# Proof By ContraPositive: if xy is odd, then x is odd and y is odd

If $$xy$$ is odd, then $$x$$ is odd and $$y$$ is odd

I was just wondering if the correct contrapositive would involve proving these three cases:

1) $$x$$ is even or $$y$$ is odd

2) $$x$$ is odd or $$y$$ is even

3) $$x$$ is even or $$y$$ is even

I'm not too sure about if the last case is necessary as in the answer to this question only the first two cases were shown. I guess what i'm trying to ask is why do we not check the last case, since it is a possible negation of $$x$$ is odd and $$y$$ is odd?

• No, @user170039 It is the statement of the form $P \land Q$, where $P$ means x is odd, and $Q$ means $y$ is odd, the negation of which is $\lnot (P\land Q) \equiv (\lnot P \lor \lnot Q)$ Mar 10, 2020 at 19:56
• @amWhy: Indeed. Nice catch. I have removed my comment.
– user170039
Mar 11, 2020 at 5:29

The counter-positive of the assertion

if $$xy$$ is odd, then $$x$$ and $$y$$ are odd

is

if $$x$$ is even or $$y$$ is even, then $$xy$$ is even.

And asserting that $$x$$ is even or $$y$$ is even is equivalent to asserting that we have one of the following possibilities:

1. $$x$$ is even and $$y$$ is odd;
2. $$x$$ is odd and $$y$$ is even;
3. both $$x$$ and $$y$$ are even.
• though we need only two cases: 1. $x$ is even (and $y$ arbitrary); 2. $y$ is even (and $x$ arbitrary) Mar 10, 2020 at 13:07