Finding a rational number in between two rationals Can anyone give me some hints or show me how to solve this problem?
Problem Find a rational number in between $\frac{9}{10}$ and $\frac{10}{11}$ which may be written in the form $\frac{m}{2^n}$ where $m$ is an integer and $n$ is a non negative integer.
I just do not know where to even start or what theorems I need to use to approach this. I would appreciate hints or guidance to solving these types of problems.

I'll show you a method that works for any base. I'll do the question but with base $5$ instead of $2$, and leave base $2$ as an exercise.
$\frac{10}{11} - \frac{9}{10} = \frac{100}{110} - \frac{99}{110} = \frac{1}{110}$, i.e. the distance between $\frac{10}{11}$ and $\frac{9}{10}$ is $\frac{1}{110}$.
$\frac{1}{5^2} > \frac{1}{110}\ $, but $\frac{1}{5^3} < \frac{1}{110}$.
So $ \exists k \in \mathbb{Z}$ such that $\frac{k}{5^3} \in (\frac{10}{11}, \frac{9}{10})$, otherwise $\exists j \in \mathbb{Z}\ $ such that both $\frac{j}{5^3} < \frac{99}{110}$ and $\frac{j+1}{5^3} > \frac{100}{110}$, contradicting the fact that $\frac{1}{5^3} < \frac{1}{110}$.
$\frac{9}{10} = \frac{112.5}{125}\ \text{and} \ \frac{10}{11} = \frac{113.6...}{125} ,\ \therefore \frac{113}{125} \ $ does the trick.
