I was reading this proof

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

I could understand until the last step.


"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."

I thought the contradiction would come up if you took the limit of $n \to \infty$, in the last inequality, but taken the limit doesn't help, as the left side goes to $0$, the middle goes to "$\infty$"$\cdot$$(p-q(\cosh \sqrt2))$, and the rigth side goes to $0$.

Also, I think the right member of the inequality is missing a $q$, so I rewrote as

$0<p-qs_n<\frac{2^n}{(2n)!} \frac{2q}{(2n+1)(2n+2)}\frac{(2n+3)^2}{(2n+3)^2-2}$

And took the limit again, as $n$ aproaches infinity, so the left side goes to $0$, the right side goes to $0$, and the middle goes to $p-q(\cosh \sqrt2)$. I need to make sure that $p-q(\cosh \sqrt2)$ is greater than $0$ ? I don't understand the last step on the proof. If someone could explain to me I'll be thankful.


The thing at the center of the inequality $$ 0<p\frac{(2n)!}{2^n}-qs_n\frac{(2n)!}{2^n}<\frac{2q}{(2n+1)(2n+2)}\frac{(2n+3)^2}{(2n+3)^2-2}\tag1 $$ is a natural number (i.e., a positive integer) for every $n$. But the RHS tends to $0$ as $n\to\infty$, which means the RHS is less than $1$ for large enough $n$. The contradiction is that you've now identified a natural number between $0$ and $1$.

EDIT: To see why the central entity $$p\frac{(2n)!}{2^n}-qs_n\frac{(2n)!}{2^n}$$ is an integer, note first that $(2n)!/2^n$ is an integer for each $n$ (an induction proof works). It follows that $s_n(2n)!/2^n$ is also an integer, since $$ s_n \frac{(2n)!}{2^n}=\sum_{k=0}^n \frac{(2n)!}{2^n}\frac{2^k}{(2k)!}=\sum_{k=0}^n{2n \choose 2k}\frac{(2(n-k))!}{2^{n-k}}. $$ Since $p$ and $q$ are assumed integers, the claim follows.

  • $\begingroup$ So there is a natural and finite $n$ that makes the RHS less than one (because the limit goes to $0$ as $n \to \infty$), and the same $n$ makes the center a natural number? which implies a natural number greater than $0$ and smaller than $1$ ? $\endgroup$ – Pinteco Apr 7 '18 at 5:13
  • $\begingroup$ Also, how me make sure the center is always a natural number? because $s_n$ can be a rational number, so maybe the center can be a rational bewteen $0$ and $1$. $\endgroup$ – Pinteco Apr 7 '18 at 5:36
  • $\begingroup$ You already had $0<p-qs_n<\frac{2^n}{(2n)!} \frac{2q}{(2n+1)(2n+2)}\frac{(2n+3)^2}{(2n+3)^2-2}$ and you accepted $p-q s_n$ to be an integer. The RHS also tends to 0. Thus everything works. $\endgroup$ – Jens Schwaiger Apr 7 '18 at 7:33
  • $\begingroup$ But $s_n$ tends to $\cosh \sqrt 2$ as $n \to \infty$. So $p-q s_n$ can be not a integer. So what I'm trying to understand is how I can make sure that $p-qs_n$ is always an integer. $\endgroup$ – Pinteco Apr 7 '18 at 20:46
  • $\begingroup$ I think there is naturals $p,q,n$ that makes $p-qs_n$ not natural. $\endgroup$ – Pinteco Apr 7 '18 at 21:01

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