Square roots

How to prove that the square root of a non perfect square is irrational.
So I guess it's showing that $\sqrt n \not\in \mathbb{Q}, (n\in \mathbb{N})$ if $n\neq k^2, k\in \mathbb{N}$.

Meh, you can do better than that, you can prove that if $a,b \in \mathbb{Z}_{>0}$ then $a^{1/b}$ is either an integer or irrational. I'll go ahead and prove that instead.

Proof 1: Let $a^{1/b} = \frac{p}{q}$ with $(p,q) = 1$ and hence $(p^b, q^b) = 1$ then $a = \frac{p^b}{q^b}$ is a contradiction.

Proof 2: Rational root theorem gives us that the rational roots of $x^b - a$ must be of the form $\frac{p}{q}$ with $(p, q) = 1$ and $q \mid 1$. So the root must be either an integer or irrational.

Proof 3: I'll let you consider other methods.

How does your proof explain why cube root of 8 is rational? So when a=8 and b =3 ?