Show that a number belongs or does not belong to a Cantor set I have trouble showing that $\frac{10}{31}$ and $\frac{45}{69}$ belong/do not belong to the Cantor set. I tried to do the following
$$\sum_{n = x}^\infty \frac{2}{3^n} = \frac{2}{3^x} + \frac{2}{3^{(x + 1)}} + \frac{2}{3^{(x + 2)}} + \dots = \frac{2}{3^x} \left[ 1 + \frac{1}{3^1} + \frac{1}{3^2} + \dots \right] = \frac{2}{3^x} \sum_{k = 0}^\infty \frac{1}{x^k} =  \frac{2}{3^x} \times \frac{1}{1- \frac{1}{3}} = \frac{1}{3^{x- 1}}$$
and then plug different values of $x = 1, 2, \dots$, but I could not seem to get these numbers. Does it mean they do not belong to the Cantor set? Please help if you can. Thank you so much.
 A: Here is the start of a long division base $3$ for the first problem:
         0.0222010
      ------------
  1011)101.0000000
        20 22
        -----
        10 010
         2 022
         -----
           2110
           2022
           ----
             1100
             1011
             ----
               120

At this point it is clear that $\frac{10}{31}$ is not in the Cantor set, since no later digits can change that $1$ into a $2$. One step earlier it wasn’t clear  from the quotient digits alone, since $(0.022201)_3$ could conceivably have continued $$(0.02220122222\ldots)_3\,,$$ which is equal to $(0.022202)_3$, which is in the Cantor set. However, if you realize that the only numbers that have two ternary expansions are rational numbers that can be written with a denominator that is a power of $3$, you can stop as soon as you get that $1$.
What we have at this point says that $(0.022201)_3<\frac{10}{31}<(0.022202)_3$, so it is in one of the deleted open intervals.
The second one is even easier.
A: Neither one is in the Cantor set. $\dfrac{10}{31} = 0.02220101110012..._{3}$, and $\dfrac{45}{69} = \dfrac{15}{23} = 0.12212110220122..._{3}$. Both has $1$ in their ternary presentation.
