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.
 A: 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:
$$
\frac{9}{10} < \frac{19(9) + 6(10)}{19(10) + 6(11)} < \frac{10}{11} \\
\color{white}{\text{empty}}\\
\frac{9}{10} < \frac{231}{256} < \frac{10}{11}
$$
A: 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.
A: 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.
A: 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.
A: You are looking for integers $m,n$ such that $$\frac9{10}<\frac m{2^n}<\frac{10}{11}$$
or
$$\frac9{10}2^n<m<\frac{10}{11}2^n.$$
That means that the left and right expression must have different integer parts,
$$\left\lfloor\frac9{10}2^n\right\rfloor\ne\left\lfloor\frac{10}{11}2^n\right\rfloor.$$
We will be on the safe side if the numbers themselves differ by at least one unit,
$$\frac9{10}2^n\le\frac{10}{11}2^n-1,$$
which gives us
$$2^n\ge\frac1{\dfrac{10}{11}-\dfrac9{10}}=110.$$
So a solution is $n=7,m=116$. Note that this is not the tightest, as $n=5,m=29$ also works.
A: Hint
Some ingredients:

*

*$1/2^n$ is positive and can be as small as you want providing $n$ is large enough.

*Then $1/2^n, 2/2^n, ...$ is an increasing sequence with constant difference between two terms.

*If an interval, like $(9/10,10/11)$ is of length larger than $1/2^n$, then one element of the sequence above has to belong to it.

*You should be done!
Note: making a drawing of all that will help you a lot to understand what happens.
