I'm trying to prove that $e^{\sqrt 2}$ is irrational. My approach: $$ e^{\sqrt 2}+e^{-\sqrt 2}=2\sum_{k=0}^{\infty}\frac{2^k}{(2k)!}=:2s $$ Define $s_n:=\sum_{k=0}^{n}\frac{2^k}{(2k)!}$, then: $$ s-s_n=\sum_{k=n+1}^{\infty}\frac{2^k}{(2k)!}=\frac{2^{n+1}}{(2n+2)!}\sum_{k=0}^{\infty}\frac{2^k}{\prod_{k=1}^{2k}(2n+2+k)}\\<\frac{2^{n+1}}{(2n+2)!}\sum_{k=0}^{\infty}\frac{2^k}{(2n+3)^{2k}}=\frac{2^{n+1}}{(2n+2)!}\frac{(2n+3)^2}{(2n+3)^2-2} $$ Now assume $s=\frac{p}{q}$ for $p,q\in\mathbb{N}$. This implies: $$ 0<\frac{p}{q}-s_n<\frac{2^{n+1}}{(2n+2)!}\frac{(2n+3)^2}{(2n+3)^2-2}\iff\\ 0<p\frac{(2n)!}{2^n}-qs_n\frac{(2n)!}{2^n}<\frac{2}{(2n+1)(2n+2)}\frac{(2n+3)^2}{(2n+3)^2-2} $$ But $\left(p\frac{(2n)!}{2^n}-qs_n\frac{(2n)!}{2^n}\right)\in\mathbb{N}$ which is a contradiction for large $n$. Thus $s$ is irrational. Can we somehow use this to prove $e^\sqrt{2}$ is irrational?


Since the sum of two rational numbers is rational, one or both of $e^{\sqrt{2}}$ and $e^{-\sqrt{2}}$ is irrational. But, $e^{-\sqrt{2}}=1/e^{\sqrt{2}}$, and hence both are irrational.

  • $\begingroup$ Uhhh, so close! Thanks for the answer, +1 $\endgroup$ – Redundant Aunt Sep 18 '15 at 17:10
  • 3
    $\begingroup$ You did all the hard work :-) $\endgroup$ – parsiad Sep 18 '15 at 17:11
  • 7
    $\begingroup$ Nice. Here is an easier one :If A,B,C are rational and $Ae^2 +Be+C=0$ then A=B=C=0. $\endgroup$ – DanielWainfleet Sep 18 '15 at 23:21
  • $\begingroup$ Why can't both numbers be rational? $\endgroup$ – DVD Sep 24 '15 at 4:09
  • $\begingroup$ @DVD: OP proves it in his post $\endgroup$ – parsiad Sep 24 '15 at 14:18

$e^{\sqrt{2}}$ is transcendental because of Lindemann–Weierstrass theorem:

If $a\neq 0$ is algebraic, then $e^a$ is transcendental.

It is written in the list of transcendental numbers.


To prove that $e^{\pm\sqrt{n}}\not\in\mathbb{Q}$, it is enough to exploit the Gauss' continued fraction for $\tanh$ and Lagrange's theorem about the periodicity of continued fractions of quadratic irrationals.
A detailed explanation of this approach (for $n=2$) is outlined in this answer.

  • $\begingroup$ you might want to look at this method math.stackexchange.com/questions/1864678/… here $\sum_{k=0}^\infty \frac{2^k}{(2k)!} = \sum_{n=1}^\infty \frac{1}{b_n}$ where $b_1 = 2, b_{n+1} = b_n \frac{n (n+1)}{2}$, and the same proof applies. But I'm not sure how to modify it for $e^{\sqrt{n}}$ $\endgroup$ – reuns Sep 28 '16 at 14:23
  • $\begingroup$ @user1952009: to exploit some Taylor expansion is another way for providing tight rational approximations of such numbers (then prove their irrationality) but I somewhat prefer the CF approach since it gives more manageable and more accurate approximations. $\endgroup$ – Jack D'Aurizio Sep 28 '16 at 14:25

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.