Question

Prove that for any integer A is valid: If A^2 is a multiple of 2, then A is also a multiple of 2 ?

Answer 1

Use the contraposition: If and only if `A->B` is true, `notB->notA` is also true.

Explanation:

You can prove the problem using contraposition.
This proposition is equivalent to:

If `A` is not a multiple of `2`, then `A^2` is not a multiple of `2.` (1)

Prove the proposition (1) and you're done.

Let `A=2k+1` (`k`: integer). Now `A` is an odd number.Then,
`A^2=(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1`
is also odd. Proposition (1) is proven and so as the original problem.