Rational Square Root of Integer is Integer

Theorem
Let $n$ be an integer.

Suppose that $\sqrt n$ is a rational number.

Then $\sqrt n$ is an integer.

Proof
Suppose that $\sqrt n = \dfrac a b$, with $a, b$ coprime integers and $b > 0$.

Then we would have $n = \dfrac {a^2} {b^2}$, i.e. $n b^2 = a^2$.

However, since $a \perp b$ and $b \mid a$, this means that necessarily $b = 1$.

That is, $\sqrt n = a$, an integer; hence the result.