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Does this proof for irrationality of sqrt(2)+sqrt(3) work?

  1. Oct 1, 2008 #1

    Simfish

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    Gold Member

    1. The problem statement, all variables and given/known data

    Prove that [tex]\sqrt{2}+\sqrt{3}[/tex] is irrational.

    2. Relevant equations



    3. The attempt at a solution

    So we know that [tex](\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2}) = 1[/tex]. But a rational number must be of the form a/b, and if (a/b)c = 1, the only number c that works (for rational numbers) is c = b/a in reduced form due to unique inverses for rational numbers. But here we have a value of c that is NOT of the form c = b/a. And so once we prove that [tex]\sqrt{2}+\sqrt{3}[/tex] is NOT [tex]\frac{1}{\sqrt{3}-\sqrt{2}}[/tex], we can only conclude that [tex](\sqrt{2}+\sqrt{3})[/tex] is irrational.
     
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  3. Oct 1, 2008 #2

    statdad

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    Homework Helper

    No, this won't do it, because by claiming that [tex] c = \sqrt 3 - \sqrt 2 [/tex] is not a rational number you are making an assumption of precisely the type of thing you are trying to prove.

    As an alternate start, try this. Assume, to the contrary, that [tex] \sqrt 2 + \sqrt 3 [/tex] is rational. Then there are integers [tex], a, b [/tex] such that

    [tex]
    \sqrt 3 + \sqrt 2 = \frac a b
    [/tex]

    Square each side

    [tex]
    3 + 2\sqrt 6 + 2 = \frac{a^2}{b^2}
    [/tex]

    Rewrite this to isolate [tex] \sqrt 6 [/tex] - you will see that you assumption says something about the type of number [tex] \sqrt 6 [/tex] is. Go from there.
     
  4. Oct 1, 2008 #3

    morphism

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    Actually I don't even understand what the OP is asking. It seems as though he/she is trying to prove this:
    which it is.
     
  5. Oct 1, 2008 #4

    Simfish

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    Gold Member

    Okay thanks to you both! Sorry I was wrong - yeah - my argument was circular (and was contingent on a wrong assumption too - one that I forgot to check). So i'll go along with your suggestion.
     
  6. Oct 1, 2008 #5

    HallsofIvy

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    Do you already know that [itex]1/\sqrt{2}[/itex] and [itex]1/\sqrt{3}[/itex] are irrational- or, equivalently, that [itex]\sqrt{2}[/itex] and [itex]\sqrt{3}[/itex] are irrational are you allowed to use those facts? If so, that simplifies the problem enormously.
     
  7. Oct 1, 2008 #6

    Simfish

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    Yeah, I already know from the book (apostol math analysis) that [tex]\sqrt{2}[/tex] is irrational. But I'm not sure whether or not I know that [tex]\sqrt{6}[/tex] is irrational or not - but this can be easily proven using the same techniques as those involved in [tex]\sqrt{2}[/tex]. I got it now - thanks :)
     
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