Find a fraction given the repeating binary expansions

Question

I can do binary expansion given a fraction just fine but the question I'm stuck on says: Find fractions for the numbers with the following binary expansion: (i) $0.00\overline{110}$ and (ii) $0.0\overline{0110}$

So I know the definition for a binary expansion is as follows: The binary expansion of a number $x_0 \in [0,1)$ is a sequence $s_0,s_1,s_2,...$ in which each $s_k$ is either $0$ or $1$ and $x_0 = \sum_{k=0}^{\infty} \frac{s_k}{2^{k+1}}$

So I tried my hand at it and I got an answer close to what I need but it's not quite there.

I figured since there is a $1$ in the $\frac{1}{8}$ and $\frac{1}{16}$ decimal place I would do $\sum_{k=1}^{\infty} \frac{1}{8^k}+\frac{1}{16^k}$ which solves to $\frac{1}{7} + \frac{1}{15} = \frac{22}{105}$

When I checked my work I got $\frac{22}{105} = 0.001101011010001...$ for its binary expansion and this is very close to what I need so I felt like I am on the right track of what to do I just messed up somewhere.

Thanks in advance for any help!

Answer 1

The rule is the same as for decimal repeating numbers, just replace $9$ with $1$; so $$ 0.00\overline{110}=\frac{110-0}{11100}=\frac{11}{1110} $$ and, in decimal, $3/14$. Similarly, $$ 0.0\overline{0110}=\frac{110-0}{11110}=\frac{11}{1111} $$ which, in decimal, is $3/15=1/5$. Note that $0.00\overline{110}=0.0\overline{011}$ and $0.0\overline{0110}=0.\overline{0011}$.

With bc I get, just for rough confirmation,

bc 1.06
Copyright 1991-1994, 1997, 1998, 2000 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type `warranty'. 
obase=2
3/14
.0011011011011011011011011011011011011011011011011011011011011011010
1/5
.0011001100110011001100110011001100110011001100110011001100110011001

Answer 2

Let $S=0.00\overline{110}_2$. Then $8S = 1.10\overline{110}_2$. Subtracting $8S - S$ gives us $7S = 1.1_2$.

Can you solve it from there?