Let us assume, to the contrary, that $\sqrt2$ is rational. Then $\sqrt2=\frac{p}{q}$ where p, q are coprime integers, $q\ne0$. Squaring: $2=\frac{p^2}{q^2} \Rightarrow p^2=2q^2$. So $p^2$ is even, hence $p$ is even. Let $p=2m$. Then $4m^2=2q^2 \Rightarrow q^2=2m^2$, so $q$ is also even. This contradicts that p, q are coprime (both even means they share factor 2). Hence our assumption is wrong, and $\sqrt2$ is irrational.