# Is $\ \Large\pi^e$ rational?

Is the number $\ \Large\pi^e$ rational?

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No­­${}{}{}{}{}$ – evil999man Apr 24 '14 at 11:49
Nobody knows for sure. Cf. mathworld.wolfram.com/IrrationalNumber.html – derpy Apr 24 '14 at 11:50
In fact I hope it is rational. If it ever turns out to be like that, then I will be shocked in a pleasant way. – drhab Apr 24 '14 at 12:32

It seems highly likely that $\pi^e$ is irrational, but no one has figured out how to prove it yet.

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What about $\pi^{e^\pi}$? – Jika Apr 24 '14 at 11:56
@Jika: $e^\pi$ is a transcedental number, look the Gelfond's Theorem and $e^\pi = (-1)^{-i}$. So, looks like the $\pi^{e^\pi}$ is irrational and transcedental. – m0nhawk Apr 24 '14 at 13:58

The answer is unknown (though it seems intuitive to assume that the result will be irrational - just because $\pi$ raised to any rational number other than 0 will be already irrational).

Though nothing is for certain. Maybe by some bizarre connection, $e$ and $\pi$ could be established to be 2 faces of the same thing, such that some non-trivial operations on them (like yours) would result in a rational number!

The challenge, fundamentally, lies in proving that $\pi$ and $e$ are algebraically independent of one another. I.e. there is no polynomial expression with coefficients which are algebraic numbers (numbers which themselves are roots of polynomials with rational coefficients) equating the 2 numbers.

For imagine if there was indeed such a polynomial that equated the 2 - then you could take the numerator of one side, divide both sides by it, and then attain 1 = … where RHS would be a "mixture" of $\pi$ and $e$.

Maybe, given that the polynomial "connections" were workable enough, you could find ways to create even more astounding equalities, such as the one you are pondering about.

As it turns out, proving the independency of the two numbers is not easy.

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Algebraic independence is not sufficient for this case. At least on present knowledge, any of the four possibilities of ($\pi$ and $e$ { are / are not } algebraically independent) × ($\pi^e$ is { rational / irrational }) might hold. – MJD Apr 24 '14 at 12:55