# Why is $e^π$ transcendental?

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.

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)?

• Where does LW say $e^a$ is algebraic if $a$ not algebraic? Feb 11 at 7:14
• The Lindemann–Weierstrass theorem says "if $a$ is algebraic then $e^a$ is transcendental". You are attempting to apply a different implication, namely "if $a$ ($\pi$ in this case) is not algebraic then $e6a$ is not transcendental". This is an example of confusing an implication for its inverse—the two are not equivalent. Feb 11 at 7:25
• @Don Thousand and Greg Martin. I just negated the statement of the lindemann weierstrass theorem but it seems like I was wrong. Feb 11 at 10:15

The Gelfond-Schneider Theorem actually says:

If:

• $$\alpha$$ and $$\beta$$ are algebraic numbers such that $$\alpha \notin \{0,1\}$$
• $$\beta$$ is either irrational or not wholly real

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."

• I think there is something wrong in your answer. The place where you mentioned the gelfond Schneider theorem, it will not be $\alpha \beta$ ,it will be $\alpha ^{\beta}$ Feb 11 at 10:17
• Yes of course you're right -- approved the edit. Feb 11 at 13:12