How to rationalize non-terminating binary number?

Question

I would like to know the way to rationalize non terminating but periodic binary numbers, such as (0.101010...) or (0.100111 100111...). I’m thinking of grouping the periodic numbers together...

Answer 1

Same we would do it for base 10.

Multiply by $2^k$ where $k$ is the period. ($0.100111100111100.... *2^6=100111.100111100.....$)

Subtract $2^kn - n$ to get terminating string. $2^6*0.100111100111100...-0.100111100111100.... = 100111.100111100..... -0.100111100111100.... = 100111$)

And divide by $2^k - 1$. $n = 0.100111100111100.... = \frac {100111}{2^6- 1} = \frac {100111}{111111}$.

(in this case $\frac {39}{63} = \frac {13}{21} = \frac 12 + \frac {15}{128} + \frac {15}{128*64} + ..... $)

Answer 2

For instance, if you want to rationalize $x=0,101010\ldots$, note that$$100x=10,101010\ldots=10+x$$and therefore $100x-x=10$. But$$100x-x=10\iff11x=10\iff x=\frac{10}{11}.$$

Answer 3

Suppose the number is $x=0.b_1 b_2 \cdots$, with $b_k \in \{0,1\}$.

If it is periodic, then there is some $n,p$ such that $(b_n,\cdots ,b_{n+p-1}) = (b_{n+mp},\cdots ,b_{n+(m+1)p-1}) $ for all $m \ge 0$.

Then $x= \sum_{k<n} b_k {1 \over 2^k} + \sum_{m\ge 0}{1 \over 2^{mp}} \sum_{j=n}^{n+p-1}b_{j} {1 \over 2^j} = \sum_{k<n} b_k {1 \over 2^k}+ {1 \over 1-{1 \over 2^{mp}}} (\sum_{j=n}^{n+p-1}b_{j} {1 \over 2^j} )$.