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Rationality
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spoudyal
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#1
Rationality

Is \pi ^ e rational or not? I know that an irrational number to an irrational power is rational (ex. :sqrt:2:sqrt:2) , but can anyone prove that this holds in this case? Good luck!
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It is confessed emongste all men, that knowe what learnyng meaneth, that besides the Mathematicalle artes, there is noe vnfallible knowledge, excepte it bee borowed of them.
Last edited by spoudyal on Sun May 15, 2005 8:27 am; edited 1 time in total 
PostPosted: Sat May 14, 2005 7:59 pm  Back to top 
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bomb
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#2
Isnt irrational to power of irrational irrational transeendatal even

Bomb
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PostPosted: Sun May 15, 2005 12:14 am  Back to top 
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pbornsztein
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#3
bomb wrote:
Isnt irrational to power of irrational irrational transeendatal even

Bomb


I don't think so : e^{i \pi} = -1.

Pierre.

PostPosted: Sun May 15, 2005 12:29 am  Back to top 
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kevinatcausa
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#4
I think the most general result known for this type of problem is what's called the Gelfond-Schneider theorem:

If b is an (edit: algebraic and) irrational number, and a is any number other than 0 and 1 which is algebraic (meaning that there is a polynomial with integer coefficients to which a is a root), then a^b must be transcendental (meaning not algebraic)

For example, taking a=b=\sqrt{2} gives that \sqrt{2}^{\sqrt{2}} must be transcendental.

However, the theorem doesn't apply with a=\sqrt{2}^{\sqrt{2}} and b=\sqrt{2} because a wouldn't be algebraic), and indeed taking a^b=\sqrt{2}^2=2.

The theorem doesn't apply directly to e^{\pi} (e isn't algebraic), but we can write e^{\pi}=(e^{\pi i})^{-i}=(-1)^{-i}, and now the theorem does hold and we get e^{\pi} is transcendental. However, I think the \pi^e case is still open

(editted due to error pointed out by Grobber...thanks!)
Last edited by kevinatcausa on Sun May 15, 2005 2:54 am; edited 3 times in total 
PostPosted: Sun May 15, 2005 2:20 am  Back to top 
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grobber
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#5
In the Gelfond-Schneider theorem b must also be algebraic (so it's algebraic and irrational, not just irrational, as you wrote, kevinatcausa).

PostPosted: Sun May 15, 2005 2:29 am  Back to top 
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spoudyal
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#6
Still Open

Hmmm ... Never knew about Gelfond-Schneider theorem. Alright well it's still open for whoever wants it. Thanks guys.
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It is confessed emongste all men, that knowe what learnyng meaneth, that besides the Mathematicalle artes, there is noe vnfallible knowledge, excepte it bee borowed of them.

PostPosted: Sun May 15, 2005 8:23 am  Back to top 
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