Find the mystery fraction Question: There is a fraction with integer numerator and integer denominator, each smaller than $100$, whose decimal expansion begins with $0.11235\ldots$. Both the numerator and the denominator are positive. What is that fraction?
I suppose I could brute force this by testing every fraction whos numerator and denominator is less than 100, but that feels like cheating and I won't learn anything from doing it; So I'll try to solve it the way it's designed to be solved.
So far, I have tried to continue adding convenient values to the end of the decimal in an attempt to simply the fractional representation
to have both the numerator and denominator be less than 100. However, this has gotten me nowhere, and I don't see another way of approaching this problem. Can someone give me a hint to get started on the right path? Any and all help will be much appreciated.
 A: Without computer, you can try Continued fractions approach. You may learn the Continued fractions method here.
First $1/0.11235 \approx 8.901=8+0.9007$
Then $1/0.9007 \approx 1.10=1+0.11$
Then $1/0.11 \approx 9$
Then you can try $\frac{1}{8+\frac{1}{1+\frac{1}{9}}}=\frac{10}{89}=0.11235...$
You may to need to try and error a few time to find a suitable fraction. You do not know how many level of continued fractions give the answer you want.
A: Another way you can try is to use the fact that
$$\frac{a}{c} < \frac{b}{d} \Longrightarrow \frac{a}{c} < \frac{a+b}{c+d} < \frac{b}{d}$$
Start with two fractions that are respectively lesser and greater than the given decimal; since you want a fraction with numerator and denominator under 100, you should start with similar fractions, e.g. $\frac{11}{100}$ and $\frac{12}{100}$.
Apply the above property successively, each time using the newly generated fraction and the one of the previous two that would 'envelop' the given decimal.
$$\begin{align}
\frac{11}{100},\frac{12}{100} &\Longrightarrow \frac{23}{200}=0.115 \\
\\
\frac{11}{100},\frac{23}{200} &\Longrightarrow \frac{34}{300}=0.11\overline{3} \\
\\
\frac{11}{100},\frac{34}{300} &\Longrightarrow \frac{45}{400}=\frac{9}{80}=0.1125 \\
\\
\frac{11}{100},\frac{9}{80} &\Longrightarrow \frac{20}{180}=\frac{1}{9}=0.\overline{1} \\
\\
\frac{1}{9},\frac{9}{80} &\Longrightarrow \frac{10}{89}=0.11235... \\
\end{align}$$
You can start with $\frac{1}{10}$ and $\frac{3}{25}$ and get the same result, though it takes a few more iterations.
Update: at @Hagen von Eitzen's suggestion, starting with reciprocals $\frac{1}{k}$ and $\frac{1}{k+1}, k \in \mathbb{N}, (\frac{1}{8}$ and $\frac{1}{9}$ would work for your given decimal) could be advantageous; while it would take more iterations in this particular case, it also keeps the numerators and denominators lower.
A: The other answers have explained how to do this in general, but I can't help but mention another approach (see also e.g. here) based on the observation that those aren't just any digits:  they're the first five Fibonacci numbers.  The recurrence relation for Fibonacci numbers implies that if $x \approx 0.11235$, then
\begin{align*}
x + 10x & \approx 0.11235 + 1.1235 \\
& \approx 1.2358 \\
& \approx 11.235 - 10 \\
& \approx 100x - 10,
\end{align*}
and thus $89x \approx 10$ and $x \approx \frac{10}{89}$.
Of course if you define $x = \sum_{i=1}^{\infty} \frac{F_i}{10^i}$ where $F_i$ is the $i$-th Fibonacci number, then all of these approximations can be turned into equalities, so that $x = \frac{10}{89}$.
