box, it is (b - 1)c- b-1; if there must be 3 at least in each box, it is (b - 1)c -
2b - 1; and so on.
The number of ways in which m odd numbers can be put together to
make n, is the same as the number of ways in which m even numbers (0
included) can be put together to make n-m; and this is the number of ways
in which m numbers (odd or even, 0 included) can be put together to make
½(n-m). Accordingly, the number of ways in which m odd numbers can be
put together to make n is the same as the number of combinations of m-1
things out of ½(n-m) + m-1, or ½(n + m)-1. Unless n and m be both even or
both odd, the problem is evidently impossible.
There are curious and useful relations existing between numbers of
combinations, some of which may readily be exhibited, under the simple
expression of mₙ to stand for the number of ways in which m things may be
taken out of n. Suppose we have to take 5 out of 12: Let the 12 things be
marked a, b, c, &c. and set apart one of them, a. Every collection of 5 out of
the 12 either does or does not include a. The number of the latter sort must
be 5₁₁; the number of the former sort must be 4₁₁, since it is the number of
ways in which the other four can be chosen out of all but a. Consequently,
5₁₂ must be 5₁₁ + 4₁₁, and thus we prove in every case,
mₙ′ = mₙ₋₁ + (m - 1)ₙ₋₁
0ₙ and nₙ both are 1; for there is but one way of taking none, and but one
way of taking all. And again mₙ and (n-m)ₙ are the same things. And if m be

greater than n, mₙ is 0; for there are no ways of doing it. We make one of
our preceding results more symmetrical if we write it thus,
2ⁿ = 0ₙ + 1ₙ + 2ₙ + ... + nₙ
If we now write down the table of symbols in which the (m + 1)th
0 1 2 3 &c.
1 0₁ 1₁ 2₁ 3₁, &c.
2 0₂ 1₂ 2₂ 3₂, &c.
3 0₃ 1₃ 2₃ 3₃, &c.
&c. &c. &c. &c. &c.
number of the nth row represents mₙ, the number of combinations of m out
of n, we see it proved above that the law of formation of this table is as
follows: Each number is to be the sum of the number above it and the
number preceding the number above it. Now, the first row must be 1, 1, 0,
0, 0, &c. and the first column must be 1, 1, 1, 1, &c. so that we have a table
of the following kind, which may be carried as far as we please:
0 1 2 3 4 5 6 7 8 9 10
11 1 0 0 0 0 0 0 0 0 0
21 2 1 0 0 0 0 0 0 0 0
31 3 3 1 0 0 0 0 0 0 0
41 4 6 4 1 0 0 0 0 0 0
5 1 5 10 10 5 1 0 0 0 0 0
6 1 6 15 20 15 6 1 0 0 0 0
7 1 7 21 35 35 21 7 1 0 0 0
8 1 8 28 56 70 56 28 8 1 0 0
9 1 9 36 84 126 126 84 36 9 1 0
10 1 10 45 120 210 252 210 120 45 10 1
Thus, in the row 9, under the column headed 4, we see 126, which is 9 ×
8 × 7 × 6 ÷ (1 × 2 × 3 × 4), the number of ways in which 4 can be chosen
out of 9, which we represent by 4-{9}.

If we add the several rows, we have 1 + 1 or 2, 1 + 2 + 1 or 2², next 1 +
3 + 3 + 1 or 2³, &c. which verify a theorem already announced; and the law
of formation shews us that the several columns are formed thus:
11 121 1331
11 121 1331
121 1331 1 4 6 4 1 , &c.
so that the sum in each row must be double of the sum in the preceding. But
we can carry the consequences of this mode of formation further. If we
make the powers of 1 + x by actual algebraical multiplication, we see that
the process makes the same oblique addition in the formation of the
numerical multipliers of the powers of x.

1+x
1+x
1+x
x + x²
1 + 2x + x²

1 + 2x + x²
1+ x
1 + 2x + x²
x + 2x² + x³
1 + 3x + 3x² + x³

Here are the second and third powers of 1 + x: the fourth, we can tell
beforehand from the table, must be 1 + 4x + 6x² + 4x³ + x⁴; and so on.
Hence we have
(1 + x)ⁿ = 0ₙ + 1ₙx + 2ₙx² + 3ₙx³ + ... + nₙxⁿ
which is usually written with the symbols 0ₙ, 1ₙ, &c. at length, thus,
n-1 n-1 n-2
(1 + x)ⁿ = 1 + nx + n 2
x² + n 2 3
x³ + &c.

This is the simplest case of what in algebra is called the binomial theorem.
If instead of 1 + x we use x + a, we get
(x + a)ⁿ = xⁿ + 1ₙaxⁿ⁻¹ + 2ₙa²xⁿ⁻² + 3ₙa³xⁿ⁻³ + ... + nₙaⁿ
We can make the same table in another form. If we take a row of ciphers
beginning with unity, and setting down the first, add the next, and then the
next, and so on, and then repeat the process with one step less, and then
again with one step less, we have the following:

1 0 0 0 0 0 0
1 1 1 1 1 1 1
1 2 3 4 5 6
1 3 6 10 15
1 4 10 20
1 5 15
1 6
1
In the oblique columns we see 1 1, 1 2 1, 1 3 3 1, &c. the same as in the
original table, and formed by the same additions. If, before making the
additions, we had always multiplied by a, we should have got the several
components of the powers of 1 + a, thus,

1 0 0 0 0
1 a a² a³ a⁴
1 2a 3a² 4a³
1 3a 6a²
1 4a
1
where the oblique columns 1 + a, 1 + 2a + a², 1 + 3a + 3a² + a³, &c., give
the several powers of 1 + a. If instead of beginning with 1, 0, 0, &c. we had

begun with p, 0, 0, &c. we should have got p, p × 4a, p × 6a², &c. at the
bottom of the several columns; and if we had written at the top x⁴, x³, x², x,
1, we should have had all the materials for forming p(x + a)⁴ by multiplying
the terms at the top and bottom of each column together, and adding the
results.
Suppose we follow this mode of forming p(x + a)³ + q(x + a)² + r(x + a)
+ s.

x³ x² x 1 x² x 1 x 1 1
p 0 0 0 q 0 0 r 0 1
p pa pa² pa³ q qa qa² r ra
p 2pa 3pa² q 2qa r
p 3pa q
p

px³ + 3pax² + 3pa²x + pa³ + qx² + 2qax + qa² + rx + ra + s
= px³ + (3pa + q)x² + (3pa² + 2qa + r)x + pa³ + qa² + ra + s
Now, observe that all this might be done in one process, by entering q, r,
and s under their proper powers of x in the first process, as follows

x³ x² x 1
p q r s
p pa + q pa² + qa + r pa³ + qa² + ra + s
p 2pa + q 3pa² + 2qa + r
p 3pa + q
p

This process[65] is the one used in Appendix XI., with the slight alteration of
varying the sign of the last letter, and making subtractions instead of
additions in the last column. As it stands, it is the most convenient mode of
writing x + a instead of x in a large class of algebraical expressions. For

instance, what does 2x⁵ + x⁴ + 3x² + 7x + 9 become when x + 5 is written
instead of x? The expression, made complete, is,

2x⁵ + 1x⁴ + 0x³ + 3x² + 7x + 9
1 0 3 7 9
2 11 55 278 1397 6994
2 21 160 1078 6787
2 31 315 2653
2 41 520
2 51

Answer, 2x⁵ + 51x⁴ + 520x³ + 2653x² + 6787x + 6994.

APPENDIX XI.
ON HORNER’S METHOD OF SOLVING EQUATIONS.

The rule given in this chapter is inserted on account of its excellence as an exercise in computation.
The examples chosen will require but little use of algebraical signs, that they may be understood by
those who know no more of algebra than is contained in the present work.
To solve an equation such as
2x⁴ + x² - 3x = 416793,
or, as it is usually written,
2x⁴ + x² - 3x - 416793 = 0,
we must first ascertain by trial not only the first figure of the root, but also the denomination of it: if it
be a 2, for instance, we must know whether it be 2, or 20, or 200, &c., or ·2, or ·02, or ·002, &c. This
must be found by trial; and the shortest way of making the trial is as follows: Write the expression in its
complete form. In the preceding case the form is not complete, and the complete form is
2x⁴ + 0x³ + 1x² - 3x - 416793.
To find what this is when x is any number, for instance, 3000, the best way is to take the first multiplier
(2), multiply it by 3000, and take in the next multiplier (0), multiply the result by 3000, and take in the
next multiplier (1), and so on to the end, as follows:
2 × 3000 + 0 = 6000; 6000 × 3000 + 1 = 18000001
18000001 × 3000 - 3 = 54000002997
54000002997 × 3000 - 416793 = 162000008574207
Now try the value of the above when x = 30. We have then, for the steps, 60 (2 × 30 + 0), 1801, 54027,
and lastly,
1620810 - 416793,
or x = 30 makes the first terms greater than 416793. Now try x = 20 which gives 40, 801, 16017, and
lastly,
320340 - 416793,
or x = 20 makes the first terms less than 416793. Between 20 and 30, then, must be a value of x which
makes 2x⁴ + x²-3x equal to 416793. And this is the preliminary step of the process.
Having got thus far, write down the coefficients +2, 0, +1,-3, and -416793, each with its proper
algebraical sign, except the last, in which let the sign be changed. This is the most convenient way when
the last sign is-. But if the last sign be +, it may be more convenient to let it stand, and change all which
come before. Thus, in solving x³-12x + 1 = 0, we might write
-1 0 +12 1
whereas in the instance before us, we write
+2 0 +1 -3 416793
Having done this, take the highest figure of the root, properly named, which is 2 tens, or 20. Begin with
the first column, multiply by 20, and join it to the number in the next column; multiply that by 20, and

join it to the number in the next column; and so on. But when you come to the last column, subtract the
product which comes out of the preceding column, or join it to the last column after changing its sign.
When this has been done, repeat the process with the numbers which now stand in the columns,
omitting the last, that is, the subtracting step; then repeat it again, going only as far as the last column
but two, and so on, until the columns present a set of rows of the following appearance:

a b c d e
f g h i
k l m
n o
p

to the formation of which the following is the key:

f = 20a + b,
g = 20f + c,
h = 20g + d,
i = e - 20h,
k = 20a + f,
l = 20k + g,
m = 20l + h,
n = 20a + k,
o = 20n + l,
p = 20a + n.
We call this Horner’s Process, from the name of its inventor. The result is as follows:

2 0 1 -3 416793 (20
40 801 16017 96453
80 2401 64037
120 4801
160
We have now before us the row
2 160 4801 64037 96453
which furnishes our means of guessing at the next, or units’ figure of the root.
Call the last column the dividend, the last but one the divisor, and all that come before antecedents.
See how often the dividend contains the divisor; this gives the guess at the next figure. The guess is a
true one,[66] if, on applying Horner’s process, the divisor result, augmented as it is by the antecedent

processes, still go as many times in the dividend. For example, in the case before us, 96453 contains
64037 once; let 1 be put on its trial. Horner’s process is found to succeed, and we have for the second
process,

2 160 4801 64037 96453
162 4963 69000 27453
164 5127 74127
166 5293
168

As soon as we come to the fractional portion of the root, the process assumes a more[67] methodical
form.
The equation being of the fourth degree, annex four ciphers to the dividend, three to the divisor, two
to the antecedent, and one to the previous antecedent, leaving the first column as it is; then find the new
figure by the dividend and divisor, as before,[68] and apply Horner’s process. Annex ciphers to the
results, as before, and proceed in the same way. The annexing of the ciphers prevents our having any
thing to do with decimal points, and enables us to use the quotient-figures without paying any attention
to their local values. The following exhibits the whole process from the beginning, carried as far as it is
here intended to go before beginning the contraction, which will give more figures, as in the rule for the
square root. The following, then, is the process as far as one decimal place:

2 0 1 -3 416793(213
40 801 16017 96453
80 2401 64037 274530000
120 4801 69000 47339778
160 4963 74127000
162 5127 75730074
164 529300 77348376
166 534358
1680 539434
1686 544528
1692
1698
1704
If we now begin the contraction, it is good to know beforehand on what number of additional root-
figures we may reckon. We may be pretty certain of having nearly as many as there are figures in the
divisor when we begin to contract—one less, or at least two less. Thus, there being now eight figures in
the divisor, we may conclude that the contraction will give us at least six more figures. To begin the
contraction, let the dividend stand, cut off one figure from the divisor, two from the column before that,
three from the one before that, and so on. Thus, our contraction begins with

0002 1 704 5445 28 7734837 6 47339778

The first column is rendered quite useless here. Conduct the process as before, using only the figures
which are not cut off. But it will be better to go as far as the first figure cut off, carrying from the
second figure cut off. We shall then have as follows:

1 704 5445 28 7734837 6 47339778(6
5445 5 7767570 6 734354
5465 7 7800364 8
5475 9

At the next contraction the column 1|704 becomes |001704, and is quite useless. The next step,
separately written (which is not, however, necessary in working), is

54 759 780036 48 734354(0

Here the dividend 734354 does not contain the divisor 780036, and we, therefore, write 0 as a root
figure and make another contraction, or begin with

54759 78003 648 734354(9
78008 5 32277
78013 4

At the next contraction the first column becomes |0054759, and is quite useless, so that the remainder of
the process is the contracted division.

7801 34)32277(4137
1072
292
58
3
and the root required is 21·36094137.
I now write down the complete process for another equation, one root of which lies between 3 and
4: it is

x³ - 10x + 1 = 0
1 0 -10 -1(3·1110390520730990796
3 -1 2000
6 1700 209000

9 0 1791 19769000
9 1 188300 743369000000
9 2 189231 172311710273000
9 30 19016300 991247447681
9 31 19025631 39462875420
9 32 1903496300 0 0 1391491559
9 33 0 1903524299 0 9 58993123
9 33 1 1903552298 2 7 0 0 1886047
9 33 2 1903560698 0 5 9 1 172835
9 33 30 0 1903569097 8 5 6 3 1515
9 33 30 3 1903569144 5 2 2 183
9 33 30 6 1903569191 1 8 8 12
9 33 30 90 1903569193 0 6 1
9 33 30 99 1903569194 9 3
9 33 31 08
09 33 31 17

The student need not repeat the rows of figures so far as they come under one another: thus, it is not
necessary to repeat 190356. But he must use his own discretion as to how much it would be safe for
him to omit. I have set down the whole process here as a guide.
The following examples will serve for exercise:

1. 2x³ - 100x - 7 = 0
x = 7·10581133.

2. x⁴ + x³ + x² + x = 6000
x = 8·531437726.
3. x³ + 3x² - 4x - 10 = 0
x = 1·895694916504.

4. x³ + 100x² - 5x - 2173 = 0
x = 4·582246071058464.
_
5. ∛2 = 1·259921049894873164767210607278.[69]

6. x³ - 6x = 100
x = 5·071351748731.

7. x³ + 2x² + 3x = 300
x = 5·95525967122398.

8. x³ + x = 1000

x = 9·96666679.

9. 27000x³ + 27000x = 26999999
x = 9·9666666.....

10. x³ - 6x = 100
x = 5·0713517487.

11. x⁵ - 4x⁴ + 7x³ - 863 = 0
x = 4·5195507.

12. x³ - 20x + 8 = 0
x = 4·66003769300087278.

13. x³ + x² + x - 10 = 0
x = 1·737370233.

14. x³ - 46x² - 36x + 18 = 0
x = 46·7616301847,
or x = ·3471623192.

15. x³ + 46x² - 36x - 18 = 0
x = 1·1087925037.

16. 8991x³ - 162838x² + 746271x - 81000 = 0
x = ·111222333444555....

17. 729x³ - 486x² + 99x - 6 = 0
x = ·1111..., or ·2222..., or ·3333....

18. 2x³ + 3x² - 4x = 500
x = 5·93481796231515279.

19. x³ + 2x² + x - 150 = 0
x = 4·6684090145541983253742991201705899.

20. x³ + x = x² + 500
x = 8·240963558144858526963.

21. x³ + 2x² + 3x - 10000 = 0
x = 20·852905526009.

22. x⁵ - 4x - 2000 = 0
x = 4·581400362.

23. 10x³ - 33x² - 11x - 100 = 0
x = 4·146797808584278785.

24. x⁴ + x³ + x² + x = 127694
x = 18·64482373095.

25. 10x³ + 11x² + 12x = 100000
x = 21·1655995554508805.

26. x³ + x = 13
x = 2·209753301208849.

27. x³ + x² - 4x - 1600 = 0
x = 11·482837157.

28. x³ - 2x = 5
x = 2·094551481542326591482386540579302963857306105628239.

29. x⁴ - 80x³ + 24x² - 6x - 80379639 = 0
x = 123.[70]

30. x³ - 242x² - 6315x + 2577096 = 0
x = 123.[71]

31. 2x⁴ - 3x³ + 6x - 8 = 0
x = 1·414213562373095048803.[72]

32. x⁴ - 19x³ + 132x² - 302x + 200 = 0
x = 1·02804, or 4, or 6·57653, or 7·39543[73].

33. 7x⁴ - 11x³ + 6x² + 5x = 215
x = 2·70648049385791.[74]

34. 7x⁵ + 6x⁴ + 5x³ + 4x² + 3x = 11
x = ·770768819622658522379296505.[75]

35. 4x⁶ + 7x⁵ + 9x⁴ + 6x³ + 5x² + 3x = 792
x = 2·052042176879605365214043401281201973460275599545541724214.[76]

36. 2187x⁴ - 2430x³ + 945x² - 150x + 8 = 0
x = ·1111...., or ·2222...., or ·3333...., or ·4444....