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 $k>2$?

  • $\begingroup$ The only way to get two is if the digits eventually stabilize to zero. You seem to know that if you're in such a case that there are indeed two. If they don't stabilize to zero, then let $\delta_n$ be the first digit that two representations differ. Then sum over $\Sigma_n^\infty 10^{-k}(\delta_k - \delta_k')$ and this will be less than one (but not zero) so mod $1$ they'll be different numbers. For a number that eventually stabilizes (say at decimal $n$), multiply by $10^n$ and just use a limit argument from either side on your interval $[0,1]$. $\endgroup$ Feb 9 at 20:36
  • $\begingroup$ Would $1.000000000.....$ with the "last" digit of $1$ after an infinite numbers of $0$ would qualify as a different expansion? $\endgroup$
    – Andrei
    Feb 9 at 20:36
  • $\begingroup$ @Andrei Does it fit the given definition of decimal expansion? If so, yes, if not, no. $\endgroup$
    – user239203
    Feb 9 at 20:40
  • $\begingroup$ @Andrei yes it would count, as $\delta=(1,0,0,0,...)$ does indeed satisfy $$\sigma(\delta)=1$$ $\endgroup$
    – clathratus
    Feb 9 at 20:43

$\cal C_k$ 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.

  • $\begingroup$ So, $\mathcal C_{3}$ is empty because there are at most $2$ ways to use the $1=0.\bar{9}$ trick? $\endgroup$
    – clathratus
    Feb 9 at 20:48
  • 1
    $\begingroup$ Yes. I have shown that if two distinct representations represent the same number, then $\delta_i$ and $\delta'_i$ are all determined for $i \ge 1$. Moreover, $\delta_0$ and $\delta'_0$ will be fixed by looking at $\lfloor x \rfloor$. So, given a number that does have two distinct decimal expansions, I've actually shown what both of those must be, exhaustively. (In particular, there's no third possibility.) $\endgroup$ Feb 9 at 20:51
  • 1
    $\begingroup$ Ah I see. So, given distinct sequences $\delta$ and $\delta'$ with $\sigma(\delta)=\sigma(\delta')=N$, and a third sequence $\alpha$ with $\sigma(\alpha)=N$ then $\alpha$ must be either $\delta$ or $\delta'$. Very clever! +1 $\endgroup$
    – clathratus
    Feb 9 at 20:55
  • $\begingroup$ What about for any base? $\endgroup$
    – AMDG
    Jul 22 at 17:02
  • $\begingroup$ @AMDG: (I'm assuming that your base is still an integer.) Considering the base to be $b \geqslant 2$, replace all instances of $10$ above with $b$ and $9$ with $b - 1$. The analogous result follows. (Note that $$\sum_{i \geqslant 1} \frac{b - 1}{b^i} = 1$$ still holds.) $\endgroup$ Jul 22 at 17:23

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