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Proof that sqrt(6)-sqrt(2)-sqrt(3) is irrational

  1. Mar 21, 2008 #1
    I want to prove that [tex]\sqrt 6 - \sqrt 2- \sqrt 3[/tex] is irrational.

    I already know that [tex]\sqrt 2+\sqrt 3[/tex] is irrational (by squaring it). I would like a proof that doesn't use a polynomial and the rational root theorem.

    Last edited: Mar 21, 2008
  2. jcsd
  3. Mar 21, 2008 #2


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    Is [itex]\mahtbb{Q}(\sqrt{2}, \sqrt{3})[/itex] Galois over Q? If so, something is rational if and only if it is equal to all of its conjugates. (The same is true if the field isn't galois... it's just that in that case, its conjugates would live in other fields)
  4. Mar 21, 2008 #3
    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 [tex]\sqrt 2 + \sqrt 3[/tex] is irrational is that, supposing it is rational, its square should also be rational. But [tex](\sqrt 2 + \sqrt 3)^2 = 5+2\sqrt6[/tex] is irrational because [tex]\sqrt 6[/tex] is irrational.

    I would like to come up with a similar proof for [tex]\sqrt 6 - (\sqrt 2 + \sqrt 3)[/tex]
  5. Mar 21, 2008 #4


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    Hi abruzzi! :smile:

    How about squaring [tex]n+\sqrt 2+\sqrt 3[/tex], for some whole number n?
  6. Mar 21, 2008 #5
    By taking n=-1 and squaring we get [tex]2\sqrt 6 - 2\sqrt 3 - 2\sqrt 2 + 6 = 6 + 2(\sqrt 6 -\sqrt 3-\sqrt 2)[/tex]

    But from this I cannot conclude anything - since knowing that number a is irrational doesn't mean that a^2 is (for example [tex]a=\sqrt 2[/tex]).

    Or am I not on the right path?
  7. Mar 22, 2008 #6


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    ooh, you're right! :rolleyes:

    ok, let's try this:

    rational + irrational = irrational.

    rational x irrational = irrational.

    Suppose √6 - √3 - √2 is rational.

    Then (1 +√3)(1 + √2) is irrational, because it is (1 + 2√6) - (√6 - √3 - √2).

    But (1 - √3)(1 - √2) is rational, because it is 1 + (√6 - √3 - √2).

    So the product (1 +√3)(1 + √2)(1 - √3)(1 - √2) is irrational.

    But it isn't - it's 2. :smile:
  8. Mar 22, 2008 #7


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    sqrt2 + sqrt3 = sqrt 6 + r, with r rational,

    implies, by squaring both sides, that 1 + r^2 = (2-2r)sqrt6. which equates a rational and an irrational.
  9. Mar 22, 2008 #8


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    … mathwonk is cool …

    :smile: ooh, mathwonk, that's much better! :smile:
  10. Mar 22, 2008 #9
    That was exactly what I was looking for, thanks!
  11. Mar 22, 2008 #10


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    Much nicer than mine
    let x=√6-√3-√2
    ->x is irrational
    let y=√323-√19-√17
    ->y is irrational
    mathwonks works for that too
    ->y is irrational
    just two was of writing m√ab
    one in Z[√a,√b]
    one in the field of fractions
    the fraction sure give nice numbers though
  12. Mar 23, 2008 #11
    Well I think if you can prove that any one of the terms is irrational, which should be easy, and if the number subtracted by it will be irrational, unless the number you subtract by is the same number or follows the same irrational pattern, which would be quite hard as it is impossible to find a pattern, thus if you were to subtract 2 irrational numbers, there should be an infinite probability that they will be irrational and rational =\.

    for example you have an irrational number

    .342526524525352325........n where n are the rest of the terms
    -.1412413431413........n then the 2 n's would cause repeating zeros, thus being rational.

    Just bringing up irrat-irrat

    But there is no way to prove that at some point the rest of the digits would be the same (aka n), yet as it goes on for ever it could happen. Very weird little thing infinity =P, I'm in very low level maths, and I just wanted to contribute so this is all I could think of =], I'm sure there is a clever way to prove it though.
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