Proving the irrationality of $e^n$
This essentially proves $e^n \ $ is irrational for all possible values of n and since $ e^{\pi^2} $ is essentially $e^{\pi\times\pi}$ and this fits the $e^n \ $ category so does that make it irrational?
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2Huh??????????????? – David C. Ullrich Dec 23 '17 at 16:10
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2$n$ is supposed to be an integer. Is $\pi^2$ an integer? – James Dec 23 '17 at 16:11
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No, this does not prove that $\mathrm{e}^{\pi^2}$ is irrational; the proof that you cite assumes that $n$ is a natural number. – Xander Henderson Dec 23 '17 at 16:11
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That link shows that $e^n$ is irrational for positive integers $n$. $\pi^2$ is not a positive integer – lulu Dec 23 '17 at 16:12
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If $n$ was arbitrary, you could write $e^{\ln 2}=2$ irrational. – Dec 23 '17 at 16:12
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Ok thanks for the help – Dec 23 '17 at 16:15
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To the original question: I, for one, don't know if the transcendence of $e^{\pi^2}$ has been shown (surely it is transcendental, but proving it is another matter). Of course, someone else may know more about it. The transcendence of $e^{\pi}$ follows from the Gelfond-Schneider Theorem. – lulu Dec 23 '17 at 16:19
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No. The link shows that $e^n$ is irrational for any natural number $n$. But $\pi^2$ isn't a natural number, so that result doesn't apply. Note that there absolutely are exponents $x$ for which $e^x$ is rational - e.g. $e^{\ln 2}=2,$ and so on.
Noah Schweber
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Well, $e^{\pi^2}$ isn't just “essentially” $e^{\pi\times\pi}$; they're the same thing. And the question to which you have posted a link is about the irrationality of the numbers of the form $e^n$, with $n$ natural. Since $\pi^2\notin\mathbb N$, no, $e^{\pi^2}$ doesn't fit this category.
José Carlos Santos
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No. The answers to that question only show that $e^n$ is irrational for any positive integer $n$. This would only apply to $e^{\pi^2}$ if $\pi^2$ were a positive integer, which it is not.
Eric Wofsey
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