decimal expansion - Writing a finite ternary as an infinite ternary with infinite number of $3^s$

A finite ternary can be written as an infinite ternary with finitely many trailing $3^s$.
We can say $$0_3.t_1t_2t_3...t_nt=0_3.t_1t_2t_3...t_n(t-1)\bar{2}$$ where $t=1,2$.

What does the $(t-1)$ bit mean?

And why is there a line above 2?

Why cant we just write $0_3.t_1t_2t_3...t_n2222...$ instead?

Answer

The line above the $2$ means that the $2$ is to be repeated infinitely often. For example, $0.10\overline{2}$ is an abbreviation for $0.1022222\ldots\;$.

The $t-1$ means exactly that. For example,

$$0_3.120\color{magenta}1=0_3.120\color{red}0\overline{2}=0_3.12002222\ldots\;,$$

and

$$0_3.120\color{magenta}2=0_3.120\color{red}1\overline{2}=0_3.12012222\ldots\;.$$

In the first example $t$ is the magenta $1$, and $t-1$ is the red $0$. In the second $t$ is the magenta $2$, and $t-1$ is the red $1$. Note that $t$ cannot be $0$, since a digit of $-1$ isn’t allowed: $t$ is the last non-zero digit of the finite ternary expansion.

We can’t simply replace the last non-zero digit by a string of $2$s because the resulting number isn’t equal to the one with which we started. Start with $0_3.1$, for instance; that’s $\frac13$. But

$$0_3.\overline{2}=\frac23+\frac2{3^2}+\frac2{3^3}+\ldots=\sum_{k\ge 1}\frac2{3^k}=\frac{\frac23}{1-\frac13}=1\;,$$

not $\frac13$. Dropping the $1$ down to $0$ and attaching an infinite string of $2$s to that, on the other hand, yields

$$0_3.0\overline{2}=\frac2{3^2}+\frac2{3^3}+\ldots=\sum_{k\ge 2}\frac2{3^k}=\frac{\frac29}{1-\frac13}=\frac{2/9}{2/3}=\frac13\;,$$

as it should.

In general we have

$$\begin{align*}
0_3.t_1t_2\ldots t_nt&=\frac{t_1}3+\frac{t_2}{3^2}+\ldots+\frac{t_n}{3^n}+\frac{t}{3^{n+1}}\\
&=\frac{t_1}3+\frac{t_2}{3^2}+\ldots+\frac{t_n}{3^n}+\frac{t-1}{3^{n+1}}+\frac1{3^{n+1}}\\
&=\frac{t_1}3+\frac{t_2}{3^2}+\ldots+\frac{t_n}{3^n}+\frac{t-1}{3^{n+1}}+\frac1{3^{n+1}}\color{red}{\sum_{k\ge 1}\frac2{3^k}}\\
&=\frac{t_1}3+\frac{t_2}{3^2}+\ldots+\frac{t_n}{3^n}+\frac{t-1}{3^{n+1}}+\sum_{k\ge n+2}\frac2{3^k}\\
&=0_3.t_1t_2\ldots t_n(t-1)\overline{2}\;,

\end{align*}$$

because the summation in red is equal to $1$. (All of the summations are just geometric series and hence easily evaluated.)