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Math Help - Integers in a logarithmic relation

  1. #1
    Newbie Demandeur's Avatar
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    Integers in a logarithmic relation

    Do there exist a,b\in\mathbb{Z}^+ such that \log(\frac{a}{b}) = \frac{1}{\sqrt{2}}\log{2}?
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    If you do a bit of manipulation [LaTeX ERROR: Convert failed]

    From this you get that you need positive integers a and b, such that

    [LaTeX ERROR: Convert failed]

    In essence you are asked if [LaTeX ERROR: Convert failed] is rational.

    What theorems about properties of numbers do you have at your disposal? Given certain theorems this shouldn't be a difficult thing to answer, although it may be hard if you have to actually find and prove your answer.
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  3. #3
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    Quote Originally Posted by Vlasev View Post
    [LaTeX ERROR: Convert failed] In essence you are asked if [LaTeX ERROR: Convert failed] is rational.
    It's almost as if you read my thought process. This was indeed what I was trying to determine - but having
    failed to make any progress on it, I figured that it was equivalent to finding the integers a and b satisfying
    the logarithmic relation in the first post, under the assumption that it might be easier to deal with it that
    way.

    What theorems about properties of numbers do you have at your disposal?
    What are the theorems? I know just the basic stuff - no more nor less advanced than what one usually
    needs to show that \sqrt{2}\notin\mathbb{Q}. But if you have a 'non-elementary' answer (or a theorem that holds the key),
    feel free to provide it - I will make the effort to understand (or apply) it.
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    If you have covered transcendental numbers before, you will know that if a number is transcendental, then it is irrational. Now, there is a theorem, the Gelfond-Schneider theorem, which characterizes numbers of the form [LaTeX ERROR: Convert failed] . It states that

    If a and b are algebraic numbers with x ≠ 0, and if y is not a rational number, then any value of [LaTeX ERROR: Convert failed] a transcendental number.

    Now, x = 2 is algebraic, since it is obviously the root of a polynomial with rational coefficients. Since y = [LaTeX ERROR: Convert failed] is irrational,then [LaTeX ERROR: Convert failed] is also irrational. Then by the G-S theorem, [LaTeX ERROR: Convert failed] must be transcendental. Hence it is irrational.

    Now, unless this was the theorem you are supposed to use, there must be some other way to prove it, without invoking this theorem.
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  5. #5
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    Hello, Demandeur!

    This problem has an even more interesting form.
    [But no, I haven't solved it yet.]



    \text{Do there exist }a,b\in\mathbb{Z}^+\text{ such that: }\;\log\left(\dfrac{a}{b}\right) \;=\; \frac{1}{\sqrt{2}}\log{2}?

    As vlasev pointed out: . \log\left(\dfrac{a}{b}\right) \:=\:\log\left(2^{\frac{1}{\sqrt{2}}}\right)

    So we have: . \dfrac{a}{b} \;=\;2^{\frac{1}{\sqrt{2}} }

    Note that: . 2^{\frac{1}{\sqrt{2}}} \;=\;2^{\frac{\sqrt{2}}{2}} \;=\;2^{\left(\frac{1}{2}\cdot\sqrt{2}\right)}   \;=\;\left(2^{\frac{1}{2}\right)^{\sqrt{2}} \;=\;\left(\sqrt{2}\right)^{\sqrt{2}}


    The equation becomes: . \left(\sqrt{2}\right)^{\sqrt{2}} \;=\;\dfrac{a}{b}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Somewhere, there is a theorem about: . \text{(irrational)}^{\text{irrational}}

    But I can't recall it nor can I find the reference.
    Does that sound familiar to anyone?
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    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Soroban View Post
    Somewhere, there is a theorem about: . \text{(irrational)}^{\text{irrational}}

    But I can't recall it nor can I find the reference.
    Does that sound familiar to anyone?
    I don't know of that theorem, but according to Wikipedia, it is unknown whether \pi^e and \pi^{\sqrt{2}} are irrational.

    Irrational number - Wikipedia, the free encyclopedia (subheading: Open questions)

    Perhaps the theorem you have in mind deals with algebraic irrationalities?
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Soroban View Post

    Somewhere, there is a theorem about: \text{(irrational)}^{\text{irrational}}
    But I can't recall it nor can I find the reference.
    Does that sound familiar to anyone?
    If m,n\notin\mathbb{Q}, it's possible that m^n\in\mathbb{Q}. (see here for proof - at the beginning of the page).
    It's interesting the proof actually uses the concerned constant, without determining whether or not it's rational.
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  9. #9
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    I seem to be suffering from dementia... I read post #4 and almost immediately forgot having read it!
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  10. #10
    MHF Contributor Also sprach Zarathustra's Avatar
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    I didn't see that... (Sorry Vlasev!)
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  11. #11
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    Quote Originally Posted by Vlasev View Post
    ...
    That's a very nice theorem, Vlasev, and a nice proof. It's really heavy theorem, though. I'm not doing this
    for a class, so I won't need prove the theorem to use it (and I really doubt I would understand the proof).
    It would be interesting if we could find a proof using contradiction or the other usual elementary techniques.
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  12. #12
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Demandeur View Post
    That's a very nice theorem, Vlasev, and a nice proof. It's really heavy theorem, though. I'm not doing this
    for a class, so I won't need prove the theorem to use it (and I really doubt I would understand the proof).
    It would be interesting if we could find a proof using contradiction or the other usual elementary techniques.

    I don't think so... this theorem is a case of 7th Hilbert's problems - Wikipedia, the free encyclopedia,so there many mathematicians that thought about it... [but maybe all of them miss something...]
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  13. #13
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    Quote Originally Posted by Also sprach Zarathustra View Post
    I don't think so... this theorem is a case of 7th Hilbert's problems - Wikipedia, the free encyclopedia,so there many mathematicians that thought about it... [but maybe all of them miss something...]
    You misunderstood me, Nietzsche. I'm not trying to find a contradiction to this well established theorem (one
    would have to be so vain to do that). I'm merely seeking an elementary way of proving that the square root
    two raised as much is irrational. That is what I mean when I talk of the hope of deriving a contradiction and
    elementary techniques, not Gelfond's theorem.
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  14. #14
    Newbie Demandeur's Avatar
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    Found this in MathWorld:
    Quote Originally Posted by MathWorld
    Both the Gelfand-Schneider constant [LaTeX ERROR: Convert failed] and Gelfond's constant [LaTeX ERROR: Convert failed] were singled out in the 7th of Hilbert's problems as examples of numbers whose transcendence was an open problem (Wells 1986, p. 45).
    So indeed my quest to find an elementary proof was in vain! It's probably what Nietzsche was trying to
    tell me as well - now that I realise it also says the same in the Wikipedia article whose link he's posted.
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  15. #15
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Demandeur View Post
    You misunderstood me, Nietzsche. I'm not trying to find a contradiction to this well established theorem (one
    would have to be so vain to do that). I'm merely seeking an elementary way of proving that the square root
    two raised as much is irrational. That is what I mean when I talk of the hope of deriving a contradiction and
    elementary techniques, not Gelfond's theorem.

    You misunderstood me, but I don't blame you, my English is very bad, sorry for that...

    By the way, who it is in your picture avatar...?
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