<s>[INST] 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?
 [/INST] 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.
</s>