I just picked up the subject of ternary expansions (actually I'm trying to gain an understanding of the cantor set for measure theory) so my knowledge is still extremely weak but I just a quick question. Let $x\in[0,1]$. Then $x=0.a_1a_2...$ where $$a_i=\mathrm{sup}\{k\in \mathbb{Z}: \frac{a_1}{3}+ \frac{a_2}{3^2}+...+\frac{a_{i-1}}{3^{i-1}}+\frac{k}{3^i}\leq x\}.$$ Using this, I understand that $2/3=0.2$ but why is it also $0.12222...$? So first of all $a_1=\mathrm{sup}\{k\in \mathbb{Z}:\frac{k}{3}\leq\frac{2}{3}\}=2$. So $a_1$ is also $1$ as well? I'm a little confused here. Thank you for your assistance!
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Have you every seen why $0.\bar{9} = 1$? It's the same general idea as that. – user141592 Oct 19 '14 at 20:39
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Oh, no I have not. I will look this up. – user23793 Oct 19 '14 at 20:40
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Yes, every number with a finite ternary representation also has an infinite ternary representation. This is the same story as with Is it true that $0.999999999\ldots = 1$?
First, convince yourself (perhaps with the help of the above thread) that within the ternary system, $0.222222\ldots =1$, meaning $$\sum_{n=1}^\infty \frac{2}{3^n} = \frac{2/3}{1-1/3} = 1$$ This immediately generalizes to $0.1 = 0.0222\ldots$, etc. Finally, $$ 0.2 = 0.1+0.1 = 0.1+0.0222\ldots = 0.1222\ldots $$ and the same can be done with every finite ternary representation.
