Apart from the trivial cases, $x=\log a$ where $a\in\mathbb{Q}$, are all values of $e^x$ irrational? Are some transcendental?

  • $\begingroup$ they are irrational by definition. $\endgroup$
    – Dunham
    Oct 24, 2014 at 13:42
  • $\begingroup$ All transcendental numbers are irrational, so I'm not sure what the role of the "or" is here. Did you mean to ask if $e^x$ is transcendental whenever $x\ne\log{a}$ for some $a\in\mathbb{Q}$? $\endgroup$
    – mdp
    Oct 24, 2014 at 13:52
  • $\begingroup$ I've updated the question. Hope that helps. $\endgroup$
    – gone
    Oct 24, 2014 at 13:57
  • $\begingroup$ @pbs Thanks, that's clearer. $\endgroup$
    – mdp
    Oct 24, 2014 at 14:13
  • $\begingroup$ Would $0$ count? $\endgroup$
    – Amad27
    Aug 11, 2015 at 19:58

2 Answers 2


$e^x$ is rational if and only if $x = \log a$ for $a \in \mathbb{Q}$. This is basically by definition since $\log x$ is defined to be the inverse of $e^x$: $e^x = a$ with $a$ rational $\iff$ $x = \log e^x = \log a$ with $a $ rational.

Similiarly, $e^x$ is transcendental if and only if $x = \log a$ for $a$ transcendental, by the same proof.


An immediate consequence of the Hermite-Lindemann Transcendence Theorem is that if $x$ is algebraic (which includes "rational") and $x\not =0$ then $e^x$ is transcendental.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.