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From: Maury Barbato on 29 Jan 2010 02:18
Hello,
it's a well-known fact that pi and e are two irrational
numbers. Now, let m, n be two nonzero integers.
Are (pi)^m and e^n independent over Q?
I think the answer is yes, but I couldn't find a proof.
Thank you very much for your attention.
My Best Regards,
Maury Barbato
From: T.H. Ray on 29 Jan 2010 02:39
Maury wrote

> Hello,
> it's a well-known fact that pi and e are two
> irrational
> numbers. Now, let m, n be two nonzero integers.
> Are {pi, (pi)^2, ..., (pi)^m, e, e^2,..., e^n}
> independent over Q?
> I think the answer is yes, but I couldn't find a
> proof.
> Thank you very much for your attention.
> My Best Regards,
> Maury Barbato

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

Tom
From: Pubkeybreaker on 29 Jan 2010 12:40
On Jan 29, 12:18 pm, Maury Barbato <mauriziobarb...(a)aruba.it> wrote:
> Hello,
> it's a well-known fact that pi and e are two irrational
> numbers. Now, let m, n be two nonzero integers.
> Are (pi)^m and e^n independent over Q?
> I think the answer is yes, but I couldn't find a proof.


Neither can anyone else.
From: Gerry on 29 Jan 2010 17:36
On Jan 30, 4:18 am, Maury Barbato <mauriziobarb...(a)aruba.it> wrote:
> Hello,
> it's a well-known fact that pi and e are two irrational
> numbers. Now, let m, n be two nonzero integers.
> Are (pi)^m and e^n independent over Q?

I'm not sure whether you mean linearly independent
or algebraically independent, but in either case
the answer is "Of course - but no one can prove it."

What might be the simplest case, the question of linear
independence of pi and e over the rationals, is equivalent
to the question of the irrationality of pi / e, which
is open.
--
GM
From: John T on 2 Feb 2010 11:48
Pi and e are shapes that are iterations I believe. So as the expand (or change shape geometrically) their values change and approach a certain value as they approach their limit. The guy on this news video can draw pi shows exactly how it works.

http://www.pnwlocalnews.com/south_king/fwm/lifestyle/82848357.html
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