Can you prove The product of two integers is even if and only if at least one integer is even?

Answer 1

To prove this, we need to be able to prove two things:

1. The product of two integers is always even if at least one integer is even
2. The product of two integers is never even if neither integer is even

First, let's prove that the product of two integers is always even if at least one integer is even:

Let's say that n is an even integer, and that k is some integer (even or odd). We want to prove that k*n has to be even. Since even is even, we can think of it as 2m, where m is some integer. In other words, n is equal to (2 + 2 + 2 + 2 ...) for however many twos it takes to get there.

We can re-write k*n as k*2m. We can factor out a 2 to get:

2*k*2m-1.

Since we are multiplying an integer quantity by two, this proves that the quantity has to be divisible by 2, which, by definition, means that the number must be even.

Now let's prove that the product of two integers is never even if neither integer is even. Let's call the two integers k and n. Since nither k and n are even, they must both be odd.

k*n can be re-written as k*n - n + n, which can be factored into (k-1)n + n.

Now we know that (k-1)n has to be even, since k-1 is even, and we have proved that a number times an even number is even above.

So now we have an even number plus and odd number. We can re-write n as 2m + 1, where m is an integer. Since (k-1)n is even, that quantity can be written as 2p, where p is an integer.

To recap so far, this means that we can say that if k and n are both odd:

k*n = (k-1)n + n = 2p + 2m + 1.

This can be re-written as 2p+m + 1. Since 2p+m is a multiple of two, it must be even, and an even number plus one must be odd, so the product of two odd numbers must be odd.