Following is a proof that $\sqrt{2}$ is irrational.

The polynomial $x^2-2$ is irreducible over $\mathbb{Z}$ by Eisenstein's criterion with $p=2$ Thus, $x^2-2$ is irreducible over $\mathbb{Q}$ by Gauss's lemma. Therefore, $x^2-2$ does not have any roots in $\mathbb{Q}$ Since $\sqrt{2}$ is a root of $x^2-2$ it must be irrational.

This method generalizes to show that any number of the form $\sqrt[r]{n}$ is not rational, where $r \in \mathbb{Z}$ with $r>1$ and $n \in \mathbb{Z}$ such that there exists a prime $p$ dividing $n$ with $p^2$ not dividing $n$