Leta = x + y - z,

thenma = mx + my - mz.

For, if a had been x + y, ma would have been mx + my. But since a is less than x + y by z, too much by z has been repeated every time that x + y has been repeated;—that is, mz too much has been taken; consequently, ma is not mx + my, but mx + my-mz. Similar reasoning may be applied to other cases, and the following results may be obtained:

m(a + b + c - d) = ma + mb + mc - md.

a(a - b) = aa - ab.
b(a - b)=ba - bb.
3(2a - 4b)=6a - 12b.
7a(7 + 2b)=49a + 14ab.
(aa + a + 1)a=aaa + aa + a.
(3ab - 2c)4abc=12aabbc - 8abcc.

55. There is another way in which two numbers may be multiplied together. Since 8 is 4 times 2, 7 times 8 may be made by multiplying 7 and 4, and then multiplying that product by 2. To shew this, place 7 counters in a line, and repeat that line in all 8 times, as in figures I. and II.

I.
A ● ●● ●● ●●
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B ● ●● ●● ●●
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II.
● ●● ●● ●●
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The number of counters in all is 8 times 7, or 56. But (as in fig. I.) enclose each four rows in oblong figures, such as a and b. The number in each oblong is 4 times 7, or 28, and there are two of those oblongs; so that in the whole the number of counters is twice 28, or 28 x 2, or 7 first multiplied by 4, and that product multiplied by 2. In figure II. it is shewn that 7 multiplied by 8 is also 7 first multiplied by 2, and that product multiplied by 4. The same method may be applied to other numbers. Thus, since 80 is 8 times 10, 256 times 80 is 256 multiplied by 8, and that product multiplied by 10. If we use the signs, the foregoing assertions are made thus:

7 × 8 = 7 × 4 × 2 = 7 × 2 × 4.
256 × 80 = 256 × 8 × 10 = 256 × 10 × 8.