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?
 A: $4n-2 = (a/b)^2$ 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!
A: The number $\sqrt{4n-2}$ is rational iff $4n-2 = a^2$ reduction mod 4 shows that this is impossible.
Here is a proof of the general fact that $\sqrt{k}$ is irrational unless $k$ is a square: Suppose $\frac{u}{v}$ is a solution to $x^2 - k = 0$, then it is an integer $i$ by Gauss lemma, but then $k = i^2$.
A: $$\sqrt{4n - 2} = \sqrt{2(2n - 1)} = \sqrt{2}\sqrt{2n - 1}$$ 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. $$\begin{align} \therefore \sqrt{2} &\neq p/q \\ \therefore \sqrt{4n - 2} &\neq p/q \end{align}$$
A: Suppose $n$ is a natural number. If $n$ is even, then $n^2 = (2k)^2 = 4k^2$ is divisible by 4. If $n$ is odd, then $n^2 = (2k-1)^2 = 4k^2-4k+1 = 1\ \textrm{mod}\ 4$.
Hence, if $n$ is a natural number, then $n^2$ is either 0 or 1 mod 4.
Thus, if $m$ is a natural number and is 2 or 3 mod 4, then $\sqrt{m}$ is not a natural number, and hence must be irrational, since $\sqrt{m}$ is either irrational or natural.
Hence, as a corollary, since $4n-2$ is 2 mod 4, then $\sqrt{4n-2}$ is irrational.
(This shows that $\sqrt{4n-3}$ is also irrational.)
