TITLE: Critiques on proof showing $\sqrt{12}$ is irrational.
QUESTION [22 upvotes]: My only exposure to proofs was in a math logic class I took in University. I was wondering if my attempt at proving that $\sqrt{12}$ is irrational is OK.
$$\Big(\frac{m}{n}\Big)^2 = 12$$
$$\Big(\frac{m}{2n}\Big)^2 = 3$$
$$m^2=3*(2n)^2$$
This implies $m$ is even and so $n$ must be odd.
The problem can be reduced to:
$$\Big(\frac{p}{n}\Big)^2 = 3$$ 
Because $n$ is odd, $p^2$ is odd, so $p$ is odd.
This implies:
$$4a+1 = 3(4b+1)$$
$$4a - 12b = 2$$
$$2a - 6b = 1$$
I'm kind of stuck at this point. I know that this can't be true but I don't know how to state it. Any critiques or suggestions? Thanks!

REPLY [2 votes]: Just call $\sqrt {12}$ a rational number. What can you say about it? $12$ must be a square of rational number.
$12= k^2$ where ($k =\dfrac{p}{q}$)
GCD$(p,q)=1$
$12= 2^2 \cdot 3=\dfrac{p^2}{q^2} \implies 3= \dfrac{p^2}{q^2 \cdot 2^2} \implies 3$ is a square. Which implies it has odd number of divisors, but it doesn't.