Irrationality of √n for non-square n

Prime factorization. $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$. Since $n$ is not a square, some $\alpha_i$ is odd. Let $p=p_i$.

Contradiction. Suppose $\sqrt n=a/b$ with $\gcd(a,b)=1$.

Square. $a^2=nb^2$, so $p\mid a^2$, hence $p\mid a$. Write $a=pk$.

Substitute. $p^2k^2=nb^2$, so $p^2\mid nb^2$, and $v_p(a^2)=2v_p(a)$ is even while $v_p(nb^2)=v_p(n)+2v_p(b)$ has parity $v_p(n)$ odd. Contradiction.

Conclusion. $\sqrt n$ is irrational.