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

Intuitive explanation

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

@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]

Intuitive counterproof

The logic in the so-called intuitive proofs (rather: naïve arguments) relies on extending known properties and algorithms for finite decimals to infinite decimals, without formal definitions or formal proof. Along the same lines:

0.9 < 1
0.99 < 1
0.999 < 1
...
hence 0.999... < 1.

I think this fallacious intuitive argument is at the core of students' misgivings about 0.999... = 1, and I think this should be in the article - but that's just me ... I know I'd need a source. I have not perused the literature, but isn't there a good source saying something like this anywhere? Nø (talk) 08:50, 29 November 2024 (UTC)[reply]

Greater than or equal to

I inserted "or equal to" in the lead, thus:

In mathematics, 0.999... (also written as 0.9, 0..9, or 0.(9)) denotes the smallest number greater than or equal to every number in the sequence (0.9, 0.99, 0.999, ...). It can be proved that this number is 1; that is,

0.999...=1.

(I did not emphasize the words as shown here.) But it was reverted by user:Tito Omburo. Let me argue why I think it was an improvement, while both versions are correct. First, "my" version it s correct because it is true: 1 is greater than or equal to every number in the sequence, and any number less than 1 is not. Secondly, if a reader has the misconception that 0.999... is slightly less than 1, they may oppose the idea that the value must be strictly greater than alle numbers in the sequence - and they would be right in opposing that, if not in this case, then in other cases. E.g., 0.9000... is not greater than every number in the corresponding sequence, 0.9, 0.90, 0.900, ...; it is in fact equal to all of them. Nø (talk) 12:07, 29 November 2024 (UTC)[reply]

I think it's confusing because 1 doesn't belong to the sequence, so "or equal" are unnecessary extra words. A reader might wonder why those extra words are there at all, and the lead doesnt seem like the place to flesh this out. Tito Omburo (talk) 13:40, 29 November 2024 (UTC)[reply]

Certainly, both fomulations are correct. This sentence is here for recalling the definition of the notation in this specific case, and must be kept as simple as possible. Therefore, I agree with Tito. The only case for which this definition of ellipsis notation is incorrect is when the ellipsis replaces an infinite sequence of zeros, that is when the notation is useful only for emphasizing that finite decimals are a special case of infinite decimals. Otherwise, notation 0.100... is very rarely used. For people for which this notation of finite decimals has been taught, one could add a footnote such as 'For taking into account the case of an infinity of trailing zeros, one replaces often "greater" with "greater or equal"; the two definitions of the notation are equivalent in all other cases'. I am not sure that this is really needed. D.Lazard (talk) 14:46, 29 November 2024 (UTC)[reply]