Proving that something is irrational I'm trying to evaluate the following claim:
$$ \sqrt{2} + \sqrt{n} $$ is irrational.
This is what I tried: 
Proof by contrapositive: Suppose $$ r = \sqrt{2} + \sqrt{n} $$
and r is rational. Then $$ \frac{m}{l} = \sqrt{2} + \sqrt{n} $$
$$\frac{m^2}{l^2} = 2 + 2\sqrt{2n} + n $$
I'm not sure where to proceed or if I'm even heading in the right direction.
Could anyone give me a tip?
 A: I don't think this is going in the right direction, because $\sqrt{2n}$ could very well be an integer. (EDIT: There is actually a proof building on this in the answer by shooting-squirrel.)
I would say this. Let $r = \sqrt{n} + \sqrt{2}$, and let $s = \sqrt{n} - \sqrt{2}$. Then we have $rs = n - 2$ and $(1/2)(r - s) = \sqrt{2}$.
Now assume for a contradiction that $r$ is rational. Since $s = (n-2)/r$, the number $s$ must also be rational. And for that reason, the number $(1/2)(r - s)$ too must be rational. But this is $\sqrt{2}$, which we know to be irrational, a contradiction.
Hence $r$ must be irrational. 
A: Case 1: $n$ is of form $2u^2$ , $u$ being an integer
Then our number is equal to $\sqrt 2 (u+1)$, clearly this is irrational
Otherwise $2n$ is not a square, hence you get contradiction in the last identity you wrote.(since $\sqrt {2n}$ is irrational)
A: Dear Alex solve your equation for $\sqrt{n}$. That is $r-\sqrt{2}=\sqrt{n}$ then by squaring the equation  and solving for $\sqrt{2}$we have $\sqrt{2}=\frac{n-2-r^2}{2r}.$ A contradiction
