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Having placed the two numbers the one under the other, I have subtracted the units from the units, saying; from 8 take 7, remains 1, which I have written underneath: now coming to the tens, and trying to take 8 from 7, I see it cannot be; then I have had recourse to the following fig, at the left, from which I have mentally borrowed one unit, which is a hundred, and consequently worth ten tens; I have joined these ten tens to the 7 of the 2d. figure, and from these 17 taking 8, I have 9 left, which I set down. (Remark that in order not to forget that I have borrowed a unit from the next figure, I have put a dot over this figure.) The 3, by this borrowing, being now worth only 2, I take 2 from 2 and nothing remains; I write 0. Proceeding on, I have said from 0 taking 4 cannot be done; I have taken a unit from the next fig. and then from 10 taking 4, I have 6 left; then 3 from 3, 0 remains; and 7 from 9, remains 2. Here a new difficulty has occurred, because having 2 to subtract from 0, which is impossible, I cannot borrow from the next fig. which is also 0. I have been obliged to reascend up to the first fig. after the cyphers; I have taken a unit from the 6, which being supposed carried on the next cypher, is worth ten; but as I wanted a unit of this ten for the next cypher, I have writen a small 9 over the O next to the 6; and proceeding along, from a similar reasoning, I have also written a small 9 over the second 0; then arriving at the last, I have said from 10 take 2, remains 8; then 9 from 9 and 0 remains; then from 9 take 0, remains 9; from 5 take 7, it cannot; borrowing a unit from the next fig. from 15-7, it leaves 8; at last from 4 nothing, remains 4. It is very evident, that 48908206091, is the difference between the two given numbers, since it is formed of the difference between each of their corresponding parts.

Subtraction is proved by Addition, viz. by adding the smaller number to the difference; for, if the ope

ration be right, this sum must necessarily be equal to the greater number.

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The same process is perfectly applicable to num

bers containing decimal parts.

First Example.

Second Example.

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Proof,

754, 3845

Proof,

843, 0000 In the 2d. example, the decimals, altogether wanting in the larger number, have been supplied by cyphers, and the operation has been performed in the usual way.

OF MULTIPLICATION.

MULTIPLICATION is an operation by which we add a number to itself, as many times as another number contains units.

The first of these numbers is called Multiplicand, the second Multiplier, and the number, resulting from the operation, is called Product. The multiplicand and multiplier are also called the factors of the product.

It follows from the definition of Multiplication, that this operation is nothing but an addition; for,

24

24

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if we have two numbers to multiply one by the other, such as 24 and 6, 24 being the multiplicand, and 6 the multiplier, the question is then, to add 24 6 times to itself, which can be executed, as it is here, by the simple rules of Addition; and in reality 144 is certainly the product asked for, since it contains 24 6 times; but it is easy to see how tedious this process would be, if the multiplier were a considerable number.

24

24

24

144

Multiplication abridges this addition. The whole art consists in the use of Pythagoras's table, which must be got by heart; this table presents the products of all simple numbers, the ones by the others. A single example will show how simple, easy, and general, the rules of multiplication become, by the means of this table.

I place the multiplicand and multiplier one under the other; units under units; tens under tens, &c. and, having drawn a horizontal line under the mul

tiplier, I successively mul

EXAMPLE. 50794 multiplicand. 6033 multiplier.

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tiply all the figures of the 304764... 4th multiplicand by each fig

do.

ure of the multiplier, which 308979902 total product. gives so many partial pro

ducts, which I write, the ones under the others, with the particular and essential attention, of placing the first figure of each product, under the figure of the multiplier from which it proceeds; then I cast up all the partial products, the addition of which, gives the total product asked for. Such is the process to be followed in all cases, of which I subjoin the explication.

To get the first partial product, I say 3 times 4 are 12, I write 2, and keep, as in addition, the ten, to

join it to the second product, which I obtain by say ing; 3 times 9 are 27, and 1 (brought from the first column) are 28: I set down 8 and carry 2; then 3 times 7 are 21, and the 2 brought forward are 23; I write 3 and carry 2; 3 times 0 is 0, and 2 carried, are 2, which I set down; finally, 3 times 5 are 15, which I write down. This being done, it is evident that this first partial product contains 3 times the multiplicand, since it is formed of each part of the multiplicand taken 3 times.

By a similar process, it is easy to show, that the second partial product, as it is written, contains the multiplicand 8 times; but we must observe, that 8 being a number of tens, indicates that we must add the multiplicand to itself, not merely 8 times, but 8 tens of times, which obliges us to render this product 10 times greater; now, this is easily done, by writing the units of this product in the row of tens, whereby the tens have become hundreds, the hundreds thousands, &c. which consequently has rendered the total number ten times greater.

What we have said respecting the product of the multiplicand by 8, being applicable to all the other partial products, shows why we must write the first figure of each partial product under the figure of the multiplier from which it proceeds.

Now, casting up all the partial products, their sum must evidently be the total product asked for. The 1st. partial product contains the multipli3 times

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the 2d. contains it

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REMARK. If both the multiplicand and multiplier, or either of them, were terminated by cyphers, the multiplication might be abridged by omitting the cyphers in both numbers; but then you must write, at the end of the product, as many cyphers as there are in both factors.

EXAMPLE.-Be it proposed to multiply 6370 by 5400; I only write The multiplication being done, I add three cyphers to the product, which gives me 34398000 for the total product of the two given numbers. In fact, by cutting off the two cyphers of

637 multiplicand.

54 multiplier.

2548

3185

34398 product.

the multiplier, I had rendered this number one hundred times smaller; then the product must contain the multiplicand one hundred times less; but, this multiplicand, having itself become ten times smaller by the suppression of the cypher, it follows, that the product contains 100 times less a multiplicand 10 times too small, then this product must be 100 times 10 times, or one thousand times too small; I must then make it one thousand times greater, which is done by the addition of three cyphers.

THE MULTIPLICATION OF DECIMALS.

The Multiplication of numbers containing Decimals, is done the same as if there were none; but afterwards, you separate, at the right of the product, with a comma, as many figures as there were decimals in the two factors,

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