In a regular, positional number system (like our decimal numbers), every rational number ends in a repeating sequence (even if it's just 0). For example, $\frac{1}{3}$ in base 10 is $0.33333...$, in base 5 it's $0.131313...$, and in base 3 it's just $0.1$.
A less common number system uses the Fibonacci sequence as its base, so the first few digit places represent $1, 2, 3, 5, 8, 13, 21$, and so on (instead of the decimal $1, 10, 100, 1000,$ etc.). In this system:
$$ \begin{aligned} 17_{10} &= 100101_F \\ 40_{10} &= 10001001_F \end{aligned} $$
This can easily be extended to digits after the radix point, so $0.1_F = \frac{1}{2}_{10}$, $0.01_F = \frac{1}{3}_{10}$, $0.001_F = \frac{1}{5}_{10}$, and so on.
$$ \begin{aligned} \frac{5}{6}_{10} &= 0.11_F \\ \frac{7}{10}_{10} &= 0.101_F \end{aligned} $$
Some numbers aren't as easy to write. Greedily adding up the first places that are smaller than the remainder of the number we're trying to write: $$ \frac{1}{4} = \frac{1}{5}+\frac{1}{21}+\frac{1}{610}+\frac{1}{1597}... = 0.001001000000101..._F \\ \frac{2}{3} = \frac{1}{2}+\frac{1}{8}+\frac{1}{34}+\frac{1}{89}... = 0.10010010100001..._F $$
Do all rational numbers end in a repeating sequence in Fibonacci coding?
(For the specific case of $\frac{1}{4}$, see here.)