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Context: I'm making this Q&A question (i.e. I write the question and an answer) because I briefly quote it as a lemma of-sorts in another question of mine. There has been some discussion of this result on that question, and I think this result is reasonably interesting enough in its own right, so we should have a question on this.

For some values of $x \in \mathbb{R}$, we can define what I'll call the "average digit" of $x$ and denote as $\theta(x)$ via

$$\theta(x) = \lim_{n \to \infty} \frac{a_1 + \dots + a_n}{n}$$

where the $a_i$ are the uniquely defined as digits in the decimal expansion of $x$, that is $x = N.a_1a_2\dots$, for some integer $N$, where we do not allow infinitely trailing $9$s (for example, for $x=1$, we have $a_i=0 \forall i$, rather than $a_i = 9 \forall i$. In other words, we are using that $1=1.000...$, not that $1 = 0.999...$)

$\theta(x)$ is defined for whichever $x$ this limit exists.

Question: for which $x \in \mathbb{Q}$ is $\theta(x)=x$?

Robin
  • 6,201

1 Answers1

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The only rational number for which $\theta(x)=x$ is $x=0$.

To prove this, first note the result that a number is rational iff its decimal expansion is periodic. That is,

$$x \in \mathbb{R} \text{ is rational} \iff \begin{align}x &= n.a_1a_2 \dots \\&= n.a_1a_2\dots a_N \overline{a_{N+1} \dots a_{N+M}} \text{ for some $a_i \in \{0,...,9\}$, $n \in \mathbb{Z}$, $N, M \in \mathbb{N}$}.\end{align}$$ (see for example this question for details.)

Thus for such a rational number $x$, its "average digit" $\theta(x)$ is always well-defined, and is just going to be the average of $a_{N+1}, \dots, a_{N+M}$. Intuitively, as $n \to \infty$ in the sum $\theta_n = \frac{a_1 + \dots + a_n}{n}$, the numerator contains approximately $\frac{n}{M}$ copies of $(a_{N+1} + \dots + a_{N+M})$, and at most $N+M-1$ other summands, so $\theta_n \to \frac{1}{M}(a_{N+1} + \dots + a_{N+M})$. (To argue this a bit more formally is just algebra: find an expression for $\theta_{N + kM+ i}$ for each $i \in \{ 0, \dots, M-1\}, k \in \mathbb{N}$, and show $\theta_{N + kM+ i} \to \frac{1}{M}(a_{N+1} + \dots + a_{N+M})$.)

Now we use a fact about rational numbers. Suppose $\frac{p}{q}$ is in lowest terms, with $q \geq 2$. Then the periodicity of the decimal expansion of $\frac{p}{q}$ is at most $q-1$. (Just consider the possible remainders when performing the long division $\require{enclose}p \enclose{longdiv}{q}$.)

This means that, using the above notation, for $x =\frac{p}{q}$ we have $M \leq q-1$. As $\theta(x) = \frac{a}{M}$ for some $a \in \mathbb{Z}$, we thus have $\frac{p}{q} = \frac{a}{M}$. But $M < q$, so we've contradicted $\frac{p}{q}$ being in lowest terms.

This shows that rational numbers with a denominator greater than or equal to $2$ cannot be fixed points of $\theta$, so we are left considering those with denominator $1$, i.e. the integers. An integer just has $a_i = 0$ for all $i > 0$, so its "average digit" is $0$. Thus the only integer equal to its "average digit" is $0$.

Hence $x = \theta(x), x \in \mathbb{Q} \implies x = 0$.

Are there solutions to $x = \theta(x), x \in \mathbb{R} \backslash \mathbb{Q}$? That is the topic of my other question.

Robin
  • 6,201