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I was wondering if there is a number $x\in\mathbb Z$ with $x\neq 0$ s.t. $\exp(x)\in\mathbb N$ and if not, why is that so?

EDIT: Forgot to exclude the $0$.

Christian Ivicevic
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  • The base of the natural log ($e$) is trancendental. As far as we know, it is very hard to prove that no weird power of $e$ is NOT trancendental, like $e^{\sqrt{37 \pi}$. It is hard to show that no non-trivial natural power is uninteresting too. – amcalde Nov 03 '15 at 02:12
  • The 'why is that so?' part is quite hard to show (depending how far you want to go). You need to show that $e$ is transcendental over $\mathbb{Q}$ which is quite a lot of work. – Sam Weatherhog Nov 03 '15 at 02:12

3 Answers3

9

Besides $0$, no. Suppose that $\exp(n)\in \mathbb{Z}$, where $n\neq 0$. Then $e=\exp(n)^{1/n}$ is algebraic over $\mathbb{Q}$, contradicting the fact that $e$ is transcendental.

Note that this can be expanded to show that, moreover, $\exp(r)$ is not rational for any non-zero rational number $r$, by following the same argument.

Hayden
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3

$$\exp(0)=1$$ $\qquad$$\qquad$$\qquad$$\qquad$

0

Suppose $e^x=y$, $x\in \mathbb Z-\{0\}$, $y\in \mathbb N$, then $e$ is the solution of $z^x-y=0$, a polynomial equation with coefficient in $\mathbb Z$, contradict with the fact that $e$ is a transcendental number.