## sqrt(2) + sqrt(3) is irrational

Prove that $\sqrt{2} + \sqrt{3}$ is irrational.

Note that $\sqrt{2}$ and $\sqrt{3}$ are algebraic over $\mathbb{Q}$ with minimal polynomials $x^2-2$ and $x^2-3$, respectively, and with conjugates $\pm \sqrt{2}$ and $\pm \sqrt{3}$. Since $\sqrt{2} - \sqrt{3}$, $-\sqrt{2}+\sqrt{3}$, and $-\sqrt{2}-\sqrt{3}$ are distinct from $\sqrt{2}+\sqrt{3}$, we have $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$.

Suppose $\sqrt{2}+\sqrt{3} \in \mathbb{Q}$. Then $\mathbb{Q}(\sqrt{2}+\sqrt{3}) = \mathbb{Q}$, and so $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}$. In particular, $\sqrt{2} \in \mathbb{Q}$– a contradiction.