General Formula for Converting Repeating Decimals to Fractions

Suppose that the decimal number is

$x=a.d_1d_2\cdots d_m \overline{d_{m+1}\cdots d_{m+p}d_{m+p}}$,

where the $d_k$ are digits, $a$ is the integer part of the number, and the vinculum (overline) indicates the repeating part of the decimal. Then

$10^mx=ad_1d_2\cdots d_m.\overline{d_{m+1}\cdots d_{m+p}d_{m+p}}\tag{1}\label{eq1}$ 

and

$10^{m+p}x=ad_1d_2\cdots d_md_{m+1}\cdots d_{m+p}d_{m+p} .\overline{d_{m+1}\cdots d_{m+p}d_{m+p}}\tag{2}\label{eq2}$

Subtract (\ref{eq1}) from (\ref{eq2}) :

$10^{m+p}x-10^mx=ad_1d_2\cdots d_m d_{m+1}\cdots d_{m+p}d_{m+p}-ad_1d_2\cdots d_m$

$\Rightarrow (10^{m+p}-10^m)x=ad_1d_2\cdots d_m d_{m+1}\cdots d_{m+p}d_{m+p}-ad_1d_2\cdots d_m$

$\Rightarrow x=\frac{ad_1d_2\cdots d_m d_{m+1}\cdots d_{m+p}d_{m+p}-ad_1d_2\cdots d_m}{10^{m+p}-10^m}$

$\Rightarrow x=\frac{ad_1d_2\cdots d_m d_{m+1}\cdots d_{m+p}d_{m+p}-ad_1d_2\cdots d_m}{(10^p-1)10^m}$


$\Rightarrow x=\frac{ad_1d_2\cdots d_m d_{m+1}\cdots d_{m+p}d_{m+p}-ad_1d_2\cdots d_m}{\underbrace{99\dots 9}_{p\text{-times}}\underbrace{00\dots 0}_{m\text{- times}}}$

$x$ can be expressed as $x=a\frac{d_1d_2\cdots d_m d_{m+1}\cdots d_{m+p}d_{m+p}-d_1d_2\cdots d_m}{99\dots 900\dots 0}$

For Example,
Rewrite as a simplified fraction.

$31.5 \overline{6349}=?$

Solution :

Let $x=31.5 \overline{6349}$. Then 
$\Rightarrow x=31\frac{56349-5}{99990}$
$\Rightarrow x=31\frac{56344}{99990}$
$\Rightarrow x=31\frac{28172}{49995}$