How to express 0.999999... recurring as a fraction without equaling 1

Question

I was wondering is there any way to express $0.999999$ recurring as an actual fraction without equaling $1$? Because I tried to convert it into a fraction following the rules for normal recurring decimals like this:

$$\begin{align}n&=0.999\dot9\\10n&=9.999\dot9\\n&=0.999\dot9\\9n&=9\\\therefore n&=9/9\end{align}$$

But as you can see the result is $9/9$ which ultimately is equal to $1$ . And I've even tried calculating it other ways like this:

$$\begin{align}1/3&=0.333\dot3\\2/3&=0.666\dot6\\\therefore3/3&=0.999\dot9\end{align}$$

But it always ends up telling me that $0.9999999... = 1$. Is there any mistake in my logic? And I also realized this applied to other recurring decimals ending in $9$. E.g: $0.5999999...=5.4/9 = 0.6$ . So is there a way to write $0.999999...$ as a fraction so you can differentiate it from $1$?

Answer 1

As you have proven by yourself, $$ 0.999999999999999999999\dot9 = 1. $$ There are no (big) logical mistakes in your post.

Because $0.99\dots$ is equal to $1$, it also cannot be another fraction.

Answer 2

Here's an alternate, more rigorous proof using series

$$ 0.9999\ldots = \frac{9}{10} + \frac{9}{10^2} + \dots = \sum_{n=1}^\infty \frac{9}{10^n} $$

This is the geometric series $9\sum \left(\frac{1}{10}\right)^n$, which has common ratio $r=\frac{1}{10}<1$. So using the formula for geometric series,

$$ 0.999\ldots = \frac{\frac{9}{10}}{1-\frac{1}{10}} = 1$$