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

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.
