In the example, No. 1, say, 7 from 6 I cannot, but 7 from 16 (for you must always borrow 10, in simple numbers, in such cases) and there remains 9, which you set down. Then 1 that I borrowed and 6 are 7; 7 from 7 and there remains 0; 3 from 8 and there remains 5; nothing from 9 and there remains 9. In order to prove it, work the two last sums by addition thus:

3671

9509

9876

The other sums are worked and proved in the same manner, taking care, in compound numbers, to attend to the difference of the value of the figures.

Thus, in No. 4, say 9 from 7 I cannot, but 9 from 19 (adding 12) and there remains 10: 1 that I borrowed and 12 are 13; 13 from 6 I cannot, but 13 from 26 (adding 20)/ and there remains 13: 1 that I borrowed and 5 are 6; 6 from 3 I cannot, but 6 from 13 (adding 10) and there remains 7 1 that I borrowed and 2 are 3; 3 from 2 I cannot, but 3 from 12, and there remains 9: 1 that I borrowed from 4, and there remains 3. Prove it as before by addition, thus:

THIS, for general purposes, is the most useful rule in Arithmetic; and therefore particular attention should be paid to the following Table, which must be learned completely by heart before any thing can be done by the pupil to advantage.

In order to understand this table the learner must multiply each figure of the first column by those of the upper row, looking for the product in that square which is in a line with the one and underneath the other. Thus if the pupil wants to find the value of 6, multiplied by 5, by looking on the line where the 5 is placed in the first column, under the 6 in the top line, the product will be found to be 30. The way therefore to learn this table, which must be

done correctly, is to go on thus; twice 1 are 2, twice 3 are 6, twice 4 are 8, twice 5 are 10, and so on through the whole,

12 24 36 .48 60 72 84 96 108 120 132 144

When quite perfect in this, the learner may proceed to

the following lesson.

EXAMPLES.

In the first of the foregoing exainples say, 5 times 5 are 25; then write 5 and carry 2 to the next, saying 5 times 6 are 30, and 2 are 32; write 2 and carry 3, saying 5 times 3 are 15 and 3 are 18; therefore 365 multiplied by 5 makes 1825. In the second example the multiplier being two figures, namely 24, begin with the 4 and go through the whole of the sum to be multiplied, as in No. 1. Then with the 2 in like manner, only observing to put the product of the first figure under the multiplying figure, as set down, and multiply on as before; when both are performed, the rule of addition must be applied to ascertain the whole product, as both are to be added, and the amount will be the sum required.

Another way of working this, and which will also prove whether the sum here stated is right, is, by multiplying by 2 and 12, because twice 12 are 24, thus:

To show, in the third example, the amount of 6 times £24 38.64d. multiply thus: 6 times 2 are 12; 12 farthings being 3 pence, carry S to the pence: then 6 times 6 are 36 and 3 are 39; 39 pence being 3 shillings and 3 pence, set down 3 and carry 3; then 6 times 3 are 18 and 3 are 21; 21 shillings being pound 1 shilling, set down and carry 1: then 6 times 4 are 24 and 1 are 25; 5 and carry 2: then 6 times 2 are 12 and 2 are 14. Hence 6 pounds weight, or 6 barrels, or 6 pieces of any article, at £24 3s. 6d. would amount to £145 1s. 3d.

When either the multiplier or the multiplicand (that is the sum multiplied) or both, contains ciphers on the right hand, set down so many ciphers as there are in both, on the right of the product; and multiply only by the re mainder, thus:

DIVISION.

As Multiplication teaches the art of finding any number when repeated so many times, so Division instructs us how often one given number is contained in another. Thus, to know how many times 6 are contained in 478654, set them down in this manner:

This is performed by saying 6's in 47, 7 times and 5 over, because 7 times 6 are 42; then placing the 5 before the next figure 8, it makes 58; 6's in 58, 9 times and 4 over; which placed before the next figure 6 makes 46; then 6's in 46, 7 times and 4 over; 6's in 45, 7 times and 3 over; 6's in 34, 5 times and 4 over; therefore 79775 and over, is the answer. In order to prove it multiply it thus:

Multiply 79775-4 the Quotient.

by

6 the Divisor.

Here 6 times 5 are 30, and 4 the remainder are 34; 4 and carry 3, and so on.

When the divisor exceeds 12, it is necessary to proceed as in the following example:

Finding that 10 twenty fives make 250; place 10 on the right and multiply the 25 by 10, as here stated, then by subtracting 250 from 265, there remains 15; so that the answer

25)265(10

250

15

is 10 times and 15 over; and in order to prove it, multiply it as before.

REDUCTION.

THE next step is to reduce sums of money, &c. into an amount of different denominations; as, for instance, pounds into shillings, pence, or farthings; years into days, hours, or minutes, &c. It is not, properly speaking, a distinct rule in arithmetic, but rather the application of the two preceding ones, namely, Multiplication and Division,

In the first of these examples, begin to multiply by 20, because 20 shillings make one pound; but as it contains a cipher on the right hand, take the 4 from the 14 shillings, and set down in its proper place; then multiply by the 2, saying twice 2 are 4 and 1 from the 14 which was left are 5, and twice 3 are 6: then multiply the 654 shillings by 12, because 12 pence make one shilling, adding the 6 from the pence to the first figure multiplied; and lastly multiply the 7854 pence by 4, because 4 farthings make one penny, adding the 3 farthings to the first figure multiplied.

In the second example, the sum is proved by divison, which is the way to ascertain any similar sum, here you begin by dividing the 31419 farthings by 4, in order to bring them into pence, thus 4's in 31, 7 times and 3 over; 4's in 34, 8 times, and 2 over; 4's in 21, 5 times, and 1 over; 4's in 19, 4 times and 3 over, which are 3-4ths of a penny, and therefore you find that in $1419 farthings, are contained" 7854 pence and 3 farthings; thus you proceed through the whole, dividing the pence by 12, because 12 pence make one shilling, and the shillings by 20, taking care to carry out as here stated the overplus that remains, which must be brought down when the answer is given.

In dividing by 20, cut off the cipher, and the last figure in the quotient, which you carry out; and divide by 2, it being more easily done; thus 2's in 6, 3 times; 2's in 5, twice and 1 over, which by carrying out to the 4, makes 14 shillings over.

These two plain cxamples will give the learner a sufficient idea of the general principle of reduction, as the same method is adopted in the reducing of weights, measures, &c.