Re: exp(sqrt(2))

From: David McAnally (D.McAnally_at_i'm_a_gnu.uq.net.au)
Date: 09/02/04 

Date: 2 Sep 2004 16:24:21 GMT

panoptes@iquest.net (Daniel W. Johnson) writes:

>Barnaby Finch <barnabyfinch@verizon.net> wrote:

>> I believe I read that if x is algebraic (and not zero), then e^x is
>> transcendental.

>http://mathworld.wolfram.com/Lindemann-WeierstrassTheorem.html

While this theorem is sufficient to demonstrate that if x is a nonzero
algebraic number, then exp(x) is transcendental, another theorem that is
also sufficient to draw the conclusion is the Hermite-Lindemann Theorem,
quoted in

http://mathworld.wolfram.com/Hermite-LindemannTheorem.html

One thing that worries me about the statement of the Lindemann-Weierstrass
Theorem in Mathworld is that the independence is asserted over the p-adic
numbers, rather than over Q, as it should be.

Another result which is also sufficient to demonstrate that if x is a
nonzero algebraic number, then exp(x) is transcendental, is the
generalized form of the Gelfond-Schneider Theorem:

        If a_1, ..., a_k, are algebraic numbers, and the values of
        log(a_1), ..., log(a_k), are such that they are linearly
        independent over Z, then 1, log(a_1), ..., log(a_k), are linearly
        independent over the algebraic numbers, for these specific values
        of the logarithms.

David

>> If e^x is algebraic (and not 1), then x is transcendental.
>> I'm sure it's possible to choose a transcendental x such that e^x is also
>> transcendental.

>The set of real x such that either x or e^x is algebraic is obviously a
>countable subset of the reals; this leaves the set of real x such that
>neither x nor e^x is algebraic as uncountable.
>--
>Daniel W. Johnson
>panoptes@iquest.net
>http://members.iquest.net/~panoptes/
>039 53 36 N / 086 11 55 W

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