Find the mystery fraction

Question

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.

Answer 1

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.

Answer 2

This is a good problem for continued fractions. To see how they work for this kind of problem, I will give another example:

Find a fraction with lowest denominator between $0.70$ and $0.72$

Let us work out the continued fractions for $0.70=7/10$ and $0.72=18/25$. For $7/10$ we start with

$\dfrac{7}{10}=0+\dfrac{7}{10}$

where $0$ is the integer part and $7/10$ is the remainder. We next take the reciprocal of this remainder:

$\dfrac{7}{10}=0+\dfrac{1}{\dfrac{10}{7}}$

and now split $10/7$ into its integer part and its remainder. This yields

$\dfrac{7}{10}=0+\dfrac{1}{1+\dfrac{3}{7}}$

where the remainder is now $3/7$. Going around once more we have

$\dfrac{7}{10}=0+\dfrac{1}{1+\dfrac{1}{\dfrac{7}{3}}}=0+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3}}}$

When the remainder is $1$ divided by a whole number, in this case $1/3$, the process terminates because the next reciprocal will be that whole number exactly, with remainder zero. So we have the last expression above as our continued fraction for $7/10$. We write this in shorthand as $[0,1,2,3]$.

Now try it with $18/25$. If you use the above procedure correctly you should get

$\dfrac{18}{25}=0+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{3}}}}=[0,1,2,1,3].$

Now compare your two continued fractions. The first three integer parts $0,1,2$ are the same but then you have $3$ for one fraction and $1+1/3$ for the other. So now, you replace these parts with the smallest whole number lying in- between, which is $2$:

$[0,1,2,2]=0+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{2}}}.$

Then simplify this last fraction, working from the bottom up:

$[0,1,2,2]=0+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{2}}}=0+\dfrac{1}{1+\dfrac{1}{\dfrac{5}{2}}}=0+\dfrac{1}{1+\dfrac{2}{5}}=0+\dfrac{5}{7},$

meaning $5/7\approx0.714$ is the required minimal denominator fraction between $0.70$ and $0.72$.

Once you are confident with this example, you can try the larger numerator and denominator involved with this problem:

$0.11235=11235/100000=2247/20000$

and

$0.11236=11236/100000=2809/25000.$

$0.11235=[0,8,1,9,...]$ and $0.11236=[0,8,1,8,...]$, so the required fraction will be $[0,8,1,9]$, which simplifies to $10/89$.

Answer 3

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.

Answer 4

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 $x = \frac{10}{89}$.