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 eleventen and one[2]
two twelveten and two
three thirteenten and three
four fourteenten and four
five fifteenten and five
six sixteenten and six
seven seventeenten and seven
eight eighteenten and eight
nine nineteenten and nine
ten twentytwo tens
twenty-onetwo tens and one fiftyfive tens
twenty-twotwo tens and two sixtysix tens
&c. &c.&c. &c. seventyseven tens
thirtythree tens eightyeight tens
&c. &c. ninetynine tens
fortyfour tens a hundredten tens
&c. &c.
a hundred and oneten tens and one
&c. &c.
a thousandten hundreds
ten thousand
a hundred thousand
a millionten 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
noughtonetwothreefourfivesixseveneightnine

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. 57fifty-seven
II. 104one hundred and four.
III. 110one hundred and ten.
IV. 2348two thousand three hundred
 and forty-eight.
V. 15906fifteen thousand nine
 hundred and six.
VI. 187004one hundred and eighty-seven
 thousand and four.
VII.3697285three 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

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.
123410
2127125
11111

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. and2 tens.
15 hund. are1 thous. and5 hund.
3 thous. are3 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
12345678910
234567891011
3456789101112
45678910111213
567891011121314
6789101112131415
78910111213141516
891011121314151617
9101112131415161718

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 = 1822 + 6 = 2819 + 8 = 27
54 + 9 = 6356 + 7 = 6322 + 8 = 30
100 - 9 = 9127 - 8 = 1944 - 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

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 20524248 16923888 13323528 9723168 6122412
2448 2882088 42841728 39241368 35641008 2808 648
684 2484 324 21244320 17643960 14043204 10442844
2880 7202520 3602160 43561800 36001440 32401080
1116 2916 756 2556 396 21963996 18363636 14763276
3312 11522952 7922592 362232 40321872 36721512
1548 33481188 2988 432 2628 72 22684068 19083708
3744 15843384 8283024 4682664 1082304 41041944
1980 37801224 3420 864 3060 504 2700 144 23404140
4176 16203816 12603456 9003096 5402736 1802376

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.

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 exceed1 unitby 1 unit.
6 tens3 tens3 tens.
7 hundreds6 hundreds1 hundred.
7 thousands4 thousands3 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.
23604999923606
20996137173524096
747126792110
48024696543217222222
1884974731415926195
987654321 12345678989

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.

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 89 1011 12
2 4 6 810 1214 1618 2022 24
3 6 9 1215 1821 2427 3033 36
4 812 1620 2428 3236 4044 48
5 1015 2025 3035 4045 5055 60
6 1218 2430 3642 4854 6066 72
7 1421 2835 4249 5663 7077 84
8 1624 3240 4856 6472 8088 96
9 1827 3645 5463 7281 9099 108
10 2030 4050 6070 8090 100110 120
11 2233 4455 6677 8899 110121 132
12 2436 4860 7284 96108 120132 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,

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

abcdef
abf dec
abcdef&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.

Now64 units are written thus64
48 tens480
24 hundreds2400
8 thousands8000

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.

Multiply1679423800872
by60000
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 ×60is1422420
23707 ×500is11853500
23707 ×4000is94828000
The sum of theseis108269869

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:

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:

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 thefirst powerof 6
6 × 6second powerof 6
6 × 6 × 6third powerof 6
6 × 6 × 6 × 6fourth powerof 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.
972944784918330048
100810160641024192512
3142987216431018339288
31631000456931644451747
555530858025171416328875
678946090521312908547069
The fifthpower of 36 is60466176
fourth50 6250000
fourth108 136048896
fourth277 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:

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

91contains 137 times,
39contains 133 times,
26contains 132 times;

therefore 91 + 39 + 26 contains 13 7 + 3 + 2 times, or 12 times.

Again, 156 is made up of 100, 50, and 6.

Now100 contains13 7 timesand 9 over,
50 contains13 3 timesand 11 over,
6 contains13 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
mmmm

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:

Therefore 156 contains 13 10 + 2, or 12 times.

Again, to divide 3096 by 18.

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 than26times; therefore,
Twice 135”3”52or twice 26
10 times 135”3”260or 10 times 26
50 times 135”3”1300or 50 times 26
472contains 18 more than21times; therefore,
4720contains 18 more than210times,
47200contains 18 more than2100times,
472000contains 18 more than21000times,
32contains 12 more than2times, and less than 3 times.
320”12”20times, ”  ” 30 times.
3200”12 ”200times, ”  ” 300 times.
32000”12”2000times, ”  ” 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 ?
320711443313352

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:

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:

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,ortwice 99 and 2
3 tens,orthree times 9 and 3
and 4or4

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:

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.

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.
96944720612
1756183136562
237964841300001836484
14002564187174840
3103144207878393900
393904064768895717874
22876792454961430467215314410

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)
mbb

If n divide a and b without remainder,

a/n = a(86)
b/nb
a/b = a(87)
cbc
ab = a × b(88)
cc
ac = a(89)
bb/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 53 times,
30, or 15 + 15 contains 53 + 3 times, or 6 times,
45, or 15 + 15 + 15 contains 53 + 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

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

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 by112 gives the remainder24,
112 divided by24 gives the remainder16,
24 divided by16 gives the remainder8,
16 divided by8 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.

Therefore8is g. c. m. of16 and 8,
(97) g. c. m. of16 and 8is g. c. m. of24 and 16,
g. c. m. of24 and 16is g. c. m. of112 and 24,
g. c. m. of112 and 24is g. c. m. of360 and 112,
Therefore8is g. c. m. of360 and 112.

The process carried on may be written down in either of the following ways:

1123603
963364
16241
16162
08

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.

EXERCISES.

Numbers. g. c. m.
619795211
5836326021
5547 1470084431849
6281326041571
28915314955
15093003093

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.

Numbers proposed.Least
common
multiple.
14, 2142
16, 5, 24240
1, 2, 3, 4, 5, 6, 7, 8, 9, 10 2520
6, 8, 11, 16, 202640
876, 86463072
868, 85452948

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—
IKLMN
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).

What are10032663819and22773399?
———,———,————
1974237102424
Answer,162 23649and  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.

|||||||
ADECFGB

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
292123 × 12723
2788 = 17 × 164 = 17
492030 × 16430
93280 = 764 × 122 = 764
13786113 × 122113
888800 = 22 × 40400 = 22
40359600999 × 40400999
95469 = 121 × 789 = 121
359784456 × 789456

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 × 524′ × 6′ × 4′2′ × 2′ × 4′
27 × 28 = 3′ × 4′ = 3′ × 2′ = ⁶/₅.
9 × 701′ × 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

⅔ is2 × 7 or ¹⁴/₂₁
3 × 7
⁴/₇ is4 × 3 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

⅒ is1 × 6 × 9 or ⁵⁴/₅₄₀.
10 × 6 × 9
⅚ is5 × 10 × 9 or ⁴⁵⁰/₅₄₀.
6 × 10 × 9
⁷/₉ is7 × 10 × 6 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 proposedreduced to a common denominator.
211 2065
356 303030
1231232824184863
37142148484848484
3456 3000400506
101001000 1000100010001000
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
234560
44 - 153 = 18329
3 4271282
1 + 8 + 3 - 4 = 1834
1010010001000
2 - 1 + 12 = 253
71391
1 + 8 + 94 = 3
2161882
163 - 97 = 93066
521881459001

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

1723907 is 17230907 ,
10000  10000 
667225 is 667225 ,
1000  1000 
and 2399 is 2300099 ,
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:

123707 ,
100000
24576 ,
10
2332210 .
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,

7is 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 , and is, 120 ,
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:

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
747091968649303432465
1 × 2 × 3 × 4 = 1
2345 5
2 × 17 = 2
174545
2 × 13 × 241 = 6266
59 7 19 7874
13 × 601 = 7813
461 11 5071
Fraction
proposed.		Square. Cube.
701491401 344472101
158 24964 3944312
140196002744000
141198812803221
35512602544738875
113127691442897

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 and 4 × 7 ,
3 × 113 × 5

and their quotient, according to the rule, is

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

4 × 5 or 20
11 × 777

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.
416341
3311189
46790747167
151101136957
781360113
5071 11 461
What are¹/₅ × ¹/₅ × ¹/₅ - ²/₁₇× ²/₁₇ × ²/₁₇ ,
¹/₅ - ²/₁₇
and⁸/₁₁ × ⁸/₁₁ - ³/₁₁ × ³/₁₁ ?
⁸/₁₁ - ³/₁₁
Answer, 559and 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

(107)a = 1 × a
ba
(108)a = ma
bma
(111)a and c are the same as ad and bc
bdbdbd
(112)a + b = a + b
ccc
a - b = a - b
ccc
(113)a + c = ad + bc
bdbd
a - c = ad - bc
bdbd
(118)a × c = ac
bdbd
(121)a divᵈ. by c or a/b = ad
bdc/dbc

125. These results are true even when the letters themselves represent fractions. For example, take the fraction

whose numerator and denominator are fractional, and multiply its numerator and denominator by the fraction

e , which gives ae/bf
f ce/df
which (121) is aedf
bfce

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

But the original fraction itself is

hence

a/b = a/b × e/f
c/dc/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 , and b
q s c

Then m + n is

p + r, or ps + qr
q s qs

and (m + n)a is

ps + qr × b, or (ps + qr)b
qs c qsc
or psb + qrb .
qsc
But this (112) is psb + qrb, which is pb + rb ,
qscqscqcsc
since psb = pb , andqrb = rb (103).
qscqcqscsc
But pb = p × b, and rb , = r × b.
qc q c sc s c

Therefore (m + n)a, or

(p + r) b = p × b + r × b.
q s c q c s c

In a similar manner the same may be proved of any other formula.

The following examples may be useful:

a × c + e × g = acfh + bdeg
b d f h
—————————————
a × e + c × gaedh + bcfg
b f d h
1 = b
1 ab + 1
a + —
b
1 = 1 = bc + 1
————————————————————
1 cabc + a + c
a +———— a +———
1 bc + 1
b +—
c

Thus,

1 = 1 = 57
—————————————————
1 8350
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,

94 is 940 , or 9400 , or 94000 (106);
101001000
3 is 30 , or 300 , or 3000 (108).
30100100010000

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 , which is 16 × 4375 , or 4375 ,
16000016 × 1000010000

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.

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,

1is less1 by 3which is not1
10than 7 70so much as10
14 1 2 1
100 7 700100
142161
1000 7 70001000
1428141
10000 7 7000010000
14285151
100000 7 700000100000
142857111
1000000 7 70000001000000
&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 70000001000000

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 tables3, 17, and 1 .
with these fractions 91 143247
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 does not differ from 1 by 1 ;
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  Answer, ³³³³³³³³/₁₀₀₀₀₀₀₀₀.
⁴/₇⁵⁷¹⁴²⁸⁵⁷/₁₀₀₀₀₀₀₀₀.
¹¹³/₃₅₅³¹⁸³⁰⁹⁸⁵/₁₀₀₀₀₀₀₀₀.
³⁵⁵/₁₁₃³¹⁴¹⁵⁹²⁹²/₁₀₀₀₀₀₀₀₀.

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.  By (115) 147326 is 147326 ;
100010001000

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 = 14732 + 6
10 10
147326 = 1473 + 2 + 6
100 10100
147326 = 147 + 3 + 2 + 6
1000 101001000
147326 = 14 + 7 + 3 + 2 + 6
10000 10100100010000
147326 = 1 + 4 + 7 + 3 + 2 + 6
10000010100100010000100000
147326 = 1 + 4 + 7 + 3 + 2 + 6
1000000 101001000100001000001000000
147326 = 1 + 4 + 7 + 3 + 2 + 6
10000000 100100010000100000100000010000000

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.
10100
2700031, 2700031,&c.
10100
2073000, 2073000,&c.
10100
3331303, 3331303,&c.
100010000

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 give14732·6when the fraction is147326
10
1473·26 147326
100
147·326 147326
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
101001000100001000001000000

The next fraction is ¹⁴⁷³²⁶/₁₀₀₀₀₀₀₀, which we find by the table to be

1 + 4 + 7 + 3 + 2 + 6
100100010000100000100000010000000

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
10100100010000100000100000010000000

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.

I. 123·4 stands for1234or 1234or 123 +4
101010
12·341234or 1234or 12 +3 + 4
10010010100
1·2341234or 1234or 1 +2 + 3 + 4
10001000101001000
·12341234 or1 + 2 + 3 + 4
10000 10100100010000
·012341234 or 1 + 2 + 3 + 4
100000 100100010000100000
·0012341234 or 1 + 2 + 3 + 4
1000000 1000100001000001000000
II. ·01003 is 1003 or 1 + 3
100000100100000
·1003 is 1003 or 1 + 3
10000 10 10000
10·03 is 1003 or 10 + 3
100100
100·3 is 1003 or 100 + 3
1010
III. ·1238 = 1 + 2 + 8 + 3
10100100010000
= ·1 + ·02 + ·008 + ·0003
= ·1 + ·0283 = ·12 + ·0083
= ·128 + ·0003 = ·108 + ·0203
= ·1003 + ·028 = ·1203 + ·008
IV. In 1234·56789
inches the 		1 is 1000inches
2 is 200
3 is 30
4 is 4
5 is ⁵/₁₀of an inch
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 , which is 1439156
1000010000

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:

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

EXERCISES.

What are1527 + 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 , which is 4624776
100000100000

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:

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 is12362 - 274·22107 + ·5;
9976·2073942 - ·00143976728;
and1·2 + ·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.
82·926875·7264570135·233088
·0173·00029929·000005177717
1·432·04492·924207
·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:

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, and1 .
800125064128
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, and3.
4933370001399072908466277
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).

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 = 1240, ·00062 = ·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, and 365,
3·141592·7182818·18349

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 × 32 × 3 × 42 × 3 × 4 × 5
1 + 1 + 1 + 1 + 1 + &c. = 2·92895.
2 3 4 5
80 + 81 + 82 + 83 + 84 + &c. = 9·88286.
8182838485

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:

22212221
12344321
88842221
6663  4442
4442    6663
2221      8884
27407142740714

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
10465242
24418898
13953656
6976828
17898·152223188

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,

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.

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.

·447161833·1662483·4641016
3·77192141·41421361732·508
37719214033166248346410160
8161744 631241418052371 
150876863316625346410160
15087681326650242487112
2640343316610392305
377213266692820
2263663173205
38332771
30106001·58373
1·68665912
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
260926099
251425136
959632
949426
12061

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,

·003178, since this is ·3178 ,
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 ,
6765676545765225

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

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)3,61,76 647)361
703 29 00 329
400)32 76 546)3276
14032 76 3276
60  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   40109)986
9000 9 81 00 00 00   9 981
100000)5 78 44 01 11804)57844
180004 72 16 00 47216
40 —————— ————
100000)5 78 44 01
180004 72 16 00
40 ——————
100000)1 06 28 01 118089)1062801
180001 06 28 01 1062801
80 —————— —————
90 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.
73441271
29929001730
641424792180089
903687890625950625
42420747482776576 205962976
13422659310152401115856201

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.

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.

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
64·348·02122185
807489·8554394
103·16227766
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 + 1) 99 , or 100 × 99 , or 4950.
22

The sum of 50 terms of the series

1 , 2 , 1, 4 , 5 , 2, &c. is ( 1 + 50 ) 50 ,
3333332

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.33841452
1,3,5, &c.2855784
2,20,38, &c.100,0001799984 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.
18090251345126892
4410329 14
545
7075600 13304106368

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

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 × bd = abd  (116) = ad,
bb
and c × bd = cbd = cb:
dd
hence (180), ad = bc.

Thus, 6, 8, 21, and 28, are proportional, since

6 = 3 = 3 × 7 = 21 (180);
844 × 728

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:—

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 , and pq = p ; hence (180),
bqqbqb
a = p , or a : q ∷ p : b .
qb

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 ,
bd

it follows (181) that ad = bc, all the proportions which follow from ad = bc, by the last article, follow also from

a = c .
bd

184. From (114) it follows that

1 + a = b + a ,
bb
and if a be less than 1,
b
1 - a = b - a ,
bb
while if a be greater than 1,
b
a - 1 = a - b.
bb
Also (122), if a + b be divided by a - b
bb
the result is a + b
a - b

Hence, a, b, c, and d, being proportionals, we may obtain other proportions, thus:

Let a = c
bd
Then (114)  1 + a = 1 + c
bd
or   a + b = c + d
bd
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

or   a = c
bd
1 - a = 1 - c , if a be less than 1,
bdb
or   b - a = d - c
bd

that is, b-a: b ∷ d-c: d
or,  a-b: b ∷ c-d: d,

if a be greater than 1.
b
Again, since a + b = c + d
bd
and a - b = c - d ( a being greater than 1)
bdb

dividing the first by the second we have

a + b = c + d
a - bc - d

or a + b : a - b ∷ c + d : c - d

and also a + b : b - a ∷ c + d : d - c,

if a 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 : mb ∷ c : md
nn
ma : nb ∷ mc : nd
a : b ∷ c : d
mmmm
a : b ∷ c : d
mmnn

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 × md, or mbc ,
nn

while that of the means is

mb × c , or mad .
nn

But since a : b ∷ c : d, by (181) ad = bc,

whence, by (180), mad = mbc, and mad = mbc .
nn
Hence a , c , and md, are proportionals.
nn

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

Then  a : b ∷ c : d
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

To prove this, let a be called x.
b
Then, since a = x, and a = c ,
bbd
it follows that c = 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

1248163264&c.
2222222&c.
3 9 27 81 243729

Which are in continued proportion, since

1 : 2 ∷ 2 : 4 2 : 2 ∷ 2 : 2
339
2 : 4 ∷ 4 : 8 2 : 2 ∷ 2 : 2
39927
&c. &c.

192. Let a, b, c, d, e be in continued proportion; we have then

a : b ∷ b : c or a = b or ac = bb
bc
b : c ∷ c : d b = c bd = cc
cd
c : d ∷ d : e c = d ce = dd
de

Each term is formed from the preceding, by multiplying it by the same number. Thus,

b = b × a (180); c = c × b;
ab
and since a = b ,  b = c
bcab
or  c = b × b.
a
Again,  d = d × c ,
c
but d = c , which is = b ;
cba
therefore, d = b × c, and so on
a
If, then, b (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:

1 + r = rr - 1
r - 1
1 + r + rr=rrr - 1
r - 1
1 + r + rr + rrr=rrrr - 1
r - 1
1 + r + rr + rrr + rrrr=rrrrr - 1
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 + r = 1 - rr
1 - r
1 + r + rr=1 - rrr
1 - r
1 + r + rr + rrr=1 - rrrr
1 - r
1 + r + rr + rrr + rrrr=1 - rrrrr
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 × a, or 1 - rrrr × a,
r - 11 - r

according as r is greater or less than unity. The first fraction is

arrrr - a , or (192) e - a .
r - 1r - 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
—————  =  3——— .
1 - ½131072

EXAMPLES.

9 terms of 1 + 4 + 16 + &c. are 87381
10 ......  3 + 6 + 12 + &c. ... 847422675
749201768035
20 ......  1 + 1 + 1 + &c. ... 1048575
2481048576

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 ,
5x
and the powers of ⅖ are 1 , 1 ,
xxxxx

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 , or (112) 1 - a .
1 - r1 - r1 - 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 - r1 - 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

1or 2,
1 - ½

as was shewn in the last article.

EXERCISES.

The limit of  2 + 2 + 2 + &c.
39
or  2(1 + 1 + 1 + &c.)is 3
39
... 1 + 9 + 81 + &c.  ...  10
10 100
... 5 + 15 + 45 + &c.  ...  8¾
749

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

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 , and d ,
pqrs

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
less
than 		and
greater
than
a + b a b and c
p + q p q r
a + b + c a + b and a c and d
p + q + r p + qp rs
a + b + c + d a + b + c and a d
p + q + r + s p + q + rp 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:

abcacebcdbef
abdacfbcecde
abeadebcfcdf
abfadfbdecef
acdaefbdfdef

The combinations of four are fifteen in number, namely,

abcdabdeacdeadefbcef
abceabdfacdfbcdebdcf
abcfabefacefbcdfcdef

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:

abcdacbdacdbabdcadbcadcb
bacdcabdcadbbadcdabcdacb
bcadcbadcdabbdacdbacdcab
bcdacbdacdbabdcadbcadcba

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

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,

abcde
agivesabacadae
b···· bcbdbe
c···· cdce
d···· de
Of these, abgivesabcabdabe
ac···· acdace
ad···· ade
bc···· bcdbce
bd···· bde
cd···· cde
ae be ce and de give none.
Of these, abcgivesabcd 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 out of 8 Answer, 28
combinations411330
can be made of 2628378
6155005

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.