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.

A B No. of times. Remainder.

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.

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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,

ifa = 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:

Leta = x + y - z,

thenma = 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 ● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

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B ● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

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II.

● ● ● ● ● ● ●

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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 proposed. Square. Cube.

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 = a × b (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,