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There are many ways to see that $0.999\ldots=1$ over the Reals (or over $\Bbb Q$ or $\Bbb C$ for that matter) like "Is it true that 0.99999…=1?", and the reasoning is easy enough:

If $x=0.\bar9$ then $10x=9.\bar{9}$ and consequenty, $9x=9.0$.

Some time ago I watched a math video which explained that the equality is not true everywhere, without mentioning what would break down, and no specific example was given, but what can go wrong, what are explicit examples?

What comes to mind are fields like the $p$-adics, the Superreals and Hyperreals. While I can follow their basic definitions or popular introductions, I have absolutely no intuition, let alone knowledge, what makes them different or not.

To make sense of $$s_n:=9\sum_{k=1}^n 10^{-k}$$ it's enough to have the arithmetic of a field. To make sense of $$0.\bar 9 := \lim_{n\to\infty} s_n$$ dunno what's needed here. A metric perhaps like $\operatorname{abs}:x\mapsto |x|$. The sequence should converge and the concept of convergence should make sense of course, and the limit should be $1$. Topologically closed is maybe too strong, but if the field is also a complete metric space that should suffice?

What I don't know is whether the above structures match (except for $\bar{\Bbb Q}_p$, which is however becond by comprehension / intuition).

And suppose the series $s_n$ from above converges to an element $0.\bar 9$ of the domain. Does this imply that $0.\bar 9 -1$ is an infinitesimal?

Maybe someone can iterate on these structures or on other domains that are worth talking about. And is it a requirement that $10^{-n}\to0$? Or may it be the case that $10^{-n}$ need not to become arbitrarily small and $0.\bar9$ still makes sense? What about Duals and split-complex?

Note: My questions are not about uniqueness of representation.

emacs drives me nuts
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  • "Topologically closed is too strong" Every topological space is closed as a subspace of itself. So it's not very strong at all. – Mark Saving Sep 24 '21 at 17:43
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    Possibly referring to non-standard analysis - models of the real numbers which include infinitesimals. – Thomas Andrews Sep 24 '21 at 17:43
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    You won't make progress in mathematics by watching youtube videos. – Peter Sep 24 '21 at 17:45
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    The Wikipedia plot summary on $0.999\ldots$ has some information, in the section on infinitesimals. – Brian Tung Sep 24 '21 at 17:46
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    There's a lot that could be said and cleared up about this question, that I don't recall seeing written elsewhere on Math.SE. If no one finds a good duplicate, I hope this stays open long enough that I/others can try writing some definitive answers so that we can point people to this question when it next comes up. – Mark S. Sep 24 '21 at 18:00
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    For the sake of a uniqueness one must not allow a decimal representation finally ending in period $9$. – Michael Hoppe Sep 24 '21 at 19:04
  • A question regarding hyperreals specifically: https://math.stackexchange.com/questions/281492/about-0-999-1 – Hans Lundmark Sep 24 '21 at 20:08
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    A statement is either true or false. It can't be 'not always true' unless it's false. – David P Sep 25 '21 at 01:48
  • @DavidP What about the statement "$3n + 1$ is prime". It is true for $n=2,4,6,10,....$ but false for $n = 1,3,5,7,8,9,......$ – fleablood Sep 25 '21 at 07:21
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    @fleablood: That's two different statements. One is "$p$ is prime for $p\in A$", the other is "$p$ is prime for $p\in B$". We could invent some "density of counter-examples" for some statements, but that doesn't touch whether an indivitual statement does hold or not. – emacs drives me nuts Sep 25 '21 at 07:55
  • But the OP isn't asking an individual statement. The OP is asking are there other numerical analytic systems in which $0.99999.... \ne 1$. What purpose does "A statement is either true or false. It can't be 'not always true' unless it's false" Why did you say that? What statement are you referring to? I don't see how that comment applies to the OPs question: Is $0.999.... = 1$ is all numerical analytic systems. "Is $0.999...=1$ is $\mathbb R$" is either true or false but that does not answer "Is there a numeric system where $0.999...\ne 1$". – fleablood Sep 25 '21 at 17:10
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    @fleablood In my mind this thread was essentially asking: When is $\neg A$ true? (but we know $A$ is true). I may have missed the jist of the post. – David P Sep 26 '21 at 00:16
  • @David P: My question was rather like: I know that $A(\Bbb R)$ is true and I can follow the reasoning. What are domains $X$ such that $A(X)$ is false, and what goes wrong? Provided it makes sense to talk about $A()$ over $X$ to start with. – emacs drives me nuts Sep 26 '21 at 06:58

1 Answers1

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I think this might help answer some of the spirit of the question, I try to give some of the reasoning in the p-adic cases which ends up having the sequence not converge in two different "flavors". I think learning more about the definition of a limit and Cauchy sequences might help you here.

Ostrowski's theorem says that every absolute value we can put on the rationals is either the real, a p-adic, or trivial absolute value. This is not entirely comprehensive, but a good start for seeing where your sequence can be seen to be valid. From here we can build up the real or any of the p-adic fields by completion with respect to this absolute value.

Specifically in the p-adic cases, we define the p-adic absolute value on rational numbers by pulling out the largest power of the prime. More precisely for $a,b, n \in \mathbb{Z}$ with $a,b$ not divisible by a prime number $p$, we can represent any rational number $x=p^n\frac{a}{b}$ and define its p-adic absolute value to be:

$$\left|p^n\frac{a}{b}\right|_p = \frac{1}{p^n}$$

Here if we take $.999...$ as a limit of the sequence $\frac{9}{10}, \frac{9}{10}+\frac{9}{100}, \frac{9}{10}+\frac{9}{100}+\frac{9}{1000}, ... $ we end up in either one of two situations.

In the 2-adics or the 5-adics, the sequence can't converge because we are adding on larger and larger terms to our sequence since their absolute values diverge as $n \to \infty$: $\left|\frac{9}{10^n}\right|_2 = 2^n$ and $\left|\frac{9}{10^n}\right|_5 = 5^n$

In the other p-adic fields we have that it doesn't converge because the sequence of terms never decays to $0$ but is still bounded above and infinitely "oscillates" around, very similar to Grandi's series. To directly see this, we can simplify the geometric series of partial sums to $9\frac{(10^{-1})^n-1}{10^{-1}-1}$ and focus on the nonconstant term for which infinitely many values, $(10^{-1})^n-1 \not \equiv 0 \mod p$.

Merosity
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  • probably a silly question but... it seems to me that the sequence of partial sums accumulates to $0$ (although it does not converge to $0$), since for any $p^k$ the multiples $n$ of $ p^{k-1} a$ (where $a$ is such that $10^a \equiv 1 \text{ mod } p$) satisfy $10^n \equiv 1 \text{ mod }p^k$. Are there any other accumulation points other than $0$? – ARG May 27 '22 at 16:48
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    The last sentence is phrased somewhat unfortunately because a) it is not literally true for $p=3$ and b) even if true, by itself it does not imply that those partial sums don't converge, only maybe that they do not converge to $0$. (Together with @ARG's observation that some subsequence of them does go to $0$, that would of course suffice.) – Torsten Schoeneberg Jul 06 '22 at 05:09
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    @ARG: Actually, for $x=\xi u$ with $\xi \in \mu_{p-1}, u \in U^{(i)}\setminus U^{(i+1)}$, the set ${x^n : n\in \mathbb N}$ is dense in $\langle \xi \rangle \cdot U^{(i)}$. So the sequence discussed here has lots of accumulation points (namely this rather big subgroup of the $p$-adic units: for all $p >3$, it contains all principal units $U^{(1)} = {y \in \mathbb Z_p: v_p(y-1) \ge 1}$). – Torsten Schoeneberg Jul 06 '22 at 05:14
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    Ah I believe I oversimplified in trying to explain originally, since in the $p$-adics a sequence is Cauchy when $|a_{n+1}-a_n|_p\to 0$. We can factor out the constants to focus on $a_n=10^{-n}-1$ to look at $|10^{-n-1}-10^{-n}|_p = |10^{-1}-1|_p|10^{-1}|_p^n$, and since $|10|_p=1$ for $p \ne 2, 5$, this fails to go to $0$ and is not Cauchy. – Merosity Jul 06 '22 at 17:24
  • @TorstenSchoeneberg thanks for the details; so it contains 0 and the principal units... For $p=3$ I guess it only means that it contains 0 and $U^{(3)}$. Is there a nice way to see the density of $x^n$ in $\langle \xi \rangle U^{(i)}$? – ARG Jul 07 '22 at 06:47
  • @Merosity sorry if my message sounded aggressive, I was not trying to say that what you said is wrong, I was just wondering what are the accumulation points are. – ARG Jul 07 '22 at 06:48
  • @ARG Oh I didn't interpret it as aggressive, and I wouldn't be surprised if I made a mistake either, I'm not perfect lol. I thought your comment back in May was good and just got distracted, which I am also currently too! – Merosity Jul 07 '22 at 17:41
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    @ARG: That for $u \in U^{(i)} \setminus U^{(i+1)}$ the $u^n$ are dense in $U^{(i)}$ follows e.g. via exp-log from the density of $\mathbb N$ in $\mathbb Z_p$. ($i \ge 1$ and $p \neq 2$.) The general result, I would think, comes from fiddling a bit with exponents and $\xi^{p-1} = 1$. --- Note that this I apply to $x=\frac{1}{10}$, but we still have the other terms, so we're actually looking at $(-10)\cdot (x^n -1)$. Of course $-10$ is a $p$-adic unit, so the accum. points of this should contain $p^2 \mathbb Z_p$ for $p=3$, $p \mathbb Z_p$ for $p \ge 7$, and sometimes translates of that as well. – Torsten Schoeneberg Jul 08 '22 at 22:17