Let $x=\sum_{k=1}^{\infty }a_{k}3^{-k}, \ a_{k}=0,1,2,$ then $f(x)=\sum_{k}b_{k}2^{-k}$, were $b_{k}=a_{k}/2$ if $a_{i}=0 \ or \ 2,i\le k$, $b_{k}=1$ if $a_{k}=1,a_{i}\not =1,i<k$ and if $a_{k}=1$, then $b_{i}=0 \ \forall i>k$.. It is easy to proof by induction. But I don't calculate $x=\frac{18}{1991}=\frac{18*(3^{90}-1)/1991}{3^{90}-1}$. These representation give expressions $a_{k}\ \forall k$. Because $a_{1}=a_{2}=a_{3}=a_{4}=0=a_{6},a_{5}=2,a_{7}=1$ we get $f(\frac{18}{1991}=2^{-5}+2^{-7}=\frac{5}{128}.$

Jutaro wrote: Let $f: \ [0,\ 1] \rightarrow \mathbb{R}$ be an increasing function satisfying the following conditions: a) $f(0) = 0$; b) $f\left(\frac {x}{3}\right) = \frac {f(x)}{2}$; c) $f(1 - x) = 1 - f(x)$. Determine $f\left(\frac {18}{1991}\right)$. We have : $f(1) = f(1 - 0) = 1 - f(0) = 1$ $f(\frac {1}{3}) = \frac {f(1)}{2} = \frac {1}{2} = > f(\frac {2}{3}) = \frac {1}{2}$ $= > f(\frac {1}{9}) = \frac {1}{4} = > f(\frac {8}{9}) = \frac {3}{4}$ Similar , we also have : $f(\frac {1}{27}) = f(\frac {2}{27}) = \frac {1}{8} ; f(\frac {7}{27}) = f(\frac {8}{27}) = \frac {3}{8}$ $f(\frac {19}{27}) = f(\frac {20}{27}) = \frac {5}{8} ; f(\frac {25}{27}) = f(\frac {26}{27}) = \frac {7}{8}$ Because f is the decreasing on $[0;1]$ , we have : $f(x) = \frac {1}{2}$ for all $x \in [\frac {1}{3};\frac {2}{3}]$ $f(x) = \frac {1}{4}$ for all $x \in [\frac {1}{9};\frac {2}{9}]$ .... $f(x) = \frac {7}{8}$ for all $x \in [\frac {25}{27};\frac {26}{27}]$ We have : $\frac {19}{3^7} < \frac {18}{1991} < \frac {20}{3^7}$ But , $f(\frac {19}{27}) = \frac {5}{8} = > f(\frac {19}{3^4}) = \frac {5}{16} = > ... = > f(\frac {19}{3^7}) = \frac {5}{128}$ $f(\frac {20}{27}) = \frac {5}{8} = > f(\frac {20}{3^4}) = \frac {5}{16} = > ... = > f(\frac {20}{3^7}) = \frac {5}{128}$ Because f is the decreasing on $[0;1]$ , we have : $f(\frac {18}{1991}) = \frac {5}{128}$