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 featured article; it (or a previous version of it) has been identified as one of the best articles produced by the Wikipedia community. Even so, if you can update or improve it, please do so.

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

Current status: Featured article

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Fractions and long division

it seems like asserting that .1111... = 1/9 is circular to proving that .999...=1. I'm not saying it's not true, but if you don't believe that .9999..=1 then why would you believe that .3333...=1/3 or .1111...=1/9 — Preceding unsigned comment added by 24.19.2.53 (talk) 05:12, 27 August 2014 (UTC)[reply]

Long division cranks out an endless sequence of threes when applied to 1/3, and an endless sequence of ones when applied to 1/9. Although neither of these are infinite series, since no-one has infinite time to carry out long division, they add plausibility to the idea of these as infinite series. Since long division is a simple procedure we learn in school, and gives the correct results for other problems, it's easier for people to believe in than more abstract procedures. -- The Anome (talk) 09:11, 27 August 2014 (UTC)[reply]

Those are absolutely infinite series, btw.

{\frac {1}{9}}=\sum _{k=1}^{\infty }({\frac {1}{10}})^{k}

, multiply by 3 for 1/3. No one may have "infinite time to carry out long division," but that doesn't mean the solution isn't provably infinite. Gnassar (talk) 09:44, 29 November 2014 (UTC)[reply]

Dedekind cuts

According the Dedekind cut section, a real number is a subset of the set of rational numbers, but a rational number is a kind of real number, which means that all rational numbers contain themselves which according to ZF is impossible. Blackbombchu (talk) 00:18, 12 May 2014 (UTC)[reply]

Technically speaking, a mathematician would say that there is a subset of the real numbers that is in one-to-one correspondence with the rational numbers, and that the real number addition and multiplication exactly correspond to the rational number addition and multiplication. In other words there is a set within the real numbers that is isomorphic to the rational numbers as ordered fields. So while the are different we choose not to distinguish them in most contexts. Thenub314 (talk) 00:44, 12 May 2014 (UTC)[reply]

More generally, given any set S, I can form the set

S'=\{\{x\}|x\in S\}

, where obviously S and S' are in bijection. So while there's an obvious mapping which sends x to {x}, neither x nor {x} contain themselves as an element. Something similar happens for the rational numbers. Huon (talk) 21:22, 12 May 2014 (UTC)[reply]

This seems to really belong in talk:Dedekind cut, as it's not a question somehow unique to 0.999.... But really, if I'm understanding the comment correctly, it has a simple answer: The Dedekind cut of a rational number is the set of all real numbers *less* than that rational number. Rational numbers, therefore, do not "contain themselves." Gnassar (talk) 09:25, 29 November 2014 (UTC)[reply]

I'm a bit late to the party here, but while it may be true that the Dedekind cut corresponding to a rational number does not contain that rational number, that actually isn't the real point here.

The real point is the one Thenub314 made above. The real number corresponding to a particular rational number, in this approach to their construction at least, is not the same as that rational number. So literally speaking, Blackbombchu's claim that "a rational number is a kind of real number" is wrong.

That said, practically no one ever speaks quite that literally, except in very peculiar contexts like this one. For almost all purposes, it's just fine to say that a rational number is a kind of real number, and in fact people would look at you funny if you said the opposite. Just the same, in this context, it's not literally true.

To keep this on-topic, does this point need to be clarified in the article? I haven't checked. --Trovatore (talk) 20:30, 28 December 2015 (UTC)[reply]

Graphical representation of the limit

Section Infinite series and sequences includes a graphical representation of the limit, but it's base-4 and doesn't represent the $\epsilon$ measure of "arbitrarily closeness". I think we can add a second graph representing the distance |x − xn| as a segment on the real line, and showing how it becomes closer to 1 than 1- $\epsilon$ as the series increases. This would provide a redundant, visual representation of the concept already explained in that section, which could help us visual thinkers. Diego (talk) 10:33, 15 October 2014 (UTC)[reply]

Subpage nominated at MfD

I have begun a discussion of this talk page's "Arguments" subpage at MfD, because it is being used as a forum in violation of Wikipedia policy. Lagrange 613 04:18, 16 October 2014 (UTC)[reply]

Update: the MfD was closed as keep, for the same reasons as before: the existence of that page is the lesser of two evils. There was, however, one very good argument there that it should really be part of the Reference Desk system: perhaps we should consider moving it to Wikipedia:Reference desk/0.999... or Wikipedia:Reference desk/Mathematics/0.999...? -- The Anome (talk) 12:54, 3 November 2014 (UTC)[reply]

No citations in the introduction

Hi, after reading the article, I saw no citations in the introduction, and only one in algebraic proof. It is a valid argument, but it is not verifiable. Please see what you can do to fix this. Thanks! The f18hornet (talk) 19:25, 20 March 2015 (UTC)[reply]

See WP:LEADCITE. The lead does not generally require citations. If there is a direct quotation or other element that requires it then yes, but otherwise citations generally can be found in the main sections of the article. Repeating them wholesale in the lead is unnecessary and can lead to excessive clutter – where for example a paragraph summarises a whole section and so is based on all the sources in that section. If there are particular problem statements then they should be checked and if necessary reworked or sourced, but tagging every paragraph as you did is excessive.--JohnBlackburne^words_deeds 19:32, 20 March 2015 (UTC)[reply]

Okay, Good to know! Thank you for helping. I will point out however, some things in the fourth paragraph of the introduction:

The equality 0.999... = 1 has long been accepted by mathematicians^{[by whom?]} and is part of general mathematical education^[where?]. Nonetheless, some students^[who?] find it sufficiently counterintuitive that they question or reject it. Such skepticism is common enough that the difficulty of convincing them of the validity of this identity has been the subject of numerous studies in mathematics education^[vague].

These are just some of the things that I saw that could have been written better. So please put these issues into consideration. Thanks! The f18hornet (talk) 20:10, 20 March 2015 (UTC)[reply]

Hi there. I've reverted your changes, because every one of these points is already addressed in the body of the article, with citations to reliable sources as and where appropriate. Which mathematicians? We name and cite a representative sample (eg. Leonhard Euler, Tom Apostol and Georg Cantor, to name just the first three I registered in a quick glance at the article). General mathematical education? See the section on mathematical education. Some students? See the section on mathematical education. Numerous studies? See the section on mathematical education, where we cite several. -- The Anome (talk) 17:32, 21 March 2015 (UTC)[reply]

Problem rendering formulas in this article

In the Fractions and long division section, the formulas that use the "math & /math" format don't seem to render correctly. For example, the first one simply appears as, "\begin{align} \frac{1}{9} & = 0.111\dots \\ 9 \times \frac{1}{9} & = 9 \times 0.111\dots \\ 1 & = 0.999\dots \end{align}". Is this a problem with this article or is it the browser's? I noticed that another article that used the "{{convert|}" template rendered correctly. Am I the only person with this problem? __209.179.0.121 (talk) 02:22, 28 December 2015 (UTC)[reply]

No problem here, using standard Firefox 43.0.1 and IE 11 on Win8.1 Pro. Formulas are perfect, both in PNG and MathML — see Preferences, Appearance, Math. - DVdm (talk) 14:53, 28 December 2015 (UTC)[reply]

Then it must be I. Time to upgrade, again. Thanks for your help. __209.179.0.121 (talk) 02:34, 29 December 2015 (UTC)[reply]

No problem. Are you still working with IE3 on Win 95?

- DVdm (talk) 10:25, 29 December 2015 (UTC)[reply]

Actually, I don't use Windoze at all, unless I'm forced to, like when I was still working. I'm one of those people who uses a Mac, and have proudly since I got my first one in 1985. Why my browser is behaving oddly is a mystery, as it no longer displays pictures, either. Again, thanks. __209.179.0.121 (talk) 16:28, 2 January 2016 (UTC)[reply]