Prove that the infinite $\sum_{\text{ p prime}}\frac{1}{2^p}$ is an irrational number.

Question

Prove that the infinite $\sum_{\text{ p prime}}\frac{1}{2^p}$ is an irrational number.

My progress:

• Suppose $$\omega = \sum_{\text{ p prime}}\frac{1}{2^p}= \frac{1}{2^2}+\frac{1}{2^3}+\dots$$
• Also let $f(x)= x^{th}$ prime

We will try to show that for $\omega$ and $\epsilon > 0,$ there is a positive integer $q$ and an integer $p$ such that $0 < |q\omega − p| < \epsilon.$

• Suppose $$\frac{p}{q}= \frac{1}{2^2}+\frac{1}{2^3}+\dots +\frac{1}{2^{f(n)}}$$
• Then we let $$q=2^{f(n)},~~p=2^{f(n)}\left(\frac{1}{2^2}+\frac{1}{2^3}+\dots +\frac{1}{2^{f(n)}}\right)$$
• So $$|q\omega − p|= |2^{f(n)}\omega - 2^{f(n)}\left(\frac{1}{2^2}+\frac{1}{2^3}+\dots +\frac{1}{2^{f(n)}}\right)|$$ $$= \frac{2^{f(n)}}{2^{f(n+1)}}+\frac{2^{f(n)}}{2^{f(n+2)}} +\dots$$ $$=\frac{1}{2^{f(n+1)-f(n)}}+\frac{1}{2^{f(n+2)-f(n)}}+\dots \le \frac{1}{2^{f(n+1)-f(n)}}+\frac{1}{2^{f(n+1)-f(n)+1}}+\frac{1}{2^{f(n+1)-f(n)+2}}\dots $$ $$= \frac{1/2^ {f(n+1)-f(n)}}{1-2^ {f(n+1)-f(n)}}=\frac{1}{2^ {f(n+1)-f(n)}-1}$$

Now, the ending which I think of is that the difference between $f(n+1)-f(n)$ can be very big. Then when we have $f(n+1)-f(n)$ to be very big, then $1/2^{f(n+1)-f(n)}< \epsilon .$ I was actually motivated by the proof of proving e irrational.

I am not sure about it. Any hints?

Answer 1

Since there are arbitrarily large gaps between consecutive primes ($n!+2$ to $n!+n$ are all composite), $\sum 2^{-p_i}$ has arbitrarily long strings of $0$'s in binary and so can not be periodic which means that it is not rational.

Answer 2

Here's a simpler proof:

By Dirichlet's Theorem on Arithmetic Progressions, for any coprime naturals $a,d$, the arithmetic progression $a+nd$ contains infinitely many primes (and conversely, $a+nd$ for $a,n,d\in\mathbb N$ can only be prime if $a,d$ are coprime). But any arithmetic progression contains infinitely many composites (let $n=ak$ for $k\in\mathbb N$).

This completes the proof, since for a number to be rational, its decimal representation in binary must eventually repeat. Letting $d$ be the period and $a$ some natural such that $a+nd$ contains at least one prime completes the proof.