Talk:0.999...

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Frequently asked questions

Q: Are you positive that 0.999... equals 1 exactly, not approximately?

A: In the set of real numbers, yes. This is covered in the article. If you still have doubts, you can discuss it at Talk:0.999.../Arguments. However, please note that original research should never be added to a Wikipedia article, and original arguments and research in the talk pages will not change the content of the article—only reputable secondary and tertiary sources can do so.

Q: Can't "1 - 0.999..." be expressed as "0.000...1"?

A: No. The string "0.000...1" is not a meaningful real decimal because, although a decimal representation of a real number has a potentially infinite number of decimal places, each of the decimal places is a finite distance from the decimal point; the meaning of digit d being k places past the decimal point is that the digit contributes d · 10^-k toward the value of the number represented. It may help to ask yourself how many places past the decimal point the "1" is. It cannot be an infinite number of real decimal places, because all real places must be finite. Also ask yourself what the value of

{\frac {0.000\dots 1}{10}}

would be. Those proposing this argument generally believe the answer to be 0.000...1, but, basic algebra shows that, if a real number divided by 10 is itself, then that number must be 0.

Q: The highest number in 0.999... is 0.999...9, with a last '9' after an infinite number of 9s, so isn't it smaller than 1?

A: If you have a number like 0.999...9, it is not the last number in the sequence (0.9, 0.99, ...); you can always create 0.999...99, which is a higher number. The limit

0.999\ldots =\lim _{n\to \infty }0.\underbrace {99\ldots 9} _{n}

is not defined as the highest number in the sequence, but as the smallest number that is higher than any number in the sequence. In the reals, that smallest number is the number 1.

Q: 0.9 < 1, 0.99 < 1, and so forth. Therefore it's obvious that 0.999... < 1.

A: No. By this logic, 0.9 < 0.999...; 0.99 < 0.999... and so forth. Therefore 0.999... < 0.999..., which is absurd.

Something that holds for various values need not hold for the limit of those values. For example, f (x)=x³/x is positive (>0) for all values in its implied domain (x ≠ 0). However, the limit as x goes to 0 is 0, which is not positive. This is an important consideration in proving inequalities based on limits. Moreover, although you may have been taught that

0.x_{1}x_{2}x_{3}...

must be less than

1.y_{1}y_{2}y_{3}...

for any values, this is not an axiom of decimal representation, but rather a property for terminating decimals that can be derived from the definition of decimals and the axioms of the real numbers. Systems of numbers have axioms; representations of numbers do not. To emphasize: Decimal representation, being only a representation, has no associated axioms or other special significance over any other numerical representation.

Q: 0.999... is written differently from 1, so it can't be equal.

A: 1 can be written many ways: 1/1, 2/2, cos 0, ln e, i⁴, 2 - 1, 1e0, 1₂, and so forth. Another way of writing it is 0.999...; contrary to the intuition of many people, decimal notation is not a bijection from decimal representations to real numbers.

Q: Is it possible to create a new number system other than the reals in which 0.999... < 1, the difference being an infinitesimal amount?

A: Yes, although such systems are neither as used nor as useful as the real numbers, lacking properties such as the ability to take limits (which defines the real numbers), to divide (which defines the rational numbers, and thus applies to real numbers), or to add and subtract (which defines the integers, and thus applies to real numbers). Furthermore, we must define what we mean by "an infinitesimal amount." There is no nonzero constant infinitesimal in the real numbers; quantities generally thought of informally as "infinitesimal" include ε, which is not a fixed constant; differentials, which are not numbers at all; differential forms, which are not real numbers and have anticommutativity; 0⁺, which is not a number, but rather part of the expression

\lim _{x\rightarrow 0^{+}}f(x)

, the right limit of x (which can also be expressed without the "+" as

\lim _{x\downarrow 0}f(x)

); and values in number systems such as dual numbers and hyperreals. In these systems, 0.999... = 1 still holds due to real numbers being a subfield. As detailed in the main article, there are systems for which 0.999... and 1 are distinct, systems that have both alternative means of notation and alternative properties, and systems for which subtraction no longer holds. These, however, are rarely used and possess little to no practical application.

Q: Are you sure 0.999... equals 1 in hyperreals?

A: If notation '0.999...' means anything useful in hyperreals, it still means number 1. There are several ways to define hyperreal numbers, but if we use the construction given here, the problem is that almost same sequences give different hyperreal numbers,

0.(9)<0.9(9)<0.99(9)<0.(99)<0.9(99)<0.(999)<1\;

, and even the '()' notation doesn't represent all hyperreals. The correct notation is (0.9; 0.99; 0,999; ...).

Q: If it is possible to construct number systems in which 0.999... is less than 1, shouldn't we be talking about those instead of focusing so much on the real numbers? Aren't people justified in believing that 0.999... is less than one when other number systems can show this explicitly?

A: At the expense of abandoning many familiar features of mathematics, it is possible to construct a system of notation in which the string of symbols "0.999..." is different than the number 1. This object would represent a different number than the topic of this article, and this notation has no use in applied mathematics. Moreover, it does not change the fact that 0.999... = 1 in the real number system. The fact that 0.999... = 1 is not a "problem" with the real number system and is not something that other number systems "fix". Absent a WP:POV desire to cling to intuitive misconceptions about real numbers, there is little incentive to use a different system.

Q: The initial proofs don't seem formal and the later proofs don't seem understandable. Are you sure you proved this? I'm an intelligent person, but this doesn't seem right.

A: Yes. The initial proofs are necessarily somewhat informal so as to be understandable by novices. The later proofs are formal, but more difficult to understand. If you haven't completed a course on real analysis, it shouldn't be surprising that you find difficulty understanding some of the proofs, and, indeed, might have some skepticism that 0.999... = 1; this isn't a sign of inferior intelligence. Hopefully the informal arguments can give you a flavor of why 0.999... = 1. If you want to formally understand 0.999..., however, you'd be best to study real analysis. If you're getting a college degree in engineering, mathematics, statistics, computer science, or a natural science, it would probably help you in the future anyway.

Q: But I still think I'm right! Shouldn't both sides of the debate be discussed in the article?

A: The criteria for inclusion in Wikipedia is for information to be attributable to a reliable published source, not an editor's opinion. Regardless of how confident you may be, at least one published, reliable source is needed to warrant space in the article. Until such a document is provided, including such material would violate Wikipedia policy. Arguments posted on the Talk:0.999.../Arguments page are disqualified, as their inclusion would violate Wikipedia policy on original research.

0.999... is a former featured article. Please see the links under Article milestones below for its original nomination page (for older articles, check the nomination archive) and why it was removed.

This article appeared on Wikipedia's Main Page as Today's featured article on October 25, 2006.

Article milestones
Date	Process	Result
May 5, 2006	Articles for deletion	Kept
October 10, 2006	Featured article candidate	Promoted
August 31, 2010	Featured article review	Kept
September 24, 2024	Featured article review	Demoted
Current status: Former featured article

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Arguments Archives: Index, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

Yet another anon

Moved to Arguments subpage

Sourcing question

Latest comment: 3 months ago5 comments3 people in discussion

Does the argument in 0.999...#Impossibility of unique representation come from somewhere? Other than that, the sourcing seems OK. XOR'easter (talk) 05:10, 29 June 2024 (UTC)Reply

The first and the second, as well as the bullet list, remain unsourced. Dedhert.Jr (talk) 06:18, 29 June 2024 (UTC)Reply

I have provisionally trimmed the passage I couldn't find support for. It wasn't technically wrong, as far as I could tell, but we aren't a repository for everything that could be said about a math topic. XOR'easter (talk) 17:47, 29 June 2024 (UTC)Reply

Probably a correct removal, but sort of a pity, since it's the only bit of actual mathematical interest.

No matter. This isn't really a math article, or shouldn't be. Mathematicians are unlikely to care about 0.999... per se. We should keep that in mind when thinking about how to present the material. I'm totally against lies to children, but I also don't see the point in making this an article about real analysis. If you understand real analysis you don't need this article. --Trovatore (talk) 21:50, 29 June 2024 (UTC)Reply

Are any of the concerns in the FA review still outstanding, then? XOR'easter (talk) 02:30, 23 July 2024 (UTC)Reply

Two representations in every positional numeral system with one terminating?

Latest comment: 4 months ago22 comments6 people in discussion

The article contains the statement

... every nonzero terminating decimal has two equal representations ... all positional numeral systems have this property.

Every positional numeral system has two representations for certain numbers, but is this necessarily true of terminating representations? A counterexample would seem to be balanced ternary: the numbers that have two representations seem to be nonterminating, e.g. 1 = 1.000..._bal3 has no other representation, but 1/2 = 0.111..._bal3 = 1.TTT..._bal3 (where T = −1) has two. Or maybe I need some coffee? —Quondum 01:56, 30 June 2024 (UTC)Reply

Well, the trouble is that "balanced ternary" is not a "usual" positional numeral system. Perhaps it might be better to write, "and this is true of all bases, not just decimal". In the end it depends whether you think Wikipedia is here for the benefit of lawyers, or just to help people understand things. Imaginatorium (talk) 04:07, 30 June 2024 (UTC)Reply

A better viewpoint is that such systems have no "terminating" representations at all, but only ones that eventually repeat the digit 0. That's beyond the scope of this article. Still, I think we should de-emphasize the notion of "terminating" representations. We don't really need to talk about them. We can say, for example, that 3.4999... is the same as 3.5000.... --Trovatore (talk) 05:36, 30 June 2024 (UTC)Reply

While I agree that "terminating representations" are a little peripheral, including that to answer the immediate question "How easy is it to find such values?" seems reasonable, although extrapolating from the example would seem obvious to us, and the phrase is adequately defined as linked. I don't feel strongly about keeping "terminating representations" or any other specific description of the class, though. I have clarified the statement in a way that fits the section 0.999... § Generalizations.

That aside, it is interesting that having multiple representations depends on the definition of a positional numeral system as having position weights and values associated with symbols that do not depend on the value of other digits; I say this, because Gray codes are remarkably close to being a positional system and (extended to a fractional part) they evidently have a unique representation for each real number. —Quondum 15:11, 30 June 2024 (UTC)Reply

Not sure quite what you mean about the Gray codes. The key point here is that representations of this sort are zero-dimensional in the product topology, which seems to be the natural one to put on them, whereas the reals are one-dimensional. So a continuous surjection from the representations to the real numbers cannot be an injection ~~with a continuous inverse~~, because otherwise it would be a homeomorphism, contradicting the previous observation about dimension. Therefore there must be reals with non-unique representations.

Maybe the representation by Gray codes you're talking about isn't continuous (or its inverse is not); would need to see what you mean. --Trovatore (talk) 18:30, 30 June 2024 (UTC)Reply

I intended that a digit incrementing reverses the interpretation of all subsequent digits – in my quest to get rid of the infinite number of digits 'rolling over' (e.g. 0.999... to 1.000...). Looking more closely, I see that my supposition was incorrect: you still get two representations for a (presumably the same) set of real values, but now a pair of equal representations differs in one digit instead of infinitely many. This leads me to wonder whether a representation composed of an infinite sequence of discrete symbols could avoid doubles, once we have the the restriction that every real number must have a representation. —Quondum 19:12, 30 June 2024 (UTC)Reply

If you put no restrictions at all on what you mean by a "representation", the answer is clearly yes, you can avoid duplicates. For example the set of all countably infinite decimal strings and the set of all reals have the same cardinality, so there's a bijection between them. You can even make that bijection pretty explicit, by playing games with (the proof of) the Schröder–Bernstein theorem.

However, if you care at all about continuity, you're going to need to deal with the dimension issue I mentioned in my previous comment. --Trovatore (talk) 19:52, 30 June 2024 (UTC)Reply

I was having difficulty framing the issue properly, and I thought about the cardinality argument after my post. I this context, all I would care about in this context is finding a simple constructive definition of a class of surjective maps from sequences of symbols to the reals for which multiple representations occur, ideally independent of the axiom of choice and it would be nice if it gave a sense of the lengths one might have to go to to avoid these multiples. The class of standard positional systems suffices, but clearly the class for which this is true is larger, and this gives room for possibly finding a class that is simpler to define. Improving on just copying the second sentence in Positional notation seems to me to be challenge. —Quondum 21:40, 30 June 2024 (UTC)Reply

In this context, a nice way to think of continuity is that, if you want finitely much information about the answer, you need only finitely much information about the input. On the "real" side, finitely much information means an open interval. On the "representation" side, it means finitely many digits. If that's true in both directions, then there must be duplicates.

I'm a little skeptical that this can (or should) be worked into this article, but it would be satisfying if it could be. --Trovatore (talk) 22:13, 30 June 2024 (UTC)Reply

I'm not of the opinion that anything more complicated than a sentence defining 'positional numeral system' is called for here to address the concerns that have been expressed, and it seems that my hope that a broader class could be defined more easily was too much. I have to confess that I do not know what you mean by 'finitely many digits': we are dealing with a countably infinite sequence of digits, each of which is an element of a finite (or possibly infinite) set of symbols. I'm afraid educating me on this is, for the moment, a lost cause, so I suggest that we just focus on language to include in the article. —Quondum 23:35, 30 June 2024 (UTC)Reply

I tend to agree that this is getting off-scope for the talk page. I'll drop a note on your talk page. I don't think it's a lost cause; I probably just haven't found the right way of explaining myself.

But as long as we're here, I do want to correct the record for the benefit of any lurkers. Turns out my maunderings about the continuity of the inverse mapping were unnecessary. As long as

The alphabet is finite (or at least there are only finitely many choices for a digit at any given position),
The mapping is continuous, and
The mapping is injective

you get continuity of the inverse mapping, and therefore a contradiction, for free. That's because the representation space is compact (by Tychonoff's theorem), so any closed set is compact. Then the continuous image of a closed set is compact and therefore closed, which in the injective case implies that the inverse map is continuous.

So if we can source it, we could say that any continuous interpretation of the representations would have to have duplicates. Is that appropriate for this article? I doubt it. This article ought to be pitched considerably lower. Anyone who understands the above argument isn't looking to understand 0.999.... --Trovatore (talk) 03:06, 1 July 2024 (UTC)Reply

Very nice. This also in some sense explains how p-adic integers can have unique expansions: the p-adic integers are compact. Tito Omburo (talk) 10:59, 1 July 2024 (UTC)Reply

Or more importantly, the p-adic integers are zero-dimensional, or it might be easier to think of it in terms of the p-adic integers are totally disconnected (not quite equivalent but it gets at the same point for our purposes).

The representations are totally disconnected whereas the reals are connected, so intuitively, to map the representations to the reals, you have to "connect something", which is where the duplicates come from. The p-adics are already totally disconnected, so the problem doesn't come up. --Trovatore (talk) 20:05, 1 July 2024 (UTC)Reply

Also, the product space

\{0,...,9\}^{\mathbb {N} }

is totally disconnected, and so is not the continuous image of a real interval. Tito Omburo (talk) 23:16, 1 July 2024 (UTC)Reply

As user:Imaginatorium point out, the statement is valid in usual positional numeral systems, but not in all non-standard positional numeral systems. Luckily, positional numeral systems redirects to a section of List of numeral systems on "Standard positional numeral systems", so I suggest we simply use that wikilink. Nø (talk) 15:32, 30 June 2024 (UTC)Reply

We should never have a redirect and its plural linking to different places, so that is not a solution. —Quondum 15:54, 30 June 2024 (UTC)Reply

I have changed that redirect to be consistent with the singular form, after verifying that there are no mainspace uses. In any event, the definition at Positional notation (essentially a weighted sum) is precisely correct for the statement as it now stands (i.e. including all nonstandard positional systems that meet this definition, with the proviso that they can represent all real numbers), and as supported by the text of the article. —Quondum 16:43, 30 June 2024 (UTC)Reply

As far as I understand, this section discusses supposed properties of all positional numeral systems. But this supposes a precise definition of a positional numeral system, and of a positional numeral system that accepts infinite strings. Without such a definition, everything is original research.

As an example, the standard $p$ -adic representation of $p$ -adic numbers is an example of a positional numeral system such that there is always a unique representation.

By the way it is astonishing that nobody mention what is, in my opinion, the main reason for which there is so much confusion with the subject of the article: it is that "infinite decimals" make a systematic use of actual infinity, a concept that is so counterintuitive that, before the 20th century, it was refused by most mathematicians. It seems that some teachers hope that kids could understand easily concept that were refused by mathematicians and philosophers a century ago. D.Lazard (talk) 16:44, 30 June 2024 (UTC)Reply

I think you're a bit off on that point, Prof. Lazard. My impression is that the rejection of the actual infinite was more in theory than in practice, and its systematic use considerably predated the 20th century, since real analysis was developing in the mid-19th century and used the actual infinite implicitly. It took Cantor to make it explicit, but the ideas of Bolzano and Cauchy and Weierstrass and Dedekind were already laying the groundwork.

That said, sure, it's a key psychological point.

ObSMBC. --Trovatore (talk) 18:53, 30 June 2024 (UTC)Reply

This is interesting, but does it apply? As I understand it (and admittedly this is outside my area of knowledge), p-adic numbers do not embed the reals. The ability to represent all reals is core to the statement that there are necessarily multiple representations. —Quondum 16:57, 30 June 2024 (UTC)Reply

Interesting point about p-adic numbers. I think the lead should mention infinity somewhere. I think the issues are resolved if the article is clear on what a "positional number system" is. I am unclear exactly what is meant. Tito Omburo (talk) 17:01, 30 June 2024 (UTC)Reply

So we have two people saying that a clear definition of a 'positional number system' is needed in the article, and I tend to agree in the context of this claim. I imagine that this can be omitted from the lead, but it might make sense in 0.999... § Generalizations. —Quondum 18:09, 30 June 2024 (UTC)Reply

Intuitive explanation

Latest comment: 1 month ago6 comments5 people in discussion

There seems to be an error in the intuitive explanation:

For any number x that is less than 1, the sequence 0.9, 0.99, 0.999, and so on will eventually reach a number larger than x⁠⁠.

If we set x = 0.̅9 then the sequence will never reach a number larger than x. 2A01:799:39E:1300:F896:4392:8DAA:D475 (talk) 12:16, 4 October 2024 (UTC)Reply

If x = 0.̅9 then x is not less than 1, so the conditional statement is true. What is the error? MartinPoulter (talk) 12:50, 4 October 2024 (UTC)Reply

If you presuppose that 0.̅9 is less than one, the argument that should prove you wrong may apprear to be sort of circular. Would it be better to say "to the left of 1 on the number line" instead of "less than 1"? I know it's the same, but then the person believing 0.̅9 to be less than one would have to place it on the number line! Nø (talk) 14:47, 4 October 2024 (UTC)Reply

What does the notation 0.̅9 mean? Johnjbarton (talk) 15:43, 4 October 2024 (UTC)Reply

It means zero followed by the decimal point, followed by an infinite sequence of 9s. Mr. Swordfish (talk) 00:24, 5 October 2024 (UTC)Reply

Thanks! Seems a bit odd that this is curious combination of characters (which I don't know how to type) is not listed in the article on 0.999... Johnjbarton (talk) 01:47, 5 October 2024 (UTC)Reply

B and C

Latest comment: 10 days ago4 comments2 people in discussion

@Tito Omburo. There are other unsourced facts in the given sections. For example:

There is no source mentions about "Every element of 0.999... is less than 1, so it is an element of the real number 1. Conversely, all elements of 1 are rational numbers that can be written as..." in Dedekind cuts.
There is no source mentions about "Continuing this process yields an infinite sequence of nested intervals, labeled by an infinite sequence of digits ⁠ $b 1, b 2 ⁠⁠, b 3, ...$ , and one writes..." in Nested intervals and least upper bounds. This is just one of them.

Dedhert.Jr (talk) 11:00, 30 October 2024 (UTC)Reply

The section on Dedekind cuts is sourced to Richman throughout. The paragraph on nested intervals has three different sources attached to it. Tito Omburo (talk) 11:35, 30 October 2024 (UTC)Reply

Are you saying that citations in the latter paragraph supports the previous paragraphs? If that's the case, I prefer to attach the same citations into those previous ones. Dedhert.Jr (talk) 12:52, 30 October 2024 (UTC)Reply

Not sure what you mean. Both paragraphs have citations. Tito Omburo (talk) 13:09, 30 October 2024 (UTC)Reply

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