# Is $2^n = p \cdot m$ solvable?

When I divide the scale between 0 and 1 continuously in halfs, are then all rational numbers (<1) covered? I.e. can every rational number $p/q (p<q)$ be written as $m/2^n$ for some $m,n \in \mathbb{N}$?

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No. For example, $1/3$ cannot be so written. Suppose that $\dfrac13 = \dfrac{m}{2^n}$ for some integers $m$ and $n$; then $3m=2^n$. But $2^n$ is not divisible by $3$ for any $n$.
The numbers that can be written in the form $\dfrac{m}{2^n}$ are called dyadic rationals.
HINT $\$ The lowest terms representation of a fraction is unique. In lowest terms, your dyadic fractions have denominator of form $\rm\:2^n\:.\:$ But there are lowest term fractions not of that form.