<s>[INST] 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.
 [/INST] Another possibility is to use the following well-known fact:
$$\frac{a}{c} < \frac{b}{d} \Longrightarrow \frac{a}{c} < \frac{a+b}{c+d} < \frac{b}{d}$$
A simple consequence of this is that
$$\frac{a}{c} < \frac{b}{d} \Longrightarrow \frac{a}{c} < \frac{ma+nb}{mc+nd} < \frac{b}{d} \quad (m,n > 0) $$
This means that for your question, we simply have to find multiples of $10$ and $11$ which add to a power of 2. The simplest one would be $1(10) + 2(11) = 32 = 2^5$, which gives
$$
\frac{9}{10} < \frac{1(9) + 2(10)}{1(10) + 2(11)} < \frac{10}{11} \\
\color{white}{\text{empty}}\\
\frac{9}{10} < \frac{29}{32} < \frac{10}{11}
$$
Trying for $64$ or $128$ just yields fractions equivalent to the above, but we can use $19(10) + 6(11) = 256 = 2^8$ to find another one:
$$
\frac{9}{10} < \frac{19(9) + 6(10)}{19(10) + 6(11)} < \frac{10}{11} \\
\color{white}{\text{empty}}\\
\frac{9}{10} < \frac{231}{256} < \frac{10}{11}
$$
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