# On irrationality of natural logarithm

Is there any rational number $r$ such that ln (r) is rational as well?

If so, what's the proof?

If proofs are too lengthy to be cointained as an answer here, I would truly appreciated any easy-to-understand references to study them.

• Does $r=1$ count? Feb 27, 2017 at 0:40

Aside from $r=1$, no. To prove it, suppose we had an example. Then we'd write $$\frac mn=e^{\frac ab}\implies e^a=\left( \frac mn \right)^b$$ But, with $a\neq 0$ this would tell us that $e$ was algebraic, which is not the case.
• @JimThio: A number is algebraic if it is the solution to a polynomial equation with integer coefficients. $e$ is not algebraic; this is well-known but not so easy to prove. See also en.wikipedia.org/wiki/Algebraic_number and divisbyzero.com/2010/09/28/the-transcendence-of-e. Feb 27, 2017 at 10:32
• @JimThio The proof that $e$ is transcendental is far from trivial, but it is an old result (Hermite, around 1875). The proof of the stronger theorem of Lindemann and Weiesrstrass, which shows the algebraic independence of $\{e^{\alpha_i}\}$ for distinct algebraic numbers $\{\alpha_i\}$ follows similar lines and is not all that much harder, so I recommend looking at that. The wiki article offers a fairly standard version, and the references provided in it are very good.