Proving that $\sqrt{2}+\sqrt{3}$ is irrational This is from Spivak.
Prove that $\sqrt{2}+\sqrt{3}$ is irrational.
So far, I have this:
If $\sqrt{2}+\sqrt{3}$ is rational, then it can be written as $\frac{p}{q}$ with integral $p, q$ and in lowest terms. 
$$\sqrt{2}+\sqrt{3}=\frac{p}{q}$$
$$2\sqrt{6}+5 =\frac{p^2}{q^2}$$
$$(2\sqrt{6}+5)q^2=p^2$$
And that's about where I get stuck. In a similar question (prove that $\sqrt{2}+\sqrt{6}$ is irrational,) I was able to show that both $p,q$ had to be even which is impossible. I obviously can't apply this trick here. Any hints?

We can solve this problem by contradiction. Suppose that it is rational, so it is a non-zero rational number such as $\frac{a}{b}$.
$(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=1\rightarrow \sqrt{3}-\sqrt{2}=\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{b}{a}$
So we conclude that $\sqrt{3}-\sqrt{2}$ is a rational number as well. 
$\sqrt{3}-\sqrt{2}=\frac{b}{a}\rightarrow 5-2\sqrt{6}=(\frac{b}{a})^2$
$\sqrt{3}+\sqrt{2}=\frac{a}{b}\rightarrow 5+2\sqrt{6}=(\frac{a}{b})^2$
Subtract the former equation from the latter one, we conclude $4\sqrt{6}=(\frac{a}{b})^2-(\frac{b}{a})^2$ 
From above equation we conclude $\sqrt{6}$ is rational, but we can prove the irrationality of $\sqrt{6}$ quite simple(It has been brought in following) which shows contradiction and hence $\sqrt{2}+\sqrt{3}$ is irrational.
proof for irrationality of $\sqrt{6}$
If it is rational then it is equal to $\frac{p}{q}$ such that $p$ and $q$ are relatively prime natural numbers. then $p^2=6q^2\rightarrow q^2|p^2\rightarrow$ $p$ and $q$ are relatively prime if and only if $q=1$ which implies 6 is a square integer which is a contradiction. 
