Exercise 60Number theory › IrrationalityHigh schoolAdded 3 Nov 2026

A square root in disguise

#irrationality#quadratic#conjugates

Statement

Show that $\sqrt{2+\sqrt2}$ is irrational.

Hint

Let $\alpha=\sqrt{2+\sqrt2}$ . Then $\alpha^2-2=\sqrt2$ , which is irrational.

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Compute. $\alpha^2=2+\sqrt2$ , so $\alpha^2-2=\sqrt2$ .

Irrationality. If $\alpha\in\mathbb{Q}$ , then $\alpha^2\in\mathbb{Q}$ , hence $\sqrt2=\alpha^2-2\in\mathbb{Q}$ , contradiction.

Conclusion. $\alpha$ is irrational.

Remark. $\alpha$ is algebraic of degree $4$ over $\mathbb{Q}$ .