Exercise 38Number theory › IrrationalityHigh schoolAdded 3 Nov 2026

Irrationality of √p for prime p

#sqrt p#prime#irrationality

Statement

Let $p$ be prime. Prove that $\sqrt p$ is irrational.

Hint

Same argument as $\sqrt2$ and $\sqrt3$ , using Euclid's lemma: $p\mid n^2\Rightarrow p\mid n$ .

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

Square. $a^2=pb^2$ , so $p\mid a^2$ , hence $p\mid a$ (Euclid's lemma since $p$ is prime). Write $a=pk$ .

Substitute. $p^2k^2=pb^2$ , so $b^2=pk^2$ , hence $p\mid b^2$ , so $p\mid b$ .

Contradiction. $\gcd(a,b)\ge p>1$ , contradiction.

Conclusion. $\sqrt p$ is irrational.