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the others, without separating it into parts except in the mind.

I say 5 from 7, there remains 2: then borrowing 10 (which must in fact come from the 3 (thousand), I say, 6 (tens) from 10 (tens) there remains 4 (tens ;) then I borrow ten again, but since I have already used one of these, I say, 2 (hundreds) from 9 (hundreds) there remains 7 (hundreds ;) then I borrow ten again, and having borrowed one out of the 3 (thousand,) I say, 5 (thousand) from 12 (thousand) there remains 7 (thousand ;) then 1 (ten-thousand) from í (tenthousand) nothing remains. The answer is 7,742 as before.

The general rule for subtraction may be expressed thus : The less number is always to be subtracted from the larger. Begin at the right hand and take successively each figure of the less number from the corresponding figure of the larger number, that is, units from units, tens from tens, fc. If it happens that any figure of the less number cannot be taken from the corresponding figure of the larger, borrow ten and join it with the figure from which the subtraction is to be made, and then subtract; before the next figure is subtracted take care to diminish by one the figure from which the subtraction is to be made.

N. B. When two or more zeros intervene in the number from which the subtraction is to be made, all, except the first, must be called 9s in subtracting, that is, after having borrowed ten, it must be diminished by one, on account of the ten which was borrowed before.

Note. It is usual to write the smaller number under the greater, so that units may stand under units, and tens under

tens, &c.

Proof. A man bought an ox and a row for 73 dollars, and the price of the cow was 25 dollars ; what was the price of the ox?

The price of the ox is evidently what remains after taking 25 from 73.

Operation.
Ox and cow 73 dollars
Cow

25 do.

Ox

48 do. It appears that the ox cost 48 dollars. If the cow cost 25 dollars, and the ox 48 dollars, it is evident that 25 and 46 added together must make 73 dollars, what they both cost.

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Hence to prove subtraction, add the remainder and the smaller number together, and if the work is right their sum will be equal to the larger number.

Another method. If the ox cost 48 dollars, this number taken from 73, the price of both, must leave the price of the cow, that is, 25. Hence subtract the remainder from the larger number, and if the work is right, this last remainder will be equal to the smaller number.

Proof of addition. It is evident from what we have seen of subtraction, that when two numbers have been added together, if one of these numbers be subtracted from the sum, the remainder, if the work be right, must be equal to the other number. This will readily be seen by recurring to the last example. In the same manner if more than two numbers have been added together, and from the sum all the numbers but one, be subtracted, the remainder must be equal to that one.

DIVISION.

So on.

IX. A boy having 32 apples wished to divide them equally among 8 of his companions; how many must he give them apiece ?

If the boy were not accustomed to calculating, he would probably divide them, by giving one to each of the boys, and then another, and so on. But to give them one apiece would take 8 apples, and one apiece again would take 8 more, and

The question then is, to see how many times 8 may be taken from 32 ; or, which is the same thing, to see how many times 8 is contained in 32. It is contained four times. Ans. 4 each.

A boy having 32 apples was able to give 8 to each of his companions. How many companions had he?

This question, though different from the other, we perceive, is to be performed exactly like it. That is, it is the question to see how many times 8 is contained 32. We take away 8 for one boy, and then 8 for another, and so on.

A man having 54 cents, laid them all out for oranges, at 6 cents apiece. How many did he buy?

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It is evident that as many times as 6 cents can be taken from 54 cents, so many oranges he can buy. Ans. 9 oranges

A man bought 9 oranges for 54 cents ; how much did he give apiece ?

In this example we wish to divide the number 54 into 9 equal parts, in the same manner as in the first question we wished to divide 32 into 8 equal parts. Let us observe, that if the oranges had been only one cent apiece, nine of them would come to 9 cents ; if they had been 2 cents apiece, they would come to twice nine cents; if they had been 3 cents apiece, they would come to 3 times 9 cents, and so on. Hence the question is to see how many times 9 is contained in 54. Ans. 6 cents apiece.

In all the above ques.ions the purpose was to see how many times a small number is contained in a larger one, and they may be performed by subtraction. If we examine them again we shall find also, that the question was, in the two first, to see what number 8 must be multiplied by, in order to produce 32; and in the third, to see what the number 6 must be multiplied by, to produce 54; in the fourth, to see what number 9 must be multiplied by, or rather what number must be multiplied by 9, in order to produce 54.

The operation by which questions of this kind are performed is called division. In the last example, 54, which is the number to be divided, is called the dividend ; 9, which is the number divided by, is called the divisor ; and 6, which is the number of times 9 is contained in 54, is called the quotient.

It is easy to see from the above reasoning, that the quotient and divisor multiplied together must produce the divi

for the question is to see how many times the divisor must be taken to make the dividend, or in other words to see what the divisor must be multiplied by to produce the dividend. It is evident also, that if the dividend be divided by the quotient, it must produce the divisor. For if 54 contains 6 nine times, it will contain 9 six times.

To prove division, multiply the divisor and quotient together, and if the work be right, the product will be the dividend. Or divide the dividend by the quotient, and if the work be right, the result will be the divisor.

This also furnishes a proof for multiplication, for if the

dend ;

quotient multiplied by the divisor produces the dividend, it is evident, that if the product of two numbers be divided by one of those numbers, the quotient must be the other num ber.

It appears that division is applied to two distinct purposes, though the operation is the same for both. The object of the first and fourth of the above examples is to divide the numbers into equal parts, and of the second and third to find how many times one number is contained in another. At present, we shall confine oui attention to examples of the latter kind, viz. to find how many times one number is contained in another.

At 3 cents apiece, how many pears may be bought for 57 cents ?

It is evident, that as many pears may be bought, as there are 3 cents in 57 cents. But the solution of this question does not appear so easy as the last, on account of the greater number of times which the divisor is contained in the dividend. If we separate 57 into two parts it will appear more easy.

57 = 30 +- 27. We know by the table of Pythagoras that 3 is contained in 30 ten times, and in 27 nine times, consequently it is contained in 57 nineteen times, and the answer is 19 pears. Ного

many barrels of cider, at 3 dollars a barrel, can be bought for 84 dollars ?

Operation.

84 = 60 +- 24 3 is contained in 6 twice, but in 6 tens it is contained ten times as often, or 20 times. 3 is contained in 24 eight times, consequently 3 is contained 28 times in 84. Ans. 28 barrels.

How many pence are there in 132 farthings ? As many times as 4 farthings are contained in 132 farthings, so many pence there are.

Operation.

132 = 120 + 12 120 is 12 tens, 4 is contained in 12 three times, consequently it is contained 30 times in 12 tens. 4 is contained 3 times in 12 units, consequently in 132 it is contained 33 times. Ans. 33 pence.

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How many barrels of flour, at 5 dollars a barrel, may be bought for 785 dollars.

Operation. 785 = 500 + 250 + 35 5 is contained in 5 once, and in 500 one hundred times. 250 is tens. 5 is con ned 5 times in 25, consequently 50 times in 250. 5 is contained 7 times in 35 units. In 785, 5 is contained 157 times. Ans. 157 barrels. How

many dollars are there in 7464 shillings ? As many times as 6 shillings are contained in 7464 shillings, so many dollars there are.

Operation. 7464 = 6000 + 1200 + 240 +- 24 6 is contained 1000 times in 6000, 200 times in 1200, 40 times in 240, and 4 times in 24, making in all 1244 times.* Ans. 1244 dollars.

It is not always convenient to resolve the number into parts in this manner at first, but we may do it as we perform the operation. In 126 days how many weeks ?

Operation.

126 = 70 + 56 Instead of resolving it in this man ner, we will write it down as follows. Dividend 126 07 Divisor

70

10 56 8 56

18 quotient

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1 observe that 7 cannot be contained 100 times in 126, ! therefore call the two first figures on the left 12 tens or 120, rejecting the 6 for the present. 7 is contained more than once and not so much as twice in 12, consequently in 12 tens it is contained more than 10 and less than 20 times. I take 10 times 7 or 70 out of 126, and there remains 56. Then 7 is contained 8 times in 56, and 18 times in 126. Ans. 18 weeks.

* Let the pupil perform a large number of examples in this manner when he first commences; as he is obliged to separate the numbers into parts, he will at length come to the common method.

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