Ternary Expansion Ambiguity

Question

I am following along and reading this notes: https://www.maths.tcd.ie/~levene/221/pdf/cantor.pdf

I am having trouble understanding why we necessarily have $e_n=d_n+1$, $d_{n+1}= d_{n+2} =···= 2$ and $e_{n+1} = e_{n+2} = ··· = 0$ when $d_n > e_n$. It would much appreciated if someone can guide me through this.

Answer 1

You'll have to play around with the inequalities a little bit to establish that.

To make life easy, assume without loss of generality $d_1 \neq e_1$. Next, notice,

$$x = \sum_{n \geq 1} \frac{e_n}{3^n} = \frac{e_1}{3} + \sum_{n \geq 2} \frac{e_n}{3^n} \geq \frac{e_1}{3}$$

And also,

$$x = \sum_{n \geq 1} \frac{d_n}{3^n} = \frac{d_1}{3} + \sum_{n \geq 2} \frac{d_n}{3^n} \leq \frac{d_1}{3} + \sum_{n \geq 2} \frac{2}{3^n} = \frac{d_1+1}{3}$$

So, we have, $d_1 < e_1 \leq d_1 + 1 \implies e_1 = d_1+1$

Answer 2

All numbers have at least one ternary expansion. For example, \begin{eqnarray} \frac{1}{2} &=& 0.11111111... \\ \frac{2}{9} &=& 0.02000000... \\ \frac{\pi}{8} &=& 0.21001211... \\ \gamma &=& 0.12012021... \end{eqnarray} However, some numbers have two ternary expansions \begin{eqnarray} \frac{2}{9} &=& 0.02000000... = 0.01222222...\\ \frac{19}{27} &=& 0.20100000... = 0.20022222...\\ \frac{100}{243} &=& 0.10201000... = 0.10200222... \end{eqnarray} Note that whenever a number has two ternary expansions, it always has one that ends in $222222...$ and one that ends in $000000...$, with the preceding digit being one larger for the $000000...$ expansion. That's what the article is talking about.

You'll also notice that such numbers are always fractions with a power of 3 in the denominator. This is no coincidence, as they're exactly the numbers with terminating ternary expansions. As for why the other expansion has $222222...$, consider that $\sum_1^\infty (2/3)^n = 1$.

Answer 3

As in the case of $\frac 13$ there are two ternary expressions. In this example $n $ as they have defined it $=1$. That is, the first place they don't agree is the first place after the decimal. ..

So, $0.e_1e_2\dots e_n=0.10\dots0$ and $0.d_1d_2\dots d_n=0.02\dots 2\dots $. Looking at this you should be able to see the $e_n's $ and $d_n's $ are as described. .. In particular, $e_1=1=0+1=d_1+1$, and after that all the $e_n's $ are $0$, and all the $d_n's $ are $2$...

It turns out this is always the situation when ambiguity occurs (that is, there is more than one expression, in ternary,for a number in the cantor set )...