Re: prove it if you can

In article <3sspfmFpsc6sU1@xxxxxxxxxxxxxx>,
Robert Low <mtx014@xxxxxxxxxxxxxx> wrote:

> Proginoskes wrote:
> > José Carlos Santos wrote:
> >
> >>[...]
> >>Let _s_ be the square root of 2. I don't know whether s^s is rational or
> >>not. If it is, then you're done. Otherwise, consider
> >>
> >> (s^s)^s = s^{s^2} = s^2 = 2.
> >>
> >>So, if s^s is irrational, you have the example that you want.
> > For the record, s^s is not only irrational, but transcendental. This is
> > a consequence of Lindemann's Theorem.
>
> How so? All I know about Lindemann's Theorem is what I
> googled for twenty seconds or so ago, and I can't see
> how it implies that. The theorem tells us that if
> a is algebraic, then exp(a) is transcendental. But
> s^s is exp(ln(s)s), and surely ln(s)s isn't algebraic.

I think it needs Gelfond-Schneider, which came after Lindemann.
If a is algebraic and not 0 or 1, and b is algebraic and irrational,
then a^b is transcendental.

http://mathworld.wolfram.com/GelfondsTheorem.html

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.

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