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tens under tens, &c. as in addition. It is also most convenient, and, in fact, frequently necessary, to begin with the units as in addition and multiplication.
Operation. Ox 47 dollars. I say first 3 from 7, and there will Cow 23 dollars. remain 4. Then 2 (tens) from 4
(tens) and there will remain 2 (tens), 24 difference. and the whole remainder is 24. A man having 62 sheep in his flock, sold 17 of them ; how many had he then ?
Operation. He had 62 sheep In this example a difficulty immedi Sold 17 sheep ately presents itself, if we attempt to
perform the operation as before ; for Ilad left 45 sheep we cannot take 7 from 2. We can, however, take 7 from 62, and there remains 55 ; and 10 from 55, and there remains 45, which is the answer.
The same operation may be performed in another way, which is generally more convenient. I first observe, that 62 is the same as 50 and 12; and 17 is the same as 10 and 7. They may be written thus :
62 = 50 + 12 That is, I take one ten from the six 17 = 10 + 7 tens, and write it with the two units.
But the 17 I separate simply into units 45 = 40 + 5 and tens as they stand. Now I can take 7 from 12, and there remains 5. Then 10 from 50, and there remains 40, and these put together make 45.*
This separation may be made in the mind as well as to write it down. Operation.
62 Here I suppose 1 ten taken from the B tens, 17 and written with the 2, which makes 12. I say
7 from 12, 5 remains, then setting down the 5, I 45 say, 1 ten from 5 tens, or simply 1 froin 5, and there remains 4 (tens), which written down shows the remainder, 45.
The taking of the ten out of 6 tens and joining it with the 2 units, is called borrowing ten.
* Let the pupil perform a large number of examples by separating them in this way, when he first commences subtraction.
Sir Isaac Newton was born in the year 1642, and he died in 1727 ; how old was he at the time of his decease ?
It is evident that the difference between these two numbers must give his age.
1600 + 40 + 2 = 1642
. In this example I take 2 from 7 and there remains 5, which I write down. But since I cannot take 4 (tens) from 2 (tens,) I borrow I (hundred) or 10 tens from the 7 (hundreds,) which joined with 2 (tens) makes 12 (tens,) then 4 (tens) from 12 (tens) there remains 8 (tens,) which I write down. Then 6 (hundreds) from 6 (hundreds) there remains nothing. Also I thousand) from 1 (thousand) nothing remains. The answer is 85 years.
A man bought a quantity of flour for 15,265 dollars, and sold it again for 23,007 dollars, how much did he gain by the bargain?
0, and it does not immediately appear where
2. I shall borrow the 10 (tens,) since there is nothing in the hundreds' place. This will be evident, how ever, if I decompose the numbers into parts.
7,000 + 700 + 40 + 2 = 7,742 The 23,000 is equal to 10,000 and 13,000; this last 19 equal to 12,000 and 1,000 ; and 1,000 is equal to 900 and 100. Now I take 5 from 7, and there remains 2 ; 60 from 100, or 6 tens from 10 tens, and there remains 40, or 4 tens; 2 hundreds from 9 hundreds, and there remains 7 hundreds ; 5 thousands from 12 thousands, and there remains 7 thousands; and 1 ten-thousand from 1 ten-thousand, and nothing remains. The answer is 7,742 dollars.
This example may be performed in the same manner as
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 ! (ten-thousand) from 1 (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, first, must be called Is 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
Proof. A man bought an ox and a cow 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.
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 48 added together must make 73 dollars, what they both cost.
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, 23. 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.
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 so on. The question then is, to see how many times 8 may be taken from 32 ; in 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 per ceive, is to be performed exactly like it. That is, it is the question to see how many times 8 is contained in 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. Ilow many did he buy?
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 dividend; 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