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.


1 Answer 1


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.

  • 2
    $\begingroup$ why e isn't algebraic? $\endgroup$
    – user4951
    Feb 27, 2017 at 9:48
  • $\begingroup$ It's most likely true, but I do not know what algebraic means and your proof needs more explanation $\endgroup$
    – user4951
    Feb 27, 2017 at 9:48
  • 7
    $\begingroup$ @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. $\endgroup$ Feb 27, 2017 at 10:32
  • $\begingroup$ @JimThio A complex number is said to be transcendental if it is not algebraic. While proving number being transcendental is hard in general, we know the set of transcendental numbers is uncountable. The proof is simply that there are countable many polynomials with rational coefficients and every nonzero polynomial has finite number of roots. Since countable union of countable sets is countable, the set of algebraic numbers is countable. So the set of transcendental numbers must be uncountable since the set of complex numbers is uncountable. $\endgroup$
    – Alex Vong
    Feb 27, 2017 at 11:05
  • 3
    $\begingroup$ @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. $\endgroup$
    – lulu
    Feb 27, 2017 at 11:29

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