Elements of arithmetic

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—

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

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:

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).

33. When the last two articles have been thoroughly studied, you will be able to find the sum of any numbers by the following process,[6] which is the same as that in (29).

Rule I. Place the numbers under one another, units under units, tens under tens, and so on.

II. Add together the units of all, and part the whole number thus obtained into units and tens. Thus, if 85 be the number, part it into 8 tens and 5 units; if 136 be the number, part it into 13 tens and 6 units (20).

III. Write down the units of this number under the units of the rest, and keep in memory the number of tens.

IV. Add together all the numbers in the column of tens, remembering to take in (or carry, as it is called) the tens which you were told to recollect in III., and divide this number of tens into tens and hundreds. Thus, if 335 tens be the number obtained, part this into 33 hundreds and 5 tens.

V. Place the number of tens under the tens, and remember the number of hundreds.

VI. Proceed in this way through every column, and at the last column, instead of separating the number you obtain into two parts, write it all down before the rest.

Example.—What is

1805 + 36 + 19727 + 3 + 1474 + 2008

•   1805
•   36
• 19727
• 3
•   1474
•   2008
• ——-
• 25053

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

34. As an exercise in addition, you may satisfy yourself that what I now say of the following square is correct. The numbers in every row, whether reckoned upright, or from right to left, or from corner to corner, when added together give the number 24156.

35. If two numbers must be added together, it will not alter the sum if you take away a part of one, provided you put on as much to the other. It is plain that you will not alter the whole number of a collection of pebbles in two baskets by taking any number out of one, and putting them into the other. Thus, 15 + 7 is the same as 12 + 10, since 12 is 3 less than 15, and 10 is three more than 7. This was the principle upon which the whole of the process in (29) was conducted.

36. Let a and b stand for two numbers, as in (24). It is impossible to tell what their sum will be until the numbers themselves are known. In the mean while a + b stands for this sum. To say, in algebraical language, that the sum of a and b is not altered by adding c to a, provided we take away c from b, we have the following equation:

(a + c) + (b - c) = a + b;

which may be written without brackets, thus,

a + c + b - c = a + b.

For the meaning of these two equations will appear to be the same, on consideration.

37. If a be taken twice, three times, &c., the results are represented in algebra by 2a, 3a, 4a, &c. The sum of any two of this series may be expressed in a shorter form than by writing the sign + between them; for though we do not know what number a stands for, we know that, be it what it may, 2a + 2a = 4a, 3a + 2a = 5a, 4a + 9a = 13a; and generally, if a taken m times be added to a taken n times, the result is a taken m + n times, or

ma + na = (m + n)a.

38. The use of the brackets must here be noticed. They mean, that the expression contained inside them must be used exactly as a single letter would be used in the same place. Thus, pa signifies that a is taken p times, and (m + n)a, that a is taken m + n times. It is, therefore, a different thing from m + na, which means that a, after being taken n times, is added to m. Thus (3 + 4) × 2 is 7 × 2 or 14; while 3 + 4 × 2 is 3 + 8, or 11.

39. When one number is taken away from another, the number which is left is called the difference or remainder. The process of finding the difference is called subtraction. The number which is to be taken away must be of course the lesser of the two.

40. The process of subtraction depends upon these two principles.

I. The difference of two numbers is not altered by adding a number to the first, if you add the same number to the second; or by subtracting a number from the first, if you subtract the same number from the second. Conceive two baskets with pebbles in them, in the first of which are 100 pebbles more than in the second. If I put 50 more pebbles into each of them, there are still only 100 more in the first than in the second, and the same if I take 50 from each. Therefore, in finding the difference of two numbers, if it should be convenient, I may add any number I please to both of them, because, though I alter the numbers themselves by so doing, I do not alter their difference.

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

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

41. If a, b, and c be three numbers, of which a is greater than b (40), I. leads to the following,

(a + c) - (b + c) = a - b.

Again, if c be less than a and b,

(a - c) - (b - c) = a - b.

The brackets cannot be here removed as in (36). That is, p- (q-r) is not the same thing as p-q- r. For, in the first, the difference of q and r is subtracted from p; but in the second, first q and then r are subtracted from p, which is the same as subtracting as much as q and r together, or q + r. Therefore p-q-r is p-(q + r). In order to shew how to remove the brackets from p -(q-r) without altering the value of the result, let us take the simple instance 12-(8-5). If we subtract 8 from 12, or form 12-8, we subtract too much; because it is not 8 which is to be taken away, but as much of 8 as is left after diminishing it by 5. In forming 12-8 we have therefore subtracted 5 too much. This must be set right by adding 5 to the result, which gives 12-8 + 5 for the value of 12-(8-5). The same reasoning applies to every case, and we have therefore,

p - (q + r) = p - q - r.
p - (q - r) = p - q + r.

By the same kind of reasoning,

a - (b + c - d - e) = a - b - c + d + e.
2a + 3b - (a - 2b) = 2a + 3b - a + 2b = a + 5b.
4x + y - (17x - 9y) = 4x + y - 17x + 9y = 10y - 13x.

42. I want to find the difference of the numbers 57762 and 34631. Take these to pieces as in (29) and

57762 is 5 ten-th. 7 th. 7 hund. 6 tens and 2 units.

34631 is 3 ten-th. 4 th. 6 hund. 3 tens and 1 unit.

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:

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.

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.