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?

$$\sqrt{4n - 2} = \sqrt{2(2n - 1)} = \sqrt{2}\sqrt{2n - 1}$$ Now for some natural number $p$ and some natural number $q$, let: $$\begin{align} \sqrt{2} &= \{p/q : p/q \text{ is simplified to lowest terms}\} \\ \implies 2 &= (p/q)^2 \\ &= p^2/q^2 \\ \implies 2q^2 &= p^2 \\ \implies p &= 2r \text{ for some natural number $r$ (simply, $r \in \mathbb{N}$)} \\ \implies 2q^2 &= (2r)^2 \\ &= 4r^2 \\ \implies q^2 &= 2r^2 \end{align}$$ However if both $q^2$ and $p^2$ are divisible by $2$, they are both even; contrary (a contradiction) to our original assumption since we already established that $p/q$ was simplified to lowest terms. $$\begin{align} \therefore \sqrt{2} &\neq p/q \\ \therefore \sqrt{4n - 2} &\neq p/q \end{align}$$
