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Thread: TA’s Challenge Problem #7

  1. #1
    Senior Member TheAbstractionist's Avatar
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    TA’s Challenge Problem #7

    “Prove” that 1 is transcendental.

    “Proof”:

    The Gelfond–Schneider theorem states that if $\displaystyle \alpha$ and $\displaystyle \beta$ are algebraic numbers such that $\displaystyle \alpha\ne0,\,1$ and $\displaystyle \beta\notin\mathbb Q,$ then $\displaystyle \alpha^\beta$ is a transcendental number.

    Hence, taking $\displaystyle \alpha=-1$ and $\displaystyle \beta=\sqrt2,$ we have that $\displaystyle (-1)^{\sqrt2}$ is transcendental.

    But $\displaystyle (-1)^{\sqrt2}=(-1)^{2\frac{\sqrt2}2}=\left((-1)^2\right)^{\frac{\sqrt2}2}=1^{\frac{\sqrt2}2}=1.$

    Hence 1 is transcendental.

    Explain what is wrong with the above “proof”.
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  2. #2
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    I guess maybe a couple of things are wrong with it, but for one, the proof manipulates things so that $\displaystyle \alpha=1,$ and then attempts to apply Gel'fond-Schneider, contrary to hypothesis.
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    You are on the right track, but you need to be more specific about what things are being “manipulated”.
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  4. #4
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    Okay, well... the correct way to evaluate $\displaystyle (-1)^{\sqrt{2}}$ would probably be to use the rule $\displaystyle z^\alpha=\exp(\alpha\log z)$, which I believe has infinitely many values when $\displaystyle \alpha$ is not a rational real number.... So the expression is multi-valued, and so we can't apply Gel'fond-Schneider?
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  5. #5
    Senior Member TheAbstractionist's Avatar
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    Gelfond–Schneider certainly applies in this case.

    But you’ve the nail on the head.
    Spoiler:
    $\displaystyle (-1)^{\sqrt2}$ is not a real number. It is complex-valued, and for complex multiplication, the exponent laws for real numbers break down – e.g. $\displaystyle \left[(-1)(-1)\right]^{1/2}\ne(-1)^{1/2}(-1)^{1/2}.$
    Last edited by TheAbstractionist; Aug 7th 2009 at 05:39 AM.
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  6. #6
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    Quote Originally Posted by TheAbstractionist View Post
    Gelfond–Schneider certainly applies in this case.

    But you’ve the nail on the head.
    Spoiler:
    $\displaystyle (-1)^{\sqrt2}$ is not a real number. It is complex-valued, and for complex multiplication, the exponent laws for real numbers break down – e.g. $\displaystyle \left[(-1)(-1)\right]^{1/2}\ne(-1)^{1/2}(-1)^{1/2}.$
    Heh... I think you've been far too generous in crediting me with providing a solution to this. My answers were clumsy all along. What I ought to have said was that Gel'fond-Schneider cannot be applied in the way that it was being applied in the OP. I could tell that something was wrong with the way the exponent was being taken, but I wasn't sure exactly what, because I've always done these "by the book" using logs, and so to be honest, I wasn't really sure where the problem was in the OP.

    Which is sort of embarrassing to me, because I am supposed to be a specialist in transcendental number theory....
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