# Rational Square Root of Integer is Integer

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## 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:

- $n b^2 = a^2$

From Number divides Number iff Square divides Square:

- $b^2 \divides a^2 \implies b \divides a$

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.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $2$: Some Properties of $\Z$: Exercise $2.14$