A somewhat unintuitive result of real analysis is that decimal expansions are not unique. For example,
$$0.99999…=1.$$
So it can be gathered that every real number has at least one base-$10$ decimal expansion, sometimes even two. But is two the maximum? Are there any real numbers with three different base-$10$ decimal expansions?
Intuitively, I would think not, but I have close to no idea how to prove it other than knowing that the easiest method would be a proof by contraction.
It may help to restrict the problem by considering the expansions of numbers only in $[0,1]$. That is because if $a\in[0,1]$ has more than two base-$10$ decimal expansions, then so does $a+x$ for all $x\in\Bbb R$. And likewise, if $x\in\Bbb R\setminus [0,1]$ has more than two base-$10$ expansions, it can be written as $x=\mathrm{sgn}(x)(\lfloor x\rfloor+a)$, where $a\in[0,1)$, and by necessity $a$ has more than two base-$10$ expansions.
To be clear, I should define what I mean by base-$10$ expansion. Let $N\in[0,1]$. Then a decimal expansion of $N$ is a sequence $\delta=(\delta_0,\delta_1,\delta_2,…)$ of integers $0\le \delta_i\le 9$ such that
$$\sigma(\delta):=\sum_{i\ge0}\frac{\delta_i}{10^i}=N.$$
Furthermore, let $\mathcal U=\{(a_0,a_1,a_2,…):0\le a_i\le 9,\, a_i\in\Bbb Z\}$ and let
$$D_N=\{\delta\in\mathcal U :\sigma(\delta)=N\}.$$
Lastly, let
$$\mathcal C_k=\{N\in[0,1]:\#(D_N)\ge k\},\qquad k\in\Bbb N$$
where $\#(S)$ is the number of elements in the set $S$.
So, is $\mathcal C_k$ empty for $k>2$?
Best Answer
$\cal C_k$ is indeed empty for $k > 2$.
Suppose that a number has two distinct decimal expansions $\delta = (\delta_0, \delta_1, \delta_2, \ldots)$ and $\delta' = (\delta_0', \delta'_1, \delta'_2, \ldots)$. Then, we must have
$$0 = \sum_{i \ge 0}\dfrac{\delta_i - \delta'_i}{10^i}.$$
Wlog, we may assume that $\delta_0 \neq \delta'_0$. (There must be some smallest $i$ for which they are unequal. By multiplying by a suitable power of $10$, we may assume that it is $i = 0$.) Furthermore, we may assume that $\delta_0 > \delta'_0$.
Thus, we have $$1 \le \delta_0 - \delta'_0 = \sum_{i \ge 1}\dfrac{\delta'_i - \delta_i}{10^i}.$$
Taking absolute value on all sides and using triangle inequality for series, we see that
$$1 \le \delta_0 - \delta'_0 \le \sum_{i \ge 1}\dfrac{|\delta'_i - \delta_i|}{10^i} \le \sum_{i \ge 1}\dfrac{9}{10^i} = 1.$$
Thus, we have equality throughout. Note that this means that $|\delta'_i - \delta_i| = 9$ for all $i$. In fact, we see that $\delta'_i - \delta_i$ must have the same sign for all $i$. This sign must, of course, be positive.
Thus, we have $\delta_0 = \delta'_0 + 1$ and $\delta_i = 0$ for all $i \ge 1$, $\delta'_i = 9$ for all $i \ge 1$.
(All the manipulations above were justified as the series converged absolutely.)
The above shows that if a number has two decimal expansions, it must actually just be a variant of $1 = 0.\bar{9}$. In particular, there are at most two decimal expansions.