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


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)?

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    $\begingroup$ Where does LW say $e^a$ is algebraic if $a$ not algebraic? $\endgroup$ Feb 11 at 7:14
  • 2
    $\begingroup$ 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. $\endgroup$ Feb 11 at 7:25
  • $\begingroup$ @Don Thousand and Greg Martin. I just negated the statement of the lindemann weierstrass theorem but it seems like I was wrong. $\endgroup$ Feb 11 at 10:15

The Gelfond-Schneider Theorem actually says:


  • $\alpha$ 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."

  • $\begingroup$ 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 $\alpha ^{\beta}$ $\endgroup$ Feb 11 at 10:17
  • $\begingroup$ Yes of course you're right -- approved the edit. $\endgroup$ Feb 11 at 13:12

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