Prove that $\sqrt 2 + \sqrt 3$ is irrational

If $\sqrt{2} + \sqrt{3}$ is rational, then so is $(\sqrt{2} + \sqrt{3})^2 = 5 + 2 \sqrt{6}$. But this is absurd since $\sqrt{6}$ is irrational.

If $\sqrt 3 +\sqrt 2$ is rational/irrational, then so is $\sqrt 3 -\sqrt 2$ because $\sqrt 3 +\sqrt 2=\large \frac {1}{\sqrt 3- \sqrt 2}$ . Now assume $\sqrt 3 +\sqrt 2$ is rational. If we add $(\sqrt 3 +\sqrt 2)+(\sqrt 3 -\sqrt 2)$ we get $2\sqrt 3$ which is irrational. But the sum of two rationals can never be irrational, because for integers $a, b, c, d$ $\large \frac ab+\frac cd=\frac {ad+bc}{bd}$ which is rational. Therefore, our assumption that $\sqrt 3 +\sqrt 2$ is rational is incorrect, so $\sqrt 3 +\sqrt 2$ is irrational.

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

$$\sqrt6=\frac{2p'}q\implies 2\mid q\;\;\;\;\text{also , and this is a contradiction}$$