A slightly variation of this question: Can every real number be represented by a (possibly infinite) decimal?
Here I restrict my discussion into infinite decimals (without ending with infinite zeros starting from somewhere along the number):
My question is a straightforward one, is this representation unique?
Firstly note the following, where $(a_i)$ are the digits of some decimal expansion: $$\sum_{i=n}^\infty a_i 10^{-i}\le \sum_{i=n}^\infty 9. 10^{-i}=10^{-(n-1)}$$ The first bit follows from a comparison of the coefficients and the second is just a geometric progression calculation — it's essentially the "0.999...=1" result.
Now suppose we have two different decimal expansions $\sum_{i=m}^\infty a_i 10^{-i}$ and $\sum_{i=m}^\infty b_i 10^{-i}$ which differ for the first time when $i=n-1$. Without loss of generality, assume $a_{n-1}<b_{n-1}$, so $a_{n-1}+1\le b_{n-1}$. We have $$\begin{align} \sum_{i=m}^\infty a_i 10^{-i}&=\sum_{i=m}^{n-2} a_i 10^{-i}+a_{n-1}10^{-(n-1)}+\sum_{i=n}^\infty a_i 10^{-i} \\ &\le\sum_{i=m}^{n-2} a_i 10^{-i}+a_{n-1}10^{-(n-1)}+10^{-(n-1)} \\ &\le\sum_{i=m}^{n-2} a_i 10^{-i}+b_{n-1}10^{-(n-1)} \end{align}$$ However, because the $a_i$s and $b_i$s match for $i \le n-2$, $$\begin{align} \sum_{i=m}^\infty b_i 10^{-i}&=\sum_{i=m}^{n-2} a_i 10^{-i}+b_{n-1}10^{-(n-1)}+\sum_{i=n}^\infty b_i 10^{-i} \\ &\ge \sum_{i=m}^\infty a_i 10^{-i}+\sum_{i=n}^\infty b_i 10^{-i} \end{align}$$ So equality of the two expansions can only possibly be achieved if $\sum_{i=n}^\infty b_i 10^{-i}=0$, which is not the case if you insist on the decimal expansions being infinite. Thus an infinite decimal expansion of a number is unique.
