# A Quick Little Proof

Prove that $mn$ is odd if and only if both $m$ and $n$ are odd. We can write this in mathamatical notation as $mn \in \mathbb O \Longleftrightarrow m,n \in \mathbb O$

Let’s first prove that if both $m$ and $n$ are odd then their product is odd. An odd number is of the form two times and integer plus one. So let $m = 2r + 1$ and $2q + 1$. Then $mn = (2r + 1)(2q + 1) = 4rq + 2r + 2q + 1 = 2(2rq + r + q) + 1$. Which is odd, by definition, it’s two time an integer — in this case the integer is $2(2rq + r + q)$ — plus one.

Now we have to prove things in the other direction. If $mn$ is odd then so are both $m$ and $n$. Let’s do a proof by contradiction. We must assume that either $m$ or $n$ is even and then prove that $mn$ is even. Assume $m$ is even. Then $mn = (2r)(2q + 1) = 4rq + 2r = 2(2rq + r)$, which is even. The same holds if n is even. We’ve completed our proof by contradiction and so have proven our original statement.