If $n$ is any positive integer, prove that $\sqrt{4n-2}$ is irrational If $n$ is any positive integer, prove that $\sqrt{4n-2}$ is irrational.
I've tried proving by contradiction but I'm stuck, here is my work so far:
Suppose that $\sqrt{4n-2}$ is rational. Then we have $\sqrt{4n-2}$ = $\frac{p}{q}$, where $ p,q \in \mathbb{Z}$ and $q \neq 0$.
From $\sqrt{4n-2}$ = $\frac{p}{q}$, I just rearrange it to:
$n=\frac{p^2+2q^2}{4q^2}$. I'm having troubles from here, $n$ is obviously positive but I need to prove that it isn't an integer.
Any corrections, advice on my progress and what I should do next?

$4n-2 = (a/b)^2$ so $b$ divides $a$.
But $\operatorname{gcd}(a,b) = 1$
so $b = 1$.
So now $2$ divides $a$
so write $a = 2k$
then by substitution, we get that
$2n-1 = 2k^2$
Left side is odd but the right side is even. Contradiction!
