I need to prove the following by contrapositive:
"$x$ and $y$ are integers, and $xy$ is even, then $x$ is even or $y$ is even"
I'm sure this question isn't very hard to solve, however my understanding of contraposition is very weak. I have only learned it recently and I do not feel like I am totally grasping the concept.
From my understanding I am almost trying to prove this by saying the opposite statement. But I feel like this is oversimplifying it.
I know that based on previous proofs an even number is in a form like $z = 2a$ an odd is the same as the even except plus 1: $x = 2a + 1$. Based on knowing this I assume I am able to use contraposition. But where do I start to use this?
Best Answer
What your are starting with is
Let $x, y$ be integers.
($xy$ is even) $\implies (x$ is even or y is even)
The contrapositive is then $\lnot(x$ is even or $y$ is even)$\implies \lnot(xy$ is even).
This means we want to prove that if $x$ is odd AND $y$ is odd, then $xy$ is odd.
Start in the standard way: Let $x = 2a +1$ and let $y = 2b +1$ where $a, b \in \mathbb Z$.
Then calculate $xy$, and represent the product as an odd integer.