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Math Help - Integers in a logarithmic relation

  1. #16
    Newbie Demandeur's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    You misunderstood me, but I don't blame you, my English is very bad, sorry for that...
    Not at all. On the contrary, your English is very good. It was my fault.
    By the way, who it is in your picture avatar...?
    The one and only one Carl Friedrich Gauss.
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  2. #17
    MHF Contributor Also sprach Zarathustra's Avatar
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    Not at all. On the contrary, your English is very good. It was my fault.
    Thank you for that...


    The one and only one Carl Friedrich Gauss.
    I'm ashamed...
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  3. #18
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    Quote Originally Posted by Also sprach Zarathustra View Post

    I'm ashamed...
    It's ok, he's young in that picture!
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  4. #19
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    I don't know whether or not x = \sqrt{2}^\sqrt{2} is rational or not, but there is an interesting proof in which this quantity figures.

    Problem: Can a^b be rational if a and b are irrational?

    Consider x = \sqrt{2}^\sqrt{2} . If x is rational then we have our example. If not, and x is irrational, then consider x^{\sqrt{2}} = \sqrt{2}^2 = 2, and we again have an example of rational a^b with irrational a, b. So whether x is rational or not, we have the required example.

    This is an example of (to me) an absolutely infuriating proof. We know either x or x^\sqrt{2} is an example of rational a^b with irrational a and b, but we don't know which, and the proof gives us no clue!

    This proof is discussed in "The Princeton Companion to Mathematics".
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  5. #20
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    The Gelfond-Schneider theorem says that [LaTeX ERROR: Convert failed] is transcendental and hence irrational. But I hear you. it's quite a nasty little existence proof!
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