elementary number theory - Proof verification: Prove $sqrt{n}$ is irrational.

Let $n$ be a positive integer and not a perfect square. Prove $\sqrt{n}$ is irrational.

Consider proving by contradiction. If $\sqrt{n}$ is rational, then there exist two coprime integers $p,q$ such that $$\sqrt{n}=\frac{p}{q},$$ which implies $$p^2=nq^2.$$
Moreover, since $p, q$ are coprime, by Bézout's theorem, there exist two integers $a,b$ such that $$ap+bq=1.$$
Thus

$$p=ap^2+bpq=anq^2+bpq=(anq+bp)q,$$ which implies $$\sqrt{n}=\frac{p}{q}=anq+bp \in \mathbb{N^+},$$ which contradicts.