Prove that $\sqrt 2 + \sqrt 3$ is irrational I have proved in earlier exercises of this book that $\sqrt 2$ and $\sqrt 3$ are irrational. Then, the sum of two irrational numbers is an irrational number. Thus, $\sqrt 2 + \sqrt 3$ is irrational. My first question is, is this reasoning correct?
Secondly, the book wants me to use the fact that if $n$ is an integer that is not a perfect square, then $\sqrt n$ is irrational. This means that $\sqrt 6$ is irrational. How are we to use this fact? Can we reason as follows:
$\sqrt 6$ is irrational
$\Rightarrow \sqrt{2 \cdot 3}$ is irrational.
$\Rightarrow \sqrt 2 \cdot \sqrt 3$ is irrational
$\Rightarrow \sqrt 2$ or $\sqrt 3$ or both are irrational.
$\Rightarrow \sqrt 2 + \sqrt 3$ is irrational.
Is this way of reasoning correct?

Hints:
Suppose there exist coprime $\,a,b\in\Bbb Z\,$ s.t.
$$\sqrt2+\sqrt3=\frac ab\implies \sqrt6=\frac{a^2}{2b^2}-\frac52=\frac{a^2-5b^2}{2b^2}$$
If you already know $\,\sqrt6\,$ is irrational then you're already done, otherwise prove it as with $\,\sqrt2\,$ , say:
$$\sqrt6=\frac pq\;,\;\;(p,q)=1\implies 6q^2=p^2\implies 2\mid p$$
and thus we can write
$$\sqrt6=\frac{2p'}q\implies 2\mid q\;\;\;\;\text{also , and this is a contradiction}$$
