Exercise 73Number theory › IrrationalityHigh schoolAdded 3 Nov 2026

Irrationality of √2 + √3

#algebraic#sum of roots#irrationality

Statement

Show that $\sqrt2+\sqrt[3]{2}$ is irrational.

Hint

If $\alpha=\sqrt2+\sqrt[3]{2}\in\mathbb{Q}$ , then $\sqrt[3]{2}=\alpha-\sqrt2$ , and cubing gives a relation involving $\sqrt2$ , forcing $\sqrt2\in\mathbb{Q}$ .

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Solution

Assume. $\alpha=\sqrt2+\beta\in\mathbb{Q}$ with $\beta=\sqrt[3]{2}$ .

Cube. $\beta^3=(\alpha-\sqrt2)^3=\alpha^3-3\alpha^2\sqrt2+3\alpha\cdot2-2\sqrt2=(\alpha^3+6\alpha)-(3\alpha^2+2)\sqrt2$ .

But. $\beta^3=2\in\mathbb{Q}$ , so $(3\alpha^2+2)\sqrt2=\alpha^3+6\alpha-2$ .

Contradiction. Since $3\alpha^2+2\neq0$ and $\alpha\in\mathbb{Q}$ , $\sqrt2\in\mathbb{Q}$ , contradiction.

Conclusion. $\alpha$ is irrational.

Irrationality of √2 + √3

Statement

Solution

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