Talk:0.999...

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.

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This article was nominated for deletion on May 5, 2006. The result of the discussion was Keep (early closure).

Question about the FAQ

Hi. The answer to the second question in the FAQ says that "...an infinite string of zeroes cannot be followed by a 1." While this is, of course, true of decimal expansions, one can have a well-ordered set of order-type $\omega +1$ consisting of countably many zeros followed by a 1. At the risk of feeding the trolls, it may be worth rephrasing the answer to that question (though I cannot immediately think of the right way to do so). Molinari 19:53, 31 October 2006 (UTC)[reply]

It might be counterproductive to try to explain things concerning ordinals and cardinals to those who refuse to accept that 1=1, but you're welcome to try. --King Bee 21:27, 31 October 2006 (UTC)[reply]

Point taken. Molinari 23:34, 31 October 2006 (UTC)[reply]

The answer also says that "0.000...1 is not a meaningful string of symbols", but then goes on to discuss exactly what this string of symbols means, i.e., an infinite string of zeros followed by a one. That notion may be self-contradictory but it is not meaningless as the discussion of its meaning clearly shows.Davkal 21:37, 31 October 2006 (UTC)[reply]

The Riemann Zeta Function might not be a meaningless string of symbols either, but it's out of reach for the people disagreeing with the fact that 1=1, and probably has no place here; just as a lengthy discussion of ordinals should not be here either. --King Bee 21:47, 31 October 2006 (UTC)[reply]

That may well be so, but I don't see what it has to do with the error referred to above, i.e., not meaningful versus self-contradictory.Davkal 22:12, 31 October 2006 (UTC)[reply]

I have no idea what you're talking about. The response in the FAQ is not contradictory. It's a response to those who do not want to grasp a concept of infinity. It succeeds by asking them questions about what they just wrote down, making them find their own flaw. --King Bee 22:23, 31 October 2006 (UTC)[reply]

It's contradictory inasmuch as it says "x is not meaningful" and then proceeds to explain exactly what x means. Davkal 22:27, 31 October 2006 (UTC)[reply]

I rewrote the second answer; the above discussion seems to indicate that it was not convincing as it should have been. I believe my contribution is less dogmatic, less confusing, and appeals more to what is consistent with definitions. I also got rid of the Archimedean property link, since that has to do with abstract algebra, and gives a lot of abstract algebra before actually saying that the reals have the property. A novice would not understand it, and, worse, might say that the real numbers don't necessarily lack the property. Calbaer 23:44, 31 October 2006 (UTC)[reply]

Hi. I wrote the original answer. The current version is fine too. What I meant by "meaningless" is that the string of symbols does not represent a real number; it is not a well-formed string in the language of the system. I was reading a book by Douglas Hofstadter called Godel, Escher, Bach and picked it up from there. Maybe I am using the terminology incorrectly. Check chapter two if you have the book, entitled "Meaning and Form in Mathematics." I like what's up now just as well. Argyrios 01:02, 1 November 2006 (UTC)[reply]

Glad to hear it. I didn't like the old version (which I must admit I thought of editing prior to Molinari's comments) because saying something is meaningless without saying why seems like a dismissal of opponents without appeal to reason. If I were convinced that 0.000...1 meant something, someone calling it "meaningless" would not change my opinion one iota. However, pointing out that the "1" must have a well-defined, finite place explains why there's no such thing as 0.000...1. A doubter might say, "Well, there should be," but he or she must in the end admit that decimal notation — as it exists — doesn't allow for 0.000...1. Since that seems to be a big stumbling block — rather than claims that 0.999... isn't a real number or that not all real numbers can be expressed in decimal notation — that should be precisely addressed. Calbaer 01:17, 1 November 2006 (UTC)[reply]

If you tell us at what value N for :

$\sum _{n=1}^{N}{\frac {9}{10^{n}}}$ is 1, then I will tell you the position of the 1 in 0.000...1. --68.211.195.82 13:45, 1 November 2006 (UTC)[reply]

There does not exist a value of N for which that expression above is equal to 1. That's what you don't understand. --King Bee 14:47, 1 November 2006 (UTC)[reply]

There is no value of N for which that expression is equal to 1, this is true. But it is also true that there is no value of N for which that expression is equal to 0.999..., either. There is not a finite number of nines in 0.999..., you cannot represent that number of nines by an integer. Maelin (Talk | Contribs) 16:08, 1 November 2006 (UTC)[reply]

Time for a comment?

I've noticed that many pages that are subject to vandalism/controversy have an HTML comment at the beginning saying something like "The content of this article is well-established. If you plan to make a significant change, please consider discussing it on the talk page." Is it time for 0.999... to have one? Confusing Manifestation 01:49, 1 November 2006 (UTC)[reply]

That's not a bad idea. However, keep in mind that that won't deter some users with a seemingly religiously held conviction that the entire article is wrong, so we should do our best to divert them to the Arguments page. Supadawg (talk · contribs) 02:04, 1 November 2006 (UTC)[reply]

That's what I thought. What about a wording like "WARNING: This article contains several proofs that 0.999... = 1. If it is your intention to try and disprove this, please see the Arguments page at http://en.wikipedia.org/wiki/Talk:0.999.../Arguments first, as most of the common objections are dealt with repeatedly there. Also, please understand that the earlier, more naive proofs are not as rigorous as the later ones as they intend to appeal to intuition, and as such may require further justification to be complete. If there is a significant flaw with a proof, however, please discuss it on the Talk page." Confusing Manifestation 02:12, 1 November 2006 (UTC)[reply]

Done. I hope this is a good wording. Supadawg (talk · contribs) 02:23, 1 November 2006 (UTC)[reply]

Proofs need to be reexamined

Moved to Talk:0.999.../Arguments#Proofs need to be reexamined by Calbaer 01:30, 2 November 2006 (UTC)[reply]

Interwiki

Is anyone else surprised that there exists a mathematics article in 14 languages such that none of them is German? Melchoir 19:15, 1 November 2006 (UTC)[reply]

I am. I suggest we get someone to translate this to the language of our Vaterland so that the Germans can start arguing over the correctness of rigorous proofs. =) --King Bee 20:32, 1 November 2006 (UTC)[reply]

Maybe Germans just get it and so are busy laughing at these silly English-speaking people and no one has bothered to translate/write and article on such a silly topic which is understood by everyone Nil Einne 10:12, 5 November 2006 (UTC)[reply]

How can it be a real number and a hyperreal?

A representation of a number can't be two different numbers in two different number sets. If you want to express a hyperreal number that is infinitely close to 1, but not one, what would you use? Just like 999/1000 doesn't = 1 in the set of integers, $0.{\bar {9}}$ shouldn't = 1 in the reals. Fresheneesz 23:00, 1 November 2006 (UTC)[reply]

Since the real numbers are vastly more common than the hyperreals and since

0.{\bar {9}}

is a well-defined real number, in order to avoid confusion between the real 0.9999... and the hyperreal one you would have to rename the hyperreal version. But I have yet to see a case where it was in doubt whether the real number or the hyperreal was meant. --Huon 23:07, 1 November 2006 (UTC)[reply]

There is no hyperreal named 0.999…, and the existence of an article here is not an invitation to start making stuff up. Melchoir 23:31, 1 November 2006 (UTC)[reply]

I'm no expert, but I believe there is indeed a hyppereal 0.999…, since the hyperreals are a superset of the reals. Inverse hyperreals are infinitesimals, so you could say that 1-e is "infinitely close" to 1 but not equal to 1, where e is some infinitesimal. But 0.999… is 1, whether it's a real or a hyperreal. But more importantly, any discussion of hyperreals is irrelevant to this article, which is a discussion of (and limited to) the decimal representation of real numbers. — Loadmaster 23:53, 1 November 2006 (UTC)[reply]

999/1000 isn't in the rationals; 0.999... is in the reals. No mathemetician uses 0.999... to mean 1-e (with e an infintesimal in the hyperreals), because that would conflict with the usual meaning of decimal expansions of real numbers. -- SCZenz 06:50, 2 November 2006 (UTC)[reply]

999/1000 isn't an element of the rationals? --King Bee 15:24, 2 November 2006 (UTC)[reply]

SCZenz probably meant "integers". The hyperreals have infinitely many infinitesimals. One of them is ε. Accordingly, there are infinitely many hyperreals infinitesimally close to 1. One of them is 1 - ε. And you still didn't explain where you got the idea that 0.999... is a hyperreal. One would rarely (if ever) use 0.999... to represent a hyperreal number (as it is a decimal expansion, and decimal expansions are reserved for reals), and one definitely does not need such a notation to represent numbers infinitely close to 1. -- Meni Rosenfeld (talk) 21:38, 2 November 2006 (UTC)[reply]

I am under the impression that 0.999… is both a real (obviously) and a hyperreal (not so obviously) because of what Wiki says about hyperreals:

The hyperreals, or nonstandard reals, (usually denoted as *R) denote an ordered field which is a proper extension of the ordered field of real numbers R and which satisfies the transfer principle.

Perhaps I'm reading it wrong, but I take that to mean every real is also a hyperreal. But whatever the case, it's irrelevant to this article, which is about the real value 0.999…. — Loadmaster 22:47, 2 November 2006 (UTC)[reply]

That's a somewhat subtle issue. You can define things in a way that the reals would be a subset of the hyperreals, but the other approach (which is clearer) is to define them as two disjoint sets. My opposition to 0.999... being a hyperreal is clear with the second approach; as for the first, Fresheneesz implied that 0.999... is a non-real hyperreal, which is flawed. Your interpretation, that 0.999... is a hyperreal by virtue of being the real number 1, is more sensible, but is not what Fresheneesz was getting at. -- Meni Rosenfeld (talk) 07:17, 3 November 2006 (UTC)[reply]

Proposal to globally change inline mathematics into sans-serif

There is currently a proposal suggesting this addition to the global CSS of Wikipedia:

span.texhtml {
 font-family: sans-serif;
}

This would mean that inline mathematics would be displayed in sans-serif rather than serif. This proposal was shot down twice before, but it seems that the strange nature of Wikipediamocratics allow for it to be suggested a third time. Maybe you would like to take a look there and partake in the discussion, since you're all active editors in the field of mathematics. —msikma <user_talk:msikma> 06:54, 2 November 2006 (UTC)[reply]

A proof of the more general concept

One interesting thing to note is that there are other ways of lexicographically representing real numbers in a fixed range (say, [0,2]) by infinite strings of digits (or bits or other alphabets), and they all have the multiple representation "problem." By lexicographically, I mean that x >= y if and only if x is after y in alphabetical (dictionary) order. In any such system, there will be no string between "0hhh..." and "1000..." where 'h' is the highest digit of the alphabet, 9 in decimal. If these two strings represent two different real numbers, their average will fail to have a represenation. Thus they must represent the same real number. An interesting example of an alternative representation is LCF ([1]), in which the representation is based on continued fractions, and the alternative representations, rather than being 0.999... and 1.000, are $0+{\frac {1}{1+0}}$ and $1+{\frac {1}{\infty }}$ . I'd add this, but I'm not sure how to do so without possibly causing additional confusion. Calbaer 20:58, 5 November 2006 (UTC)[reply]

A counterproof?

Is this a valid argument?

Tan(89.999999....) = infinity

Tan(90) = undefined

Savager 16:56, 8 November 2006 (UTC)[reply]

No. --King Bee 17:04, 8 November 2006 (UTC)[reply]

To be precise, the reason it's invalid is that it assumes that, if

\lim _{x\uparrow 90}x=90

, then

\lim _{x\uparrow 90}\tan(x)=\tan(90)

. This is untrue; if something were always equal to its limit given a parameter that is equal to its limit, there would be no need for calculus! Calbaer 17:40, 8 November 2006 (UTC)[reply]

Student stories

Many of the examples of students misunderstanding this concept are apocryphal and not referenced. It seems that they should be rewritten or removed.