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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 $1$?

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    $\begingroup$ Why do you expect to have $0.99\ldots \ne 1$? $\endgroup$ Aug 11 '17 at 13:35
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    $\begingroup$ @Ruby: that is a common misbelief. $0.\overline{9}$ and $1$ are two representations of the same number. The decimal representation is not unique: $$\sum_{n\geq 1}\frac{9}{10^n} = 1.$$ $\endgroup$ Aug 11 '17 at 13:37
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    $\begingroup$ @Ruby: $5.\overline{9}=6$, simply. $\endgroup$ Aug 11 '17 at 13:39
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    $\begingroup$ There is lots of good stuff about this at en.wikipedia.org/wiki/0.999... . Btw the fact that this equality holds is not intuitive so don't feel bad if it came as a surprise. I also remember feeling uneasy about it when I first saw it. $\endgroup$
    – EHH
    Aug 11 '17 at 13:41
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    $\begingroup$ If they are not equal, what is a real number between them? "Because they are not equal" is not an argument, it is an intuition. Your intuition is wrong. $\endgroup$ Aug 11 '17 at 13:48
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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.

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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,

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

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  • $\begingroup$ Wow this seems pretty confusing.... $\endgroup$
    – yt.
    Aug 11 '17 at 13:54
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    $\begingroup$ @Ruby he's just basically saying that:$$0.9=1-0.1\\0.99=1-0.01\\0.999=1-0.001\\\vdots\\0.\overline{999}=1$$ $\endgroup$ Aug 11 '17 at 13:58
  • $\begingroup$ @SimplyBeautifulArt Is the line over the 9's also showing it's recurred? $\endgroup$
    – yt.
    Aug 11 '17 at 14:25
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    $\begingroup$ @Ruby More useful than the dot if you want something like $0.\overline{12}=0.121212\dots$ $\endgroup$ Aug 11 '17 at 14:31
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    $\begingroup$ @Ruby Because$$\lim_{n\to\infty}\frac1{10^n}=0$$ $\endgroup$ Aug 11 '17 at 14:32

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