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Am I right in thinking this is not possible since both are known to be transcendental?

Also, $e^{i\pi}+1=0$ suggests this is not possible - we can not isolate $e$ or $\pi$ from this since it involves taking a log at some point, thus "cancelling" $e$.

pshmath0
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    This is unknown at the moment. They both are transcendental, but it is unknown if they are algebraically independent. – M Turgeon Aug 30 '12 at 11:50
  • But doesn't $e^{i\pi}+1=0$ show them to be algebraically independent? – pshmath0 Aug 30 '12 at 11:55
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    No. It simply shows that there exists a non-algebraic relation between them. – M Turgeon Aug 30 '12 at 11:57
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    So there could be an algebraic relation that exists that we just don't know of. Ok. – pshmath0 Aug 30 '12 at 11:59
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    Note that $e$ and $2e$ are both known to be transcendental, and despite that it is possible to express each in terms of the other. – Gerry Myerson Aug 30 '12 at 12:50
  • I don't suppose you would accept $e=(-1)^{1/(\pi i)}$ as an algebraic expression for $e$ in terms of $\pi$? – Gerry Myerson Aug 30 '12 at 12:52
  • @Gary Myerson I don't think so since then we would have $$e=(-1)^{i(-1/\pi)}=e^{i(-1/\pi)\log(-1)}=e^{i(-1/\pi)i\pi}=e,$$ so we just get back what we started with. – pshmath0 Aug 31 '12 at 09:11
  • "we just get back what we started with." Isn't that true, even when there is a "legitimate" way to express one number algebraically in terms of another? If you express 2 in terms of 3 by $2=3-1$ then $2=(2+1)-1=2+(1-1)=2$. You have an example where you can't do this? – Gerry Myerson Aug 31 '12 at 12:45
  • Yes, but what you are saying above is that $e=e^{i(−1/\pi)i\pi}$, so in fact you're expressing $e$ in terms of $e$ and $\pi$, not $\pi$ alone. – pshmath0 Aug 31 '12 at 16:14

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The answer is: This is unknown at the moment. They both are transcendental, but it is unknown if they are algebraically independent. If you want to know more about this, you can read this Mathoverflow thread, or this Wikipedia article on Schanuel's conjecture.

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M Turgeon
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