# $e$ and $\pi$ Spoof

It is known that that $\pi$ and $e$ are transcendental. $e$ was proven to be transcendental by Hermite in 1873, and $\pi$ by Lindemann in 1882. $e^{\pi}$ is transcendental from the Gelfond's Theorem. It is still not known if $e^e,$ $\pi ^\pi$ or $\pi ^e$ are transcendental. However, below is a proof that $\pi ^e$ is rational. It comes from E. Barbeau's Mathematical Fallacies, Flaws, and Flimflam which is a collection of his columns *The College Mathematical Journal*. The proof has been contributed by Christian Counts.

### Theorem

$\pi ^e$ is rational.

### Proof

First off, for a rational $r,$ $\log_{\pi}r$ is irrational because, otherwise, the equation $\pi ^s=r,$ with rational $r,$ would have a rational solution which is impossible given that $\pi$ is transcendental.

Now suppose, if possible, that $\pi ^e \not\in \mathbb{Q}.$ Then, for all $r\in\mathbb{Q},$ $\pi ^e\ne r,$ so that $e=\log_{\pi}\pi ^e\ne\log_{\pi}r$ because the logarithm is a 1-1 function. Therefore, as was observed, $e\ne k$ for $k\not\in\mathbb{Q}.$ In other words, $e$ is rational. But this is a contradiction. It follows that $\pi ^e$ is rational as claimed.

### References

- E. J. Barbeau, Mathematical Fallacies, Flaws, and Flimflam, MAA, 2000, p. 10

|Contact| |Front page| |Contents| |Algebra| |Up|

Copyright © 1996-2018 Alexander Bogomolny

65278010 |