1
$\begingroup$

This question already has an answer here:

I know that this is a common question, but I want to see your take on this. The way I approached this was by contraposition (i'm still new with the proofs but any help would be appreciated, even if it's by contradiction.)

Assume $n$ is odd, thus $n^2$ is odd as well.

Since $n$ is odd then $n=2k+1$ for some integer $k$.

Then $n^2= (2k+1)^2 = 4k^2+4x+1 \rightarrow 2(2k^2+2k)+1.$ Which clearly shows that $n^2$ is odd.

$\endgroup$

marked as duplicate by Dietrich Burde, Ethan Bolker, Community Mar 9 '18 at 16:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3
$\begingroup$

There's a flaw in your proof. That's when you write “Assume $n$ is odd, thus $n^2$ is odd as well.” That's exactly what you are supposed to prove! Eliminate that sentence, and your proof will be just fine.

$\endgroup$
  • $\begingroup$ thanks ! i'll do that right now. $\endgroup$ – Killercamin Mar 9 '18 at 16:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.