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== Yet another anon ==
'''Previous discussions:'''
''Moved to ] subpage''


== Intuitive explanation ==
*] (] to ])
*] (] to ])
*] (] to ])
*] (] to ])


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. ] (]) 12:16, 4 October 2024 (UTC)
:If x = 0.̅9 then x is not less than 1, so the conditional statement is true. What is the error? ] (]) 12:50, 4 October 2024 (UTC)
::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! ] (]) 14:47, 4 October 2024 (UTC)
:What does the notation 0.̅9 mean? ] (]) 15:43, 4 October 2024 (UTC)


::It means zero followed by the decimal point, followed by an infinite sequence of 9s. ] (]) 00:24, 5 October 2024 (UTC)
== Rules of engagement ==
:::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... ] (]) 01:47, 5 October 2024 (UTC)


== B and C ==
Dear all,


@]. There are other unsourced facts in the given sections. For example:
I assume all of us are well-meaning people, wanting a good result for this article. That makes me ask the following favor to anonymous contributors:
* 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 ], labeled by an infinite sequence of digits ⁠{{math|1=''b''<sub>1</sub>, ''b''<sub>2</sub>⁠⁠, ''b''<sub>3</sub>, ...}}, and one writes..." in Nested intervals and least upper bounds. This is just one of them.
] (]) 11:00, 30 October 2024 (UTC)


: The section on Dedekind cuts is sourced to Richman throughout. The paragraph on nested intervals has three different sources attached to it. ] (]) 11:35, 30 October 2024 (UTC)
* Please make an account. It is not productive for us to deal with a person who always uses a different IP address. It is impossible for us to keep in touch with you this way. Making an account will take you five secods. Just choose an imaginary username which has nothing to do with your real name, and a password. No more. But so much gained.
::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. ] (]) 12:52, 30 October 2024 (UTC)
:::Not sure what you mean. Both paragraphs have citations. ] (]) 13:09, 30 October 2024 (UTC)


== Intuitive counterproof ==
* Please sign your posts. Use four tildas for that, like this: <nowiki>~~~~</nowiki>. I would really, really ask you to give it a try. You leave a lot of unsigned comments and nobody can tell which is what you wrote, and which is somebody else, and which is another anonymous user.


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:
I truly appreciate you taking your time to read all this, and do us a the small favors I asked which will take you a very small amount of time (not infinitesimal, but close :) and will make it so much more pleasant for us to have constructive discussions. Sincerely, ] (]) 06:04, 18 December 2005 (UTC)
* 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? ] (]) 08:50, 29 November 2024 (UTC)


== Greater than or equal to ==
As an aside, a discussion about this very topic has been hotly debated by an on-line society for people with a high IQ. Keep it going SM! {{unsigned|68.148.229.166|06:30, 2005 December 18}}


I inserted "or equal to" in the lead, thus:
What?! You mean Hardy is discussing this without me?! How can this be? I too have a high IQ (over 140) -not that it means anything. The IQ concept was developed by a '''human''' who I believe had to wipe his arse every time he had a BM. Do you think Hardy wipes or washes? I '''wash only'''. Toilet paper is for those who have bad hygienne and stinky butts. Although I don't care to0 much for Islam, this is something we could learn from them. Wait, I think the ancient Greeks invented the bidet if I am not mistaken? Oops, I think Melchoir is going to censor this when he wakes up. Hopefuly a few people will get to read it before he does and have a good laugh if nothing else. Not that it's any of my business but I am infinitesimally (singular ONLY) curious, do you wash or wipe Oleg? ] 14:14, 18 December 2005 (UTC)
:In ], '''0.999...''' (also written as '''0.{{overline|9}}''', '''0.{{overset|.|9}}''', or '''0.(9)''') denotes the smallest number greater than '''''or equal to''''' every ] in the sequence {{nowrap|(0.9, 0.99, 0.999, ...)}}. It can be proved that this number is{{spaces}}]; that is,
:: <math>0.999... = 1.</math>
(I did ''not'' emphasize the words as shown here.)
But it was reverted by ]. 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. ] (]) 12:07, 29 November 2024 (UTC)
: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. ] (]) 13:40, 29 November 2024 (UTC)
::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. ] (]) 14:46, 29 November 2024 (UTC)
:::Could you point to where the values of decimals are defined in this way - in wikipedia, or a good source? I can eassily find definitions in terms of limits, but not so easily with inequality signs (strict or not).
:::I think the version with strict inequality signs is weaker in terms of stating the case clearly for a skeptic. ] (]) 17:45, 30 November 2024 (UTC)
:::Agree that both versions are correct. My inclination from years of mathematical training is to use the simplest, most succinct statement rather than a more complicated one that adds nothing. So, I'm with Tito and D. here. ] (]) 18:24, 30 November 2024 (UTC)
::::I think many mathematicians feel that "greater than or equal to" is the primitive notion and "strictly greater than" is the derived notion, notwithstanding that the former has more words. Therefore it's not at all clear that the "greater than" version is "simpler". --] (]) 03:13, 1 December 2024 (UTC)
:::The general case is "greater than or equal to", and I would support phrasing it that way. I think we don't need to explain why we say "or equal to"; just put it there without belaboring it. --] (]) 03:06, 1 December 2024 (UTC)


== Image ==
I am not going to argue about infinitesimal. As far as I am concerned, this (notice I will not even consider the plural form - it makes no sense whatsoever to me) does not exist and you most certainly cannot prove that x+x+x+... < 1 for an infinite number of terms in x unless x is zero. You can only show in a similar manner to Rasmus that n*10^(-n) < 1 using induction. Now as I stated, one can show that 0.999... < 1 and that 0.999... = 1 depending on how you approach the proof. I believe that this anomaly exists because 0.999... is not a finitely represented number. This is a problem with the decimal system and all other radix systems. What this means is that the Archimedean property applies only to reals that can be finitely represented. Rasmus's proof is no better than the induction proof that 0.999... < 1 since it uses a result of induction to arrive at the conclusion that nx < 1. Rasmus justifies his argument by stating that because the Archimedean principle cannot be applied, x must be zero.
{{Discussion top|There is no consensus to remove the image, and a rough consensus to keep it. ] (]) 21:42, 10 January 2025 (UTC) }}
However, I maintain that the Archimedean property can only be applied to finitely represented numbers. For x > 1/n, n can take on the value of a suitable natural number but not infinity. To use the fact that 0.999... has an upper bound in a proof such as Rasmus's defeats the purpose. So what should be believed? I think that 0.999... should be considered less than 1 because it has to be considered in the context of the decimal system. If the full extent of 0.999... were known, there would be no problem with the Archimedean property or any of its corollaries. ] 18:49, 21 December 2005 (UTC)


The image included at the top of this article is confusing. Some readers may interpret the image to mean that 0.999... represents a sequence of digits that grows over time as nines are added, and never stops growing. To make this article less confusing I suggest that we explicitly state that 0.999... is not used in that sense, and remove the image. ] (]) 17:31, 1 January 2025 (UTC)
== Infinitesimals ==


: I do not see how this is confusing. The caption reads: "Stylistic impression of the number 0.9999..., representing the digit 9 repeating infinitely" - nothing remotely like "sequence... that grows over time". I cannot see how one could meaningfully add a comment that "0.999..." is not used in a sense that has not even been mentioned. Of course lots of people are confused: that is the reason for the article, which in an ideal world would not be needed. ] (]) 04:29, 2 January 2025 (UTC)
I have not yet given up on the "infinitesimal" proof, especially since anon agrees there are no infinitesimals in the field of real numbers. (Compare the 0:16 post of 18 December 2005. By the way, meaningful concepts of infinitesimals in larger number sets can be found in ], part 1, which is written on a rather basic level.)
::If a sequence of digits grows over time as nines are added, and never stops growing, it is reasonable to conclude that the digit nine is repeating infinitely. ] (]) 18:14, 2 January 2025 (UTC)

:::Yes, notation 0.999... means that the digit nine is repeating infinitely. So, the figure and its caption reflect accurately the content of the article. ] (]) 18:28, 2 January 2025 (UTC)
0.999... and 1 are real numbers. Let x = 1-0.999... be their difference. Note that I do not care whether I can give a decimal representation for x. Note also that I do not make claims about the existence of numbers between 0.999... and 1. All I claim is that I can subtract, and that the difference of two real numbers is again a real number. And that is due to the fact that the real numbers are a ].
::::When we use the word repeating we should expect that some people will think we are referring to a process which occurs over time, like the operation of a ]. ] (]) 22:03, 2 January 2025 (UTC)
By definition, a number y ≠ 0 is an ] if every sum |y|+...+|y| of finitely many terms is less than 1, no matter how large the finite number of terms. We agreed such a thing does not exist in the reals. Now form any sum |x|+...+|x| of finitely many terms, say n terms. Obviously, |x| = 1-0.999... < 1/n. Thus, |x|+...+|x|<1. So if x were greater than 0, it would satisfy the definition given above. That can't be, since we agreed that no real number is an infinitesimal. Our only way out is x=0. Thus, 0.999...=1.
:::::You can think of this as a "process" if you like. 0.9999... means the limit of the sequence . Of course in mathematics nothing ever really "occurs over time", though I suppose you could consider it a kind of algorithm which if implemented on an idealization of a physical computer with infinite memory capacity might indefinitely produce nearer and nearer approximations. –]&nbsp;] 22:20, 2 January 2025 (UTC)

::::::I think you are going in a very productive direction. We should explain to readers how what they might think we mean, "occurring over time", relates to what we actually mean. ] (]) 00:43, 3 January 2025 (UTC)
If there are problems with this proof, please be precise in denoting them.--] 00:00, 20 December 2005 (UTC)
:::::::I personally think that would be distracting and not particularly helpful in the lead section. There is further discussion of this in {{alink|Infinite series and sequences}}, though perhaps it could be made more accessible. –]&nbsp;] 03:42, 3 January 2025 (UTC)

::::::::Yes, I agree that detailed discussion does not belong in the lead section. I personally think that the image is distracting and not helpful. In the lead section we can simply state that in mathematics the term 0.999... is used to denote the number one. We can use the rest of the article to explain why. ] (]) 16:23, 3 January 2025 (UTC)
The proof would be '''flawless''' if your definition of infinitesimal is true. Only problem is it is not true because it '''does matter''' how large the finite number of terms become. If you feel comfortable that the sum of these terms will always be less than 1, how is it that you do not feel the same way about the sum of 9/10+9/100+9/1000+... ? I can make the same statement here, i.e. for finitely many terms, this sum will always be less than 1. So what?! ] 02:54, 20 December 2005 (UTC)
::::::::: Except that it's not true that 0.999... denotes the number one. It denotes the least number greater than every element of the sequence 0.9, 0.99, 0.999,... It's then a theorem that the number denoted in this way is equal to one. ] (]) 16:31, 3 January 2025 (UTC)

::::::::::It also denotes the least number greater than every number which is less than one, just as 0.333...denotes the least number greater than every number which is less than one-third. That's why we say it denotes 1/3, and why we also say that the one with 9s denotes 1. ] (]) 17:39, 3 January 2025 (UTC)
: Hmmm, I had sort of given up on this discussion, but one last try: Let x = 1-0.999... as above. Consider the set S = {x, 2x, 3x, 4x, ...} = {nx|n in N}. (If you don't agree with me setting x = 1-0.999... , just consider the set S = { (1-0.999...), 2(1-0.999...), 3(1-0.999...), ... } = { n(1-0.999... ) | n in N} instead ). Could you answer these questions?
::::::::::@], notice that @] just wrote above "we also say that the one with 9's denotes 1". The description "the least number greater than every element of the sequence 0.9, 0.99, 0.999,..." does describe the number one, just as does "the integer greater than zero and less than two". ] (]) 18:21, 3 January 2025 (UTC)
:# Does S have an upper bound?
::::::::::: This is an incorrect use of the word "denotes". Denotes an equality by definition, whereas one instead has that 0.999... and 1 are ''judgementally'' equal. For example, does "All zeros of the Riemann zeta function inside the critical strip have real part 1/2" denote True or False? ] (]) 18:56, 3 January 2025 (UTC)
:# Does S have a least upper bound?
::::::::::::I think you are inventing this - please find reliable sources (dictionaries and things) to back up your claimed meaning of "denote". ] (]) 04:55, 9 January 2025 (UTC)
:# If S has a least upper bound, what is it? If you can't give an exact answer, can you give an interval (ie. 0.5 < sup S < 1)?
::::::::::I agree that it is better to write that the term is used to denote the number one, rather than that the term denotes the number one. ] (]) 20:06, 3 January 2025 (UTC)
:# If 1-0.999... != 0, then 1/(1-0.999...) must be a real number. Can you describe the properties of 1/(1-0.999...)? For instance is there any natural number n, so that 1/(1-0.999...) < n ?
:::::::::::Its not "used to denote". It is a mathematical theorem that the two terms are equal. ] (]) 20:46, 3 January 2025 (UTC)
: ] ] 07:23, 20 December 2005 (UTC)
::::::::::I think we can make this issue very clear. Assume that x equals the least number greater than every element of the sequence 0.9, 0.99, 0.999,... . Applying the theorem we learn that x = 1. Substituting 1 for x in the opening sentence of this article we have: In mathematics 0.999... denotes 1. If we also insist that 0.999... does not denote 1, we have a contradiction. ] (]) 18:45, 4 January 2025 (UTC)

:::::::::::You have redefined the word "denote" to mean precisely the same as "is equal to", which is confusing and unnecessary. It's better to just say "is equal to" when that's what you mean, so that readers are not confused. –]&nbsp;] 18:56, 4 January 2025 (UTC)
::Answers:
::::::::::::I agree that redefining the word denote would be confusing and unnecessary. I simply defined a variable x to be equal to a number, the least number. ] (]) 20:04, 4 January 2025 (UTC)
1. S does not have an upper bound therefore it cannot have a least upper bound.
:::::I'm in agreement with @] and @] on this. The image does not suggest a process extended over time, and it correctly reflects the (correct) content of the article, so there is no need to remove it. I'm not persuaded that people will interpret "repeating" as purely temporal rather than spatial. If I say my wallpaper has a repeating pattern, does this confuse people who expect the wallpaper to be a process extended over time? (Are there people who think purely in firearm metaphors?) ] (]) 17:30, 3 January 2025 (UTC)
So questions 2 and 3 are not relevant.
::::::Consider the number 999. Like the wallpaper, it contains a repeating pattern. That pattern could be defined over time, one nine at a time. Or it could be defined at one time, using three nines. ] (]) 18:27, 3 January 2025 (UTC)

:Is it OK if I go ahead and edit the article, keeping in mind all the concerns which have been raised with my proposed changes? ] (]) 17:56, 8 January 2025 (UTC)
4. 1/(1-0.999...) is a real number. Properties: all we can say is that it is a
::Can you be more specific about which changes you want to implement? ] (]) 20:32, 8 January 2025 (UTC)
very large indeterminate number comparable with infinity. There is no natural
:::The first change would be to remove the image. ] (]) 15:06, 9 January 2025 (UTC)
number n, so that 1/(1-0.999...) < n.
::::I'm confused, @]. Where in the above discussion do you see a consensus to remove the image? You have twice said the image should be removed, and I have said it should stay. No matter how many times you express it, your opinion only counts once. Other users have addressed other aspects of your proposal. Do you sincerely think the discussion has come to a decision about the image? ] (]) 13:47, 10 January 2025 (UTC)

::No. I do not think there is agreement on removing the image. (I don't personally think it is spectacularly good, but the argument for removing it appears to me to be completely bogus.) ] (]) 04:57, 9 January 2025 (UTC)
So now you are going to conclude that since 4 is true that 1/(1-0.999...) is not a real number - yes? What about 1/(3.15-pi)? Is there a natural number n so that 1/(3.15-pi) < n ? ] 13:09, 20 December 2005 (UTC)
:::The term 0.999... is literally a sequence of eight characters, just as y3.p05&9 is. Yet, the term itself implies meaning. I think confusion about the term can be reduced simply by acknowledging different meanings the term might imply. It does imply different meanings to different people. We can respect everyone, including children who are not willing to simply accept everything a teacher tells them. We can do our best to help everyone understand what we mean when we use the term. ] (]) 15:32, 9 January 2025 (UTC)

:::For example, if a child thinks that by 0.999... we mean a sequence of digits growing over time, and the child objects when told that the sequence of digits is equal to one, we can respond by saying something like the following: You are correct that a growing sequence of digits does not represent one, or any number, because the sequence is changing. We don't mean that 0.999... represents a changing or growing sequence of digits. ] (]) 16:12, 9 January 2025 (UTC)
::: All members of S are of the form nx. Consider one such member, nx. As we saw a month ago, we can choose a natural number <math>m > \log_{10} (n) </math>, and use the fact that <math>0.999...>\sum_{i=1}^{m}\frac{9}{10^i}</math> to show that <math> x = 1-0.999... < 1 - \sum_{i=1}^{m}\frac{9}{10^i} = \frac{1}{10^m} < \frac{1}{10^{\log_{10} (n)}} = \frac{1}{n} </math>, and thus <math> nx < n \frac{1}{n} = 1 </math>. So all members of S are less than 1, yet you claim it has no upper bound?
:::We don't mean a changing or growing sequence of digits. That is what it is confusing to say that we mean a repeating sequence of digits. ] (]) 16:15, 9 January 2025 (UTC)
:::As for 4, you claimed that you accepted Planet Maths definition of the Archimedean property (or was that another anon?). You don't feel this is a contradiction? Planet Math claims: "Let xbe any real number. Then there exists a natural number n such that n>x".
:::What we mean is a number. ] (]) 16:18, 9 January 2025 (UTC)
:::And finally, of course there is a natural number n so that 1/(3.15-pi) < n. pi<3.142, so 1/(3.15-pi) < 1/(3.15-3.142) = 125.
::::This article is about the meaning of 0.999... '''in mathematics''' not about the possible meanings that people may imagine. If people imagine another meaning, they have to read the article and to understand it (this may need some work), and they will see that their alleged meaning is not what is commonly meant. If a child objects to 0.999... = 1, it must be told to read the elementary proof given in the article and to say which part of the proof seems wrong. ] (]) 16:58, 9 January 2025 (UTC)
:::] ] 15:28, 20 December 2005 (UTC)
:::What do we mean by the term number? A number is a measure, not a sequence of digits. We may denote a number using a sequence of digits, but we don't always. Sometimes we denote a number using a word, like one. Sometimes we use a phrase such as: the least number greater than any number in a certain sequence. We may use a lowercase Greek letter, or even notches in a bone. ] (]) 16:44, 9 January 2025 (UTC)

::::By the term "number", we mean a number (]). It is difficult to define a number, and this took several thousands years to mathematicians to find an acceptable definition. A number is certainly not a measure, since a measure requires a ] and numbers are not associated with any measurement unit. The best that can be said at elementary level is something like "the natural number three is the common property of the nines in 0.999..., of the consecutive dots in the same notation, and of the letters of the word ''one''". ] (]) 17:20, 9 January 2025 (UTC)
You have just proved that 0.999... < 1: nx < n*1/n = 1 => x < 1. If x is a real number greater than 0, there exists a natural n such that 0 < 1/n < x or nx > 1. So if x = 0 (which is what you would require for having 0.999... = 1) then no n exists such that nx > 1. Hence x must be greater than 0 and if x is greater than 0, then 0.999... must be less than 1. Now do the right thing and delete this garbage article. ] 17:35, 20 December 2005 (UTC)
:::::I see. A number is not a measure, but it is used to measure. Thanks. ] (]) 17:40, 9 January 2025 (UTC)

:::::A number is a value used to measure. ] (]) 17:42, 9 January 2025 (UTC)
:Could you clarify that argument please? In the above I showed how for all natural numbers n, nx < 1. If you accept that "If x is a real number greater than 0, there exists a natural n such that 0 < 1/n < x or nx > 1", the conclusion must be that x is not a real number greater than 0.
:::The caption on the image is: Stylistic impression of the number 0.9999..., representing the digit 9 repeating infinitely.
:You also didn't comment on the Archimedean property (are you the same person as ]?)
:::The caption can be understood to mean that the term 0.999... '''''is''''' a zero followed by a decimal point followed by the digit 9 repeating infinitely, which meaning is distinct from the meaning that 0.999... '''''denotes''''' the number one.
:] ] 22:28, 20 December 2005 (UTC)
:::If we retain the caption, we may communicate to readers that we mean that 0.999... '''''is''''' a repeating sequence, which sequence '''''denotes''''' the number one. That doesn't work because repeating sequences themselves cannot be written completely and and therefore cannot be used to notate.

:::0.999... '''''is''''' notation. The purpose of this article should be to help others understand what it denotes. If it denotes a repeating sequence of digits, then we should say so in the lead sentence. ] (]) 18:32, 9 January 2025 (UTC)
You showed nx < 1. The Archimedean property says there exists an n s.t. nx > 1. There is no x that satisfies nx < 1 and nx > 1. So how do you reach the conclusion that x = 0? You are looking only at nx < 1 and thus drawing the conclusion that x must be zero? Okay, let me try to understand what you are saying:
:::: How does the first sentence of the article not explain that notation? The meaning of the notation is the smallest number greater than every element of the sequence (0.9,0.99,...). ] (]) 18:39, 9 January 2025 (UTC)

:::::Because it does not make sense to say that the sequence is repeating, because all the nines have not already been added, and at the same time to say that the sequence represents a number, because all the nines have already been added. It is confusing because it is contradictory.
The Archimedean property shows the relationship between a natural number n and a number x greater than 0 such that nx > 1. This means that x and n must be greater than zero. Your proof demonstrates that a number x and some natural number n have the property that nx < 1. The only n that satifies this is n=0 for otherwise x must be zero. How do you associate the Archimedean property with your proof? They both state different facts. So what I am trying to say is this: if you are to draw any conclusion that is backed by the Archimedean property, then your proof must result in a form that resembles it, i.e. nx > 1 and not nx < 1. You cannot arrive at the conclusion that nx < 1 and then state by the Archimedean property that x is not a real number greater than 0. By demonstrating that x (1-0.999...) < 1/n for any n, you have proved conclusively that x is greater than 0 because for whatever 1/n you give me, I can always find an x that is smaller. This x is greater than zero and sounds very real to me. 00:26, 21 December 2005 (UTC)
:::::When we say that the sequence is repeating, people who are not trained in mathematics will likely assume that we mean that all the nines have not already been added, and therefore that the sequence is changing and therefore, does not represent a number. Which, I believe, is why the subject of this article is not more widely understood. ] (]) 19:05, 9 January 2025 (UTC)

:::::: I think I understand part of the confusion, which I've hopefully tried to correct with an edit. The notation 0.999... refers to a ], a concept which had not been linked. There is a way of associating to any decimal expansion a number as its value. For the repeating decimal 0.999..., that number is 1. ] (]) 19:09, 9 January 2025 (UTC)
:What I showed before were that for x=1-0.999... and ''all natural numbers'' n: nx<1. (I actually only wanted to use it for showing that S had an upper bound, since you had earlier rejected the application of the Archimedean property). Since the Archimedean property state that for all real x>0, there exists a natural number n, so that nx>1, we have a contradiction unless x ''is not'' a real number greater than 0.
:::::::I like the edits. Because the least number is one, the meaning of the lead sentence can be understood to be that 0.999... is a recurring decimal whose value '''''is defined as''''' one. The notation below should match. Instead of <math>0.999... = 1</math>, we should write <math>0.999... \ \overset{\underset{\mathrm{def}}{}}{=}\ 1</math>. ] (]) 19:40, 9 January 2025 (UTC)
:I can't make much sense of your last argument. You claim that "x (1-0.999...) < 1/n for any n" => "x is greater than 0"? I assume the parenthesis is just a clarification and not a multiplication, so that it is actually (for all n in N: x < 1/n) => (x > 0) ? Your argument for this seems to imply that you can change the x as you go?! Anyway x=-1 (or even x=0) is a counterexample, which, frankly, you ought to have been able to see for yourself.
:::::::: No. The point is that the notation has a definition which is a standard one for repeating decimals of this form. It is a ''theorem'' that this number is one, but that is not the definition. ] (]) 19:47, 9 January 2025 (UTC)
: ] ] 07:32, 21 December 2005 (UTC)
:::::::::I agree. What you are saying agrees with what I am saying. It is a theorem that the least number is one, not a definition. The notation has a standard definition which defines the notation to be equal to the least number, whatever that least number is.

:::::::::#Given that the notation is defined to be equal to the least number
Fine. I see your argument now. It's always been confusing because for any 1/n, I can always find an x that is smaller but not zero. In an earlier discussion, you maintained that the induction proof was incorrect because it does not show P(infinity). Do you realize that one can say the same to you regarding this argument? You may say that the lowest x one can find is zero but then you are assuming P(infinity) is true. So although your argument is valid, you have not shown P(infinity). It seems to me that one can show equally well by induction that 0.999... < 1 and using your method that 0.999... = 1. How can 0.999... be both less than and equal to 1? This is strange... ] 13:14, 21 December 2005 (UTC)
:::::::::#And given a theorem that the least number does equals one

:::::::::#Therefore the notation is defined to be equal to a number which does equal one.
:Well, the difference is that I don't need to go to the limit. To use the Archimedean property, I only need to show that for all ''finite'' natural numbers n: nx<1. I do not need to show that "<math> \infty x < 1 </math>" (whatever meaning one would assign to that statement). ] ] 14:07, 21 December 2005 (UTC)
:::::::::#Note that it does not follow from the givens that the notation is equal to one, or that the notation is equal to the least number.

:::::::::] (]) 20:23, 9 January 2025 (UTC)
One can say exactly the same for the induction proof, i.e. only need to show that 0.9999xn < 1. Same thing. Let me get one thing straight: you are also saying that if the Archimedean property does not apply, then x cannot be a real number, right? If your answer is 'yes', then the Archimedean property only applies to finitely represented reals in any radix system. 0.999... is not finitely respresented. ] 14:20, 21 December 2005 (UTC)
::::::::This is not correct, but I feel like we're talking in circles here. Cf. ]. {{pb}} Let me try one more thing though. If we wanted a more explicit ''definition'' of 0.999..., we might use mathematical notation and write something like <math display=block>0.999\ldots \ \stackrel{\text{def}}{=}\ \sum_{k=1}^{\infty} 9 \cdot 10^{-k} = 1.</math> This is discussed in the article in {{alink|Infinite series and sequences}}. –]&nbsp;] 02:58, 10 January 2025 (UTC)

:::::::::Can you see that the summation is a process which must occur over time, and can never end? Do you notice that k cannot equal 1 and 2 at the same time? However, if we insist that the summation does occur all at once, then we affirm that k does equal 1 and 2 at the same time. We affirm that we do intend contradiction. If so, then we should clearly communicate that intention. ] (]) 15:14, 10 January 2025 (UTC)
:Rasmus is going to say that
::::Please stop misusing the word ''denotes'' when you mean "is equal to". It's incredibly confusing. –]&nbsp;] 20:57, 9 January 2025 (UTC)
:* the Archimedean property is a property for the entire set of numbers, and it applies.
:::::I agree that the difference between the two is critical. I've tried to be very careful. ] (]) 21:13, 9 January 2025 (UTC)
:* x is indeed a real number, but not one that is greater than 0.
:::::I don't know if this will help at all, but it may. I think that we have been preoccupied with what infinity means, and have almost completely ignored what it means to be finite. We don't even have an article dedicated to the subject. So, I have begun drafting one: ]. ] (]) 00:00, 10 January 2025 (UTC)
:After all, what we are trying to show is just x=0. If you are now willing to sacrifice the Archimedean property for your brand of "real" numbers, you will probably agree that yours are not what mathematicians usually call the real numbers.
:I think the problem here is that there are two levels of symbol/interpretation. The literal 8-byte string "0.999..." is a "symbol for a symbol", namely for the infinitely long string starting with 0 and a point and followed by infinitely many 9s. Then that infinitely long symbol, in turn, denotes the real number 1.

:It's also possible that people are using "denote" differently; I had trouble following that part of the discussion. But we need to be clear first of all that when we say "0.999..." we're not usually really talking about the 8-byte string, but about the infinitely long string. --] (]) 05:06, 10 January 2025 (UTC)
:Concerning the definition of ]s I gave above: That was the Misplaced Pages definition; I just copied it. If you don't believe that definition to be correct, look it up in, say, ].
::This is also a weird use of "denote", in my opinion. For me, the word ''denote'' has to do with notation, as in, symbols that can be physically written down or maybe typed into a little text box. For example, the symbol {{tmath|\pi}} denotes the ]. The symbol {{tmath|1}} denotes the number ]. The mathematical expression {{tmath|1= ax^2 + bx + c = 0}} denotes the general ] with unknown coefficients. {{pb}} An "infinitely long string" is an abstract concept, not anything physically realizable, not notation at all. From my point of view it doesn't even "exist" except as an idea in people's minds, and in my opinion it can't "denote" anything. But again, within some abstract systems this conceptual idea can be said to equal the number 1. –]&nbsp;] 07:05, 10 January 2025 (UTC)
:Finally, of course 0.999...9 with a finite number of nines is less than one - by 10^{-n}, if n is the number of 9's. Now if you truly were going to use a limit argument for the case of an infinite number of nines, then the difference between 1 and .999... would have to be <math>10^{-\infty}</math>, whatever that is. I personally do not endorse the following reasoning, but you might still find it interesitng: In order to show that x:=1-0.999... is an infinitesimal or zero, I can also show that x+x+x+... < 1 for an infinite number of terms: For every natural n, n*10^{-n} < 1. Thus, by your own methods, <math>\infty\cdot 10^{-\infty}<1</math>. Thus, even with a stronger (and more strange) definition and with your methods of reasoning, x is an infinitesimal or zero, and infinitesimals don't exist. Thus, x=0 and 0.999...=1. What now? If you still doubt that x is either zero or an infinitesimal, please give a definition of infinitesimal you are willing to accept (keeping in mind that in the reals, there are no infinitesimals). --] 17:15, 21 December 2005 (UTC)
:::While you can't physically ''use'' infinitely long notation, I don't see why it should be thought of as "not notation at all". Heck, this is what ] is all about. In my opinion this is the clearest way of thinking about the topic of this article &mdash; it's an infinitely long numeral, which denotes a numerical value, which happens to be the real number 1. <small>The reason I keep writing "the real number 1" is that this is arguably a distinct object from the natural number 1, but that's a fruitless argument for another day. </small> --] (]) 07:15, 10 January 2025 (UTC)

::'''''The''''' infinitely long string. The one that is not growing over time because it already has all of the nines in it, and because it is not growing can be interpreted as a number. The one that is repeating, because it does not at any specific instance in time have all the nines yet. That one? The one that is by definition a contradiction? ] (]) 15:56, 10 January 2025 (UTC)
I am not going to argue about infinitesimal. As far as I am concerned, this (notice I will not even consider the plural form - it makes no sense whatsoever to me) does not exist and you most certainly cannot prove that x+x+x+... < 1 for an infinite number of terms in x unless x is zero. You can only show in a similar manner to Rasmus that n*10^(-n) < 1 using induction. Now as I stated, one can show that 0.999... < 1 and that 0.999... = 1 depending on how you approach the proof. I believe that this anomaly exists because 0.999... is not a finitely represented number. This is a problem with the decimal system and all other radix systems. What this means is that the Archimedean property applies only to reals that can be finitely represented. Rasmus's proof is no better than the induction proof that 0.999... < 1 since it uses a result of induction to arrive at the conclusion that nx < 1. Rasmus justifies his argument by stating that because the Archimedean principle cannot be applied, x must be zero. However, I maintain that the Archimedean property can only be applied to finitely represented numbers. For x > 1/n, n can take on the value of a suitable natural number but not infinity. To use the fact that 0.999... has an upper bound in a proof such as Rasmus's defeats the purpose. So what should be believed? I think that 0.999... should be considered less than 1 because it has to be considered in the context of the decimal system. If the full extent of 0.999... were known, there would be no problem with the Archimedean property or any of its corollaries. ] 18:51, 21 December 2005 (UTC)
:::"By definition a contradiction". Huh? What are you talking about? If you can find a contradiction in the notion of ], you're wasting your time editing Misplaced Pages. Go get famous. --] (]) 18:25, 10 January 2025 (UTC)

::::Above, I just described P and not P, a contradiction. ] (]) 19:26, 10 January 2025 (UTC)
:We agree that I can't show x+x+x+... < 1 for infinitely many summands unless x=0. But I definitely can "show" that (1-0.999...)+(1-0.999...)+(1-0.999...)+... < 1 for infinitely many summands. Let's do it step by step:
:::::Um. No. You didn't. I would explain why but in my experience this sort of discussion is not productive. You're wandering dangerously close to the sorts of arguments we move to the Arguments page. --] (]) 21:13, 10 January 2025 (UTC)
:* 1-0.9 = 0.1 < 1
::::I'm not wasting my time. I believe in Misplaced Pages. ] (]) 19:33, 10 January 2025 (UTC)
:* (1-0.99)+(1-0.99) = 0.02 < 1
::::We look to famous people to tell us what to understand? ] (]) 19:40, 10 January 2025 (UTC)
:* (1-0.999)+(1-0.999)+(1-0.999) = 0.003 < 1 ...
::::I see Misplaced Pages as a great place for people to learn about and evaluate the ideas of people who, over time, have become famous for their ideas. ] (]) 20:04, 10 January 2025 (UTC)
:Similarly, for every n, the sum of n terms of the form (1-0.999...9) (n nines) is less than 1. Thus,
::::The fact of the matter is that if any theory logically entails a contradiction, then that theory is logically inconsistent. If we accept logical inconsistency as fact, then we can save everyone a lot of time by saying so. ] (]) 20:19, 10 January 2025 (UTC)
:* (1-0.999...)+(1-0.999...)+(1-0.999...)+... < 1 for infinitely many summands (using methots not endorsed by me).
::I suggest that we address each of the following in our article:
:Thus, we have 1-0.999... = 0. To be precise, this "proof" is not mathematically rigorous (that's why I employ all these quotation marks), but it is just as good as the "induction proof" claimed to show that 0.999...<1. If one of these "proofs" is correct, then so is the other. Thus, in a way, I have disproved the induction proof, since using its methods leads to a contradiction. --] 19:42, 21 December 2005 (UTC)
::#The 8-byte term

::#(0.9, 0.99, 0.999, ...)
I am not sure this method works but this is not relevant to what I said. Anyway, I agreed one could prove this but I stated that it is a result of induction. I said that Rasmus's proof is also a result of induction. ] 19:51, 21 December 2005 (UTC)
::#The least number

::#The growing sequence
: Actually, none of the proofs here are using ]. Induction is a special technique to show that a statement is true for all natural numbers. An induction-proof is easily recognized by being split into two parts: The basis (showing that the statement is true for n=0) and the inductive step (showing that if the statement is true for n, it is also true for n+1). Neither mine, nor your or Huons have this form.
::#The contradiction
: As for infinitesimals, they are not defined by any infinite sums. Rather they are defined by the property that any ''finite'' sum is less than 1. Ie. if for all natural numbers n, nx<1, we say that x is an infinitesimal. But it is just a name, if you don't like it, we can call them for very-small-numbers. We also found out that if 0.999...<1 then 1/(1-0.999...) is a <s>illimited number</s> very-large-number. The existence of very-small- and very-large-numbers is in contradiction with the Archimedean property and the Least-upper-bound-property. What we call the real numbers is (uniquely) characterized by being a complete ordered field that has the LUB-property. You can of course define another set of numbers without the LUB-property and choose to call them the real numbers (but then you need to convince everybody else to follow your naming-convention...). Let us take a look at your options:
::] (]) 17:11, 10 January 2025 (UTC)
:You can follow Fred Richman and use the ''Decimal Numbers''. Then by definition 0.999...<1, but you not only lose LUB-property, you also lose negative numbers (since 0.999... + 0.999... = 1.999... = 1 + 0.999..., 0.999... doesn't have an unique additive inverse) and division (since there is no ''Decimal Number'' x, so that x(1-0.999...)=1).
:::There is no contradiction. There is no growing sequence. 0.999... is indeed infinitely long, and = 1. ] ] 21:14, 10 January 2025 (UTC)
:You can extend the real number field with <s>infinitesimals</s> very-small-numbers (using the ]). The only important property you lose is the LUB-property (and thus the Archimedean property). Of course you cannot express these numbers using decimal numbers, and most people don't like the concept of very-small-numbers. Also, even here there is no real reason not to define 0.999...=1. But these numbers are actually rather interesting. They are called the ] and are the subject of ].
{{Discussion bottom}}
:] ] 22:17, 21 December 2005 (UTC)

Actually all the proofs here use induction. You have also used induction whether you like it ot not. How did you arrive at x < 1/n ? You started by assuming a finite sum and then continued to show how it is always less than 1. This is true and it proves that 0.999... is '''always less than 1'''. However, you took this result and then tried to explain it away with the Archimedean property. You are defining real numbers using the Archimedean property. What you don't seem to understand is that the Archimedean property does not allow for very small numbers or very large numbers. It just so happens 0.999... is a number that is misunderstood because it is very close to 1. However in the decimal system, there is no way to represent many numbers exactly, so you resort to the LUB property to reach conclusions about numbers. In my opinion it is just as easy to have 0.999... < 1. You do not lose any of the properties you mentioned. Look, you could easily define the real number system in terms of the decimal system if you were to use finite representation of numbers. In fact, this is how we use the decimal system. By doing so, all the properties of the real numbers hold, including very small and very large numbers and the Archimedean property.] 23:01, 21 December 2005 (UTC)

:Rasmus and I do not use induction, since we do not use that the result holds for n in order to show it is true for n+1. Rasmus arrived at x<1/n by an explicit calculation; he did not use x<1/(n-1) in order to show it. (By the way, x was taken as 1-0.999... How does x<1/n imply 0.999...<1 ?)
:Using only finite decimal representations as numbers sacrifices several properties:
:* Without division (such as 1/3, which has no finite representation, if I understand you correctly), the reals are no longer a field.
:* Completeness is also lost, since there are Cauchy sequences of finitely represented numbers whose limit has no fintite representation.
:* That's not one of Rasmus' properties, but surely an orthogonal triangle with two sides of length one should have a third side whose length is again a real number? After all, to the ancient greek mathematicians numbers were objects of geometry, be it lengths, areas or volumes.
:Finally, then our whole discussion would be rather empty, since 0.999... is not finitely represented and would not even be a real number (whether it is less than 1 cannot be answered using a set which does not even contain 0.999...). Probably I misunderstood the statement about "defining the real number system using only finite representations of numbers"; please clarify it. --] 00:13, 22 December 2005 (UTC)

Very well. I concede that Rasmus's proof is quite solid. I can't argue against it. I still don't think you can rule out the design of the decimal system contributing to these anomalies, i.e. you have Rasmus's proof on the one hand and a simple proof by induction on the other hand that says exactly the opposite. Maybe you should include both proofs in an article about whether 0.999... equals 1 or not. Remember the decimal system is a model of the reals. ] 01:39, 22 December 2005 (UTC)

== Explanation for removal of good-faith insertion ==

An anonymous IP added a proof idea in the "Elementary" section that essentially duplicated a proof in the "Advanced section". While a good-faith edit (thanks), it doesn't seem to offer any improvement. The "squeeze play" idea is a good one, and does sometimes appear in informal explanations. The point here is that we already cover the same ground more rigorously. --]<sup>]</sup> 18:56, 10 January 2006 (UTC)

== The debate resumes - Problems with Rasmus's Proof ==

The article is still biased and incorrect. It shows only one view of this problem - Rasmus's proof that 0.999... = 1. This is an induction proof contrary to what Rasmus states. Rasmus also states that the difference between his proof and the opposing induction proof is that he does not need to run the limit to infinity. Well, the opposing proof does not require this either. To say that 0.999... > Sum (i=1 to m) 9/10^i is a result of induction. Is this still true if we run m through to infinity? The answer is no. All of Rasmus's remaining proof is based on this first induction result. Seems like Rasmus's proof is not as solid as once thought. I am inclined to have 0.999... < 1. ] 18:55, 11 January 2006 (UTC)

:That 0.999... is greater than <math>\sum_{i=1}^m\frac{9}{10^i}</math> for all natural numbers m might indeed be proved by induction; one can also prove it directly. I assume Rasmus believed it to be obvious. And indeed this result becomes wrong if we proceed to the limit for m tending to infinity. But Rasmus need not do that. On the other hand, the "opposing proof" states quite analogously that <math>\sum_{i=1}^m\frac{9}{10^i}<1</math> for every finite m (which is undisputed), and then "shows", supposedly by induction, that the same must be true for the limit - just the argument which we now see to be false. That "proof" makes an even more general (false) claim; it can be found in the ].
:Still, this gave me just another idea for a proof that 0.9999...=1. I will stop making any assumptions about the nature of 0.9999... but the following:
:* For every natural number m, <math>\sum_{i=1}^m\frac{9}{10^i}<0.9999...</math>. Here I make no claims about limits.
:Now we use two facts of analysis:
:* Let a_m be a convergent sequence of real numbers with <math>\lim_{m\to\infty}a_m=a</math>. Let b be any real number with a_m < b for all m. Then <math>a\leq b</math>. Note that not necessarily a<b; a proof should be found in almost every undergraduate textbook on analysis.
:* We have <math>\lim_{m\to\infty}\left(\sum_{i=1}^m\frac{9}{10^i}\right)=1</math>. This is a statement about limits only; it does not depend on (or even use) the definition or properties of 0.9999... A proof would be a slight generalisation of the proofs used to show convergence of the geometric series.
:Now let us combine these facts. Take <math>a_m=\sum_{i=1}^m\frac{9}{10^i}</math>, b=0.9999... Then by using the second fact of analysis, the sequence of the a_m's converges to a=1. Now using the first fact of analysis and the assumed property of 0.9999..., we conclude that <math>1\leq 0.9999...</math>. Since 0.9999... is not greater than 1, we conclude 1=0.9999... --] 14:10, 20 January 2006 (UTC)

By the second fact on analysis, the sequence of the a_m's converges to b=1 and not a=1 as you have written. So you end up concluding that <math>0.999...\leq1</math> and not <math>1\leq 0.999...</math>. So in essence you were unable to show that it is equal. Finally since we can show by induction that <math>\sum_{i=1}^m\frac{9}{10^i}<1</math> for any m (except infinity), it would be correct to say that <math>0.9999... < 1</math>. ] 22:17, 22 January 2006 (UTC)

:I'm sorry, but I can't follow any of your arguments. I defined a to be the limit of the a_m's. Since you agree that the a_m's converge to 1, we have a=1. By comparison, b is a number which is larger than all the a_m; 0.9999... satisfies that condition.
:On the other hand, I agree to <math>\sum_{i=1}^m\frac{9}{10^i}<1</math> for any m (except infinity). But why does that imply anything about 0.9999..., which has an infinite number of nines? By comparison, I could state: <math>\sum_{i=1}^m\frac{9}{10^i}</math> is rational for any m (except infinity). Is 0.9999... still rational? If so, please give numerator and denominator; if not, then why should "being less than 1" be preserved if "being rational" is not?
:By the way, please create an account. --] 19:51, 23 January 2006 (UTC)

No, since the a_ms converge to 1, you have that b=1, not a=1. You do not know what the last
a_m is, do you? Rationality is preserved with 0.999... < 1. There is no problem with any facts of analysis. ] 11:52, 24 January 2006 (UTC)

:Excuse me? a is defined to be the limit of the a_m's. Thus, when the a_m's converge to 1, we have a=1. Or, as a formula: <math>a:=\lim_{m\to\infty}a_m=1.</math> Why should a not be 1? Why does b even enter this part of the discussion?
:The "last a_m" comment is rather confusing. There is no "last a_m"; how should I know it? Why should I even want to know it?
:Finally, if 0.9999... really is rational, give numerator and denominator. --] 13:13, 24 January 2006 (UTC)
:Oops, I was in a hurry and forgot to sign in; that was me. --] 15:07, 24 January 2006 (UTC)

It matters a lot that b enters the discussion because you have made several erroneous assumptions based on b being the LUB: you stated that a_m < b for all m. Now since b=1 it must follow that any of the a_ms are less than 1 for a_m < b. You are getting yourself horribly muddled up. I did not say 0.999... is rational. I said there is nothing irrational about 0.999... < 1. Please do not misquote me. ] 17:35, 24 January 2006 (UTC)

:I made no assumptions based on anything being an LUB. I don't even talk or care about LUBs here. All I assumed about b was that it is greater than all of the a_m's. I did not claim b=1; but indeed 1 would fit that description, and if I had chosen b=1, then I would have proved 1 being less than or equal to 1, which is correct. But 2, for example, would also fit that description, and if I had chosen b=2, then I would have proved 1 being less than or equal to 2 (which is, of course, also correct). But 0.9999... is also greater than all of the a_m's (or do you doubt that? It was the one property of 0.9999... I assumed). Thus, b=0.9999... is a valid choice. And then I prove that 1, which is the limit of the a_m's, is less than or equal to 0.9999...; that's what I wanted to show.
:On the other hand, concerning rationality: You said, more or less: "All of the a_m's are less than 1; thus, 0.9999... is also less than 1." That is the best "proof" we have for 0.9999... being less than one, but it contains a serious gap when you say that a property which is shared by all the a_m's (being less than 1) must also be shared by 0.9999... Now I chose another property all of the a_m's have in common - being rational numbers. By an analogous argument, 0.9999... would also have to be a rational number, would it not? If you claim this kind of reasoning is sound, then you claim 0.9999... to be a rational number, and you should be able to give numerator and denominator. If, on the other hand, you do not believe that this kind of reasoning is sound for the property "being a rational number", then why should it be true for "being less than 1"?
:I am sorry for misinterpreting your remark about rationality of statements instead of numbers, but I had assumed it to be an answer to my argument about rational numbers (which, maybe, was a bit too short itself).
:Finally, I would ask you to be more precise. I made erroneous assumptions? Maybe, but which ones? You say b=1? Why, if I never stated it to be so? --] 23:31, 24 January 2006 (UTC)

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? view · edit 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 Misplaced Pages 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 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 0.000 1 10 {\displaystyle {\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 = lim n 0. 99 9 n {\displaystyle 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)=x/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 . . . {\displaystyle 0.x_{1}x_{2}x_{3}...} must be less than 1. y 1 y 2 y 3 . . . {\displaystyle 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, i, 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 0 + f ( x ) {\displaystyle \lim _{x\rightarrow 0^{+}}f(x)} , the right limit of x (which can also be expressed without the "+" as lim x 0 f ( x ) {\displaystyle \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 {\displaystyle 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 Misplaced Pages 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 Misplaced Pages policy. Arguments posted on the Talk:0.999.../Arguments page are disqualified, as their inclusion would violate Misplaced Pages policy on original research.

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)

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)
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! (talk) 14:47, 4 October 2024 (UTC)
What does the notation 0.̅9 mean? Johnjbarton (talk) 15:43, 4 October 2024 (UTC)
It means zero followed by the decimal point, followed by an infinite sequence of 9s. Mr. Swordfish (talk) 00:24, 5 October 2024 (UTC)
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)

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 ⁠b1, b2⁠⁠, b3, ..., 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)

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

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? (talk) 08:50, 29 November 2024 (UTC)

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. {\displaystyle 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. (talk) 12:07, 29 November 2024 (UTC)

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)
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)
Could you point to where the values of decimals are defined in this way - in wikipedia, or a good source? I can eassily find definitions in terms of limits, but not so easily with inequality signs (strict or not).
I think the version with strict inequality signs is weaker in terms of stating the case clearly for a skeptic. (talk) 17:45, 30 November 2024 (UTC)
Agree that both versions are correct. My inclination from years of mathematical training is to use the simplest, most succinct statement rather than a more complicated one that adds nothing. So, I'm with Tito and D. here. Mr. Swordfish (talk) 18:24, 30 November 2024 (UTC)
I think many mathematicians feel that "greater than or equal to" is the primitive notion and "strictly greater than" is the derived notion, notwithstanding that the former has more words. Therefore it's not at all clear that the "greater than" version is "simpler". --Trovatore (talk) 03:13, 1 December 2024 (UTC)
The general case is "greater than or equal to", and I would support phrasing it that way. I think we don't need to explain why we say "or equal to"; just put it there without belaboring it. --Trovatore (talk) 03:06, 1 December 2024 (UTC)

Image

The following discussion is closed. Please do not modify it. Subsequent comments should be made in a new section. A summary of the conclusions reached follows.
There is no consensus to remove the image, and a rough consensus to keep it. Mr. Swordfish (talk) 21:42, 10 January 2025 (UTC)

The image included at the top of this article is confusing. Some readers may interpret the image to mean that 0.999... represents a sequence of digits that grows over time as nines are added, and never stops growing. To make this article less confusing I suggest that we explicitly state that 0.999... is not used in that sense, and remove the image. Kevincook13 (talk) 17:31, 1 January 2025 (UTC)

I do not see how this is confusing. The caption reads: "Stylistic impression of the number 0.9999..., representing the digit 9 repeating infinitely" - nothing remotely like "sequence... that grows over time". I cannot see how one could meaningfully add a comment that "0.999..." is not used in a sense that has not even been mentioned. Of course lots of people are confused: that is the reason for the article, which in an ideal world would not be needed. Imaginatorium (talk) 04:29, 2 January 2025 (UTC)
If a sequence of digits grows over time as nines are added, and never stops growing, it is reasonable to conclude that the digit nine is repeating infinitely. Kevincook13 (talk) 18:14, 2 January 2025 (UTC)
Yes, notation 0.999... means that the digit nine is repeating infinitely. So, the figure and its caption reflect accurately the content of the article. D.Lazard (talk) 18:28, 2 January 2025 (UTC)
When we use the word repeating we should expect that some people will think we are referring to a process which occurs over time, like the operation of a Repeating firearm. Kevincook13 (talk) 22:03, 2 January 2025 (UTC)
You can think of this as a "process" if you like. 0.9999... means the limit of the sequence . Of course in mathematics nothing ever really "occurs over time", though I suppose you could consider it a kind of algorithm which if implemented on an idealization of a physical computer with infinite memory capacity might indefinitely produce nearer and nearer approximations. –jacobolus (t) 22:20, 2 January 2025 (UTC)
I think you are going in a very productive direction. We should explain to readers how what they might think we mean, "occurring over time", relates to what we actually mean. Kevincook13 (talk) 00:43, 3 January 2025 (UTC)
I personally think that would be distracting and not particularly helpful in the lead section. There is further discussion of this in § Infinite series and sequences, though perhaps it could be made more accessible. –jacobolus (t) 03:42, 3 January 2025 (UTC)
Yes, I agree that detailed discussion does not belong in the lead section. I personally think that the image is distracting and not helpful. In the lead section we can simply state that in mathematics the term 0.999... is used to denote the number one. We can use the rest of the article to explain why. Kevincook13 (talk) 16:23, 3 January 2025 (UTC)
Except that it's not true that 0.999... denotes the number one. It denotes the least number greater than every element of the sequence 0.9, 0.99, 0.999,... It's then a theorem that the number denoted in this way is equal to one. Tito Omburo (talk) 16:31, 3 January 2025 (UTC)
It also denotes the least number greater than every number which is less than one, just as 0.333...denotes the least number greater than every number which is less than one-third. That's why we say it denotes 1/3, and why we also say that the one with 9s denotes 1. Imaginatorium (talk) 17:39, 3 January 2025 (UTC)
@Tito Omburo, notice that @Imaginatorium just wrote above "we also say that the one with 9's denotes 1". The description "the least number greater than every element of the sequence 0.9, 0.99, 0.999,..." does describe the number one, just as does "the integer greater than zero and less than two". Kevincook13 (talk) 18:21, 3 January 2025 (UTC)
This is an incorrect use of the word "denotes". Denotes an equality by definition, whereas one instead has that 0.999... and 1 are judgementally equal. For example, does "All zeros of the Riemann zeta function inside the critical strip have real part 1/2" denote True or False? Tito Omburo (talk) 18:56, 3 January 2025 (UTC)
I think you are inventing this - please find reliable sources (dictionaries and things) to back up your claimed meaning of "denote". Imaginatorium (talk) 04:55, 9 January 2025 (UTC)
I agree that it is better to write that the term is used to denote the number one, rather than that the term denotes the number one. Kevincook13 (talk) 20:06, 3 January 2025 (UTC)
Its not "used to denote". It is a mathematical theorem that the two terms are equal. Tito Omburo (talk) 20:46, 3 January 2025 (UTC)
I think we can make this issue very clear. Assume that x equals the least number greater than every element of the sequence 0.9, 0.99, 0.999,... . Applying the theorem we learn that x = 1. Substituting 1 for x in the opening sentence of this article we have: In mathematics 0.999... denotes 1. If we also insist that 0.999... does not denote 1, we have a contradiction. Kevincook13 (talk) 18:45, 4 January 2025 (UTC)
You have redefined the word "denote" to mean precisely the same as "is equal to", which is confusing and unnecessary. It's better to just say "is equal to" when that's what you mean, so that readers are not confused. –jacobolus (t) 18:56, 4 January 2025 (UTC)
I agree that redefining the word denote would be confusing and unnecessary. I simply defined a variable x to be equal to a number, the least number. Kevincook13 (talk) 20:04, 4 January 2025 (UTC)
I'm in agreement with @Imaginatorium and @D.Lazard on this. The image does not suggest a process extended over time, and it correctly reflects the (correct) content of the article, so there is no need to remove it. I'm not persuaded that people will interpret "repeating" as purely temporal rather than spatial. If I say my wallpaper has a repeating pattern, does this confuse people who expect the wallpaper to be a process extended over time? (Are there people who think purely in firearm metaphors?) MartinPoulter (talk) 17:30, 3 January 2025 (UTC)
Consider the number 999. Like the wallpaper, it contains a repeating pattern. That pattern could be defined over time, one nine at a time. Or it could be defined at one time, using three nines. Kevincook13 (talk) 18:27, 3 January 2025 (UTC)
Is it OK if I go ahead and edit the article, keeping in mind all the concerns which have been raised with my proposed changes? Kevincook13 (talk) 17:56, 8 January 2025 (UTC)
Can you be more specific about which changes you want to implement? MartinPoulter (talk) 20:32, 8 January 2025 (UTC)
The first change would be to remove the image. Kevincook13 (talk) 15:06, 9 January 2025 (UTC)
I'm confused, @Kevincook13. Where in the above discussion do you see a consensus to remove the image? You have twice said the image should be removed, and I have said it should stay. No matter how many times you express it, your opinion only counts once. Other users have addressed other aspects of your proposal. Do you sincerely think the discussion has come to a decision about the image? MartinPoulter (talk) 13:47, 10 January 2025 (UTC)
No. I do not think there is agreement on removing the image. (I don't personally think it is spectacularly good, but the argument for removing it appears to me to be completely bogus.) Imaginatorium (talk) 04:57, 9 January 2025 (UTC)
The term 0.999... is literally a sequence of eight characters, just as y3.p05&9 is. Yet, the term itself implies meaning. I think confusion about the term can be reduced simply by acknowledging different meanings the term might imply. It does imply different meanings to different people. We can respect everyone, including children who are not willing to simply accept everything a teacher tells them. We can do our best to help everyone understand what we mean when we use the term. Kevincook13 (talk) 15:32, 9 January 2025 (UTC)
For example, if a child thinks that by 0.999... we mean a sequence of digits growing over time, and the child objects when told that the sequence of digits is equal to one, we can respond by saying something like the following: You are correct that a growing sequence of digits does not represent one, or any number, because the sequence is changing. We don't mean that 0.999... represents a changing or growing sequence of digits. Kevincook13 (talk) 16:12, 9 January 2025 (UTC)
We don't mean a changing or growing sequence of digits. That is what it is confusing to say that we mean a repeating sequence of digits. Kevincook13 (talk) 16:15, 9 January 2025 (UTC)
What we mean is a number. Kevincook13 (talk) 16:18, 9 January 2025 (UTC)
This article is about the meaning of 0.999... in mathematics not about the possible meanings that people may imagine. If people imagine another meaning, they have to read the article and to understand it (this may need some work), and they will see that their alleged meaning is not what is commonly meant. If a child objects to 0.999... = 1, it must be told to read the elementary proof given in the article and to say which part of the proof seems wrong. D.Lazard (talk) 16:58, 9 January 2025 (UTC)
What do we mean by the term number? A number is a measure, not a sequence of digits. We may denote a number using a sequence of digits, but we don't always. Sometimes we denote a number using a word, like one. Sometimes we use a phrase such as: the least number greater than any number in a certain sequence. We may use a lowercase Greek letter, or even notches in a bone. Kevincook13 (talk) 16:44, 9 January 2025 (UTC)
By the term "number", we mean a number (the word is not the thing). It is difficult to define a number, and this took several thousands years to mathematicians to find an acceptable definition. A number is certainly not a measure, since a measure requires a measurement unit and numbers are not associated with any measurement unit. The best that can be said at elementary level is something like "the natural number three is the common property of the nines in 0.999..., of the consecutive dots in the same notation, and of the letters of the word one". D.Lazard (talk) 17:20, 9 January 2025 (UTC)
I see. A number is not a measure, but it is used to measure. Thanks. Kevincook13 (talk) 17:40, 9 January 2025 (UTC)
A number is a value used to measure. Kevincook13 (talk) 17:42, 9 January 2025 (UTC)
The caption on the image is: Stylistic impression of the number 0.9999..., representing the digit 9 repeating infinitely.
The caption can be understood to mean that the term 0.999... is a zero followed by a decimal point followed by the digit 9 repeating infinitely, which meaning is distinct from the meaning that 0.999... denotes the number one.
If we retain the caption, we may communicate to readers that we mean that 0.999... is a repeating sequence, which sequence denotes the number one. That doesn't work because repeating sequences themselves cannot be written completely and and therefore cannot be used to notate.
0.999... is notation. The purpose of this article should be to help others understand what it denotes. If it denotes a repeating sequence of digits, then we should say so in the lead sentence. Kevincook13 (talk) 18:32, 9 January 2025 (UTC)
How does the first sentence of the article not explain that notation? The meaning of the notation is the smallest number greater than every element of the sequence (0.9,0.99,...). Tito Omburo (talk) 18:39, 9 January 2025 (UTC)
Because it does not make sense to say that the sequence is repeating, because all the nines have not already been added, and at the same time to say that the sequence represents a number, because all the nines have already been added. It is confusing because it is contradictory.
When we say that the sequence is repeating, people who are not trained in mathematics will likely assume that we mean that all the nines have not already been added, and therefore that the sequence is changing and therefore, does not represent a number. Which, I believe, is why the subject of this article is not more widely understood. Kevincook13 (talk) 19:05, 9 January 2025 (UTC)
I think I understand part of the confusion, which I've hopefully tried to correct with an edit. The notation 0.999... refers to a repeating decimal, a concept which had not been linked. There is a way of associating to any decimal expansion a number as its value. For the repeating decimal 0.999..., that number is 1. Tito Omburo (talk) 19:09, 9 January 2025 (UTC)
I like the edits. Because the least number is one, the meaning of the lead sentence can be understood to be that 0.999... is a recurring decimal whose value is defined as one. The notation below should match. Instead of 0.999... = 1 {\displaystyle 0.999...=1} , we should write 0.999...   = d e f   1 {\displaystyle 0.999...\ {\overset {\underset {\mathrm {def} }{}}{=}}\ 1} . Kevincook13 (talk) 19:40, 9 January 2025 (UTC)
No. The point is that the notation has a definition which is a standard one for repeating decimals of this form. It is a theorem that this number is one, but that is not the definition. Tito Omburo (talk) 19:47, 9 January 2025 (UTC)
I agree. What you are saying agrees with what I am saying. It is a theorem that the least number is one, not a definition. The notation has a standard definition which defines the notation to be equal to the least number, whatever that least number is.
  1. Given that the notation is defined to be equal to the least number
  2. And given a theorem that the least number does equals one
  3. Therefore the notation is defined to be equal to a number which does equal one.
  4. Note that it does not follow from the givens that the notation is equal to one, or that the notation is equal to the least number.
Kevincook13 (talk) 20:23, 9 January 2025 (UTC)
This is not correct, but I feel like we're talking in circles here. Cf. WP:LISTEN. Let me try one more thing though. If we wanted a more explicit definition of 0.999..., we might use mathematical notation and write something like 0.999   = def   k = 1 9 10 k = 1. {\displaystyle 0.999\ldots \ {\stackrel {\text{def}}{=}}\ \sum _{k=1}^{\infty }9\cdot 10^{-k}=1.} This is discussed in the article in § Infinite series and sequences. –jacobolus (t) 02:58, 10 January 2025 (UTC)
Can you see that the summation is a process which must occur over time, and can never end? Do you notice that k cannot equal 1 and 2 at the same time? However, if we insist that the summation does occur all at once, then we affirm that k does equal 1 and 2 at the same time. We affirm that we do intend contradiction. If so, then we should clearly communicate that intention. Kevincook13 (talk) 15:14, 10 January 2025 (UTC)
Please stop misusing the word denotes when you mean "is equal to". It's incredibly confusing. –jacobolus (t) 20:57, 9 January 2025 (UTC)
I agree that the difference between the two is critical. I've tried to be very careful. Kevincook13 (talk) 21:13, 9 January 2025 (UTC)
I don't know if this will help at all, but it may. I think that we have been preoccupied with what infinity means, and have almost completely ignored what it means to be finite. We don't even have an article dedicated to the subject. So, I have begun drafting one: Draft:Finiteness. Kevincook13 (talk) 00:00, 10 January 2025 (UTC)
I think the problem here is that there are two levels of symbol/interpretation. The literal 8-byte string "0.999..." is a "symbol for a symbol", namely for the infinitely long string starting with 0 and a point and followed by infinitely many 9s. Then that infinitely long symbol, in turn, denotes the real number 1.
It's also possible that people are using "denote" differently; I had trouble following that part of the discussion. But we need to be clear first of all that when we say "0.999..." we're not usually really talking about the 8-byte string, but about the infinitely long string. --Trovatore (talk) 05:06, 10 January 2025 (UTC)
This is also a weird use of "denote", in my opinion. For me, the word denote has to do with notation, as in, symbols that can be physically written down or maybe typed into a little text box. For example, the symbol ⁠ π {\displaystyle \pi } ⁠ denotes the circle constant. The symbol ⁠ 1 {\displaystyle 1} ⁠ denotes the number one. The mathematical expression ⁠ a x 2 + b x + c = 0 {\displaystyle ax^{2}+bx+c=0} ⁠ denotes the general quadratic equation with unknown coefficients. An "infinitely long string" is an abstract concept, not anything physically realizable, not notation at all. From my point of view it doesn't even "exist" except as an idea in people's minds, and in my opinion it can't "denote" anything. But again, within some abstract systems this conceptual idea can be said to equal the number 1. –jacobolus (t) 07:05, 10 January 2025 (UTC)
While you can't physically use infinitely long notation, I don't see why it should be thought of as "not notation at all". Heck, this is what infinitary logic is all about. In my opinion this is the clearest way of thinking about the topic of this article — it's an infinitely long numeral, which denotes a numerical value, which happens to be the real number 1. The reason I keep writing "the real number 1" is that this is arguably a distinct object from the natural number 1, but that's a fruitless argument for another day. --Trovatore (talk) 07:15, 10 January 2025 (UTC)
The infinitely long string. The one that is not growing over time because it already has all of the nines in it, and because it is not growing can be interpreted as a number. The one that is repeating, because it does not at any specific instance in time have all the nines yet. That one? The one that is by definition a contradiction? Kevincook13 (talk) 15:56, 10 January 2025 (UTC)
"By definition a contradiction". Huh? What are you talking about? If you can find a contradiction in the notion of completed infinity, you're wasting your time editing Misplaced Pages. Go get famous. --Trovatore (talk) 18:25, 10 January 2025 (UTC)
Above, I just described P and not P, a contradiction. Kevincook13 (talk) 19:26, 10 January 2025 (UTC)
Um. No. You didn't. I would explain why but in my experience this sort of discussion is not productive. You're wandering dangerously close to the sorts of arguments we move to the Arguments page. --Trovatore (talk) 21:13, 10 January 2025 (UTC)
I'm not wasting my time. I believe in Misplaced Pages. Kevincook13 (talk) 19:33, 10 January 2025 (UTC)
We look to famous people to tell us what to understand? Kevincook13 (talk) 19:40, 10 January 2025 (UTC)
I see Misplaced Pages as a great place for people to learn about and evaluate the ideas of people who, over time, have become famous for their ideas. Kevincook13 (talk) 20:04, 10 January 2025 (UTC)
The fact of the matter is that if any theory logically entails a contradiction, then that theory is logically inconsistent. If we accept logical inconsistency as fact, then we can save everyone a lot of time by saying so. Kevincook13 (talk) 20:19, 10 January 2025 (UTC)
I suggest that we address each of the following in our article:
  1. The 8-byte term
  2. (0.9, 0.99, 0.999, ...)
  3. The least number
  4. The growing sequence
  5. The contradiction
Kevincook13 (talk) 17:11, 10 January 2025 (UTC)
There is no contradiction. There is no growing sequence. 0.999... is indeed infinitely long, and = 1. Hawkeye7 (discuss) 21:14, 10 January 2025 (UTC)
The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion. Categories: