$e$ and $\pi$ are rather peculiar numbers. It turns out that, in addition to being irrational numbers, they are also transcendental numbers. Basically, a number is transcendental if there are no polynomials with rational coefficients that have that number as a root.

Clearly, $p(x) = (x-e)(x-\pi)$ is a polynomial whose roots are $e$ and $\pi$, so its coefficients cannot all be rational, by the definition of transcendental numbers. Expanding that expression, we get

$$(x-e)(x-\pi) = x^2 - (e+ \pi)x + e\pi$$

This means that $1, -(e+\pi), e\pi$ cannot all be rational. If all the coefficients were rational, we would have found a polynomial with rational coefficients that had $e$ and $\pi$ as roots, and that has been proven impossible already. Hermite proved that $e$ is transcendental in 1873, and Lindemann proved that $\pi$ is transcendental in 1882. In fact, Lindemann's proof was similar to Hermite's proof and was based on the fact that $e$ is also transcendental.

In other words, at most one of $e+\pi$ and $e\pi$ is rational. (We know that they cannot both be rational, so that's the most we can say). Are there any more conditions required for this proof to be correct?

  • 1
    $\begingroup$ What are you trying to prove? $\endgroup$ – Alice Ryhl Jan 7 '15 at 16:47
  • $\begingroup$ Using that $e$ and $\pi$ are transcendental, to prove irrationality of either $e+\pi$ or $e\pi$ seems like overkill to me, but the proof looks sound. $\endgroup$ – Henrik - stop hurting Monica Jan 7 '15 at 16:56
  • $\begingroup$ I think I'm missing something. What prevents $e+π$ and $eπ$ from both being irrational at the same time? $\endgroup$ – nicks Jan 17 '15 at 11:17

We do not really need to use the fact that both $\pi$ and $e$ are trascendental numbers. If both $e\pi$ and $\pi+e$ would be rational numbers, then $e$ would be a quadratic irrational, so its continued fraction would be eventually periodic due to Lagrange's theorem. However, the continued fraction of $e$ is well-known:

$$ e = [2;1,2,1,1,4,1,1,6,1,1,8,\ldots] $$ and its coefficients are unbounded.


The proof is correct. For a nit-pick, you should say you have a non-zero polynomial.

Note that it would even suffice to know that at least one of $\pi$ and $e$ is transcendental.

  • $\begingroup$ Are you certain that knowing "at least one of $\pi,e$ is transcendental" is sufficient? What if both were transcendental and just happened to sum and multiply to rationals? $\endgroup$ – abiessu Jan 7 '15 at 16:56
  • $\begingroup$ @abiessu what I meant to say is that for your argument to work you'd only need to know one of them is transcendental. $\endgroup$ – quid Jan 7 '15 at 17:00

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