Exercise 40Number theory › IrrationalityHigh schoolAdded 3 Nov 2026

Irrationality of √3

#sqrt3#irrationality#mod 3

Statement

Prove that $\sqrt3$ is irrational.

Hint

Same structure as $\sqrt2$ , but use divisibility by $3$ instead of parity.

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Contradiction. Suppose $\sqrt3=p/q$ with $p,q$ coprime.

Square. $p^2=3q^2$ , so $3\mid p^2$ , hence $3\mid p$ (since $3$ is prime). Write $p=3k$ .

Substitute. $9k^2=3q^2$ , so $q^2=3k^2$ , hence $3\mid q$ .

Contradiction. $3\mid p$ and $3\mid q$ contradict $\gcd(p,q)=1$ .

Conclusion. $\sqrt3$ is irrational.