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?
 A: 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 $0. \overline{23} = 23/99$
$0. \overline{123} = 123/999 = 61/333$
A: 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 $2$).
A: 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.
A: 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 $0.\overline{25} = \frac{25}{99}$.
