Converting repeating decimal to fraction How do I convert $0.297$ to a fraction, if the $2$ and $9$ are repeating? The non-repeating number is in the middle, so I am not sure how to proceed from here. 
 A: Let me give you a general idea:
Let $x = I.n_1n_2\ldots n_t r_1r_2\ldots r_k r_1 r_2 \ldots$
Where $I$ is the integer part of $x$ i.e. $I = \lfloor x \rfloor$. The $n_i$ are the numbers before the repeating block and $r_j$ are the numbers in the repeating block.
Then:
$$10^tx =  In_1n_2\ldots n_t. r_1r_2\ldots r_k r_1 r_2 \ldots$$
(Now the decimal part is all repeating).
Note that the decimal part now is:
$$r_1·10^{-1}+r_2·10^{-2} + \ldots + r_k·10^{-k} + r_1·10^{-(k+1)} + r_2·10^{-(k+2)} + \ldots =\\ = r_1\left(10^{-1} + 10^{-(k+1)} + \ldots\right) + r_2\left(10^{-2} + 10^{-(k+2)} + \ldots\right) + \ldots +r_k\left(10^{-k} + 10^{-2k} + \ldots\right) = r_1·10^{-1}\left(1 + 10^{-k} + \ldots\right) + r_2·10^{-2} \left(1 + 10^{-k} + \ldots\right) + r_k·10^{-k}\left(1 + 10^{-k} + \ldots\right) = 10^{-k}[r_1r_2\ldots r_k]\frac{1}{1-10^k} = \frac{[r_1r_2\ldots r_k]}{\underbrace{9\cdots9}_{k}\underbrace{0\cdots0}_{k}}$$
Where $[a_1a_2\ldots a_n]$ means concatenating the digits $a_j$.
Now we want to add up the integer part of $10^t x$:
$$10^tx = [In_1n_2\ldots n_t] + \frac{[r_1r_2\ldots r_k]}{\underbrace{9\cdots9}_{k}\underbrace{0\cdots0}_{k}} = \frac{(10^{2k}-10^k)[In_1n_2\ldots n_t]+[r_1r_2\ldots r_k]}{\underbrace{9\cdots9}_{k}\underbrace{0\cdots0}_{k}}$$
Therefore:
$$x = \frac{(10^{2k}-10^k)[In_1n_2\ldots n_t]+[r_1r_2\ldots r_k]}{\underbrace{9\cdots9}_{k}\ \underbrace{0\cdots0}_{k+t}}$$
A: $0.297292929...=0.297292929...\times \frac{1000}{1000}=\large\frac{297.2929...}{1000}$
$=297+\large\frac{29}{100}+\frac{29}{10000}+\frac{29}{1000000}+...$
$=297+29(\large\frac{1}{100}+\frac{1}{10000}+...)$
Sum of the bracket terms in GP$ = \large\frac{1}{100} \left[ \frac{1-(\frac{1}{100})^n}{1-\frac{1}{100}} \right] $
When $n \rightarrow \infty$, Sum = $\large\frac{1}{99}$
$\therefore 0.2972929...= ( 297+\large\frac{29}{99})\frac{1}{1000}$
$\therefore 0.2972929...=\large\frac{29432}{99000}$
A: Note that parenthesized "phrases" should be considered to repeat ad infinitum.  
$.297(29)=.(92)-.632$.  
$.(92)={{92}\over{{10}^2}-1}={{92}\over{99}}$.  
$.632={{632}\over{1000}}={{79}\over{125}}$.  
So, the sought-after fraction $={{92}\over{99}}-{{79}\over{125}}=
{{92·125-79·99}\over{99·125}}={{11500-7821}\over{12375}}={{3679}\over{12375}}$.
A: If $n = 0.2972929\ldots$ then $1000n = 297.2929\ldots$ and $100000n = 29729.29\ldots$ Hence:
$$\begin{eqnarray*}
100000n - 1000n &=& (29729.29\ldots) - (297.2929\ldots) \\ \\
99000n &=& 29432 \\ \\
n &=& \frac{29432}{99000} = \frac{3679}{12375}
\end{eqnarray*}$$
