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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.

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

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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 $a \notin \mathbb{Q}$

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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.

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  • $\begingroup$ Wish the downvoter had left a comment why. $\endgroup$ – dxiv May 15 '17 at 16:28
  • $\begingroup$ P.S. Somebody must really hate this answer to downvote it (again without comment) months later. Sorry to disappoint, but I am not going to delete it since it is mathematically correct, and it is different from the other posted answers. My take is that it's always beneficial to be aware of different lines of attack for the same problem, just in case the more direct approaches may not work sometimes. $\endgroup$ – dxiv Jul 21 '17 at 21:08
  • $\begingroup$ Maybe they were trying to get you out of the hole. If so, I think it's working. $\endgroup$ – Mr. Brooks Jul 24 '17 at 20:38

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