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, I, 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 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 (85). 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 The same reasoning applies to every case, and we have therefore

85 for the value of 12 -(8—5).

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—13 x

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. 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. I 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. I 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. I thous. 2 hund. 7 tens and 14 units,*

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

Those numbers which have been altered are put in italics.

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. I 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

o ten-thous. o thous. 8 hunds. I ten and 8 units;

the two first numbers are nothing, and the rest is

8 hundreds, I 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.

I. What is 18337 +149263200—6472902 ?— Answer 142808635. What is 1000-464+ 3279—646 ?—Answer 3169.

II. Subtract

64+76+144-18 from 33−2+ 100037.— Answer 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.

and D.

V. There are four places on the same line in the order A, B, C, From A to D it is 1463 miles; from A to c it is 728 miles ; How far is it from A to B, from B From A to B 146, from в to c 582,

and from B to D it is 1317 miles. to c, and from c to D?-Answer. 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 в can be no longer subtracted. Find how many times в can be subtracted from a, and what is the last remainder.

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.

17

17

17

17

17

85

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 is nothing more than the addition of a number of the same Here 17 is called the multiplicand, and 5 is called the

quantities. 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 enor