Exercise 1031Logic, sets and proofs › Proof techniquesHigh schoolAdded 3 Nov 2026

Proving an equivalence

#equivalence#iff#proof

Statement

Prove that for any integer $n$ , $n^2$ is even if and only if $n$ is even.

Hint

Prove both directions: $n$ even $\Rightarrow$ $n^2$ even, and $n^2$ even $\Rightarrow$ $n$ even.

Show solutionHide solution

Solution

$\Rightarrow$ (n even implies $n^2$ even). If $n=2k$ , then $n^2=4k^2=2(2k^2)$ is even.

$\Leftarrow$ (n² even implies $n$ even). Contrapositive: if $n$ is odd, then $n=2k+1$ and $n^2=4k^2+4k+1=2(2k^2+2k)+1$ is odd. So $n^2$ even implies $n$ not odd, i.e. $n$ even.

Conclusion. $n^2$ even $\iff$ $n$ even.

Proving an equivalence

Statement

Solution

More problems in Proof techniques