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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" $e$.

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    This is unknown at the moment. They both are transcendental, but it is unknown if they are *algebraically independent*.2012-08-30
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    But doesn't $e^{i\pi}+1=0$ show them to be algebraically independent?2012-08-30
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    No. It simply shows that there exists a non-algebraic relation between them.2012-08-30
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    So there could be an algebraic relation that exists that we just don't know of. Ok.2012-08-30
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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.2012-08-30
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    I don't suppose you would accept $e=(-1)^{1/(\pi i)}$ as an algebraic expression for $e$ in terms of $\pi$?2012-08-30
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    @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.2012-08-31
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    "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?2012-08-31
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    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.2012-08-31

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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.