Let $n,m\in N0$ and $\sqrt{n}\not\in Q.$ Show that $\sqrt{n} + \sqrt{m}\not\in Q$

Question

Let $n,m\in N0$ and $\sqrt{n}\not\in Q.$ Show that $\sqrt{n} + \sqrt{m}\not\in Q$

My first thought was to prove it via Contradiction.

So, suppose $\sqrt{n} + \sqrt{m}\in Q$

And since $\sqrt{n}\not\in Q$, we'll have whether

1.: $\sqrt{m}\not\in Q$

2.: $\sqrt{m}\in Q$

Would this be a reasonable approach in the first place?

And also how could/should I continue or approach this?

Answer 1

Assuming $\sqrt{n}+\sqrt{m}\in\mathbb Q$;

If $m$ is a square then $\sqrt{m}\in \mathbb Q$ and $\mathbb Q $is closed under addition, i.e. $\sqrt{n}+\sqrt{m}-\sqrt{m}=\sqrt{n}\in\mathbb Q$ contradiction.

If $m$ is not a square then since $\mathbb Q$ is closed under division $\mathbb Q \ni\frac{n-m}{\sqrt{n}+\sqrt{m}}=\sqrt{n}-\sqrt{m}\in \mathbb Q$,

which means, $\frac12\left(\sqrt{n}+\sqrt{m}+\sqrt{n}-\sqrt{m}\right)=\sqrt n\in\mathbb Q$ contradiction again.