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Does there exist any positive integer $n$ such that $e^n$ is an integer ?

I was in particular trying to prove $\log 2$ is irrational; now if it is rational, then there are relatively prime integers $p,q$ both positive such that $\log2 =p/q$ that is $e^p=2^q$ is an integer. I wanted to reach a contradiction.

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    $\begingroup$ No since the number $e$ is transcendental. $\endgroup$ – user63181 Feb 2 '15 at 14:45
  • $\begingroup$ @SamiBenRomdhane: I don't know about transcendental ; can you please give a proof ? $\endgroup$ – user210961 Feb 2 '15 at 14:46
  • $\begingroup$ The proof $e$ is transcendental is probably too long to put in an answer here. "Transcendental" means it is not a solution of a polynomial equation with integer coefficients. In particular, $e$ is not a root of an equation of the form $X^n-m$ as marwalix says. $\endgroup$ – GEdgar Feb 2 '15 at 14:48
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    $\begingroup$ Probably. Take the proof that $e$ is irrational, adapt it so show $e^m$ is irrational for integer $m$. $\endgroup$ – GEdgar Feb 2 '15 at 14:51
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    $\begingroup$ @GEdgar : Well I know the irrationality prove of $e^m$ but here I only need $e^m$ is not an integer , isn't there any easier proof? $\endgroup$ – user210961 Feb 2 '15 at 14:53
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No $e$ is transcendantal and if there were integers $n$ and $m$ such that $e^n=m$ and therefore would be algebraic over $\mathbb{Q}$ as a root of the polynomial $X^n-m$. Contradiction.

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Assuming that for some couple $(n,m)$ of natural numbers we have $e^n=m$, then $\tanh(n)\in\mathbb{Q}$.

Since $\tanh z=\frac{d}{dz}\log\cosh z$, a continued fraction representation for $\tanh n$ is given by:

$$ \tanh n = \cfrac{1}{\frac{1}{n}+\cfrac{1}{\frac{3}{n}+\cfrac{1}{\frac{5}{n}+\cfrac{1}{\frac{7}{n}+\cfrac{1}{\frac{9}{n}+\ldots}}}}} \tag{1} $$ and it is not difficult to prove that such expression cannot be equal to some finite standard continued fraction. This leads to $\tanh n\not\in\mathbb{Q}$, hence to $e^n\not\in\mathbb{N}$ as wanted.

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