Is $\sqrt{2}+\sqrt{3}$ rational or irrational? I have proven that the sum of any two irrational numbers is not always irrational and that a rational plus an irrational is irrational, however I am not sure how to prove this specific case.
I was thinking along the lines of:
$\sqrt{2}+\sqrt{3}=n$   where $n$ is an element of rationals. By squaring both sides
$$(\sqrt{2}+\sqrt{3})^2 = n^2$$
$$7 + 2\sqrt{2}\sqrt{3} = n^2$$
I have proven the sum of a rational and irrational is irrational yet it is too long and complicated to prove that root two and root three make root six. Moreover, it will be an additionally difficult to prove that two times root six is irrational so I was wondering if there was a quicker less complex way to finish this proof?

The novelty in this question is the request for a proof that doesn't involve knowing (or proving) the irrationality of $\sqrt6$.  So here's a proof that only assumes the irrationality of $\sqrt2$ and/or $\sqrt3$.
If $\sqrt2+\sqrt3$ is rational, then so is $\sqrt3-\sqrt2$, since $(\sqrt2+\sqrt3)(\sqrt3-\sqrt2)=3-2=1$.  But then so is 
$$\sqrt2={(\sqrt2+\sqrt3)-(\sqrt3-\sqrt2)\over2}$$ 
(and also $\sqrt3=((\sqrt2+\sqrt3)+(\sqrt3-\sqrt2))/2$).
