# Prove that for every integer $x$, if $x^2 - 2x + 7$ is even, then $x$ is odd. (Contrapositive)

Prove that for every integer $x$, if $x^2 - 2x + 7$ is even, then $x$ is odd.

(By contrapositive)

Assume x is even, we will prove $x^2-2x+7$ is odd.

Then there exists some integer k where $x=2k$

Then $(2k)^2 - 2(2k) + 7 = 4k^2 - 4k + 7 = 2(2k^2 - 2k) + 7$

I'm not sure if I'm doing this properly, I thought that I should get 2K + 1, but I'm getting 2k+7?

• Looks good to me. You can say $2\!\left(2k^2-2k+3\right)+1$ instead. – robjohn Oct 15 '17 at 1:57
• Please don't delete this post. Others may benefit from it. – robjohn Oct 15 '17 at 2:01
• Its still fine if you are getting $2k +7$ because that is odd for all integers also. – I'mAnAccountantIKnowAlotOfMath Oct 15 '17 at 2:19

## 3 Answers

Your proof is correct, and you applied the word 'contrapositive' correct as well (switch the hypothesis and conclusion, and take the negative of both). Nice job!

Expanding on what @robjohn said, since $2(2k^2-2k)$ is clearly even, then the parity of the entire thing depends on $7$ (because $even + even = even$, and $even + odd = odd$). We can write $7$ as $2(3)+1$, which shows that the sum is odd. Alternatively, we can go an extra step and write it as $2(2k^2-2k + 3) + 1$, which is also odd.

Direct proof (in case you´re interested):

$x^2-2x+7$ is even so $x^2-2x=x(x-2)$ is odd, so both $x$ and $x-2$ are odd, because only

odd$\cdot$odd gives odd.

Because $$(x-1)^2+6$$ is even, which says $$(x-1)^2$$ is even, which gives $$x-1$$ is even, which says that $x$ is odd.

Or let $x$ be even.

Thus, $$x^2-2x+7=(x-1)^2+6$$ is odd, which is contradiction, which says that our assuming was wrong.

Id est, $x$ is odd.