Are there any other numbers like $0.999\ldots$?

In a manner similar to how the value $1$ can be represented as $0.(9)$ too, are there any other values that exhibit this property when represented in base 10?

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Every number that has a terminating decimal expansion has a decimal expansion with a tail of $9$s. E.g., $0.5= 0.499999\cdots$. –  Arturo Magidin Sep 17 '11 at 22:47
You should try to understand limits and infinite series if you want to know what such representations even mean in the first place. –  anon Sep 17 '11 at 22:50
only infinitely many... –  user12205 Sep 17 '11 at 23:00
Since only examples between $0$ and $1$ have been posted so far: $2100 = 2099.999\ldots$ –  Rahul Sep 17 '11 at 23:04
Jeeze, why edit this century old question? and also another question invovling 0.999999999999999.... –  Lost1 Nov 21 '13 at 13:20

Any number that ends in an infinite series of $9$'s is equal to the number changing all the $9$'s to $0$'s and incrementing the previous place by $1$. So $0.5=0.4999\ldots , 0.1328=0.132799999\ldots$ etc.

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Every repeating decimal can be represented as a fraction.

Example: Represent $0.\overline{25}$ as a fraction.

First, let $x = 0.\overline{25}$.

Next, multiply both sides of the equation by a power of ten to move the decimal place after the first repeat. In this case, I should choose 100 and get $$100x = 25.\overline{25}.$$

Notice that since there are an infinite number of 25's after the decimal place in $0.\overline{25}$, moving two of them in front of the decimal still leaves an infinite number of them after the decimal. That means we can write that as $$100x = 25 + x.$$

Now we can solve this for $x$.

\begin{align*} 99x &= 25\\ x &= \frac{25}{99} \end{align*}

So, $x$ is both equal to $0.\overline{25}$ and $\frac{25}{99}$. That must mean $0.\overline{25} = \frac{25}{99}$.

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This isn't really what my question was about, but now I know where the rules to convert repeating decimal fractions to ordinary ones come from. :) –  Paul Manta Sep 17 '11 at 23:01
@Paul, then I guess my answer isn't really about your question either! Hope it helps, in any case. –  The Chaz 2.0 Sep 17 '11 at 23:03
@Paul I interpreted your question as "Are there any other repeating decimals that can be represented as fractions?" Was that not what you intended? –  Austin Mohr Sep 17 '11 at 23:06
@Austin I know all rational fractions can be represented as both decimal and ordinary fractions. (I also know that if an ordinary cannot be precisely represented as a decimal, that is a limitation of the number base, not a property of the value.) My question was about numbers that, from an infinitezimal point of view, should not represent the same value, but which in actuality they do. –  Paul Manta Sep 17 '11 at 23:13

To generalize Austin's answer (and since I don't know what "this property" is, exactly):

The number $0.\overline{a_1a_2...a_n}$ is equal to $\frac{ a_1a_2 ... a_n}{99\dots9}$, where there are $n$ nines in the denominator.

So $0. \overline{23} = 23/99$

$0. \overline{123} = 123/999 = 61/333$

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And I can't, for the life of me, figure out how to fix that LaTex. I intended for there to be 99....9 in the denominator (which could be more precisely represented as nine times a sum of powers of ten...) –  The Chaz 2.0 Sep 17 '11 at 23:00
Or a power of ten minus one. –  anon Sep 17 '11 at 23:05
Thanks, Brian! And @anon, let's not get into another discussion a lá math.stackexchange.com/q/11/7850 :) –  The Chaz 2.0 Sep 17 '11 at 23:12
The number $0.\overline{a_{p-1}a_{p-2}\ldots a_{1}a_{0}}$ is the sum of a geometric series $$0.\overline{a_{p-1}a_{p-2}\ldots a_{1}a_{0}}=\dfrac{N}{10^{p}}+\dfrac{N}{10^{2p}}+\cdots =\dfrac{N/10^{p}}{1-10^{-p}}=\dfrac{N}{10^{p}-1},$$ where $$N=10^{0}a_{0}+10^{1}a_{1}+\cdots +10^{p-1}a_{p-1}.$$ –  Américo Tavares Sep 17 '11 at 23:19
I seem to recall reading somewhere (therefore it's true!! (?)) that Johannes Kepler proposed a base-3 numeral system with three digits: $0$, $1$, and $-1$. In that system, the number $1/2$ can be represented in two different ways: $$1.,\ -1,\ -1,\ -1,\ \ldots,$$ and $$0.,\ 1,\ 1,\ 1, \ \ldots\ .$$ And similarly for every binary rational number (i.e. rational number whose denominator is a power of $2$).