Show that $e^{\sqrt 2}$ is irrational

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.

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

• 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 $e^{\sqrt{n}}$ – reuns Sep 28 '16 at 14:23
• @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. – Jack D'Aurizio Sep 28 '16 at 14:25