TITLE: Repeating Decimals
QUESTION [5 upvotes]: I'm just wondering how do we simplify repeating decimals into a fraction in general?
Like, for example,
$$0.5656\dots$$
$$0.12424\dots$$
$$4.23777\dots$$
Thanks!

REPLY [2 votes]: Let's say your number have a form
$$
N = I_1I_2\ldots I_n.D_1D_2\ldots D_m \left(P_1 P_2 \ldots P_k\right)
$$
where $I_1I_2\ldots I_n$ is an integer part, $D_1D_2\ldots D_m$ is non repeating part of the decimal, and $P_1P_2\ldots P_k$ is repeating part of the decimal. So, it's most general form of such number. For example
$$
N = 455.01102(979665) = 455.01102\ 979665\ 979665\ 979665 \ldots
$$
so
$$
I_1 = 4;\quad I_2 = 5; \quad I_3 = 5 \\
D_1 = 0; \quad D_2 = 1; \quad D_3 = 1; \quad D_4 = 0; \quad D_5 = 2 \\
P_1 = 9; \quad P_2 = 7;\quad P_3 = 9; \quad P_4 = 6; \quad P_5 = 6; \quad P_6 = 5
$$
Then, you can rewrite is as
$$
N = I_1I_2\ldots I_n.D_1D_2\ldots D_m + 0.\underbrace{000\ldots 0}_{\text{$m$ times}}\left(P_1 P_2 \ldots P_k\right) = \\
= I_1I_2\ldots I_n.D_1D_2\ldots D_m + 10^{-m} \times 0.\left(P_1 P_2 \ldots P_k\right)
$$
So, you really have to consider $P = 0.\left(P_1 P_2 \ldots P_k\right)$ part only.
As it was suggested, multiply the whole thing to $10^k$ and subtract initial number, so
$$
10^kP-P = \underbrace{999\ldots 9}_{\text{$k$ times}}P = P_1P_2 \ldots P_k.(P_1P_2\ldots P_k)-0.(P_1P_2\ldots P_k) = P_1P_2\ldots P_k$$
so you can find now
$$
P = \frac {P_1P_2\ldots P_k}{\underbrace{999\ldots 9}_{\text{$k$ times}}}
$$
and finally
$$
N = I_1I_2\ldots I_n + \frac {D_1D_2\ldots D_m}{10^m} + \frac {P_1P_2\ldots P_k}{10^m\underbrace{999\ldots 9}_{\text{$k$ times}}}
$$