Recent content by abruzzi

That was exactly what I was looking for, thanks!

By taking n=-1 and squaring we get 2\sqrt 6 - 2\sqrt 3 - 2\sqrt 2 + 6 = 6 + 2(\sqrt 6 -\sqrt 3-\sqrt 2) But from this I cannot conclude anything - since knowing that number a is irrational doesn't mean that a^2 is (for example a=\sqrt 2). Or am I not on the right path?

Sorry, but I have no idea about groups, rings, fields, etc. I am looking for a basic proof. For example the proof that I have that \sqrt 2 + \sqrt 3 is irrational is that, supposing it is rational, its square should also be rational. But (\sqrt 2 + \sqrt 3)^2 = 5+2\sqrt6 is irrational because...

I want to prove that \sqrt 6 - \sqrt 2- \sqrt 3 is irrational. I already know that \sqrt 2+\sqrt 3 is irrational (by squaring it). I would like a proof that doesn't use a polynomial and the rational root theorem. Thanks.