Prove that $\sqrt{2} +\sqrt{n}$ is irrational Suppose $\sqrt{2} + \sqrt{n}$ is rational,
so it can be represented like so: $\sqrt{2} + \sqrt{n} = {x \over y}$
then  $\sqrt{2} = {{x \over y} - \sqrt{n}}$
so $\sqrt{2} = {x - y\sqrt{n}\over y}$
but $\sqrt{2}$ is irrational, which leads to a contradiction.
Does this proof make sense?
edit: fixed some mistakes

The polynomial $$\left(x^2-\left(\sqrt{2}-\sqrt{n}\right)^2\right)
   \left(x^2-\left(\sqrt{2}+\sqrt{n}\right)^2\right) = x^4-2(n+2)\,x^2+(n-2)^2$$
has $\sqrt2+\sqrt n$ as one of its roots. Any rational divisor would have to be an integer factor of $(n-2)^2$ by the rational roots theorem.
But $\sqrt 2+\sqrt n$ is not an integer, indeed, if $\sqrt 2+\sqrt n=a\in\mathbb Z$, then $n=a^2+2-2a\sqrt 2$, implying that $2a\sqrt 2\in\mathbb Z$.
