Prove $\sqrt{2} + \sqrt{5}$ is irrational How do you prove that $\sqrt{2} + \sqrt{5}$ is irrational?  
I tried to prove it by contradiction and got this equation:  $a^2/b^2 = \sqrt{40}$.

If $\sqrt{2}+\sqrt{5}\in\mathbb{Q}$, i.e., $ \sqrt{2}+\sqrt{5}=\frac{b}{a}$ for $a,b\in\mathbb{Z} \mbox{ and } a\neq 0$ then $ 2-5=\frac{b}{a}(\sqrt{2}-\sqrt{5})$ and
$$
\sqrt{2}-\sqrt{5} =\frac{-3\cdot a}{b}
$$
Here $b$ is necessarily greater than zero by cause $\sqrt{2}+\sqrt{5}>0$.Finally we get,
$$
\sqrt{2}=\frac{1}{2}\left( (\sqrt{2}+\sqrt{5})+(\sqrt{2}-\sqrt{5}) \right)= \frac{1}{2}\left( \frac{b}{a}+\frac{-3\cdot a}{b} \right)\in\mathbb{Q}.
$$
But this is a contradiction since $\sqrt{2}$ is not rational.
