Determining whether certain rationals are in the Cantor set

Question

I am trying to prove that $9\over13$ is in the Cantor set while $13 \over17$ is not.

I am having a hard time constructing a geometric series that works for $9\over13$. This is the only method I have used before for affirming membership in the Cantor set, not sure if there's another approach that would work better in this scenario.

For proving $13 \over 17$ is not in the Cantor set, I cannot think of a way other than to identify the iteration at which the interval containing $13\over17 $ gets cut. That's surely not the way to go about this, but I'm drawing a blank.

Answer 1

Hint: a positive real $0 \leq x \leq 1$ is in the Cantor set iff it has a ternary (base 3) expansion whose digits consist only of 0’s and 2’s.

Answer 2

Write $13$ and $9$ in base three and divide:

       0.2002002...
     ___________
 111)100.0000000
      22 2
      ----
         1000
          222
         ----
            1000
             222
            ----
               1

At this point it’s pretty clear that the fraction is $0.\overline{200}$ when written as a ternary decimal, so it must be in the Cantor set, since the expansion has no $1$s. Of course it doesn’t hurt to check that

$$\frac9{13}=\frac2{3^1}+\frac2{3^4}+\frac2{3^7}+\ldots\,,$$

but that’s straightforward:

$$\sum_{n\ge 0}\frac2{3^{3n+1}}=\frac23\sum_{n\ge 0}\frac1{27^n}=\frac23\cdot\frac{27}{26}=\frac9{13}\,.$$

You can take the same approach with $\frac{13}{17}$.