Prove that $\sqrt{2} + \sqrt{n}$ is irrational when $n\in\mathbb{N}$

Question

Prove that $\sqrt{2} + \sqrt{n}$ is irrational $\forall \ n \in \mathbb{N}$.

I have tried though contradiction but can't seem to come up with an answer

I did this where for $\frac ab$, $a \in \mathbb{N}, b \in \mathbb{Z}$ and in reduce form, and $0 \not \in \mathbb{N}$

$$\sqrt 2 + \sqrt n = \frac ab$$ $$\frac {2+n}{\sqrt 2 - \sqrt n} = \frac ab$$ $$\frac {(2+n)(\sqrt 2 + \sqrt n)}{2 + n} = \frac ab$$ $$\frac {(2+n)(a)}{(2-n)(b)}=\frac ab$$ $$\frac {2+n}{2-n}=1$$ $$2+n=2-n$$ $$2+2n=2$$ $$n=0$$ $$n \in \mathbb{N} \therefore n \not =0$$ $$\therefore \sqrt{2} + \sqrt{n} \not \in \mathbb{Q}$$

However this seems clearly flawed as even if $n = 0$ it still would be irrational, and that it subsitiuting $\sqrt 2 + \sqrt n$ for $\frac ab$ seems pretty incorrect.

More just a desperate attempt, pretty confident its not even close.

Answer 1

Suppose $\sqrt{2}+\sqrt{n}\in\Bbb Q$. Since the ratio of rationals is rational, $\frac{2-n}{\sqrt{2}+\sqrt{n}}=\sqrt{2}-\sqrt{n}\in\Bbb Q$. Averaging, $\sqrt2\in\Bbb Q$, which contradicts a well-known result.

Answer 2

The numerator in your second line should be $(\sqrt{2} + \sqrt{n})(\sqrt{2} - \sqrt{n}) = 2 - n$. ... and then your fourth line is a tautology.