We define the set
$S=\{x: \exists c>0 \text{ such that}\ |x-j/2^k| \geq c/2^k, \ \forall j\in\mathbb{Z},\forall k\in \mathbb{N}_0 \}$,
where $N_0$ is the set of natural numbers with $0$. Clearly, this set does not contain any dyadic rational, namely a rational of the form $j/2^k$.
I want to prove that this set contains for example the number $1/3$. After doing some experiments I found out that for all $j \in \mathbb{Z}, k \in \mathbb{N}_0$ it should be true that
\begin{equation} |1/3 - j/2^k| \geq 1/(3\cdot 2^k) \end{equation} I tested the above inequality for pretty large $k,j$ using code.
How can I prove this inequality for all $j, k$ ?
Moreover, I think all rationals of the form $a/b$ so that $b$ contains a prime factor other than $2$ should belong to the set $S$. How can I show this claim?
If $2^{k}-1 <3j < 2^{k}+1$ then $3j=2^{k}$ which is a contradiction. Hence either $2^{k}-1 \geq 3j$ or $3j \geq 2^{k}+1$ . In the first case $\frac j {2^{k}} \leq \frac 1 3 -\frac 1 {32^{k}}$ and in the second case $\frac j {2^{k}}\geq \frac 1 3 +\frac 1 {32^{k}}$. Your inequality follows from these.