On base-b expansion of nonnegative reals

[psie]

TL;DR Summary

It has probably been asked numerous times before, yet I couldn't find anything complete enough that rigorously proves that every nonnegative real number has a base-$b$ expansion.

In Folland's real analysis text, twice (so far) he drops a very similar statement about base-$b$ expansions of nonnegative reals.

The first time is when discussing the proof of $\operatorname{card}(\mathcal{P}(\mathbb N))=\mathfrak c$, where he says that every positive real number has a base-$2$ decimal expansion. The existence of such expansion is crucial for the proof. The other time is when discussing the Cantor set, where he says that every real number in $[0,1]$ has a base-$3$ decimal expansion which is unique unless the number is of the form $p3^{-k}$ for integers $p,k$. Some have said "Yes, but this is due to $0.\overline{9}=1$." But I don't think this explains, or rather proves, why the expansion is unique (it does not prove existence either).

I'm looking for lecture notes, books, anything you could imagine citing yourself if you'd want the reader to learn more about existence and uniqueness of base-$b$ expansions of nonnegative reals. If it's possible to give a treatment of this as reply to this post, I'd be happy to accept it.

 

[fresh_42]

You cannot achieve uniqueness, e.g. $1=0.\overline{9}.$ One way to define real numbers is through equivalence classes modulo zero-Cauchy sequences. So every real number is a representative of such a class. Is your question why there always exists a sequence $(a_n)_{n\in \mathbb{Z}} \subseteq \{0,1,\ldots,b-1\}$ with almost all members with negative indices are zero such that a given real number $r$ equals
$$
r=\sum_{k=0}^\infty a_{k-n}b^{-k+n}\;?
$$
The point with every real number is that we need a definition for real numbers. The equation above is already a rational Cauchy-sequence and the real number $r$ is in my definition simply the limit that we add to the rationals. I guess, things become a bit more technical if we use Dedekind cuts instead.

Say we use the definition by those equivalence classes. Then the question reduces to whether every rational number can be expressed this way. That's probably some boring work to do but not difficult.

 

[psie]

You cannot achieve uniqueness, e.g. $1=0.\overline{9}.$ One way to define real numbers is through equivalence classes modulo zero-Cauchy sequences. So every real number is a representative of such a class. Is your question why there always exists a sequence $(a_n)_{n\in \mathbb{Z}} \subseteq \{0,1,\ldots,b-1\}$ with almost all members with negative indices are zero such that a given real number $r$ equals $$r=\sum_{k=0}^\infty a_{k-n}b^{-k+n}\;?$$

Yeah, I think that's my question more or less. I wonder if

1. Why there exists a sequence $(a_n)_{n\in\mathbb Z}\subset\{0,1,\ldots,b-1\}$ with finitely many members being nonzero for nonpositive indices such that $r=\sum_{k=-\infty}^\infty a_kb^{-k}$? (I think if we do not allow $a_k=b-1$ for infinitely many positive indices, one can get a unique expansion.)
2. If we take the example with the Cantor set and $[0,1]$, why the expansion is unique (or why it isn't) for numbers not of the form (or of the form) $x=p3^{-k}$ where $p,k$ are integers such that $x\in[0,1]$?

The equation above is already a rational Cauchy-sequence and the real number $r$ is in my definition simply the limit that we add to the rationals.

I don't understand this sentence. Yes, the terms in the sum are rationals. Do you mean the terms form a rational Cauchy sequence? What do you mean by "the limit that we add to the rationals"?

Dedekind cuts would be interesting I'm looking at Rudin's book, and he uses Dedekind cuts to construct the reals, though I lean more towards construction via Cauchy sequences.

 

[fresh_42]

Yeah, I think that's my question more or less. I wonder if

1. Why there exists a sequence $(a_n)_{n\in\mathbb Z}\subset\{0,1,\ldots,b-1\}$ with finitely many members being nonzero for nonpositive indices such that $r=\sum_{k=-\infty}^\infty a_kb^{-k}$? (I think if we do not allow $a_k=b-1$ for infinitely many positive indices, one can get a unique expansion.)

We need to allow infinitely many digits. This means we have to cope with equivalences anyway. Avoiding infinitely many $b-1$ appears artificial to me. And it is not obvious that this solves the problem. What about the number $\pi \cdot 0.\overline{9}$? The moment we use equivalence classes we fail uniqueness: $a$ cannot be distinguished from $a+ \lim_{n \to \infty}b^{-n}.$

1. If we take the example with the Cantor set and $[0,1]$, why the expansion is unique (or why it isn't) for numbers not of the form (or of the form) $x=p3^{-k}$ where $p,k$ are integers such that $x\in[0,1]$?

I'm not sure I understand this question. What do Cantor sets have to do with it?

I don't understand this sentence. Yes, the terms in the sum are rationals. Do you mean the terms form a rational Cauchy sequence? What do you mean by "the limit that we add to the rationals"?

I meant that
$$
r=\lim_{m \to \infty}\underbrace{\left(\sum_{k=0}^m a_{k-n}b^{-k+n}\right)}_{=a_m}
$$
which is a Cauchy sequence $(a_m)_{m\in \mathbb{N}}\subseteq \mathbb{Q}$ of rational numbers and the given real number $r$ is its limit. "Add to the rationals" was sloppy for the construction of real numbers by equivalence classes of Cauchy sequences of rational numbers modulo those Cauchy sequences that have $0$ as their limits. The construction is basically to take $\mathbb{Q}$ and a topological completion via Cauchy sequences, i.e. expand $\mathbb{Q}$ by those limits. The details take a few lines, about five pages in my book: from an Archimedean ordered field via topological completion of the rational numbers to the real numbers.

 

[psie]

Avoiding infinitely many $b-1$ appears artificial to me.

I probably mistyped. I meant to say that the sum will converge only if $a_k$ is nonzero for a finite number of nonpositive indices $k$. And we won't get something like $0.1999\ldots$ provided $a_k\neq b-1$ for an infinite number of positive indices. So e.g. $0.1999\ldots$ is not allowed, but $0.2$ and $0.1919\ldots$ is.

I'm not sure I understand this question. What do Cantor sets have to do with it?

Well, every number of the form $x=p3^{-k}$ has two ternary expansions; one with $a_j=0$ for $j>k$ and one with $a_j=2$ for $j>k$. If $p$ is not divisible by $3$, then one of these expansions will always have $a_k=1$ and the other will have $a_k=0$ or $2$. If we then always choose the latter expansion, we see that \begin{gather*}a_1=1\iff \frac13<x<\frac23 \\ a_1\neq 1\text{ and } a_2=1\iff \frac19< x<\frac29\text{ or }\frac79<x<\frac89.\end{gather*}The Cantor set is then simply all $x\in[0,1]$ that have a base-$3$ expansion with $a_j\neq 1$ for all $j$.

 

[fresh_42]

I probably mistyped. I meant to say that the sum will converge only if $a_k$ is nonzero for a finite number of nonpositive indices $k$. And we won't get something like $0.1999\ldots$ provided $a_k\neq b-1$ for an infinite number of positive indices. So e.g. $0.1999\ldots$ is not allowed, but $0.2$ and $0.1919\ldots$ is.

Good luck with a general proof! Maybe you can do it with Dedekind cuts, but if we consider real numbers as limits of rational Cauchy sequences then it is impossible per construction: we cannot distinguish between
$$
r=\lim_{n \to \infty}a_n \text{ and }r=\lim_{n \to \infty}(a_n+b_n) \text{ if }\lim_{n \to \infty}b_n=0\quad (*).
$$
That's why we need equivalence classes of those limits modulo rational Cauchy sequences that converge to zero. But if already the sequence $r=\lim_{n \to \infty}a_n$ isn't unique, how to deduce uniqueness in any basis? Dedekind cuts are closer to the visual imagination of the real number line so maybe they can provide a particular representation that you then can call unique. But even then you will need to deal with the fact that there is more than one possible way to end up in a gap $r\in\mathbb{R}\backslash\mathbb{Q}.$

Your question made me curious and I looked it up in two more, more basic books for undergraduate students. One defined the real numbers as Dedekind cuts and didn't bother the b-adic representations, and the other one introduced b-adic numbers and didn't define real numbers and replaced them by an axiom: $\mathbb{R}$ is complete. The proposition in it states that every real number can be represented by a b-adic number (and therefore that every real number is a limit of a Cauchy sequence of rational numbers). The proof is by induction. They say in the induction step that the representative $0\leq a_n\leq b-1$ is unique, but I didn't see where they deal with $1=0.\overline{9}$ and they did not demand uniqueness in the proposition statement. They only proved that every b-adic number is a Cauchy sequence. In any case, Cauchy sequences play a central role and thus the difficulty $(*)$ kicks in that leads to an equivalence relation.

Uniqueness is in my opinion an unnecessary gimmick that would need so many technical details and in my mind artificial additional conditions that it is just not worth it. We don't even do this when it is all about representatives. We write $G/ H=\{g\cdot H\,|\,g\in H\}$ and do not care about a specific system. And although we usually take $\mathbb{Z}_n=\{0,1,\ldots,n-1\},$ nobody cares if someone uses
$$
\mathbb{Z}_n=\left\{-\lfloor n/2\rfloor ,\ldots, \lfloor n/2\rfloor \right\}
$$
instead.

 

[fresh_42]

I saw (Google search: uniqueness + "b-adic" + pdf) that computational algorithms have to deal with this problem, e.g. https://www.academia.edu/11131877/b_Digital_Sequences. So it's most likely more a problem in computer science than it is for mathematics. The induction proof I mentioned in my previous post basically considers $1$ and $0.\overline{9}$ as two different numbers, i.e. avoids the question of uniqueness.

 

[fresh_42]

The induction proof in my book via nested intervals may solve the problem with $0.\overline{9}.$ Given a real number, in this case $1,$ we start with $1\cdot (10)^0$ in the first and final step, so $0.\overline{9}$ doesn't even occur. The b-adic representation is then unique because the digits always fall into a uniquely determined interval. However, this is a bit bogus since we haven't said what "Given a real number ..." actually means. $\displaystyle{9\cdot\lim_{n \to \infty}\sum_{k=1}^n 10^{-k}}$ undoubtedly exists and is a real number that is already in a $10$-adic representation. In this case we would have started with a $0$ and had to add infinitely many $9s.$ What is with cases where we cannot compute the limit beforehand? We would have to prove that cases $0.\overline{9}$ are the only ones where we cannot decide whether to start with $1$ or with $0.$