How do I prove if a sum of two specific irrational numbers is irrational?

Question

Prove that $\sqrt 2 + \sqrt 6$ is irrational. (Note that, in general, the sum of two irrational numbers could be rational.)

I tried attempting to use proof by contradiction but I'm unsure of how to go from even there. I have no clue other than that. Please help.

Answer 1

If it is rational, its square is also rational. But this would imply that $\sqrt{3}$ is rational.

Answer 2

Side hint: let $\,x=\sqrt{2}+\sqrt{6}\,$, then $x^2=8+4 \sqrt{3}$ and $(x^2-8)^2=48 \iff x^4-16x^2+16=0\,$.

By the rational root theorem, any rational root of the latter polynomial would have to be an integer. But $\,1 \lt \sqrt{2} \lt 3/2\,$ and $\,2 \lt \sqrt{6} \lt 5/2\,$, so $\,3 \lt \sqrt{2}+\sqrt{6}\lt 4\,$, thus $\,x\,$ lies strictly between two consecutive integers, and therefore cannot be an integer itself.

Answer 3

You could try this: If $a = \sqrt 2+\sqrt 6$ is rational then so is $\frac{1}{a} = \frac{\sqrt 6 - \sqrt 2}{4} \Rightarrow \sqrt 6 - \sqrt 2 \in \mathbb{Q}$

Subtracting, we get $\sqrt 2$ is rational, which is false. Hence $a \notin \mathbb{Q}$