the title and the tags might be correct for the question.
In the wikipedia page about gelfond's constant I saw that $e^{π}$ is transcendental and the proof was
$e^{π}=(e^{iπ})^{-i}=(-1)^{-i}$
Since $-1$ and $-i$ is algebraic and $-i$ is not rational, by the Gelfond Schneider Theorem $e^{π}$ is transcendental.
Now let's take the lindemann weierstrass theorem. According to the theorem, $e^a$ is transcendental if a is algebraic
In this case where $π$ is transcendental so by the lindemann weierstrass theorem $e^π$ cannot be transcendental so it is algebraic.
Which contradicts our first result.
I just wanna know that what am I missing(I am quite sure about it) or did I just found a contradiction in maths(I don't think so)?
Best Answer
The Gelfond-Schneider Theorem actually says:
If:
then $\alpha^{\beta}$ is transcendental.
Just thought I'd put that in there because technically speaking $-i$ is not actually "irrational" as such.
Apart from that, yes, from your argument $e^\pi$ is transcendental.
Lindemann-Weierstrass Theorem says:
Let $a$ be a non-zero algebraic number.
Then $e^a$ is transcendental.
It does not say that if $a$ is transcendental, then $e^a$ is not transcendental.
That would be like saying:
"If $a$ is even then $2 a$ is even. Therefore if $a$ is odd then $2 a$ is odd."