$\sqrt{6}+\sqrt{3}$ is not rational proof I want to prove $\sqrt{6}+\sqrt{3}$ is not rational; here is my attempt:
Assume for the sake of contradiction that $\sqrt{6}+\sqrt{3}$ is rational. Then $(\sqrt{6}+\sqrt{3})^2$ must also be rational. Since $$(\sqrt{6}+\sqrt{3})^2=9+2\sqrt{6}\sqrt{3}=9+2\sqrt{2}\sqrt{3}\sqrt{3}=9+6\sqrt2,$$
we see $9+6\sqrt2$ must be rational. But, since $\sqrt{2}$ is not rational we have a contradiction and hence $\sqrt{6}+\sqrt{3}$ is not rational.
Is it correct?

Looks more or less ok, I would just add one more step.
You correctly prove that if $\sqrt 6+\sqrt 3$ is rational, then $9+6\sqrt2$ is also rational.
I would now add one more step and say that this also means that because $\sqrt{2} = \frac{9+6\sqrt{2} - 9}{6}$, that therefore, $\sqrt{2}$ is also rational, and only then move on to the conclusion.
It's a minor thing, but I would rather be a little too clear than skip an important step. Also, you can then completely explicitly write that
$$\sqrt{2} = \frac{(\sqrt{6} + \sqrt{3})^2 - 9}{6}$$
and make the connection between the two numbers even more explicit.
