Suppose \(a\) and \(b\) are odd. That is, \(a=2k+1\) and \(b=2m+1\) for some integers \(k\) and \(m\text{.}\) Then
Therefore \(ab\) is odd.
Assume that \(a\) or \(b\) is even - say it is \(a\) (the case where \(b\) is even will be identical). That is, \(a=2k\) for some integer \(k\text{.}\) Then
Thus \(ab\) is even.
Suppose that \(ab\) is even but \(a\) and \(b\) are both odd. Namely, \(ab = 2n\text{,}\) \(a=2k+1\) and \(b=2j+1\) for some integers \(n\text{,}\) \(k\text{,}\) and \(j\text{.}\) Then
But since \(2kj+k+j\) is an integer, this says that the integer \(n\) is equal to a non-integer, which is impossible.
Let \(ab\) be an even number, say \(ab=2n\text{,}\) and \(a\) be an odd number, say \(a=2k+1\text{.}\)
Therefore \(b\) must be even.