Is the set $A$ the same as the set of rational numbers?

Question

$$A = \bigcup_{n=1}^\infty \left\{\frac{a}{2^n} : a \in \Bbb Z \right\}$$

How do I prove or disprove that an arbitrary rational eg. $\frac{2}{7} \in A$

I cannot determine whether an arbitrary rational belongs to the set. I tried by determining which power of $2$ is divisible by $7$ without any result.

So is $A = \Bbb Q$ ? I am asking this because apparently $A$ is dense in $\Bbb R$.

Answer 1

If $\frac{1}{3}\in A$, then there exists $a\in\mathbb{Z}$ and $n\in\mathbb{N}$ such that $\frac{1}{3}=\frac{a}{2^n}.$ But then $2^n=a\times 3$, and thus $3$ divides $2^n$. Since $3$ is a prime number, then by Euclid's lemma it divides $2$, which is false.

Answer 2

Note that the set $A$ does NOT include all rational numbers.

As a paradigm, take $\cfrac{1}{3}$.

Even if $\cfrac{a}{2^n}$ is reducible, $2^n$ never becomes $3$ after any cancellation (since $3$ does not divide $2^n$). So, indeed, $\cfrac{1}{3} \not \in A$