Another proof method to try out if a direct proof isn't the best recourse is a proof by contrapositive.
If the statement you are trying to prove is of the form of an implication $p \to q$, instead of proving that $p$ implies $q$ (the direct way,) you can prove that $\neg q$ implies $\neg p$. That is, you will prove that $\neg q \to \neg p$, the contrapositive implication, is true.
This proof method is valid since we showed earlier that $p \to q$ is equivalent to $\neg q \to \neg p$.
Theorem. If $n^2$ is even, then $n$ is even, too.
Proof. We will prove the contrapositive of this statement, i.e., "if $n$ is not even, then $n^2$ is not even".
Note that saying "not even" is the same as saying "odd", so the contrapositive can be re-written as "if $n$ is odd, then $n^2$ is odd". We already proved this very statement! See the proof on slide 35. This means that "if $n^2$ is even, then $n$ is even" is true.
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