Identity 1

First, a word for what we've been writing. Two algebraic expressions are equal if they give the same value whenever the letters are replaced by numbers. A statement that two expressions are equal, such as a(b + 8) = ab + 8a or (a + 1)(b − 1) = ab + b − a − 1, is called an identity.

Now the general question. If a goes up by m and b goes up by n, what happens to ab?

Take (a + m) as one term and distribute: (a + m)(b + n) = (a + m)b + (a + m)n = ab + mb + an + mn.

Identity 1: (a + m)(b + n) = ab + mb + an + mn. The increase is an + bm + mn.

Read the right-hand side carefully. It's the sum of the product of each term of (a + m) with each term of (b + n): a with b, a with n, m with b, m with n. Four products, one for each pairing. The picture shows it: a big rectangle of (a + m) rows and (b + n) columns splits into four rectangles of areas ab, an, mb and mn.

Now the useful part. Integers obey distributivity too (x(y + z) = xy + xz for any integers), so the identity holds when the letters stand for negative numbers as well. You never need a separate rule for decreasing.

Check it. Earlier we found (a + 1)(b − 1) = ab + b − a − 1 by working it out separately. Write it as (a + 1)(b + (−1)) and put m = 1 and n = −1 into Identity 1: ab + 1 × b + a × (−1) + 1 × (−1) = ab + b − a − 1. The same answer, from the one identity.

Likewise:

• (a + u)(b − v) = ab + ub − av − uv;
• (a − u)(b + v) = ab − ub + av − uv;
• (a − u)(b − v) = ab − ub − av + uv.

The signs are looked after by the ordinary rules for multiplying integers (a minus times a minus is a plus).

Try it on a calendar. In any 2 × 2 square of dates, call the top-left date a; the others are a + 1, a + 7 and a + 8. The diagonal products are a(a + 8) = a² + 8a and (a + 1)(a + 7) = a² + 8a + 7. They always differ by 7, whichever square you pick.