Finding a rational number in between two rationals

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:

$$ \frac{9}{10} < \frac{19(9) + 6(10)}{19(10) + 6(11)} < \frac{10}{11} \\ \color{white}{\text{empty}}\\ \frac{9}{10} < \frac{231}{256} < \frac{10}{11} $$

Express the fractions in binary:

$ \frac{9}{10} = 0.111001100\cdots{}_2 $

$ \frac{10}{11} = 0.111010001\cdots{}_2 $

So try $ 0.11101000{\color{red}0}{}_2 = \frac{29}{32} $.

This is the solution with smallest denominator.

To keep the arithmetic simple, let's look for a number of the form $\frac{m}{2^n}$ between $\frac{1}{11}$ and $\frac{1}{10}$. Then just subtract it from $1$ for your answer. So we have: $$\frac{1}{11}<\frac{m}{2^n}<\frac{1}{10}$$ which is the same as $$10m<2^n<11m$$ Now look for $m$ such that the interval $(10m,11m)$ contains a power of $2$. You don't have to look for long.