Repeating decimals (sic) in bases other than 10

[Dustinsfl]

Prove that $.0222\ldots$ (base 3) $= .1$ (base 3) $= \frac{1}{3}$ (base 10).First, we will show $.0222\ldots$ (base 3) $= \frac{1}{3}$ (base 10).
\begin{alignat*}{3}
2\left(\frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \cdots\right) & = & 2\sum_{n = 2}^{\infty}\left(\frac{1}{3}\right)^n\\
& = & \frac{2}{9}\sum_{n = 0}^{\infty}\left(\frac{1}{3}\right)^n\\
& = & \frac{2}{9}\frac{1}{1 - \frac{1}{3}}\\
& = & \frac{1}{3}
\end{alignat*}

I am having trouble with the second part.

 

[MarkFL]

Re: base 3

This may not be what you have in mind, but this puts me in mind of a simple proof in base 10 that:

$\displaystyle 0.\bar{9}=1$

So, we have in base 3 (trinary):

$\displaystyle x=0.0\bar{2}$

Multiply by 3 in trinary:

$\displaystyle 10x=0.\bar{2}=0.2+x$

$\displaystyle 2x=0.2$

$\displaystyle x=0.1$

 

[CaptainBlack]

Re: base 3

This may not be what you have in mind, but this puts me in mind of a simple proof in base 10 that:

$\displaystyle 0.\bar{9}=1$

So, we have in base 3 (trinary):

$\displaystyle x=0.0\bar{2}$

Multiply by 3 in trinary:

$\displaystyle 10x=0.\bar{2}=0.2+x$

$\displaystyle 2x=0.2$

$\displaystyle x=0.1$

Your purported proof is invalid in any base in that it applies rules valid for finite series to infinite series without justifying their validity.

CB

 

[Dustinsfl]

How I solved it was with the fact that 3 in base 3 is 10.
$$
\frac{1}{10} = .1
$$