# Find the fraction where the decimal expansion is infinite?

Find the fraction with integers for the numerator and denominator, where the decimal expansion is $0.11235.....$

The numerator and denominator must be less than $100$.

Find the fraction.

I believe I can use generating functions here to get $1+x+2x^2+3x^3+5x^4+.....$, but I do not know how to apply it.

• What's the next digit, $8$? – barak manos Dec 20 '14 at 20:21
• Good rational approximations can be found using partial fractions. – Mark Bennet Dec 20 '14 at 20:31
• @CameronBuie It was not given. – Math is Life Dec 20 '14 at 20:33
• @barakmanos: Not sure. It was not given. – Math is Life Dec 20 '14 at 20:34
• Is the fractional term supposed to be the simple stringing together of the Fibonacci numbers, i.e., 0.112358132134...? – Mico Dec 21 '14 at 8:56

Like you hint, we can observe that the value $$0.11235955056\ldots = \sum_{k = 1}^{\infty} 10^{-k} F_k,$$ where $$F_k$$ is the $$k$$th Fibonacci number (with the convention that $$F_0 = 0$$, $$F_1 = 1$$), is simply the value of the series $$F(x) := \sum_{k = 1}^{\infty} F_k x^k = x + x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 + 13x^7 + \cdots$$ at $$x = \frac{1}{10}$$.

Hint Using the defining recurrence relation $$F_{k + 2} = F_{k + 1} + F_k$$ (and the above convention) gives that $$F(x)$$ satisfies $$F(x) = x + x F(x) + x^2 F(x).$$

Rearranging gives that on the open interval of convergence of the series, which turns out to be $$(-1/\phi, 1/\phi)$$---where $$\phi$$ is the Golden Ratio, and which in particular contains the value $$\frac{1}{10}$$ of interest)---we have $$F(x) = \frac{x}{1 - x - x^2}.$$ Thus, the series has value $$F\left(\tfrac{1}{10}\right) = \frac{\left(\frac{1}{10}\right)}{1 - \left(\frac{1}{10}\right) - \left(\frac{1}{10}\right)^2} = \color{#bf0000}{\boxed{\frac{10}{89}}} .$$ Since this is a rational number, its decimal expansion repeats: $$0.\overline{1123595505617977528089887640449438202247191}.$$

• @FlybyNight yes, that behavior is obvious and expected -- 1, 1, 2, 3, 5, 8, 13 and then the 8 and 13 overlap to give a 9 in the decimal expansion in place of the 8. – oldrinb Dec 20 '14 at 21:22

Continued fractions: \begin{align} \frac{11235}{100000}&=\frac{2247}{20000}\\ &=\cfrac{1}{8+\cfrac{2024}{2247}}\\ &=\cfrac{1}{8+\cfrac{1}{1+\cfrac{223}{2024}}}\\ &=\cfrac{1}{8+\cfrac{1}{1+\cfrac{1}{9+\cfrac{1}{13+\cfrac{2}{17}}}}}\\ &=\cfrac{1}{8+\cfrac{1}{1+\cfrac{1}{9+\cfrac{1}{13+\cfrac{1}{8+\cfrac{1}{2}}}}}}\\ &=[0;8,1,9,13,2] \end{align} The first convergent is $1/8$; the second convergent is $$\cfrac{1}{8+1}=\frac{1}{9};$$ the third convergent is $$\cfrac{1}{8+\cfrac{1}{1+\cfrac{1}{9}}}=\cfrac{1}{8+\cfrac{9}{10}}=\frac{10}{89};$$ The fourth convergent can be computed to $131/1166$ (see this continued fraction calculator)

An exhaustive search (by computer) of all fractions with numerator and denominator $< 100$ shows there is only one whose decimal representation starts 0.11235: $$\frac{10}{89}$$

The decimal expansion continues $\ldots 955056179775\ldots$ (I haven't bothered finding aprogram that can calculate enough that we can see the period), that's completely impossible to guess- I would call the question poorly worded.

• To save anyone from having to reach for a calculator, the first many digits of the expansion goes $0.1123595505617977\dots$. The question of if there is a better method than brute force remains. We could clearly see that for it to work for some $\frac{p}{q}$, we would need $8.5p < q<9p$ and so $p\leq 11$, reducing cases to test to less than 18 – JMoravitz Dec 20 '14 at 20:37
• @Henrik: I am guessing that the decimal representation would represent Fibonacci numbers, but I'm not sure.. – Math is Life Dec 20 '14 at 20:39
• The period is $44$ since $m = 44$ is the smallest positive integer such that $10^m \equiv 1 \pmod{89}$. WolframAlpha gives the decimal expansion to enough digits. – JimmyK4542 Dec 20 '14 at 20:40
• @MathisLife: I would be surprised if the decimal representation of the Fibonacci numbers were rational. – Cameron Buie Dec 20 '14 at 20:40

These are fractions $f$ with $11235/100000\le f<11236/100000.$ A direct, if slow, method for counting these would be $$\sum_{d=1}^{100}\left\lfloor\frac{100000d}{11235}\right\rfloor-\left\lfloor\frac{100000d}{11236}\right\rfloor$$

or

sum(d=1,100,100000*d\11235-100000*d\11236)