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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!!!

Arctic Char
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Tony
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  • Welcome to MSE. Write your attempts in detail so you can get a good hint –  Oct 18 '20 at 02:56
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    Welcome to the community. This is a good question! Unfortunately it will have to be closed as it has already been asked here and we do not maintain duplicate questions. For more on our expectations, you can read how to ask a good question. At any rate, I wish the best of luck to you, and please continue to contribute to our wonderful site! – gen-ℤ ready to perish Oct 18 '20 at 03:09
  • The statement holds more generally, see for example also this duplicate. I agree with the above comment, and you will be successful if you include a bit of searching before posting (see the right margin, for example). – Dietrich Burde Oct 18 '20 at 08:23

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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.

Sarvesh Ravichandran Iyer
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  • Thanks a lot. This is really helpful! In your first proof, you did a lot of algebraic manipulations that are clearly understandable but might not be intuitive to me if I were to do this proof alone. Am I right that there are various different ways to manipulate √6 - √2 - √3 as long as I can get rid of two of the square roots and only preserve one (so that I can show that such a square root is both rational and irrational under my original assumption that √6 - √2 - √3 is rational)? – Tony Oct 18 '20 at 03:23
  • Also, the first proof works because √6 = √2 * √3. If the questions were to prove that √7 - √2 - √3 is irrational, it wouldn't work. Is there always going to be some manipulations that will allow us to prove that the sum of three irrational square roots is irrational? – Tony Oct 18 '20 at 03:35
  • Or, maybe the first proof will not be a general valid way to prove the irrationality of the sum of three irrational square roots? It only works in some specific circumstances. – Tony Oct 18 '20 at 03:36
  • In the meantime, however, the rational root theorem and the second methodology is always something that I can resort to? – Tony Oct 18 '20 at 03:37
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    @Teresa Lisbon, +1, very good two methods. – sirous Oct 18 '20 at 05:10
  • @Bourbon You can always resort to the second method. As you also mention, $\sqrt 7 - \sqrt 2 - \sqrt 3$ is going to be more difficult ; I don't think the first method would be workable. – Sarvesh Ravichandran Iyer Oct 18 '20 at 10:29