The problem with trying a direct proof is that it will be hard to separate \(a\) and \(b\) from knowing something about \(a+b\text{.}\) On the other hand, if we know something about \(a\) and \(b\) separately, then combining them might give us information about \(a+b\text{.}\) The contrapositive of the statement we are trying to prove is: for all integers \(a\) and \(b\text{,}\) if \(a\) and \(b\) are even, then \(a+b\) is even. Thus our proof will have the following format:
Let \(a\) and \(b\) be integers. Assume that \(a\) and \(b\) are both even. la la la. Therefore \(a+b\) is even.
Here is a complete proof:
Let \(a\) and \(b\) be integers. Assume that \(a\) and \(b\) are even. Then \(a = 2k\) and \(b = 2l\) for some integers \(k\) and \(l\text{.}\) Now \(a + b = 2k + 2l = 2(k+l)\text{.}\) Since \(k + l\) is an integer, we see that \(a + b\) is even, completing the proof.
Note that our assumption that \(a\) and \(b\) are even is really the negation of \(a\) or \(b\) is odd. We used De Morgan's law here.