I'm trying to prove that any real number has at most two decimal expansions. I have an idea of how to prove it, but I'm not fully sure if every step of my proof works. Here is my attempt.
Let $x \in \mathbb{R}$. If $x$ has only one decimal expansion, there is nothing to prove, so it suffices to show that given any two distinct decimal expansions of $x$, one is terminating in a string of $9$'s, the other in a string of $0$'s. So let $$ \sum\limits_{i=0}^\infty a_i 10^{-i}, \; \sum\limits_{i=0}^\infty b_i 10^{-i}, $$ where $a_0$ and $b_0$ are allowed to be any integer, but $a_i, b_i$ are digits in $\{0, 1, \ldots, 9\}$ for every $i \geq 1$. Let $N$ be the least nonnegative integer for which $a_i \neq b_i$. Without loss of generality, we assume that $a_N < b_N$. We then have $$ x = \sum\limits_{i=0}^{\infty} a_i 10^{-i} = \sum\limits_{i=0}^{N-1} a_i 10^{-i} + a_N 10^{-N} + \sum\limits_{i=N+1}^{\infty} a_i 10^{-i}, $$ where we interpret $\sum\limits_{i=0}^{N-1} a_i 10^{-i}$ as $0$, summing zero numbers, if $N = 0$ (e.g., if the first decimal expansion is $0.999999 \ldots$ and the second is $1.000 \ldots$). As $a_i = b_i$ for $i < N$, we have $$ x = \sum\limits_{i=0}^{N-1} b_i 10^{-i} + a_N 10^{-N} + \sum\limits_{i=N+1}^{\infty} a_i 10^{-i}. $$ Now, as $a_i \leq 9$ for $i \geq N + 1$, we have $$ \sum\limits_{i=N+1}^{\infty} a_i 10^{-i} \leq \sum\limits_{i=N+1}^{\infty} 9 \cdot 10^{-i} = \frac{9 \cdot 10^{-N-1}}{1 - 10^{-1}} = \frac{9 \cdot 10^{-N-1}}{\frac{9}{10}} = 10^{-N}. $$ Therefore, we have: \begin{align*} x & \leq \sum\limits_{i=0}^{N-1} b_i 10^{-i} + a_N 10^{-N} + 10^{-N} \\ & = \sum\limits_{i=0}^{N-1} b_i 10^{-i} + (a_N + 1) 10^{-N} \\ & \leq \sum\limits_{i=0}^{N-1} b_i 10^{-i} + b_N 10^{-N} \\ & \leq \sum\limits_{i=0}^{N-1} b_i 10^{-i} + b_N 10^{-N} + \sum\limits_{i=N+1}^{\infty} b_i 10^{-i} \\ & = \sum\limits_{i=0}^{\infty} b_i 10^{-i} \\ & = x. \end{align*} The result is to conclude that $x \leq x$, which we know to be an equality. Therefore, every time we've written down "$\leq$," this is in fact an equality. The first line is an equality if for all $i \geq N + 1$, $a_i = 9$. The third line is an equality if and only if $a_N + 1 = b_N$. The fourth line is an equality if and only if $b_i = 0$ for every $i \geq N + 1$, in which case we're adding an infinite right tail of zeros. That is, we get $x = x$ if the following three conditions hold: $$ a_N + 1 = b_N, \; a_i = 9 \forall i \geq N + 1, \; b_i = 0 \forall i \geq N + 1, $$ which is precisely equivalent to stating that the first decimal expansions terminates in an infinite string of $9$'s, the second in an infinite string of zeros. As these are the only possible, distinct decimal expansions for $x$, we conclude that $x$ has at most two decimal expansions.
How does this proof look? Are there any leaps of logic?
