Using the technique proof by cases, show that
"For integers $x$ and $y$, if $xy$ is odd, then $x$ is odd and $y$ is odd."
There are solutions online that go through every combination of odd and even values. (Case 1: $x$ is even and $y$ is even, Case 2: $x$ is even and $y$ is odd, Case 3: $x$ is odd and $y$ is even … etc).
I don't think this is the proper way of doing a proof by cases because the cases should be in terms of what is given in the hypothesis ($xy$ is odd) instead of the conclusion we are trying to prove.
However, the only thing I can think of is Case 1: $xy$ is odd, but there is no other case which makes it more of a direct proof. What is the proper way to prove this?
Best Answer
The proof is perfectly fine. I suspect your confusion comes from the idea that to show $p\rightarrow q$ by breaking into cases, we have to break $p$ itself up into cases and discuss. That is not really true. In particular, the statement $p\rightarrow q$ is equivalent to its contrapositive $\neg q\rightarrow\neg p$, so we can just as well break into cases regarding $\neg q$ (the negation of $q$) instead.
In particular, we want to show if $xy$ is odd then $x,y$ are odd. We may just as well show that if one of $x,y$ is even (i.e. they are not both odd), then $xy$ is even (the contrapositive statement), which you can do for example by breaking into three cases: when $x$ is even and $y$ is odd, then $x$ is odd and $y$ is even, or when $x,y$ are both even. Of course, there's no necessity to do that here, since it is easy to observe that if one of the factors of $xy$ is even then it automatically is too.