TITLE: Proof that at most one of $e\pi$ and $e+\pi$ can be rational
QUESTION [12 upvotes]: $e$ 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?

REPLY [7 votes]: 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.