Re: prove it if you can


Gerry Myerson wrote:
> 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

Yes. I got those two mixed up.

--- Christopher Heckman

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