This question is Problem 13(b) in Spivak's Calculus. As stated, I haven't come up with a way to prove the irrationality of $\sqrt6 - \sqrt2 - \sqrt3$ (or, in general, the sum of any three irrational square roots).
It is straightforward to prove that the sum of any two square roots (e.g. $\sqrt2 + \sqrt3$) is irrational. One can assume it is rational, square it, then find that it is irrational and prove by contradiction. I have tried to square $\sqrt6 - \sqrt2 - \sqrt3$ twice but still can't find any contradiction.
Could anyone please give me an example of how to solve this type of question? Thanks so much!!!
If this is rational, so is $\sqrt 6 - \sqrt 2 - \sqrt 3 + 1 = (1-\sqrt 2)(1-\sqrt 3)$. This is not equal to zero, so the following is also rational : $$ \frac{-1 \times -2}{(1-\sqrt 2)(1-\sqrt 3)} = \frac{(1-\sqrt 2)(1+\sqrt 2)(1-\sqrt 3)(1+\sqrt 3)}{(1-\sqrt 2)(1-\sqrt 3)} = (1+\sqrt 2) (1+\sqrt 3)\\ = 1+\sqrt 2 + \sqrt 3 + \sqrt 6$$
where I used the difference of squares identity to factorize $-1 = 1^2-2$ and $-2 = 1^2-3$.
This implies that $\sqrt 2 + \sqrt 3 + \sqrt 6$ is rational. From the start , $\sqrt 6 - \sqrt 2 - \sqrt 3$ is rational. When you add these, something goes wrong.
One can also use the rational root theorem : find a polynomial with integer coefficients that $\sqrt 6 - \sqrt 2 - \sqrt 3$ satisfies, and show that this polynomial doesn't have rational roots using the theorem.
For that ,let $x = \sqrt 6 - \sqrt 2 - \sqrt 3$. Then $x-\sqrt 6 = -\sqrt 2 - \sqrt 3$. Squaring : $$ x^2+6 - 2x\sqrt 6 = 5+2 \sqrt 6 \implies x^2 +1 = (2+2x)\sqrt 6 $$
squaring again : $$ x^4+2x^2+1 = 24(x^2+2x+1) \implies x^4 -22x^2 - 48x -23 = 0 $$
is a polynomial satisfied by $x$. This doesn't have rational roots, by checking that each of $\pm 1 , \pm 23$ are not roots.