# Rational Square Root of Integer is Integer

Jump to navigation Jump to search

## 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:

$n = \dfrac {a^2} {b^2}$

That is:

$n b^2 = a^2$
$b^2 \divides a^2 \implies b \divides a$

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

That is, $\sqrt n = a$, an integer.

Hence the result.

$\blacksquare$