Assuming that $\sqrt{2}=\frac{p}{q}$ where $p,q\in\mathbb{N}$.
That implies that $p^2=2q^2$, and since $p$ is a natural number then the right hand side $2q^2$ must be a square of natural number and hence $2=n^2$ where $n\in\mathbb{N}$.
Since no natural number satisfies the equation $2=n^2$, then $\sqrt{2}$ can't be rational number.
Is this a valid proof?