Two different decimals representing the same number Under what condition two different decimals represent the same number? Is it "Two different decimals represent the same number if and only if one of them has 9 as its non-terminating decimal digit? For example, $0.\bar 9$ and $1$, or $3.278\bar 9$ and $3.279$.
Is it caused by the system of representing the real numbers or is it an inherent property of numbers?
 A: I don't have a definitive answer, but here are my thoughts about this. For me, there are two different kinds of representations for numbers.
Numbers as equivalence classes
If we consider the representation of rationals as fractions, any rational number has infinitely many different representations and we handle this by saying when two representations refer to the same number. For this specific case we have for example $a / b$ is equal to $c / d$ if $a \cdot d = b \cdot c$. From here, we prove that this definition of equality makes sense (it is an equivalence relation) and then we can choose a unique representative from each class as the fraction $a / b$ where $a$ and $b$ are coprime and $b > 0$.  Something similar happens if we define the integers as pairs of naturals. In this case, having multiple representations for the same number seems a normal fact of life, it holds indiscriminately for all numbers and can be avoided by adding contrraints to choose a unique preferred representation.
Notice that for this case, if we want to work only with the unique representations, the addition and multiplication become more complicated, because we need to add a simplification step at the end to obtain the correct representation.
Other representations
The decimal representation is different because it does not treat all numbers the same:

*

*some numbers have a unique representation

*some numbers have multiple representations
*** the numbers that have multiple representations have exactly 2 representations.**

It's really the third point above that I find most interesting, because it makes the choice of one representation from the two seem a bit arbitrary.
Also, this is not the only representation with this behavior. Something similar happens for continued fraction representation, where each rational number has exactly two, related, representations, while irrational numbers have just one.
If we consider the binary expansion instead of the decimal one, the numbers which have two representations are what is known as the dyadic rationals (those whose denominator is a power of 2). And the distinction of dyadic rationals from other rationals appears also in the theory of surreal numbers, but in a different way: dyadic rationals have a finite representation as surreal numbers. The moment we want to extend the set with more numbers, we get at the same time all the remaining rationals, all the irrational numbers, as well as stranger things like the infinitesimal neighbors of each dyadic rational (but not those of other rationals!).
So I guess there might be something intrinsic about the way we represent numbers which inevitably distinguishes some numbers.
A: Let $r$ be a real number which admits two different decimal representations. By substracting the integral part of $r$, we may assume that $r\in (0,1]$. Then
\begin{align} 
r &= 0.a_1a_2a_3\cdots , \\
 &= 0.b_1b_2b_3\cdots. 
\end{align}
where $a_i, b_j$ are in $\{0,1, \cdots, 9\}$. Let $k\in \mathbb N$ be the first instance where $a_k \neq b_k$. That is,
\begin{align} 
r &= 0.a_1a_2a_3\cdots a_{k-1} a_k \cdots , \\
 &= 0.a_1a_2a_3\cdots a_{k-1} b_k  \cdots 
\end{align}
Let's assume $b_k > a_k$. Then we claim that
$$ b_k = a_k +1$$
and $a_j = 9, b_j = 0$ for all $j >k$.
To see this, note that
\begin{align}
r&= 0.a_1a_2a_3\cdots a_{k-1} b_k b_{k+1} b_{k+2}  \cdots\\
 &\ge 0.a_1a_2a_3\cdots a_{k-1} b_k  000\cdots \\
&=  0.a_1a_2a_3\cdots a_{k-1} (b_k- 1) 999\cdots \\
&\ge  0.a_1a_2a_3\cdots a_{k-1} a_k 999\cdots \\
&\ge 0.a_1a_2a_3\cdots a_{k-1} a_k a_{k+1} a_{k+2} \cdots = r.
\end{align}
Thus we must have equality in each $\ge $ in the above equations. This is the same as saying that $b_k = a_k+1$, and $b_j = 0$, $a_j = 9$ for all $j>k$.
Thus the only real numbers which admits two decimal representations are of the form you described.
