Talk:0.999.../Arguments/Archive 3

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Fractional proof?

One could claim that in this case the proof is not that 0.999... = 1, but rather use this as an inequality which notes that 1/3 does not equal 0.3333... This view only notes that the decimal system is not and never will be as precise as representing value in fractions, because it must rely on the notion of infinity to represent a concrete, non-infinite value represented by a fraction. The number "1" is a finite number, and the same goes with any non-infinite decimal that in turn has a representative fraction. The point is that fractions and infinite decimals are not equal to each other. We can only fathom a representative notion of what they mean, but in truth 0.999... does not equal 1, 0.333... does not equal 1/3, etc. It just proves there is no finite decimal equivalent that represents those fractions. After all, the decimal system is based off the even number "10", which limits its accuracy in representing some values. The same goes for if the system would be based off of another number, be it 9, 8, 7, etc. It just proves the limits of our chosen quantification system.

It's incorrect. ⅓ ≠ 0.333... therefore: 3/3 ≠ 0.999...; 3/3 = 1 Is not that true? (side note: are fractions such as ⅓ and ⅔ imaginary numbers, as the decimals go on infinitely?) Jerr 00:32, 11 October 2006 (UTC)

1/3 = 0.333... by simple long division, or you can use the division algorithm to be more formal. How exactly does 1/3 ≠ 0.333...? Supadawg (talk • contribs) 02:39, 11 October 2006 (UTC)
⅓ ≠ 0.333... because 0.333...(3) ≠ 1. However, ⅓(3) = 1. I am hereby assuming (Whether incorrectly or correctly) that 0.999... ≠ 1. Of course, this goes against all that the article is saying, so IF the article is correct it should be kept that way. Jerr 02:29, 13 October 2006 (UTC)
Well, if you assume that 0.999... ≠ 1 (and that decimal arithmetic otherwise works as expected), then ⅓ ≠ 0.333.... While these statements are consistent with each other, they are not consistent with the rest of mathematics, and if you dig deeper, you find that they're both false. 0.333… does equal one-third, and 0.999… does equal one. (This is all assuming the standard definitions). Melchoir 03:12, 13 October 2006 (UTC)
You're using circular reasoning. First you said that 1 ≠ 0.999... because ⅓ ≠ 0.333..., and then you said that ⅓ ≠ 0.333... because 0.333...(3) ≠ 1. Can you explain specifically how ⅓ ≠ 0.333...? Supadawg (talk • contribs) 23:28, 13 October 2006 (UTC)
I suppose that since there is no end to 0.999... (ugh, headache) that it is such a tiny, infinitely tiny, part that it should equal one. After the 13th decimal place modern calculators show that 0.3333333333333(3) = 1. However, they also say that 0.333333333333(3) = 0.999999999999. Since the 13th place of 3 in 0.333... is so incredibly small and the difference between 0.9999999999999 and 1 is almost zero, the two therefore can be considered the same number. I withdraw my previous statement; however, I do still believe that the fractional proof is confusing. Jerr 17:01, 14 October 2006 (UTC)
I think you may be making a fundumental mistake here. The number 0.9999999999 (1 - 10-10) is very close to 1, a physicist or an engineer might not care about the difference, and a calculator might round it up. A mathematician would never say that it is equal to 1. On the other hand, the number 0.\bar{9} is 1. They're not "very close" and there's no "tiny difference" between them. They're equal.
I sort of agree with you about the fractional proof though. It relies on several non-trivial properties of real numbers without including their proofs, and thus is only relevant if it has already been agreed that these properties are already established. The proofs later in the article are more readily applicable to the fundumental definitions of real numbers. -- Meni Rosenfeld (talk) 18:46, 14 October 2006 (UTC)
Imaginary numbers are the square roots of negative real numbers. The most famous one is i, with i²=-1. (We also have (-i)²=-1, of course.) Linear combinations of real and imaginary numbers form the complex numbers. They have nothing to do with either fractions (those are rational numbers) or 0.9999... Yours, Huon 08:41, 11 October 2006 (UTC)
Whilst we hope Jerr learns the flaw in his reasoning soon and improves his understanding of 0.999..., his question does show a fairly common problem. The fractional proof just isn't convincing, because one of its premises is that ⅓ = 0.333..., and for a lot of people that's at least as dubious as the claim that 0.999... = 1. When I was young, I asked my father why there were so many threes in 0.333..., and he said it was because if you took 0.3, there was still a bit to make it a third, so you tried 0.33, but there was still a bit more to go, so you tried 0.333, etc. This kind of understanding gives the impression that 0.333... is just the slightest bit short of ⅓, but it's close enough for all purposes - leading to the similar misconception about 0.999... and 1. Is there a way we can address this issue? Maelin 23:54, 13 October 2006 (UTC)
I looked through several articles for some proof of 1/3 equaling 0.333..., but came up empty, so we'd have to create it from scratch. I remember learning through long division: you just kept adding threes at the top, so it was easy to see that it went on forever. Supadawg (talk • contribs) 00:54, 14 October 2006 (UTC)
Well, ideally the first section would be supported by the second. I've been thinking about a good way to achieve that by reworking the second section's intro, but so far I'm not sure what to say. Melchoir 04:48, 14 October 2006 (UTC)
This "proof", along with the "Algebra proof", is not rigourous. I think they are still useful to have - the point is that whilst many people are confused by 0.9..., they do not tend to share the same confusion for other recurring decimals - so this proof shows that either they have to accept 0.9... as equal to 1, or they have to say that recurring decimals can never be used to represent fractions (or to put it another way, that there exist rationals which cannot be represented by a decimal expansion).
Perhaps it would be better to emphasise that these are not rigourous proofs, but ways of helping people understand nonetheless? At one time they are labelled "elementary" and "advanced" proofs I believe, I don't know if that was better. Mdwh 00:11, 16 October 2006 (UTC)
As pointed out by several previous editors, the "fractional proof" is circular reasoning (begging the question). The idea that 0.333... is exactly equal to 1/3 is just as unproven as the idea that 0.999 is exactly equal to 3/3. The so-called "fractional proof" should be removed if the article is to have any credibility. Dagoldman 06:49, 25 October 2006 (UTC)
Assuming the "fractional proof" is not original research. I cannot see why it should not be there. I do agree as a mathematician that it is circular reasoning, and more a demonstration of the relationships between different "infinite recursion" mathematical proofs. Ansell 08:13, 25 October 2006 (UTC)

A structure for the skeptics

Thinking about what Jerr and Maelin write above, it occurs to me that some people might have something like this structure in mind: Suppose we consider formal strings of decimal digits (but with no negative numbers; this will be important) and define addition and multiplication on them by looking at sums and products of their initial segments, and taking the limit in the topology of pointwise convergence. That is, to get the nth digit of a+b, you find long enough initial segments of a and b that the nth digit of the sums of those initial segments, or any longer ones, will always be the same.

It's not instantly obvious that these operations are well-defined, but in fact they are; this isn't hard to see. Call one of these numbers "terminating" if it eventually reaches an infinite string of zeroes. Then the sum and product of a and b are terminating if and only if both a and b are terminating; this follows from monotonicity of addition and multiplication. Now it's easy to see that the digits of the sum/product of a and b are precisely the digits of their usual sum/product, disambiguating between ...000... and ...999... by the rule in the last sentence. (This doesn't work if you allow negative numbers.)

The usual laws of commutativity, associativity, and distributivity hold (see this by noting that the answer is terminating if and only if all inputs are terminating). However most nonzero numbers have no multiplicative inverse (none of them have an additive inverse), and neither the additive nor the multiplicative cancellation law holds. Another inconvenience is the structure's topology, which is that of the Cantor space, not so good for calculus.

Still, this is a well-defined structure in which affirmations like Jerr's make some sort of sense, and maybe that ought to be acknowledged. Is there a standard name for it? --Trovatore 07:16, 15 October 2006 (UTC)

Sounds to me like the commutative semiring of "decimal numbers" from 0.999...#Breaking subtraction. Richman doesn't seem to be aware of a name for it, nor for that matter any prior research on it at all. Of course, he may have simply failed to look hard enough. Melchoir 07:28, 15 October 2006 (UTC)

You all appear to be so proud of your new article?

I read the old article and the new article. Frankly, the new article is still full of false claims so I don't know what you are boasting about? None of the properties you claim as proof (including Archimedian, limits, etc) are violated by not having 0.999... = 1. You also state falsely that many textbooks support your argument - how can you be so calloused? Why is it that you don't publish the very debate in the arguments section that illustrates clearly what you are saying is wrong? Finally, it is not possible to have more than one representation in any radix system, yet you again claim falsely this is a common property across all radix systems. No matter how much you try to push your point of view, in the end, truth will prevail. It is a falacy that 0.999... = 1. 198.54.202.254 10:33, 23 October 2006 (UTC)

Welcome to mathematics, where proof is everything and wild claims made in hostile tones achieve nothing. If you dispute the claims in this article, please post the mathematical arguments with which you hope to refute them. In particular, please provide a proof of your claim, "it is not possible to have more than one representation in any radix system," since I claim the following as a counterexample: 3*2, 6. Maelin 11:47, 23 October 2006 (UTC)

No wild claims here. I am asking you to include parts of the debate in the Wikipedia article that illustrate the error of one of your editors' proofs (Rasmus). You have conveniently left this out. This 'proof' by Rasmus is the best attempt by any of your editors and/or contributors and you do not include it - why? As for your example, it is pitiful - where did you learn that 3*2 is a valid radix representation? Do you know what a radix representation is? 198.54.202.254 08:07, 24 October 2006 (UTC)

Since you originally asked for "representations in a radix system" and not "radix representations", I considered it valid. Since you have now clarified what you mean, I withdraw it as a counterexample. However, the only proof of uniqueness of radix representations that I could find with a quick Google search applied only to natural numbers. Again, I ask that you please provide a proof of your claim that all real numbers have at most one radix representation. Maelin 08:28, 24 October 2006 (UTC)

So you can't read too well either. I said: "Finally, it is not possible to have more than one representation in any radix system,..." - what makes you think that your response was valid? There is no ambiguity in this sentence. It states exactly what I mean. There is no need to provide a proof that radix representation is unique - this was designed in the architecture of a radix system. Any representation that is not the same as another representation (excluding trailing zeroes) is a different number. The onus is on Wikipedia to demonstrate that 0.999... equals 1. The best attempts have proved that this is untrue. I am not here to debate the issue with you. I asked why it is that Wikipedia left out the most 'convincing' proof that was shown to be false. Go back and read my original question. This time, try to think before you answer. 198.54.202.254 15:17, 25 October 2006 (UTC)

Sorry, but that "I don't need to provide proof of my claims" nonsense immediately disqualifies you, completely and utterly, from mathematical discussion. You must ALWAYS be able to provide proofs of your claims if you are asked to. Proof is the very foundation of mathematics. We have provided an article with several proofs that 0.999... = 1. You have thus far failed to illustrate how these proofs are all fallacious, and in addition, you have declared that you have a disproof which rests on a claim that I can't find supported anywhere with Google, and for which you have failed to deliver a proof when requested. If you can't prove your claims, you can't use them to support your arguments. Maelin 15:31, 25 October 2006 (UTC)

As I thought, you did not understand what I was writing and it is doubtful you shall. I am asking why Wikipedia did not include the debate between Rasmus and a past contributor who clearly refuted Rasmus's proofs. Read the archives! Read them and you will find the error of Rasmus's so-called proof. Then publish that section as a part of your article so that readers can decide for themselves. You falsely claim that the mathematical community supports you! Provide proof that this is so. List the names of your contributors and mathematicians. List the sources where you found this information that supports your false claims. You have not done so! You have not published any valid proofs. Where is your evidence??? 198.54.202.254 15:42, 25 October 2006 (UTC)

We do not publish original research, including "debates" that occur in our talk pages. The article has an extensive References section, complete with names of authors. I could provide more, but it's not necessary. Melchoir 15:47, 25 October 2006 (UTC)

If you do not publish original research, you should delete this article because it is original research. The reference section you have provided in no way supports your arguments. I believe that Rasmus is one of your editors? If yes, then why not publish his proof as part of the article? Are you afraid that readers will see how invalid your argument is? You talk about the construction of reals as if only you know anything about mathematics. You display an arrogance that immediately points to ignorance rather than any valid mathematics. Again, why don't you publish Rasmus's proof seeing you believe it is true? 198.54.202.254 06:19, 26 October 2006 (UTC)

I'm finding it extremely difficult to follow the flow of the debate that was going on in the archives. Could you please give a brief summary of why you, or the 'previous contributor' or whoever, believes the proof to be invalid? I can't see what's wrong with saying that 0.999... is equal to the sum of an infinite series, that the sum of the infinite series is 1, and therefore that 0.999... = 1. Perhaps I'm misunderstanding the point. Whatever, please enlighten me. Justdig 17:14, 25 October 2006 (UTC)

I have no intention of debating this with you. Your archives have ample evidence of others who have tried to enlighten you. It appears you do not wish to be enlightened. 198.54.202.254 06:19, 26 October 2006 (UTC)

So basically your argument goes like a less tactful version of this: "You have problems in your article, because it presents as fact something that is wrong. I refuse to explain why I claim it is wrong, instead, you can go looking for the justifications. You should find in the archives, somewhere, some sort of important proof that I refuse to explain, and this will, in some way that I refuse to explain, solve all our problems. Good luck. I am not going to help you do any of the things that I have so rudely demanded."
You seem to have serious trouble with the idea that if you want to reject theorems in mathematics as false, you actually have to put in some legwork for it; you can't expect to tell people that they're wrong and demand they go and prove it for you. If you think this Rasmus' proof is so important, post a link to it and explain what it is that you want us to see about it. Stop demanding that we do all the work for you. Maelin 07:08, 26 October 2006 (UTC)
You killed one small troll. You gain 25 xp and a stone club. --dreish~talk 16:51, 26 October 2006 (UTC)
That's no small troll. I suspect this is the same troll that has been plaguing this article for a year or so - if that's true, he knows we are right, so no logical argument will make him change his mind. He just enjoys making other people suffer. I therefore beg every sane editor here not to feed the trolls. -- Meni Rosenfeld (talk) 17:03, 26 October 2006 (UTC)
Meni, it's easy to call someone a troll and not have to refute any further arguments. I guess this is how Wikipedia is run. It is not a democracy or a peoples encyclopedia, instead it is the world according to the likes of Meni, Melchoir and what they think the mathematics community agrees to. I suggest you drop the 'troll' word. It's gone stale. 198.54.202.254 05:41, 29 October 2006 (UTC)
Tell you what. If you can provide a solid, rigourous disproof of 0.999... = 1, within the real numbers, using formal mathematical definitions and a standard, formal proof structure, then instead of calling you a troll we will sing your praises. Start a new section for it on this page. It should not be hard to do - if, as you claim, 0.999... != 1, then a proof by contradiction should not be difficult, especially not for a graduate like yourself. Don't forget that every mathematical proof has to stand up under peer scrutiny. Good luck! Maelin (Talk | Contribs) 08:42, 2 November 2006 (UTC)

Confusion

So yes, I am one of those students who resists .999... = 1. But I have some reasonable gripes with it. Mainly, my observation is that all "proofs" seem to rely completely on assuming that .999... does in fact extend to infinity. I understand this. But how can you reliably compute numbers using application of infinity? Based on my understanding of limits, the limit of a function does not actually equal the value. For example, if you find the limit of f(x) = (x-2)^2 / (X-2) as x approaches 2, it can be easily reduced to (x-2) and defined as 0. But that just means that as X approaches 2, then f(2) approaches zero, but F(2) != 0. Could someone clarify for me? —The preceding unsigned comment was added by 24.208.14.58 (talk • contribs) 19:50, 15 October 2006 (UTC).

You are confusing the limit of a function (your f(x) example) with the limit of a sequence (0.9999... as limit of the sequence (0.9, 0.99, 0.999, ...). This limit is defined without using infinity - see limit of a sequence. (So is the limit of a function, and that limit may indeed differ from the value at the point considered, but that's irrelevant here.) Yours, Huon 20:11, 15 October 2006 (UTC)
Before you can decide whether 0.999... = 1 is a true statement, you have to define what .999... means. If you want to impose the obvious strict ordering on decimal expansions, you run into problems. This is because then 0.999... < 1, and additionally 0.9999999999 < 0.999..., no matter how many nines are on the left half. This makes it impossible to have a set of real numbers that work like we would like them to. For example, now we can't say that limits are unique, for the sequence 0.9, 0.999, 0.9999, ... has two different limits, 0.999... and 1. These are both limits because the difference between the terms of the sequence and the value gets arbitarily small, whether you use 0.999... or 1. Also, as the article points out, even something as simple as subtraction has problems. What is 1 - 0.999...? You can invent something like 0.000...1, but now you're opening up a huge can of worms. What is that? Is it a decimal expansion? Can you now write 0.111...222...34343434...? How do you do arithmetic on these things?
An important point is that the notion of 0.999... is just a curiosity. There is no need for it. Typically, the real numbers are not even defined using decimal expansions. The definitions imply that you cannot have a real number with the properties you seem to want 0.999... to have. Our choices are:
  • Not define 0.999...
  • Define 0.999... in a manner such that 0.999... = 1.
  • Define 0.999... differently. Then either 0.999... is a not a real number, or else the real numbers fail to behave nicely.
Since there's no compelling reason to have 0.999... the third choice is not good. It is much better to have a system that is useful and has a small quirk.
Essentially, you could very well work out all the issues that arise when you try to have 0.999 < 1. But you would probably find the resulting system more unsettling than the one you get if you just say 0.999... = 1. Also, it would not be what is called the "real numbers", since what you want is impossible in the real number system. The article does, I think, make it clear that 0.999... = 1 is dependent on the nature of the set of real numbers, and that if you cook something else up (like the "decimal numbers" discussed in the article), you might very well have 0.999... < 1.
I think a lot of time is spent arguing about the notion 0.999... itself and what it means, generating lots of fruitless discussion. I think a better way to consider this question is to forget about 0.999... and instead consider the properties of real numbers. Once you have those hammered out, what to do with 0.999... falls out naturally. Eric119 21:01, 15 October 2006 (UTC)
The number 0.9... extends to infinity by definition. The value of an infinite sum is equal to the limit, again by definition. See Recurring decimal. Mdwh 00:12, 16 October 2006 (UTC)
How is .999... a real number? Isn't it undefined? I wouldn't say 1-0.999... could even exist because "0.999..." isn't properly defined. To me, all of the proofs are still either faulty or just say "this works by definition". The conclusion can't just restate the premise. I need some full clarification.
It isn't undefined. There are several ultimately equivalent definitions: 0.999… can be the sum of a certain series, the limit of a certain sequence, the least upper bound of a certain set of real numbers, the unique common element of a certain set of real intervals, the union of a certain set of Dedekind cuts, or the equivalence class of a certain Cauchy sequence. All of these definitions for 0.999… lead to it equaling 1. Melchoir 19:40, 16 October 2006 (UTC)
I read all of those definitions and I think I understand them properly. Am I wrong then in saying that 0.999... is defined as 1 only when taken, through various methods, to an infinite process or sum? I understand completely that as you take 0.999... further and further, there is eventually no discernable difference between 1 and 0.999... . But this does not convince me, mathematically, that they are equal. I am still not seeing any proof that does anything other than use limits to infinity, parlor-trick algebra, or "cuz we said so". Am I STILL missing something or is my feeble college-student brain just supposed to give up and accept it without any empirical proof?
"0.999…" is an infinite decimal. It doesn't have to be "taken" anywhere; it has infinitely many digits to begin with. Before one applies any definitions or manipulations, it already goes to infinity, and that cannot be escaped. Every possible way of making a real number from this decimal results in the number 1. Only one of these definitions involves an infinite sum, and none of them involves a "process" because there is no such thing as a process in mathematics. So... if you won't accept arithmetic, algebra, calculus, analysis, or set theory, what kind of a proof will you accept? Is there any mathematical theorem at all that you can believe? Melchoir 23:20, 16 October 2006 (UTC)
I think you are confusing "0 . followed by a lot of 9s" and the number "0.9 recurring". This article is about the latter, and as Melchoir says, it doesn't have to be "taken" anywhere. Mdwh 23:54, 16 October 2006 (UTC)
Sorry, I meant, the methods ending in 0.999...=1 involve infinity somewhere. So "0.9 recurring" and "Before one applies any definitions or manipulations, it already goes to infinity" are what I have problems with. What I would be looking for is something concrete. None of these can be done by hand, they are all concepts involving infinity. I just disagree with it being taught as a fact instead of as a concept which involves infinity. And I do completely understand the concept of 0.999...=1. I also understand the concept of Gabriel's Horn, but you couldn't convince me it's actually possible for it to exist, just like I wouldn't take you seriously if you told me you had 1.999... sisters.
I don't think this argument is going anywhere.... :(
Oh, well of course it doesn't physically exist; few things in mathematics do. There's no physical meaning to the hundredth digit of pi, but that doesn't stop people from memorizing it anyway. Sequences and their limits, the real numbers themselves... basically nothing in real analysis is "real". As such, analysis is not open to concrete demonstrations; all you can do is reason with the abstractions and hope that, at the end of the day, the structures you're dealing with form useful models for reality. But the models don't influence the structures. And if some of the results make intuitive sense, so much the better; if some of them don't, too bad. Melchoir 15:00, 17 October 2006 (UTC)
This is just incredible! Melchoir states that nothing in real analysis is real. Now here's a paradox. Why don't we call it unreal analysis? Would this not be a more appropriate name? 198.54.202.254 15:11, 28 October 2006 (UTC)
I would like to further stress out that there is nothing wrong with infinity. You seem to repeatedly imply that "infinite" is somehow synonymous with "fallacious". That is completely groundless. There are mathematical ways to provide finite (and thus accessible to humans) arguments dealing with infinite objects. In fact, very little of mathematics deals with finite objects (even the fundumental natural numbers are infinite in quantity). So, once again: The theorem "0.999... = 1" is a prefectly valid fact which involves infinity. The article presents several (finite!) essentially rigorous proofs of this equality. And I do indeed have 1.999... siblings, a brother and a sister. No problem there either. -- Meni Rosenfeld (talk) 15:18, 17 October 2006 (UTC)

No its not equal

It's just a fun paradox that occurs with our current numerical system. Simply put: the actual value represented by the figuers: '.999...' and '1' is not the same. It's just that in our current model to represent the abstract concepts and values that we call 'numbers', we can play with some symbols (and other abstract concepts) to make it look that way.

It's a flaw in our current model, that's all. it's the same flaw that makes us represent values like π(pi) like irrational numbers that continue to infinity. We all know the dimeter of a circle is a of a fixed length, but in our number system representing the ratio of that diameter to it's circunference get's us a number that continues on to infinity. Does that means that π is an infinite number? that the diameter is infinite? no. It's just a flaw in our current system that lead to interesting representations of things.Nnfolz 12:06, 16 October 2006 (UTC)

No, you're wrong. The set of real numbers is uncountable, so there is no way to assign a distinct finite sequence of symbols taken from a finite set to every real number. Any representation capable of handling every real number must involve expressions of an infinite nature - and there's nothing wrong with it. The decimal representation has a "feature", which you might call a disadvantage or a flaw, that some real numbers have 2 representations - such as 1 which can be represented as 1.000... or 0.999... . I do suggest you read the real number article so you may have a basic understanding of what they are. -- Meni Rosenfeld (talk) 13:24, 16 October 2006 (UTC)
Huh? In what way the previous parragraph adresses my statements? Please re-arrange the ideas into more understandable english sentences (i don't mean to sound like a smart a$$ or disrespetfull) because i'm having a lot of trouble trying to figure out what you just said.Nnfolz 17:05, 16 October 2006 (UTC)
Nnfolz, you're confusing something being infinite and something having an infinite representation. Different aspects of something might have different properties. For example, the diameter of a circle has finite length, but there are infinitely many points in it.
Meni Rosenfeld's points appears to be that it is impossible to represent every real number with a finite length description. Hence the "flaw" you describe is unavoidable. Eric119 17:20, 16 October 2006 (UTC)
No i'm not confusing both things. I'm aware of what you state. My answer to your comment then would be a big: "'So?" because you still repeating what I just implied/said and not making an attemp to refute what I posted.Nnfolz 18:02, 16 October 2006 (UTC)
You haven't presented any argument, so there's nothing to refute. --Trovatore 19:21, 16 October 2006 (UTC)
I'm not even gonna bother answering that.Nnfolz 17:13, 17 October 2006 (UTC)
Did you even bothered reading what I wrote? "the actual value represented by the figuers: '.999...' and '1' is not the same. It's just that in our current model to represent the abstract concepts and values that we call 'numbers', we can play with some symbols (and other abstract concepts) to make it look that way." In other words the values represented by those figures is NOT the same, but can be made to look like equal if one plays around with some simbols and notions within the actual number system. Try substracting 1-0.999... to see what you come up with.Nnfolz 20:00, 16 October 2006 (UTC)
Maybe you want to check the Other number systems section in the main article. You can define a number system where it does not hold, but it requires the familiar rules of arithmatic are broken. Thats the magic of mathematical proof, we can start from the axioms of the real numbers and by a set of logical steps arrive at the result. If you assume the non equal position and work back it implies that some axiom do not hold so the set your working with is not what we call the real numbers. --Salix alba (talk) 20:25, 16 October 2006 (UTC)
In response to your comment on subtraction, here is an excerpt of the main article:
c = 0.999…
10c = 9.999…
10c − c = 9.999… − 0.999…
9c = 9
c = 1
Also, using the simple subtraction method, 1-0.999… proceeds infinitely towards 0, reaching 0 at infinity. Since 0.999… by definition has an infinite number of nines (a static number, not a process!) subtraction is not broken. I fail to see how decimal representation is an arbitrary concept, which can be easily manipulated to produce false results. If it didn't represent reality, it would be useless. Also, don't ask a question in a mathematical discussion and then get mad when you don't understand the response. Do the research so you can. Supadawg (talk • contribs) 20:58, 16 October 2006 (UTC)
Of course I saw the substraction example. It just an equation with an ambigouss result (i forgot the name of the correct term). I've seen them a million times. Playing with symbols you can actually get some results like 1=2 1=0 and the like. That doesn't mean they are.Nnfolz 17:13, 17 October 2006 (UTC)
It is correct algebra, so how is the result ambiguous? Just saying it's incorrect doesn't make it so; you have to prove it. So far all you've done is say that you're right and everyone else is wrong, with absolutely no evidence. We're not just playing with symbols; these are mathematical concepts, with a firm basis in reality. As a side note, here is the (incorrect) proof you mentioned[1]:
a = x
a+a = a+x
2a = a+x
2a-2x = a+x-2x
2(a-x) = a+x-2x
2(a-x) = a-x
2 = 1
Where's the error? The expression a-x equals 0, and dividing by zero is undefined. If this is a weakness of the current system, then tell me what 1/0 should be. Supadawg (talk • contribs) 22:20, 17 October 2006 (UTC)
(Before someone says infinity, I should point out that the fallacy is more about trying to divide 0 by 0 than trying to divide 1 by 0. In other words, 0 can't be cancelled.) Melchoir 22:56, 17 October 2006 (UTC)
Just to make sure everything's on the table, if we want to enable 1/0 etc. we'd probably use the real projective line or the Riemann sphere; If we want to enable 0/0 we'll have to resort to wheel theory (where it's not equal to 1, so still cannot be cancelled). -- Meni Rosenfeld (talk) 13:08, 18 October 2006 (UTC)
No discussion here. We agree compleatly. thxs for presenting an example where logical and otherwise standard rules of alebra don't aply. I was having trouble making up one myself.Nnfolz 16:53, 18 October 2006 (UTC)
You seem to have a preconceived notion that 1 and 2 are different. Yet you've just seen a demonstration that 1 = 2. Why, exactly, don't you believe it? Melchoir 17:16, 18 October 2006 (UTC)
No, you misunderstand. 0/0 is undefined in the real numbers, not just in that algebraic example. This isn't just playing with symbols; it's meaning. If there is, as you say, a logical result of 0/0, what is it? Supadawg (talk • contribs) 22:31, 19 October 2006 (UTC)
No one is claiming that pi is an infinite number, or that the diameter of a circle is infinite, so your argument is a strawman one. However, numbers which have infinite decimal expansions exist and are well-defined in the real numbers - these include 0.3..., 0.9... and pi. As for the subtraction 1-0.999..., the answer is 0. What is your answer for that? If they are not the same, you should be able to tell use the non-zero answer. Mdwh 21:14, 16 October 2006 (UTC)
The substraction is .00000....1 but since the number would be placed at infinity theres is no reason to actually calculate it.Nnfolz 17:13, 17 October 2006 (UTC)
What a great number. What happens if you square it? Melchoir 17:21, 17 October 2006 (UTC)
There is no such real number as ".00000....1". If you think otherwise, please define it. What does "theres is no reason to actually calculate it" mean? Mdwh 02:32, 18 October 2006 (UTC)


To be clear, we CAN define 0.999... without the use of this ellipsis (...) notation that seems to cause so much trouble. A more formal definition of 0.999... uses summation notation, and I've put it just below. If you claim that 1 - 0.999... = 0.000...1, then you should be able to define 0.000...1 in some way, since the definition for 0.999... does not extend to represent 0.000...1 in any obvious way. -Maelin 06:21, 18 October 2006 (UTC)
\sum_{k=1}^\infty {9 \over 10^k}
I mean that it is not practical since infinity has no limit placing a '1' at the end of infinity would be a contradiction and useless (the same with the 9's).
This is exactly the point. One of the properties of the real numbers is that for any two numbers that are not equal, there exists another real number, not equal to either of them, which lie between them. In this case, we can't construct a number between 0.999... and 1. In fact, such a number does not exist. This means that the numbers 0.999... and 1 cannot be different - that is, they are equal. -Maelin 00:09, 19 October 2006 (UTC)
Nnfolz, I repeat, you have given no argument whatsoever. You have made only a bland assertion. There is nothing to refute. --Trovatore 21:10, 16 October 2006 (UTC)
[Edit conflict] Yes, I was indeed trying to say that no finite representation of real numbers exist - in response to your use of "our current system". It's not our current system that represents finite real numbers as infinite objects, it's any system.
My impression from the phrasing of your question is that you are an enthusiastic and capable thinker, but lack formal mathematical education (and I don't mean to condescend). This makes discussing real mathematical issues difficult. If you can provide us with a mathematical argument for your claim we would be happy to respond, but if you only provide us with your intuitive notions of these concepts there's very little we can do.
Your main claim can be considered correct if interpreted as "the symbol 0.999... can, in structures other than the real numbers, be reserved for an object different from 1". But, as pointed out by Salix alba, this idea is already covered in the article. Also, it is beside the main point, which is: The symbol 0.999... is usually taken to represent a certain real number, and this number is 1. -- Meni Rosenfeld (talk) 21:13, 16 October 2006 (UTC)
I will ignore the comments about my mathematical education (wich you know nothing about) and lead you guys on to onsider this other example: is 6.999... exactly equal to 7?Nnfolz 17:13, 17 October 2006 (UTC)
That would be under 0.999...#Generalizations. Melchoir 17:19, 17 October 2006 (UTC)
I apologize if I have offended you. I do believe my comments were appropriate, but feel free to ignore them. About your latest comments: We are all discussing here the real numbers, which are a field, and therefore subtraction is always defined - In contrary to your latest statements. And yes, 6.999... is indeed 7. Now, since the article already does more than enough to prove the fact that 0.999... = 1, it is up to you to try to prove that 0.999... ≠ 1. So far you have not done that. -- Meni Rosenfeld (talk) 20:28, 17 October 2006 (UTC)
That's exactly it. I cannot disprove it using any kind of mathemathical proof because the whole issue is 'designed' that way. I cannot pretend to know how to prove my point using mathematical notation cuz is only 'possible' on the shallower levels so i'll just back down from thios whole issue for the time being. I've heard to many arguments I can't refute so if/when I can i'll be back here. Anyways, you can expect to se me around other pages of ismilar subjects learning more about them. C-ya around.Nnfolz 16:53, 18 October 2006 (UTC)
It's true that the real numbers are designed in such a way that ultimately forces 0.999… = 1. But that design is not originally motivated by 0.999…. It's actually motivated by a small number of very reasonable specifications: we want a number system in which one can do arithmetic and which forms a connected line. By contrast, the only known number system in which 0.999… < 1 is not only blatantly contrived for that very purpose, it has awkward, unintuitive properties and appears to be useless. Ideas from the "shallower levels" of mathematical thought can often be formalized, and they're worth exploring for instructive purposes -- just not very far. Until you get a better feel for how mathematics works, it's more productive to stick with the same structures everyone else uses, even if you don't like one of the results.
Anyway, some of your fellow Wikipedians would be interested to know what similar subjects you turn up, but I for one don't intend to stalk you! So, feel free to ask questions at Wikipedia:Reference desk/Mathematics. Melchoir 18:00, 18 October 2006 (UTC)
One of the most delightful things in the study of mathematics (in my opinion, anyway) is finding situations in which perfectly reasonable premises and valid arguments lead to surprising or counterintuitive results. Often, the mind rejects the conclusion to begin with, but then after some more consideration of the proof and discussion with others, a beautiful moment of understanding occurs. Another popular example of this, where the truth strongly contradicts intuition, is the Monty Hall problem. Some other editors may know of more. I encourage you to continue looking into this issue, and wish you luck in finding better understanding of the problem. -Maelin 00:09, 19 October 2006 (UTC)
Banach-Tarski Paradox (or other unintuitive result regarding measure theory) seems like the best one available: what could be more reasonable than requiring there be a way to extend "length" to a notion for every set of real numbers that is translation-invariant and additive? Hilbert thought it was possible35.11.50.219 16:56, 20 October 2006 (UTC)

What exactly is your reasoning that this isn't true, as opposed to there being a "flaw" with the number system? You just state over and over that the number system is flawed with no evidence that actually stands up to any kind of scrutiny. Is it inconceivable that a slightly counter-intuitive result is in fact true, rather than the alternative of the exceptionally powerful and useful number system being wrong? You state yourself that it's impossible to show that 0.999... = 1 is incorrect using the number system BECAUSE the number system is flawed. Well, I ask you, then how is it not true? The ideas of 0.999 and 1 don't exist outside of the number system, they are abstract ideas that exist solely within the number system. Since they, therefore, have no way of being incorrect outside of the number system, and are correct within the number system, there is nothing wrong, and no flaw. Justdig 17:55, 25 October 2006 (UTC)

I've never been able to understand these "proofs"

(moved from Talk:0.999...) No doubt, this has been hashed out many times before. Anyone who isn't sick to death of the thing, please explain this to me: It seems to me that, while the limit of .999..., is 1, of course, .999... itself (i.e. the sum of that infinite series) cannot claim to represent 1, or any real number.

1)The 1/3 proof simply begs the question. I would say that .333... does not represent 1/3 either, for the same reason. 2)The x times 10 - x, has the inherent flaw that for any given number of 9s, there will always be one extra 9 after the decimal point in x, than in 10x. How is this different from the famous proof that "zero" = one: (1+-1)+(1+-1)...=1+(-1+1).... The flaw there is that no matter how far you go, you can't cut off the end of the parentheses--isn't this the same for the "last" 9? Or, if there is no "last nine", why in the false proof is there a "last -1"? 3)Finally, what seems to me to be a proof that .999... is distinct from 1:

  1. 1 - .999...=.000...1 (a 1 after infinite zeros=an infinitesmal quantity)
  2. .000...1 1/infinity (an infinitesmal quantity)
  3. If 1=.999... then 1 - .999... = 1 - 1 = 0
  4. If .000...1 = 1/infinity, then infinity times .000...1 = infinity/infinity
  5. If .000...1 = 0 (step 3) infinity times .000...1 = infinity x 0 = 0
  6. But infinity/infinity does not equal zero.

I am not claiming that this is right, but which step is wrong? --judahh

Step 1 - "I would say that .333... does not represent 1/3 either, for the same reason" You are wrong. Just do the long division by hand and it's quite obvious that an infinitely repeating .333.. is equal to 1/3. Raul654 05:45, 23 October 2006 (UTC)

I wouldn't say that's a satisfactory answer, since it isn't at all obvious that long division works exactly for infinitely many digits. So I'll try to answer as well:
0) The sum of an infinite series is defined to be the limit of the sequence of partial sums. Likewise, 0.999... is the limit of a certain sequence (namely: 0, 0.9, 0.99, 0.999, ...). Since that limit is obviously 1, 0.999... is 1.
1) It can be proven that 0.333... = 1/3, but this is not done in the article (so if you question that, focus on the other proofs).
2) There is no last nine. The flaw with the proof you mention is the false assumption that infinite sums are associative. You can't play around with paranthesis in infinite sums (unless certain conditions are met, which is not the case here).
3) The Archimedean property of the real numbers says (essentially) that there is no positive infinitesimal real number. Therefore, your hypthetical "0.000...1" does not exist - the result of subtraction is 0. This is easy to see if you understand that the decimal expansion is a sequence of digits - for every natural number n, there is a digit at place n. There isn't a "last" digit, or a digit at the ωth place (ω is the least infinite ordinal number), or at the (ω+1)th place... In the number you mention, the 1st digit is 0, the 2nd digit is 0, the 3rd digit is 0... It is no other than 0.
-- Meni Rosenfeld (talk) 09:38, 23 October 2006 (UTC)
Point 1 we're dismissing, and I'll accept your point two. Where I'm still confused is point 3. Let me ask you this: Is 1/infinity 0? If it is, then I'm not sure what is wrong with my "counterproof" of point 3.--jh151.202.111.202 23:39, 23 October 2006 (UTC)
Oh, and about the "I've never been able to understand these proofs" part; once you understand what a cauchy sequence is, what an equivalence relation and equivalence class are, how the real numbers are constructed from cauchy sequences of rational numbers, and the fact that 0.999... is simply the equivalence class of the sequence (0, 0.9, 0.99, 0.999, ...), you will be able to prove yourself that 0.999... = 1, let alone understand the proof. The linked articles provide all the information (and you can follow other links as necessary). -- Meni Rosenfeld (talk) 10:13, 23 October 2006 (UTC)
K, I'll have a look.151.202.111.202 23:46, 23 October 2006 (UTC)
The flaw with your -1+1-1+1-... proof is that your series, which can be expressed as \sum_{i=1}^\infty (-1)^i is not convergent. It does not converge to a value in the limit. However, 0.999..., which can be expressed as \sum_{i=1}^\infty {9 \over 10^i} DOES converge - in fact, it converges to 1. Maelin 11:39, 23 October 2006 (UTC)
I'll accept that point, too. The truth is, I realized you would be able to poke holes in my analogy. I'll tell you what really bothers me and other people who don't get this. (It's not that we don't realize that .999...[infinite nines] is not the same as .9...[a big finite number of nines]. That's an extremely patronizing assumption which people seem to make a lot.) It's that adding 90% of something is by definition not the same as adding a hundred percent of something. So even if the infinite sum converges on one, how can it possibly be one, if what is being added are things which by definition do not make it to one. NOTE: I'm not saying this reasoning is correct--but it's what I haven't understood. The one thing I can pin down that clearly seems to disprove this theorem to me is my third point.--jh151.202.111.202 23:46, 23 October 2006 (UTC)
There are way too many pronouns here for such a subtle point. You say "even if the infinite sum converges on one, how can it possibly be one". What, exactly, is "it"? Be careful! Melchoir 00:12, 24 October 2006 (UTC)
How can that infinite sum possibly be one, if every single one of the infinite addends of that sum is a quantity which by definition is not enough to make the total sum one.151.202.111.202 00:46, 24 October 2006 (UTC)
What is "that infinite sum"? Melchoir 00:48, 24 October 2006 (UTC)
Sorry, I meant .999... --jh151.202.111.202 00:53, 24 October 2006 (UTC)
.999... is not an infinite sum. Melchoir 01:09, 24 October 2006 (UTC)
OK, you know what I mean. It's the number which is the solution to the infinite sum \sum_{i=1}^\infty {9 \over 10^i}--you said that yourself. I don't see how the solution to that sum can be equivalent to one.
Close. There are no solutions or infinite sums here; there is an infinite series, and the sum of that infinite series. Do you know what it means for a number to be the sum of an infinite series? Melchoir 05:37, 24 October 2006 (UTC)
Infinite sum was an expression that other people used here. If it's not the correct one, fine, that doesn't make a difference. I believe I know what it means for a number to be the sum of an infinite series; I may be confused about the implications of that, of course. Why are you playing semantics?--jh129.98.212.74 14:37, 24 October 2006 (UTC)
Because there are many mathematical objects surrounding the notation "0.999…". Some of them equal 1, and some of them do not. If you do not see the subtle differences between these things, you are more likely to fall into the trap of mentally fixating on the things that do not equal 1 (without noticing that 0.999… is not defined to mean one of them). You claim that you know what it means for a number to be the sum of an infinite series. No offense, but I think you don't know what it means. Can you explain it? Melchoir 17:30, 24 October 2006 (UTC)
I'm not sure what there is to say. My understanding of an infinite series is that it is a set that contains an infinite number of objects, which can be specified individually by application of a formula. These objects are arranged in order, since they're a series, but for adding purposes, order doesn't matter. If you (hypothetically) all these objects together (if they're numbers) you will get the sum. Obviously, such addition can never be "completed". That is why I can't see how the concept of .999... can be viewed as a real number. Now, if you wanted to say that for purposes of your own, you will define the result infinite sum by its limit, there wouldn't be much to say, but such a definition doesn't seem to follow from the original definition of what an infinite sum is. (It's probably obvious that my background in math is not particularly extensive.)--jh151.202.111.202 18:04, 24 October 2006 (UTC)
Please, the less you say the phrase "infinite sum", the happier you'll be. Obviously no one can compute an infinite sum. But the sum of an infinite series is defined as the limit of the sequence of its partial sums. No, this definition does not follow from any earlier definitions; that why we need to make it a definition in the first place! Hundreds of years ago, mathematicians argued over infinite series, and some of them arrived at comically bad results. In hindsight, the confusion could have been avoided if they'd bothered to define what they were talking about. A good definition doesn't have to be intuitively obvious; it just has to be useful. There is an extensive body or research on infinite series and their sums... and it all works. Melchoir 18:33, 24 October 2006 (UTC)
"So even if the infinite sum converges on one, how can it possibly be one" - An infinite sum is by definition equal to the limit of the sum. So if you agree it converges to 1, then by definition the infinite sum, and 0.999..., is equal to 1. Mdwh 00:52, 24 October 2006 (UTC)
That "by definition" confuses the issue, I think. The only definition of an infinite sum I'm working with is the empirical one of the sum of an infinite series of numbers which are specified by given criteria--in this case each one is 90% of the previous one. If that definition leads to .999... being equal to one, great; if it doesn't, then redefining it won't help matters. I don't understand how the empirical definition leads to .999... being one.151.202.111.202 00:57, 24 October 2006 (UTC)
But that is the definition. It's how many mathematicians define infinite sum. If you don't accept that 0.9... is equal to the limit of the series {0.9, 0.99, 0.999, ...}, then the number 0.9... is meaningless. Perhaps you can define it in a different way if you like, but that's got nothing to do with what we mean by 0.999... in this article. Mdwh 02:42, 24 October 2006 (UTC)
If you can't prove that .999...equals one from the basic definition (i.e. a description of what terms are being added together), it doesn't seem like you're saying very much when you say that .999...=1 "by the definition you've chosen to use". If that's really the case, then there isn't much to say. (It doesn't bother me to say that .999... is meaningless, if that means that it doesn't refer to a real number.)151.202.111.202 04:57, 24 October 2006 (UTC)
Not exactly. He did not assume that the series converges, but rather he "proved" that it converges to 1 and that it converges to 0. The flaw in his proof is, as I mentioned, the assumption of associativity. -- Meni Rosenfeld (talk) 12:17, 23 October 2006 (UTC)
Concerning the "counter-proof" (which I made more readable), the problem is in steps 4 to 6: Multiplication by infinity, similarly to division by zero, is usually not well-defined. Even if it is in most cases (with a result of infinity), still 0/0, infinity/infinity, and 0*infinity are not defined. Consider for example 2*(oo/oo). Should that equal (2*oo)/oo = oo/oo ?
But surely infinity*0 can be clearly defined as 0. An infinite number of 0s add up to zero; conversely, 0 infinities, are simply nothing at all.--jh151.202.111.202 23:24, 23 October 2006 (UTC)
As an aside, even if 1-0.9999... and 1/infinity were both infinitesimals (and there are none in the standard reals), why should they be the same infinitesimal? --Huon 14:44, 23 October 2006 (UTC)
For (1) and (2), yes, these are not rigorous proofs. The point about (1) is for those people who do accept that 0.3... recurring equals 1/3 (I presume they must exist, because there seem to be far greater numbers of people who dispute 0.9... = 1 compared with those who dispute other fractions). If you want a rigorous proof, then check out the later proofs. As for your proof, you'll have to define ".000...1" rigorously before we can continue. And as for the rest of the stuff, would you like to show us how you are defining those operations with "infinity"? What would you say 10 - 9.9... is equal to? Surely it is still 0.0...01 in your number system? But is that equal to 1 / infinity, or 10 / infinity? Mdwh 22:36, 23 October 2006 (UTC)
Both you guys are right--I wasn't very rigorous. I am looking at an infinitesmal not as a number, but as a concept, and therefore I am looking at any infinitesmal as the same thing. This may well be fuzzy, but can you tell me how you view 1/infinity? If it is 0, then it seems to run into the problem I raised.--jh151.202.111.202 23:49, 23 October 2006 (UTC)
1/infinity is not defined in the reals. What you can say is that the limit of 1/N, as N tends to infinity, is equal to 0. Mdwh 00:29, 24 October 2006 (UTC)
That is exactly what I would say--except that I would apply it to .000...1 as well, and to .999...(the limit in that case being one). I guess I see where you are distinguishing, though; thanks.151.202.111.202 00:51, 24 October 2006 (UTC)
Apologies for being technical, but this seems like the right thing here:
Definition :\sum_{k=1}^{n}a_k = S if and only if for every real number ε such that ε > 0, there exists a natural number N such that if n is a natural number greater than N, then |\sum_{k=1}^{n}a_k - S| < \epsilon.
This, and nothing else, is what we mean we talk about the sum of an infinite series. As for decimal expansions, The notation 0.a1a2a3a4... is taken to represent the number \sum_{k=1}^{\infty}\frac{a_k}{10^k}. So, we see that 0.999... = \sum_{k=1}^{\infty}\frac{9}{10^k}, which can be easily shown to be, according to the definition above, equal to 1. That's really all there is to it.
By the way, it is legitimate to leave decimal expansions ending with infinite 9's undefined (though this creates more complications than it solves) - the important part is that 0.999... cannot represent any real number other than 1.
I'll agree to that. What I think I'm gathering from all your responses is that the definition of .999... as 1 is a mathematical convenience, not the logical result of adding an infinite number of elements (a concept which is hard to grasp in a concrete way anyhow).151.202.111.202 18:16, 24 October 2006 (UTC)
As for your other arguments: There is no element "infinity" in the set of real numbers, and this is the set to which we restrict our discussion. But suppose there was. You agree that 1 - 0.999... = 0.000...1, that 0.000...1 = 1 / infinity, and that 1 / infinity = 0. You agree, then, that 1 - 0.999... = 0. Unless I have misinterpreted your claims, how can you disagree that 0.999... = 1?
The part you misinterpret (or maybe I miswrote) is that I agree that 1/infinity is 0. I would call 1/infinity an "infinitesmal". What I said was that if it was 0, that would lead to a contradiction when it is multiplied by infinity, with result of undefined if you talk about ∞/∞, and result of 0, if you talk about ∞*0. You disagree with that below, though.151.202.111.202 18:16, 24 October 2006 (UTC)
As for 0 * ∞, that depends entirely on the context. It is not in general equal to 0 (arithmetic of limits would not work if it was). -- Meni Rosenfeld (talk) 16:13, 24 October 2006 (UTC)
Could you explain that? Doesn't the concept of a*b mean a + a + a...b number of times? And isn't \sum_{k=1}^{\infty}0_k equal to 0? Isn't that what you yourself agree to, when you consider the infinite decimal expansion of 1.000...?
Looked at the other way, isn't a zero-quantity of anything, be it 1 or ∞, equal to zero without any regard to what that number is? (actually I'm starting to see how you could hedge if you define these operations by going backwards: 0*∞ = 1*∞ - 1*∞. How would you deal with the first point, though?)151.202.111.202 18:16, 24 October 2006 (UTC)
What Meni meant by arithmetic of limits: If you have two convergent sequences a_n, b_n with limits a and b, then you would like the product sequence a_n*b_n to converge to a*b. For sequences of real numbers converging to real limits, that is correct. Now if you add infinity, it gets difficult. For example take a_n=n, b_n=1/n. Those sequences converge, the a_n's to infinity and the b_n's to 0. But the product sequence is the constant sequence (1, 1, 1, ...), which converges to 1, not 0. You can easily construct other sequences converging to 0 and infinity so that the product converges to whatever you like (including zero and infinity themselves).
In general, your definition of multiplication as just "adding the correct number of times" is a little tricky. For integers, all is fine and well. But what would, for example, e*pi be? e+e+e+... pi times? What are "pi times"? Thus, you usually define multiplication for rational numbers, then define reals as limits of cauchy sequences of rational numbers (more precisely, not as limits, but as equivalence classes) and define the multiplication of reals by the very arithmetic of limits explained above. That includes quite a little work I just omitted, but that's the general idea. Yours, Huon 19:34, 24 October 2006 (UTC)
I think a lot of loose ends can be tied by addressing a subtle issue, that of extending definitions. Often we find that a certain phenomenon occurs frequently and has some general importance. For example, if we have 3 boxes of 6 apples each, the total is 18 apples; if we have 7 boxes of 5 apples each, the total is 35 apples. We can generalize this by defining multiplication in such a way that for every natural numbers a and b we have a*b = a+a+...+a, b times. We can then say that if we have a boxes of b apples each, the total is a*b apples; and we can study the properties of this new entity, "multiplication", in a way that will help us calculate apples (and of course - any other phenomenon where multiplication is applicable).
We can then notice that what we have defined is actually just a part of something greater. For example, suppose rational numbers are defined in some way. We can then ask, "what does 4.7 * 3.5 equal?". One approach is, "it doesn't; multiplication is only defined for natural numbers". While technically true in the context in which it is said, we know in hindsight that this approach would not lead us very far. Another approach is to extend the definition of multiplication to include all rational number, in a way that will preserve as much of the logic and properties of our original definition. As we all know, in this particular case, there is one possibility which fulfills anything we could expect from this operation. So we define multiplication in the new way, and subsequently treat the original definition as just a "landmark" on the way for the more general definition.
Unfortunately, in some cases extending a definition can be desired but impossible to do in a way that is completely satisfactory. Sums are an example. We can define the sum of 2 real numbers, and we can easily extend it to 3, 4 or 5 numbers, but what about infinitely many numbers? We cannot calculate addition infinity times, so there is no single result which follows logically from the original definition. So again we are faced with a dilemma. Will we give up on defining "infinite sums"? A possibility, but evidently, not a very fruitful one. So we do what we always do: Extend the definition in a way which will preserve as much as possible of the original. And this extension is exactly what I have quoted above. Is this everything we could have hoped for? Apparently not, and your criticism of the concept is a proof of that. But this is all we can achieve. Refusing it essentially means discarding half of mathematics; Accepting it means living with a few curiosities, such as the fact that we can add as many terms as we'd like of the series \sum_{n=1}^{\infty}\frac{1}{n^2}, and never reach π2/6; and yet the sum of the infinite series is mysteriously equal to π2/6;.
This relates to the 0 * ∞ issue as well; No doubt, 0 times any real number equals 0. But when we try to extend multiplication (or any other arithmetical concept) to include infinite quantities, we are faced with very difficult choices of what to keep and what to reluctantly discard. Each choice defines a different structure the properties of which we can investigate. Some (indeed, many) lead to 0 * ∞ = 0; others do not. Since we have not decided here on a particular structure with infinite quantites as the subject of our discussion, we cannot take that for granted.
I suppose recommended reading for some insight into infinite quantities is Infinity, Real projective line, Hyperreal numbers, Ordinal numbers, Cardinal numbers. I hope that ultimately, you will have greater knowledge as a consequence of this discussion! -- Meni Rosenfeld (talk) 21:24, 24 October 2006 (UTC)


These very few thoughts might help. Limits normally mark that point which you nearly but never quite touch. So if you took the limit of f(x)=1/x as x goes to infinity this thing naturally gets closer and closer to zero. Now consider if you have f(x)=x^2. In this case you could take 2 limits as x goes to zero. The first would approach from positive numbers and the other from negative numbers. In both cases the limit equals 0. There is a rule which says that if the all the limits for a point are the same irrespective of where you come from then f(x) EQUALS the limit at that point. This is exactly the kind of thing that is happening here. 0.999... equals one exactly because infinity means continually adding extra little nines till you don't need to any more. The expression itself is a limit. If you are lost you probably need to go and read (and practise) some university level maths (they are not that hard). Trying to apply your own logic usually just makes you feel stupid, when a maths genius explains the real answer!. —The preceding unsigned comment was added by 213.202.131.74 (talk • contribs).

Actually, that's wrong. The "rule" you quoted is only correct if f is continuous. What you meant to say is that if f is not defined a priori at a point, we can extend its definition, in an attempt to make it continuous, setting it equal to the limit. -- Meni Rosenfeld (talk) 09:54, 25 October 2006 (UTC)

No Way!

This is ridiculous. Of course it's not equal to 1. One is one. If you have .9, you're pnly .1 away from 1. But you don't have .1; you have .09. Now all you need is .01, but you have to make do with .009, et cetera. It's the same as travelling half the distance between two points.WikiManGreen 00:35, 25 October 2006 (UTC)

Neither this article, nor its sources, ever claim that any finite truncation of 0.999… equals 1. In fact, some of the proofs directly exploit the exact amount by which they fall short of 1. Melchoir 00:57, 25 October 2006 (UTC)
After seeing the "Algebra proof", I see that using the rules of our number system produces a result that is consistent with what I've been taught about math. However what I walk away with is not that 0.999... == 1 but rather that our number system is imperfect and is capable of producing imperfect results. (0.999... is not equal to 1.)
A perfect example of intuitionism defeating rational thought in the human mind ("it's not my reasoning that's faulty, it's the system"). Nevertheless, standard mathematics such as this is used by actual mathematicians every day, and I haven't heard them complain. Supadawg (talk • contribs) 02:07, 25 October 2006 (UTC)
"it's not my reasoning that's faulty, it's the system" - pot calling the kettle black —The preceding unsigned comment was added by 203.173.148.254 (talk • contribs) .
Indeed, 0.9 is not equal to one. 0.99 is not equal to one. 0.999 is not equal to one. If you made a list of all such numbers, not a single one of them would be equal to one. So why do we claim that 0.999... is equal to one? Simple. 0.999... is not in that list. No number that has the form of a zero, a point, and then some finite number of nines, is equal to one, not matter how many nines you have. But 0.999... is not of that form. It's a different number to any of them, a number that has a property none of the finitely long 0.999..9 numbers has - it's equal to one, as the proofs in the article show. Maelin 02:22, 25 October 2006 (UTC)

If you take 1/3 that equals 0.333..., so if x = 1/3 then x = 0.333, now if you take 3*x, then that is BOTH 3*(1/3) which is one, AND 3 * 0.333... thus if 3* 1/3 is one then 3 * 0.333... is one. justinphd 03:30, 25 October 2006 (UTC)

1 = 1

1 = 1

0.999... = 0.999...

1 + 2 = 1 + 2
4 - 1 = 4 - 1
Therefore, 4 - 1 ≠ 1 + 2 Supadawg (talk • contribs) 02:44, 25 October 2006 (UTC)

wrong. where is your evidence that this is true? through the whole statement, you did not provide any evidence that one side was unequal to the other, or exclusively equal to another, except for the end, which is false. Xparasite9 15:51, 27 October 2006 (UTC)

Was that a response to unsigned or to Supadawg? In any case, technically unsigned has only made true statements, but has implied the conclusion 0.999... ≠ 1, which is just as fallacious as Supadawg's sarcasting conclusion 4 - 1 ≠ 1 + 2. -- Meni Rosenfeld (talk) 16:54, 27 October 2006 (UTC)
Oh. I get it now. I thought that whole thing was stated by Supadawg. Now I see the satire.Xparasite9 20:00, 30 October 2006 (UTC)

Probably another proof

I just probably find another proof of my theory that .999... isn't equal to 1. I was reading the recent modifications of the article and I that is says: lim n->inf 1/(10^n)=0 but that isn't true because lim->inf (1/10^n)*10^n = 1 and 0 multiplied by any number doesn't equals 1.... also in my thinking i also got to this .333 =/= 1/3 <--> .999 =/= 1 That means .333 isn't equal to to 1/3 only and only if .999 isn't equal to 1. You can't prove that .999... = 1 with with the afirmation that .333... is 1/3 because one affirmation depends of the other =P. Even if you can threat .333 as 1/3 because the difference is minimum the difference is still in there. Well that's just what i think... but i'm sure that what i said about 10^n is right. I hope you can understand my english. --Rensato 04:06, 25 October 2006 (UTC) just a little edit I was thinking... again... and i came to the conclution.... that.... You could possibly say that inf/0=1 O_o that would be the only way to say that.... lim n->inf 1/(10^n)=0

Sorry I'm not any math genious... i don't know if i'm getting this wrong but i want to put my part in showing the truth to the people =). --Rensato 04:13, 25 October 2006 (UTC)

Sorry, but this is a little off. Lim n->inf 1/(10^n) and Lim n->inf (1/10^n)*10^n are completely different functions. Because Lim n->inf (1/10^n)*10^n = Lim n->inf 1. The limit of a constant is the constant itself, so Lim n->inf (1/10^n)*10^n = Lim n->inf 1 = 1. However, Lim n->inf 1/(10^n) is totally different. The denominator contains an exponentially increasing function, so as n tends to infinity, 10^n gets gigantic. So the denominator get very large, which means 1/10^n gets very small. So Lim n->inf 1/(10^n) approaches 0 as n approaches infinity. Tachyon01 05:23, 25 October 2006 (UTC)

It's not quite right to say that 0 multipled by any number is never anything but 0 when dealing with limits. For example, another well known result is that \lim_{x\rightarrow 0}\frac{\mathrm{sin}\ x}{x}=\lim_{x\rightarrow 0}\mathrm{sin}\ x\cdot\frac{1}{x}=1. And here the dreaded \infty makes no appearance! Unmasked 16:38, 25 October 2006 (UTC)

The limit proof

Limits, such as shown at limit of a sequence and limit of a function, work as a value is approached, but not at the value. Why is it, then, that the limit proof used in the article can be applied at .999...? -- kenb215 talk 05:18, 25 October 2006 (UTC)

Because 0.999… is defined as the limit of the relevant sequence. You can't evaluate a sequence at infinity, so we don't even try to do so. Melchoir 05:57, 25 October 2006 (UTC)

another problem students have

as a student recently exposed to this proof my problem was not any of those mentioned but rather my interpretation of .999... as equalling 1 minus an infinitesimal

and 1-x does not equal 1 when x =any non zero number

1–x≠1 if x≠0

I totally realize I’m wrong but if you could explain what part of my premise is wrong it would be appreciated (and if this is a common problem perhaps it should be included in the article

Hey,
I will try to help. Please ignore the troll below.
Try thinking about it this way: For any number n which is greater than 0, 1-n can be evaluated to find a number that is not the same as 0.999...
So if you try to solve the equation 1-n = 0.999..., what is the only value of n that works in this equation? The only answer is 0. If you don't think that's true, then name the number. What is n specifically, if not zero? Some people who try to dispute that 0.999... = 1 might try to invent a number like "0.000...1." But this is not a meaningful string of symbols. It is impossible to have a 1 "after" an infinite string of zeroes, because the zeroes never end and the 1 does not exist anywhere. Argyrios 05:42, 25 October 2006 (UTC)
Another way to look at the "0.000...1" idea is this: take the "ones" place, 0.000...1, and assign that the number 1. Take the tenths place, 0.000...1, and assign that the number 2. Assign 3 to the hundredths place, etc, so that you can refer to any specific place on the infinite string with a number, n. For any value of n, then, the decimal in that place is a 0, and not a 1. There exists no number n for which the value in the place represented by n is a 1. Since the one does not exist in any specific place value, no matter how high you go, it does not exist in the string and it is meaningless to indicate that it does. You really just have an infinite string of zeroes, 0.000...
Helpful? Argyrios 06:08, 25 October 2006 (UTC)


But if .000...1 is not a real number then what is an infinitesimal? to me f(x)=x as x approaches 1 never reaches one but does reach every number prior to 1 including .999... if 1-an infantasimal does not equal .999... then what does it equal
to make my last point differntly when y=x on the interval x is greater than zero and x is less than one then y equals every number up to one but not one. this means y equalls (among other things) 9(1/10^1)+9(1/10^2)+9(1/10^3)....9(1/10^n) where n as an arbitrarily large number. if n is infinity (as would be included on the interval) and .999... equals 1 then this function would equal one on an interval when it exclusivly does not.
its not that i doubt or do not understand the proofs its that is see conflicting reasons so if you could clarify where i'm wrong in either the first statment or this one it would be much appreciated
Infinitesimals do not exist in the real numbers. The set of real numbers has the Archimedean property, which basically means "no infinitely small numbers". Infinitesimals are sometimes used as a justification for the dy/dx notation, but this arises because that's how we interpreted derivatives in the past. We no longer use the infinitesimal-based definition, but the notation is helpful because it gives several operations intuitive behaviour. Maelin 06:38, 25 October 2006 (UTC)
So this is true in the system of real numbers using Archimedean property? I don't think anyone would have a problem with that. However the idea that any non-real or non-Archimedan math doesn't or can't exist I think is what's seen as more difficult.--T. Anthony 08:40, 25 October 2006 (UTC)
No one has said that non-archimedean structures do not exist. Obviously they exist. But they're not what the main body of this article is about. -- Meni Rosenfeld (talk) 09:58, 25 October 2006 (UTC)
But don't you think it's telling that someone can get that impression from this article? It should really be a great place to introduce the ideas of infinitesimals and non-archimedean systems clearly rather than the current obsession with real numbers and proving an unintuitive equality. It makes the article seem like it's more interested in demonstrating a trick than in actually informing. (Not to criticise you meni, you're doing a good job with all this explaining, thanks).

You're not wrong; the proofs are faulty. See "This article is an unfortunate embarrassment to Wikipedia" above for this same discussion.

-Dworkin--76.1.39.216 05:08, 25 October 2006 (UTC)

Flaw in decimal proof

c = 0.999…

10c = 9.999…

10c − c = 9.999… − 0.999…

9c = 9

c = 1

The flaw in this proof is in the third line. We cannot simply assume that an infinite sequence multiplied by ten will simply add 9. It doesn't. You can see this by replacing .999... with 1-m. m being a hypothetical "smallest positive number." What you get is not 9, but 9-11m.

m is a real number. It has a place on the number line, next to 0. If m is unachievable, then neither is .333..., which disproves .333... = 1/3. Why would .333...3 be valid, but not .333...4? Algr 07:33, 25 October 2006 (UTC)

The action of multiplying a decimal by 10 comes straight out of elementary arithmetic, and if you don't believe it, it's immediately justified in the sum formalism, the limit formalism, or the Cauchy sequence formalism. Also, there is no smallest positive number. It's not just hypothetical; it's non-existent. Melchoir 07:50, 25 October 2006 (UTC)
Elementary arithmetic requires adding a zero to the end after you shift the digits to the left, so citing it proves that at some point a 9 is not canceled, and the result is not 9. Algr

How about (1 - 0.999...) as the smallest positive number. Davkal 07:55, 25 October 2006 (UTC)

That's what I am saying. "The smallest positive number" is certainly a more reasonable conjecture then "the square root of negative one", or "the ratio of a circle's circumference to its diameter". Algr
What's reasonable about it? If ε is "the smallest positive number", then what is ε/2? A smaller positive number? Zero? Undefined? I'm not saying this isn't workable, just that it's very unreasonable, unlike the good ol', heavily studied and rigorously defined real numbers and complex numbers. -- Meni Rosenfeld (talk) 10:01, 25 October 2006 (UTC)
If you find &epsilon to be less reasonable then i, then it is just because you haven't been taught it. That isn't logic, but just conforming to authority. Algr 17:56, 25 October 2006 (UTC)
I assume you mean the second line not the third? [Arnie/NL]
Yes, it could be taken that way. I guess it is up to what you assume 9.999... means.


Algr, you are correct, line 2 is not a legitimate calculation. The trick is a simple one, but it passes most people undetected: by multiplying by 10 then subtracting 9 they are truncating the infinitesimal--which happens to be the difference between 0.999... and 1. Unsound mathematics. --JohnLattier 05:30, 3 November 2006 (UTC)

What's the difference between the set of nonnegative integers {0, 1, 2, 3, ...} and the set of positive integers {1, 2, 3, 4, ...}? Would you say it's {0}, or {0, ∞ + 1}? --dreish~talk 15:04, 3 November 2006 (UTC)

Why am I wrong?

Imagine that all the numbers that exist, starting from zero, were placed one after the other in order of ascending value. Surely the number that would appear one place below 1 in that list would be 0.999... And since 0.999... appears prior to 1 in a list of ascendding value it must be less than 1. Why am I wrong? 82.35.70.17 07:06, 25 October 2006 (UTC)

Well, unfortunately, the first step is wrong. The real numbers can't be listed in ascending order. In fact, they can't be listed at all! See Cardinality of the continuum. The impossibility of an ascending list is easier to see than the general statement, though: if you have any two adjacent elements on the list, where does their average go? Melchoir 07:10, 25 October 2006 (UTC)

Ok, place these numbers in order of value: 0, 0.1, 38, 0.125, 1, 0.333..., 0.999..., 2, 5, 7.

0, 0.1, 0.125, 0.333..., 0.999..., 1, 2, 5, 7, 38. No? Davkal 07:19, 25 October 2006 (UTC)

That list is nondecreasing, to be sure, but it isn't strictly increasing, because two of its items are equal. Melchoir 07:22, 25 October 2006 (UTC)

That would seem to a bit question begging. What's just below 1 on the continuum? Or is there a little hole in the continuum just below 1? Or, if the continuum can be collapsed, maybe this is a proof that 0=1000 (or anything else for that matter). Davkal 07:28, 25 October 2006 (UTC)

There isn't anything "just below" 1 or any other real number. You can't shoehorn the continuum into either a list indexed by the integers, or a single collpsed point. It's a much more interesting structure than either of those. And, in fact, there are ordered sets even more interesting than the continuum; see Long line (topology) for example. Mathematical structures are often surprisingly rich, to the point where easy methods of investigation, like listing, are inadequate. Melchoir 07:34, 25 October 2006 (UTC)

Suppose x is ``just below 1 on the continuum". What's between x and 1? Nothing, since x was ``just below 1"? What about y=(x+1)/2, the average of x and 1, positioned halfway between them? But then x was never ``just below 1 on the continuum", y was. Worse still, there's still z = (y+1/2) and so on... Asking what's ``just below 1 on the continuim" is like asking how far you can walk without going 1 kilometer, or how long you can wait without waiting for a whole day. If you haven't gone 1 Km yet, you can always walk *a little bit further* and still not have gone 1 Km. If you haven't waited for a day yet, you can wait for *just a little bit* more, and stil not have waited for a full day. It's very similar to the dichotomy paradox / race course paradox from Zeno's paradoxes. There isn't a distance that's closest to 1km or an interval of time that's closest to 1 day. You can always find distances and times in between. Similarly, the continium is so named because it's continuous. There are always averages, midpoints, numbers in between. There isn't a smallest nonzero number. There isn't a number that closer to 1 than any other. Between any two different real numbers, there are always infinitely more. The continium is continuous. 210.54.148.98 21:26, 25 October 2006 (UTC)

Mr. 82.35.70.17, you aren't wrong if you are defining every number to be a point of infinitesimal dimension, which I'm sure you are. --JohnLattier 05:32, 3 November 2006 (UTC)

My thoughts

There are 2 planets with a zillion cows on each planet. Only cows, nothing else but grass and water for them to drink. On one planet, you vaporize .99999999% (repeating) of all the cows. On the other planet, you vaporize 100%, or all the cows. You must kill and eat the remaining cow from each planet. A ha! One will have 0 cows because you vaporized them all, the other will have at least 1 cow because you can't vaporize a partial cow. Thus 1 does not equal .9999999 (repeating). Perhaps in most mathematical equations, but not in reality, particularly when each factor is turned into a %. Or when the numbers actually stand for something. Which is, if i'm not mistaken, the point of math. In other words, you take a test, you get a .999999% and your friends get's em all right. It's not the same. You hit a gillion baseballs + 1 in a row. You friend hits a gillion but doesn't hit that last one. Not the same. You just have to think outside of your mathematical boxes, erm.. cubes. Have fun!


Note- if .999(repeating) is really = to 1, then 1.9999999 (repeating) would have to be equal to 2 etc... this is one of those polemicized math arguments made for people who like to show off, when the simplest explanation will work, the two are simply not =

If you vaporize only an integer number of cows, and you don't vaporize all of them, and you convert the fraction vaporized into a percent, do you really think you'll get 99.999…%? Melchoir 08:16, 25 October 2006 (UTC)
BTW, as discussed on this page 1.9999... does in fact equal to 2 etc. Nil Einne 19:18, 26 October 2006 (UTC)


If you vaporise "only" 99.999...% of the cows, you'll have no cows left, not one. (And even if you disbelieve the concept that 0.999...=1, you'd still only have an infinitesably small bit of dead cow left, which ain't much use.) --Dweller 08:46, 25 October 2006 (UTC)

Really? Unless I vaporize 100% of the cows then there will be some remaining cow left. I can't, nor do I have to define what % of the total number of cows that single cow represents but the honest answer is that 100% is not = 99.9999(repeating)% You missed my baseball analogy. Inifity + 1 is always greater than infinity.

This isn't true either, again, you either vaporize or not vaporize a cow. You don't get to choose a fraction of a cow, that's making up your own rules. Vaporize is just meant to be a means anyways, you can't kill a % of a cow. I used cows because they can adequately represent something that cannot be split into a % by someone who wishes to insert infinity in the middle.

Call me crazy, but I've always been of the firm belief that cows are not real numbers. Stebbins 14:08, 25 October 2006 (UTC)
Er infinity + 1 is greater then infinity? Do you know what infinity is? How exactly do you have something greater then infinity? If something can be greater then infinity then there is no infinity! Nil Einne 19:21, 26 October 2006 (UTC)

The cows merely represent real, whole numbers. the .999 and 1 are multipliers applied to the total sum of cows, a billion, zillion, or infinite number of cows. If the intergers are applied there will be a difference, hence they're not the same.

LOL @ Stebbins. As to the cow situation, the argument is flawed as soon as we realize that we are dealing with merely a zillion (in other words finite number of) cows. It is hard for anybody to get a sense of infinity (rightly so) and in an ideal world we could break it down to simple visualizations involving farm animals. Sadly, even the biggest number you can think of (I like to use 10 to the google factorial since it's fun to say) is nothing compared to infinity. Thus, we must forgo our human tendency to try and rationalize (unfortunate word choice) this concept. Either trust the proofs based on mathematics, or ask for a better explanation of their foundations. Please do not try and explain it (or disprove it) in any real world setting; it'll make your head explode. -bobby 14:41, 25 October 2006 (UTC)

You again, Fail to think out side the box. It's quite easy to imagine infinity, it is perhaps easier to imagine infinity than most other finite numbers. Picture 2,435,059 cows. Can't? Ok, picture infinite cows? Easier yeah? You are kidding all of us if you think your brain can get a hold of 10 to the google factorial but not infinity, I ask you what the difference is? You are still deep, buried within the box. Your mind falling back on your factorials and equations. If a simple concept like understanding that .999(repeating) is not = 1 makes your head explain you're obviously too wrapped up in preconceived mental barries.

The fact is I did explain it in a world setting. Use both numbers as multipliers on whole numbers representing finite, absolute numbers. You don't get to divide the cows, the cows represent the smallest decimal that divides .999 and 1, its a metaphor for what your mathematical lexicon can't express on paper- but I've easily proven with a rediculous example about vaporizing cows. It has to do with symbolism and representation. I also resent your directing me in not explaining this simply, and finally to you all who use this as a launching pad for needless, fruitless pride and elitism. Some mathematical concepts, like many of those in quantum physics for example can only be described with words, not with conventional mathematical symbology. I propose that this is one of those concepts. Anything less than 100% is not = 100%.

Nobody is arguing with that. What people are saying is that 99.9...% actually isn't less than 100%. Your cow argument would be stronger if you had an infinite number of cows rather than a zillion. If you have only a zillion, there is absolutely no question that vaporising 99.9...% of them will leave you with no cows at all. If you have an infinite number of cows, then what you have left is up for debate based on the number system you are using.
You're completely right. I think the above editor is having a problem visualising infinity (despite his/her claims) because s/he appears to think there's something larger then infinity Nil Einne 19:24, 26 October 2006 (UTC)
I'd like to point out that in contexts such as cardinal numbers, there are different magnitudes of infinities. So \aleph_0 is infinite, and yet \aleph_1 is greater than it. -- Meni Rosenfeld (talk) 19:36, 26 October 2006 (UTC)
I know, it's another one of those things I love to baffle people with. (Although, I think baffling people with 0.999..=1 is meaner). Of course at that stage you have to be clear that you're using very specific (although reasonable) definitions of "greater than". When will transfinite numbers or cantor diagnalisation be a featured article so we can have all this fun again?

Approaching it differently

Rather than complain about the proofs, I have a different complaint about the .99... issue. What if you wanted to represent a number infinitely close to 1, but not equal to 1? like 1+dx say (I know not exactly the samething, but therein lies the problem). Another example, if you had the number .\bar{00}1, meaning something along the lines of \lim k \rightarrow \infin (1/k).

See, my issue is that this all comes down to mathematical semantics. Its not the math that determines whether .999.. = 1 or not - its humans. Some of us simply define it to be so, and some of us don't like that. Just like there are some that don't care about the intricate possibilities in physics because we can't measure the outcomes - and some of us are sticklers for details and stuff that actually makes sense. If you asked "whats the smallest positive number" I could tell you .\bar{00}1 - because thats what I take those symbols to mean. Of course, someone else could come along and say "well of course, because .00000..1 equals zero, and zero is the smallest positive number". And thus you see it all depends on definitions - semantics. Bottom line is its ridiculous to define such a thing and impose that definition on people. If you tell people zero is positive, they're going to argue - whether its an important issue or not. Same thing with .999.. *it really doesn't matter* - but since school book make it seem as if it does, students resist.

From a practical standpoint, .9999.. will always be 1, but from a more philosophical standpoint, thats were infintesimals come into play. It might not be the standard meaning of "infintesimal" that I mean, but I mean it all the same. Fresheneesz 07:52, 25 October 2006 (UTC)

The semantics thing is a cop-out. There are important mathematical structures here that are independent of the symbols we use to describe them. And it's important for people to use the same definitions, or else we're all just talking. Melchoir 08:13, 25 October 2006 (UTC)

I think Freesheneesz makes some very good points. The article at present has a smug (0.999... = 1, bet you didn't know that) feel to it. This is then followed by all the explanations about why people are wrong who think that 0.999... is less than one, when there is a perfectly ordinary and straightforward (not just philosophical) sense in which, despite the proofs, it is actually less than one. The point being that we are talking about a mathematical decision taken for mathematical benefits and not some real thing out there in the world that was discovered to be such-and-such a way contrary to popular belief.Davkal 08:17, 25 October 2006 (UTC)

What sense would that be? Melchoir 08:20, 25 October 2006 (UTC)

The sense in which anything that starts 0.9 is always going to be less than 1 becaus e it it wasn't it would just be 1.Davkal 08:23, 25 October 2006 (UTC)

What about 0.9A? Melchoir 08:28, 25 October 2006 (UTC)

I fail to see what relevance the latest version of itunes has for this point.Davkal 08:50, 25 October 2006 (UTC)

0.9A = 1. But "0.9A" starts with a "0". Melchoir 08:55, 25 October 2006 (UTC)

Yes, and 0.9 x 25 > 1, but that's obviously not what was meant when I said that, in a perfectly straightforward sense, anything that begins 0.9 is less than one otherwise it would just be 1.Davkal 08:58, 25 October 2006 (UTC)

So does "straightforward" mean "with unstated and possibly unexplored exceptions", then? Melchoir 09:10, 25 October 2006 (UTC)

No, it doesn't have any actual exceptions if taken as a starting point, but it does have exceptions if other starting points (perhaps less mathematically problematic in the long run) are taken. We are talking about a decision, but if the decision to treat 0.999... as less than 1 had been taken then 0.999... would be less than 1 in exactly the same way it is now equal to it. Consider the following proof:

0.9 < 1

0.99 < 1

0.999 < 1

0.999... < 1 Davkal 09:34, 25 October 2006 (UTC)

But... But... there is somewhere you can't get to that proves that you are wrong!! Ansell 09:36, 25 October 2006 (UTC)


I did get there. Look, the last line of the proof! Davkal 09:41, 25 October 2006 (UTC)

That's not a proof, it's a leap of faith. At best, you are proposing that the lexicographical order be taken as a definition, rather than a meaningful statement about the real numbers. This is already done at 0.999...#Breaking subtraction. The resulting number system is frankly a piece of trash. Is it a good idea to interpret our symbols as denoting elements of this number system? No, it isn't. And no one does. Melchoir 09:47, 25 October 2006 (UTC)
0.9 is a rational number.
0.99 is a ratinal number.
0.999 is a rational number.
Is 0.9999... a rational number? If so, give enumerator and denominator. If not, why should it share the property of being less than 1 with 0.9, 0.99, 0.999 if it doesn't share the property of being rational? Yours, Huon 09:50, 25 October 2006 (UTC)
For 0.999..., the numerator is 1 and the denomenator is 1. The ratio is therefore 1/1. It's a rational number. DB 10.31.2006


If it's not a rational number then why should it share the property of being equal to 1 with rational numbers like, say, 1. And what price the proofs in the article now, making heavy use of rational numbers and the like. Yours. Davkal 09:58, 25 October 2006 (UTC)

It happens to be rational, since it's equal to 1. But those who claim it's not 1 should be hard-pressed to give enumerator and denominator. --Huon 10:16, 25 October 2006 (UTC)
Yes. Exactly. As Fresheneesz said: What if you wanted to represent something infinitely close to 1, but not equal to 1? It's 0.999... Any other suggestions or possibilities? I would change the "infinitely close" to "infinitesimally close". And I would add that we're talking about LESS than 1. But otherwise, Fresheneesz hit the nail on the head. And I agree with Davkal that the article has a very smug tone. Which is hilarious, because the "fraction proof" and "Infinite series and sequences proof" are so obviously based on circular reasoning. And the "algebra proof" is based on multiplying and subtracting "0.999... numbers", when it is obvious on closer examination (try adding "0.999... numbers" and you'll see the flaw when you try to figure out where to put the "8"), that you cannot carry out most operations on "0.999... numbers". I'm open to a valid proof. What is it? Dagoldman 09:59, 25 October 2006 (UTC)
They're all valid proofs, but some of them require more background than others. If you refuse to accept any background results at all, try the proofs at 0.999...#The real numbers. Melchoir 10:08, 25 October 2006 (UTC)


The fraction proof relies on treating 0.333... as equal to 1/3, which is simply the same point as 0.999... equalling one. How can this possibly not be circular.

As the article says, if you read it, the equality 0.333… = 1/3 can be separately justified by long division. If you are at a point in your education where you no longer think of long division as infallible, you need to move on past that proof. Melchoir 10:17, 25 October 2006 (UTC)
"Maybe a number which is infinitesimally close to 1, but not equal to 1, will magically appear if we close our eyes and wish really hard?"
That may be a useful approach to other areas in human life, but it is useless in mathematics. In mathematics, you state what it is that you wish to explore by choosing axioms and definitions, and then investigate what logically results from them. The hopes and dreams of the mathematician are meaningless at that second step - he must stick to whatever he has chosen. Some choices lead to useless sturctures; others lead to very useful ones.
A certain structure, known as the real numbers, turns out to be extremely useful in every branch of theoretical and applied mathematics. But its usefulness comes from a very specific collection of "design choices"; any slight change made would immediately render the resulting structure mostly useless. So we could modify the structure to satisfy any one of our whims, but that would destory everything else that the real numbers have to offer.
And here we are faced with a difficulty, an unfortunate mathematical fact which can be proven rigorously: There is no positive infinitesimal real number (and of course, there are no two different real numbers which are infinitesimally close). The fact that we wish for such a number to exist alters nothing. So what do we do? We can say, "real numbers are wrong! Burn them!". But what about that "extremely useful" bit mentioned earlier? We can try to tweak the definition of real numbers to include infinitesimals. But that is essentially like the previous option, since what we will get will no longer be the wonderful real numbers. Or we can simply accept this fact, and understand that this is the small sacrifice we must make in order to have an otherwise magnificent structure at our disposal. And this is the approach taken by the main body of this article.
This does not mean to say we cannot study other structures where positive infinitesimals exist and 0.999... might be different from 1. But where will that lead us? As Melchoir has mentioned, some of these structures are good-for-nothing "trash". Others are potentially useful, but not quite as much as the real numbers - so there's no sense in making them the primary subjects of our discussion.
This leads us to the final conclusion: The most useful number system we can explore is the real numbers. In this system, 0.999... = 1. This is what this article is about, and that's all there is to it. -- Meni Rosenfeld (talk) 10:29, 25 October 2006 (UTC)

Maybe if the article was hedged about with the type of caveats and provisos included above about usefulness, choices, decisions etc etc. that result in making 0.999... equal to 1 according to the system currently used for it's utility, then people would not feel that they were being presented with some obvious nonsense as if it was the 100% true version of reality absol-bloody-lutely. As things stand the article simply presents 0.999... as equal to 1 as if the latter was the case. And that, clearly, is not all there is to it. Davkal 11:12, 25 October 2006 (UTC)

I can agree that there is still some more work to be done with the article. However, my explanations above have little to do with the article itself. Most people, whether they are aware of it or not, when they speak or write about "numbers", they are actually referring to real numbers.
Even in intermediate-level mathematical texts, it is usually regarded implicit that we investigate real numbers and not some other structure. This is no more arbitrary than defining 2*3 = 6 rather than 2*3 = 5 - we have a choice, but it is quite clear what the sensible choice is. So, you could say that the moment a number appears in this article, it is agreed that we're dealing with real numbers, unless explicitly stated otherwise. Explanations of why this, and no other, is the adopted standard, do not belong in this article. In the same way, it is undertsandable to have an article say "1 + 1 \neq 0" without explicitly mentioning that we are not dealing with the integers modulu 2 or the like, or "the sum of interior angles in a triangle is 180°" without explicitly mentioning that we are dealing with Euclidean geometry and presenting the history of non-Euclidean geometries.
By the way, the article does more than what is necessary in light of the above. It does explicitly mention that we discuss real numbers, and it does mention the existence of other structures where 0.999 \neq 1.
As for the allegedly condescending tone... Well, feel free to edit it while maintaining mathematical accuracy! -- Meni Rosenfeld (talk) 11:49, 25 October 2006 (UTC)
I really don't think this is true. The fact that people have such difficulty with 0.999... = 1 pretty much proves that this isn't true. Real numbers are a formalism that does not capture what people think of when they talk about numbers. You say that real numbers are the most useful ones, and since the intuitive understanding of number is very fuzzy and vague, you may even be right, but it's definitely not the normal understanding of numbers that normal people have.
Nope, that only proves that people don't always know what they are talking about. I'd best most people will agree that if a<b, then a<(a+b)/2<b. But many of those people will also say that 0.999... is the largest number less than 1 (a contradiction), and then also fail to specify what the largest number less then π is. -- Meni Rosenfeld (talk) 15:41, 26 October 2006 (UTC)
Whether or not people have trouble accepting the conclusions is irrelevant. Many things in math are counterintuitive. For quite some time nobody would believe that any real numbers were irrational. For other examples, take a look at [Gabriel's Horn] and [Euler's Identity]. As for people's "normal understanding of numbers," there are plenty flaws other than that mentioned by Meni Rosenfeld. For example, 0.999...#Breaking subtraction points out that f(x)=x-k would no longer be single-valued. --Epee1221 04:40, 28 October 2006 (UTC)

Wow. What arguments can someone come up with to show that the usefulness of the real numbers would be diminished by adding infintesimals? Before that question is answered, one thing I can add to the discussion - semantics *are* whats at issue. It can be argued that .999... isn't a real number. Its definately not irrational, and so by asserting that its a real number, you really assert that it is equal to 1. Because if its not irrational, its rational - which makes the only possibility 1/1. Also, with the 1/3 = .333.. proof - the assertion is that 1/3 does in fact = .333 . People have written that "this can easily be shown with long division". However, it is in fact *not* possible to show that with long division, as you cannot reach the number .333... with long division. But the person doing the proof assumes that since 1/3 can't be represented by a finite number of 3s, it must be represented with an infinite number of 3s - thus .333... . But why assume this? Why assume that since a finite number of 3s doesn't work, an infinite many of them must work? My view is that the assumption simply comes out of laziness and lack of rigor. Fresheneesz 18:36, 25 October 2006 (UTC)

If you don't like the proof involving 1/3 = 0.333..., why not try the three or four other proofs mentioned in the article?
Infinitesimals are not real numbers. 153.104.208.67 05:13, 26 October 2006 (UTC)
Long division can be defined rigorously even for fractions like 1/3, and can be shown to give 0.333... in this case (we do not "assume" anything, every meaningful statement must be proven rigorously). If this is not done in long division, it's a problem with that article. However, doing this is harder than proving that 0.999... = 1 in other ways, so instead switch your attention to the Talk:0.999.../Arguments#A new proof for the skeptics thread, where I've started very close to the fundumentals. -- Meni Rosenfeld (talk) 10:20, 26 October 2006 (UTC)
As for your first question, all of standard analysis would crumble. And no, non-standard analysis does not solve this, since its interesting results are ultimately about the standard real numbers, which you are so easily willing to discard. -- Meni Rosenfeld (talk) 15:41, 26 October 2006 (UTC)
Its true that infintesimals are not real numbers. I'd argue that .999... isn't a real number either. I'm just going to declare this whole argument ridiculous. The person that decided to teach this to highschoolers was a grade A pompous ass. The debate is useless, and meaningless, because we are simply talking about what we define to be so. We say that making .999.. a real number (and thus being = to 1 as a direct result) makes things easier, or more consistant. But it would be just as consistant to say that .999... isn't a real number, but is part of the set of hyperreal numbers. What you people don't realize is that .999... is NOT math, its a symbol that was artificially defined - no matter how "natural" that definition may sound to some. Fresheneesz 03:20, 27 October 2006 (UTC)
How is 0.999... not math? It represents a number equal to 9/10+9/100+9/1000+..., which is equal to 1. That's obvious from the definition of the decimal representation of number and the value of an infinite geometric series, which are both math. -- Schapel 03:32, 27 October 2006 (UTC)
Every real number is a decimal expansion, and every decimal expansion is a real number. Therefore, 0.999... is a real number. Maelin 09:22, 27 October 2006 (UTC)
I'd replace "every real number is a decimal expansion" with "every real number has a representation or two as a decimal expansion". And no, there is no separate definition to deal with 0.999... - every property of 0.999... (such as being equal to 1) is derived from the general definition and properties of decimal expansions (and it's unfortunate if decimal representation does not discuss them properly). And where did you get the idea that 0.999... is a hyperreal number? That borders on nonsense, and please read again Talk:0.999...#A Solution we all can agree on.3F as well as my latest comment there. -- Meni Rosenfeld (talk) 09:49, 27 October 2006 (UTC)

This article is an unfortunate embarrassment to Wikipedia

Moved from main talk page. --Brad Beattie (talk) 12:22, 25 October 2006 (UTC)

Whereas mathematicians are trained in logic, etymology is sadly sometimes a total mystery. Had the authors of this article bothered to note that the 'infinitesimal' is equivalent to 1 rather than asserting that it is equal as is obviously the case by definition, then this article would have received the attention that it is due: very little.

Also, it is inherently novice to try to brow-beat opposition out of hand by refering to them as "students" or "junior". Mathematics is a science, not a political or religious arena:

Although the 'infinitesimal' converges on the number 1, thus being equivalent, it is not equal so long as the theoretical subtrahend, t, exists where 1 - t = 0.\bar{9}

-Dworkin, --76.1.39.216 02:25, 25 October 2006 (UTC)

It's true that the decimals 0.999… and 1.000…, left unevaluated, are equivalent but not equal. But this is a vacuous observation. You could just as well say that the arithmetic expressions 1 + 1 and 3 - 1 are equivalent, or that the fractions 4/2 and 6/3 are equivalent. Okay, true enough. Nonetheless, the numbers they represent are equal, exactly and identically.
And t = 0. Melchoir 02:35, 25 October 2006 (UTC)


Again, t approaches 0 asymptotically. Thus it converges on 0 and is equivalent but not equal. It's not congruent but you could use "approximately equal" if you're hung on using the word 'equal'. (Similarly, a square is a rectangle but a rectangle is not always a square.) The essence of an asymptote is that it approaches, but never touches. t never equals 0. 0.\bar{9} never equals 1.

You're close, but no cigar. Try using the word 'equivalent' and see how little debate remains.

Note: To assert that all math books support this claim is completely untrue. Of the dozens of university and secondary level books that I have, not one asserts this claim. Converge, approach and equivalent but not equal. Equal is an unwise use of notation in this case. Rounding up doesn't count; maybe the dot above the '=' sign is missing from authors relying too heavily on the works of others but not copying correctly. Furthermore, not one of the offered proofs gives a viable reason for making that tiny, yet important step to enable the use of 'equals' rather than 'approximately equals'. Numerical evaluation would yield the value of 1 for 0.\bar{9} for all practical purposes but does not retro-define mathematical notation.

-Dworkin--76.1.39.216 04:46, 25 October 2006 (UTC)

0.999... = 1. t in your example = 0. You're close, but no cigar. What part of the proof do you dispute?
Great article, by the way, and as the above illustrates evidently still necessary.
By the way, the third bullet point under "skepticism in education" appears to address your confusion, as evidenced by your use of the word "approaches." There is additional dicussion in that section about the common misconception of the sequence as a process rather than a value. Hope that helps. Argyrios 05:27, 25 October 2006 (UTC)


One last time until I see a point to your argument. As noted above, I dispute the part that is missing in every proof. - - A point is a theoretical construct. It is defined as a dimensionless geometric object with no properties except location. You can cut the distance between 0.\bar{9} and 1 as often as you like but they are asymptotic. Unless you insert a step into your 'proof' which proves that the last point diappears, then it still exists. The case study with two options under "skepticism in Education" is an assertion, not a proof. Either way, the second assertion is not the case. - - If your assertion is true, then you contend that 'approximately equal', 'converges upon' and 'approaches' are all identical to 'equals'. There is a missing step, but you'll never get there using an asymptotic 'infinitesimal'. You can ignore it; but the 'point' is still there.

There is no confusion; only lack of proof in all of the listed 'proofs'.

-Dworkin--76.1.39.216 05:46, 25 October 2006 (UTC)

I truly am sorry that you do not understand the proofs. If nothing else will convince you to stop sabotaging the efforts of others to understand this mathematical concept by making false assertions about mathematics in response to their sincere queries, perhaps WP:OR will. That 0.999... = 1 is long- and well-settled fact among mathematicians. If you are near a college campus, try finding a math professor and try to tell him that he doesn't have it right and you do. Get it published; then come back. (Of course, this will never happen.) Your own original research does not belong on wikipedia. Regards, Argyrios 05:54, 25 October 2006 (UTC)
Alternatively, find a published article in an academic journal disputing that 0.999~ = 1. You won't be able to; no journal would publish it. I can see that I am never going to convince you that you are wrong if you haven't already accepted it, so I am just trying a different tactic here to settle the debate.
I've been on the losing end of WP:OR arguments myself, sure I was right but unable to show that my idea was accepted outside of my own head. Wikipedia is a reflection of consensus knowledge. You just don't have the authorities to back up your position. We have mathematicians. You have confused high-school students. Game over. Argyrios 06:22, 25 October 2006 (UTC)
There is no asymptotic behaviour here. There is no 'behaviour' at all. We are talking about individual numbers. Points on the real line, not functions, not sequences. Saying something like "numbers x and y are asymptotically close" has no meaning in the real numbers. We are talking the number 0.999... and the number 1. They are mathematically equal. Not just "almost equal", not "equivalent but not equal". Equal. Settling for weak true statements because strong true ones cause arguments is not the sort of thing we do on Wikipedia. Maelin 06:27, 25 October 2006 (UTC)

Wow. OK, if you insist on brow-beating rather than proof then I suggest that you check the link on the main page:

http://qntm.org/pointnine

Toward the bottom, the author who agrees with you states in his Q&A:

Q: "My mate/my dad/my mathematics teacher/Professor Stephen Hawking told me that 0.9999... and 1 were different numbers."

A: They were wrong. In science, credentials are as worthless as intuition (above). Proof is everything.

So, now that we've dropped names (like Stephen Hawking on my side) and brow-beat each other sufficiently, can we get back to the missing step in your 'proof'? As your comrade says, "credentials are as worthless as intuition." But hey, at least we know that the WP:OR wasn't violated.

-Dworkin--76.1.39.216 06:48, 25 October 2006 (UTC)

There is no missing step in the proof, and it is not my own proof. I can't force you to understand it if you are unable to do so. The truth of the proposition does not depend on your consent. Unless you plan on changing the article repeatedly and then getting banned for vandalism, which is fine by me, I am going to end this conversation.
Unlike in science and math, credentials are very valuable in nonfiction writing, and essential on wikipedia. And the Stephen Hawking reference above was a joke.
Argyrios 07:40, 25 October 2006 (UTC)
Yes, the article is an embarrassment. I don't know if I would specifically blame wikipedia. Anyway, I saw three "proofs". Two of them are so obviously circular reasoning. The "fraction proof" says "0.333... is exactly equal to 1/3, so 0.999... is exactly equal to 1." But what is the proof that 0.333... is exactly equal to 1/3? There is no proof. Another "proof" is based on assuming that an infinitesimally small number is exactly equal to 0. But that is the same issue, and is not proven. The "algebra proof" is based on illegal operations on the "0.999... numbers", which don't support the normal mathematical operations for the most part. Try adding 0.999... and 0.999... together. It doesn't work. Meanwhile, it's obvious to anyone familiar with limits that a number infinitesimally close to another number is still infinitesimally different. Dagoldman 10:30, 25 October 2006 (UTC)
I wish you people would read the other comments on this page before you go making the same mistakes, because I'm getting sick of saying this. The real numbers are an Archimedean field. There are no infinitesimal quantities in the real numbers. Every real number is finitely small and finitely large. Every two nonequal real numbers x and y have a finite, nonzero, real distance between them. Every two real numbers x and y that do not have a finite, nonzero, real distance between them are equal. Maelin 13:18, 25 October 2006 (UTC)
Okay, but why does .999... even exist? By requiring an infinite sequence doesn't that go against Archimedean property, which states "Roughly speaking, it is the property of having no infinite elements"? Isn't an infinite sequence of 9s de facto an infinite element?--T. Anthony 22:06, 25 October 2006 (UTC)
No, it's two different kinds of infinite. 0.999... is infinite in length, when considered as its decimal expansion. That is perfectly allowable in the real numbers - in fact, every irrational number has an infinitely long decimal expansion. The Archimedean property implies that there are no elements that are infinite or infinitesimal in size. There are no elements that are infinitely big nor elements that are infinitely small. However, 0.999... is neither of these; it's just infinitely long, and that's perfectly allowable. Maelin 00:05, 26 October 2006 (UTC)

Dworkin, you appear to understand the topic quite well. I applaud you. --JohnLattier 05:24, 3 November 2006 (UTC)

This article is just wrong

Moved from main talk page. --Brad Beattie (talk) 12:23, 25 October 2006 (UTC)

1) This article is just wrong. The LIMIT of 0.999... is 1. But that does not mean that 0.999... is exactly equal to 1. The article is based on a misunderstanding of limits and equality. 0.9 is not exactly equal to 1. 0.99 is not exactly equal to 1. 0.999 is not exactly equal to 1. No matter how many times you add a 9, the number is never EXACTLY equal to 1. It's as close to 1 as you want to get, but never gets there.

2) Also, the article title is "0.999..." and that was originally the topic. But the topic is now "0.999... is exactly equal to 1".

I noticed the article was previously nominated for deletion. I would wholeheartedly suggest it be deleted. Dagoldman 06:35, 25 October 2006 (UTC)

These criticisms are addressed in the Skepticism section. The number 0.999... is not a process. It is not a function. It does not have a limit, it does not move anywhere. It is just a single point on the real line. That point is coincident with 1, i.e. they are equal. Yes, 0.9, 0.99, 0.999, any such number with a finite number of nines, no matter how many, is less than 1. But 0.999... is not any of them. It has an INFINITE number of nines. It's true that when you write 0.9, that number is not equal to 1, and no matter how many nines you append, the number is never equal to 1. But it's never equal to 0.999..., either. 0.999... is the limit of the sequence {0.9, 0.99, 0.999, ...} and that limit is, as I'm sure you agree, 1. Maelin 06:44, 25 October 2006 (UTC)
You say "it does not have a limit". Then, you say "0.999... is the limit of the sequence {0.9, 0.99, 0.999, ...} and that limit is, as I'm sure you agree, 1". You just contradicted yourself. Which is it? Does it have a limit or not? I agree with your second statement. 0.999... does have a limit. The limit is 1. But that's not what the article says. I read the Skepticism section. The students SHOULD be skeptical. The article is wrong, and based on a misunderstanding of limits. Math is an exact science. The article is non-rigorous, an example of math misused. By the way, "It is just a single point on the real line" is just more circular reasoning. Dagoldman 07:17, 25 October 2006 (UTC)
Look more closely. 0.999... is a limit. It doesn't have a limit. "To be" and "to have" are really basic verbs; I should think we all understand the difference between them? Melchoir 07:27, 25 October 2006 (UTC)
Yes, you said 0.999... "is" the limit. I misread your comment (it would never enter my head that someone could say such an outlandish statement), and you did not contradict yourself. But where do you get the idea that 0.999... IS a limit? Isn't this just more circular reasoning? If you assume that 0.999... IS a limit, then of course it is equal to 1. But 0.999... is NOT a limit. It is an infinitely repeating decimal. It HAS a limit. You can get as close as you want to 1, but you'll never get there. Dagoldman 07:45, 25 October 2006 (UTC)
Look, definitions are not optional. 0.999… is a limit. If you want to use the phrase "0.999…" to mean something inconsistent with its given definition, this is of course your personal right, but it will interfere with your ability to communicate with the rest of humanity. I can choose to call the color of lava "blue" and then claim "The sky is not blue". Now is that statement really a valuable insight? Melchoir 08:05, 25 October 2006 (UTC)
The argument about colors applies to you just as much as it does to me. You haven't added anything. Dagoldman 18:26, 25 October 2006 (UTC)

If 0.999... is not equal to 1, then what consists of 1? It seems to me like the concept of .999... is as empty as 1 is. But I'm reasoning philosophically, of course. I think the mistake here is to think of 1 as something finite and 0.999... as its opposite, the infinite. But that is not true. 1 consists of infinitesimal elements making it as indefinable as 0.999... Just musing.Moonwalkerwiz 07:37, 25 October 2006 (UTC)

That's an interesting point. By what I was thinking 1 would come out as "an infinity of infinitesimals" with .999... coming out as "infinity -1 of infinitesimals", but as "infinity -1" is still infinity they'd be the same. Interesting, it kind of relates to the conceptual difficulty of thinking that a number minus one equals itself as that wouldn't work in real numbers. (Getting us back to square 1)--T. Anthony 22:03, 25 October 2006 (UTC)
A number can't *have* a limit. Only expressions with variables can have a limit. The number .999... (or better yet .\bar{9}) is not a number with an arbitrary number of 9's more than 3, it is a number with infinite 9s. Fresheneesz 08:13, 25 October 2006 (UTC)
The conceptualisation of the number however, requires a thorough and rigourous definition and understanding of the concept of infinity, which is itself only well explained using limits IMO. Ansell 08:50, 25 October 2006 (UTC)
Just about every professional mathematician, historical and contemporary, you find will disagree with you Dagoldman, maybe your mathematical abilities surpass theirs though. Brentt 08:33, 25 October 2006 (UTC)
Maybe the professional mathematicians do disagree. But yes, I think they are wrong. I am not a professional mathematician. However, in my defense, I never made less than 100 on a math test in high school or college (sometimes there was extra credit). And that includes advanced calculus and number theory at Yale. I didn't go into math because the mathematicians I met seemed strange. Maybe that was a mistake, because I do love math. Anyway, back to the problem, I think a source of much of the misunderstanding here is the notation. 0.999... seems a terrible (not useful) notation to me. It's a representation of a concept. But you can't do anything with it. It leads to people multiplying and subtracting "0.999... numbers", which is not demonstrated to be valid. And it leads to bizarre statements such as "0.999... is a limit". 0.999... is a representation of an infinitely repeating decimal. This infinitely repeating decimal is properly defined, manipulated, and notated as a geometric series, which is what it really is. Forget about 0.999... notation for a second. Would anyone disagree that the series for this infinitely repeating decimal has a limit of 1? I think limits (and calculus) are very difficult for most people to understand on a fundamental level. I just happened to see this as the "article of the day". I have no idea how many high-level mathematicians would contribute to this article. But maybe zero, because the article is so obviously non-rigorous with it's inclusion of the circular-reasoning "fraction proof" that it doesn't even pass the smell test. The "fraction proof" assumes 0.333... is exactly equal to 1/3 to "prove" that 0.999... is exactly equal to 1. The "proof" under "Infinite series and sequences" is also obviously circular reasoning. It assumes that an infinitesimally small amount is equal to zero to "prove" that 0.999... is equal to 1. The "algebra proof" is not circular reasoning, but is also bogus, because the 0.999... notation is not a number that you can multiply by 10 and and then subtract. Try adding 0.999... plus 0.999... one or more times and see what happens. It doesn't work. Where do one or more 8's (9 + 9) go? Infinitesimally small differences cannot be just swept under the rug. They exist, just as much as large differences. A way to deal with the infinitesimally small is to define a limit, which is my whole point. 0.999... is as close to 1 as you want, but is infinitesimally different from 1. Three bogus proofs. Where's the beef? Dagoldman 09:34, 25 October 2006 (UTC)
This is not a mathematical question, but I'm curious about your thinking. Is 0.999… a real number or not? Melchoir 10:30, 25 October 2006 (UTC)
I would agree 0.999... is a real, irrational number. I think most mathematicians would agree that the concept of irrational numbers and exactly where to put them on the number line is not easy for to grasp. You probably have read how irrational numbers horrified and mystified Pythagoras. The diagonal of a 1 x 1 square has an exact length. Yet it's not a rational number, so how can it have an exact length? This was unacceptable to Pythagoras. Is it any easier for us to grasp? A possible argument for "0.999... equals 1" might be: \sqrt{2} ends up "exactly equal" to some "diagonal exact length", just like 0.999... ends up exactly equal to 1. But what would the diagonal "exact length" be? Nobody knows. This is a fundamental difference between 0.999... and \sqrt{2}. 0.999 is either exactly 1 or has a limit of 1. \sqrt{2} seems to be just \sqrt{2}. You can't say it equals any other number, or has a limit. So I don't think the other irrational numbers get us any closer to resolving the 0.999... issue. Dagoldman 11:30, 25 October 2006 (UTC)
Do you believe that there are non-zero, infinitesimal, real numbers? Melchoir 11:47, 25 October 2006 (UTC)
Do you? What's your point? Dagoldman 18:27, 25 October 2006 (UTC)
Given your intelligence, I would love to read what you come up with for the value and properties of 0.999...². That's my favorite conundrum for those who argue that 0.999... ≠ 1. Is its square equal to itself or not? You say 0.999... is irrational, so it is a real number -- does it have the same properties all real numbers must have? --dreish~talk 18:26, 26 October 2006 (UTC)
Here is where the discussion of infinitesimals is going. the field of real numbers is an Archimedean field -- it does not contain infinitesimals. If .999... is a real number, then 1-.999... is also a real number, and cannot be an infinitesimal. So either there is a real number between .999... and 1 or .999... is equal to 1. As for .999... as a limit, what would you say is the limit, as n approaches infinity, of the summation from 1 to n of 9 * .1n?--Epee1221 05:05, 28 October 2006 (UTC)

No, we do not. The point is that if you bear with us, we might find out your exact error. But I'll approach it differently - see the new thread I'm starting as we speak at the bottom of this page. -- Meni Rosenfeld (talk) 20:52, 25 October 2006 (UTC)

0.999... = 1

Moved from main talk page. --Brad Beattie (talk) 12:26, 25 October 2006 (UTC)

This statement is absurd and lies. 0.999... is less than 1 the same way 1.999... is less than 2 and 999,999,999 is less than 1,000,000,000. How could two different numbers could be the same, this is pure nonsense and intrinsically false. Shame On You 08:24, 25 October 2006 (UTC)

Neither "0.999..." nor "1" are numbers; they are some black pixels on your computer screen. However, in the language used by mathematicians and other people using math and numbers, they both represent numbers. Using the standard construction of real numbers, and the standard interpretation of the three dots, they both represent the number 1 (I assume everyone now interpret that symbol the way I intended it). There is more than ne way to construct the real numbers, but constructions that imply 0.999... < 1 have other counter-intuitive aspects instead.--Niels Ø 08:48, 25 October 2006 (UTC)
It is not saying that. The difference between 999,999,999 and 1,000,000,000 is a different matter because the nines end. I'm unclear why you can't express 1 as 1.00... along with .999... then neither of them ends so are unequal. In addition to that infinity apparently does not exist using the Archimedean system on which this is based. As infinity an infinitessimals are thus both made unreal I can see how this is true as defined for use in the Archimedian system of real numbers. I just confused as to this being "true" in some larger sense and in some respects I think so is the article.--T. Anthony 09:10, 25 October 2006 (UTC)
Hi people. 1 is the same as 1.00, I mean the values are the same, it's just the way you write it (the same way 1.00 = 1.000). However a number comparison implies to compare ended (determined) numbers, intrinsically an unfinished number (like 0.999...) represents an undefined value, so what is undefined (what is not terminated cannot be determined, French language's ethymology makes it clear: terminé/déterminé) cannot be measured, hence not compared. So 0.999... = 1 is false. What do you think? Shame On You 09:38, 25 October 2006 (UTC)
But 0.999... is defined. In the article. Several times, in fact. Melchoir 09:49, 25 October 2006 (UTC)
I would think .9999... is the same as .9999999999.... as both imply the same sequence. Likewise 1 is the same as 1.00000. In addition to that the proofs require that infinity does not exist for real numbers as that is what is said on the article on Archimedean property. Therefore the infinite string of .999... is simply an impossibility because if there is no infinite how can you have an infinite string? So is this article saying that .999... just doesn't exist and is essentially 1? (I might actually accept that)--T. Anthony 21:53, 25 October 2006 (UTC)
I'd also like to add that on reflection I think the article as written is clear enough on what it means not to be much of a problem. The problem I had was the soundbite on the page where it's featured isn't able to encapsulate that and makes it sound more simplistic than is really intended.--T. Anthony 21:58, 25 October 2006 (UTC)
Really? I'll figure it then. Paradoxes are hard to imagine. Shame On You 09:53, 25 October 2006 (UTC)
"Shame On You" is right. The emperor has no clothes. The "proofs" offered in the article are so obviously flawed by circular reasoning (0.333... equals 1/3; therefore 0.999... equals 1) and carrying out operations on unproven mathematical operations (10 * 0.999... - 0.999... ???) (try 0.999... + 0.999... and see what happens) that the article does not pass the smell test. Meanwhile, it is obvious to anyone familiar with the concept of limits that a number infinitesimally close to 1 is still infinitesimally different from 1. Dagoldman 10:11, 25 October 2006 (UTC)
If the proofs are so obviously flawed, surely you can produce a reference that says so? Or is it obvious only to you, and why might that be? Melchoir 10:14, 25 October 2006 (UTC)
What's the proof that "0.333... is exactly equal to 1/3"? If nobody can demonstrate that independently of some other "fact" in this circular web, such as "0.666... is exactly equal to 2/3", then by definition the "fraction proof" is flawed. Are you saying that you think the "fraction proof" is NOT circular reasoning? By the way, others on this page have pointed out the same obvious flaw. I'm not the only one. Dagoldman 11:47, 25 October 2006 (UTC)
Yes, that is not a good proof. As the article mentions, some of the proofs are more rigourous than others. That is not rigourous, and it hinges upon a large assumption that people will accept 1/3 = 0.333.... Some students will accept it, and they will accept the fraction proof because they don't consider that 1/3 = 0.333... is in question. However, it is NOT a rigourous proof. The rigourous proofs are the ones further down in the article, and if you want to dispute the key thrust of the article, you should be disputing those ones. Maelin 13:27, 25 October 2006 (UTC)
Different ways of understanding for different people I suppose. To me, the fraction proof is by far the most convincing, most of the others use things that I think normal people would find suspicious, but long division is something I trust, and you can see from long division that 8/9 is 0.888.. and 1/9 is 0.111..., and you can clearly see that adding together is a simple process that would continue for ever, producing 0.999... or 9/9 or 1. It's just long division and addition.
I'm so glad that someone finally said that! In fact, historically, repeating decimals arose not as sums of geometric series or limits or suprema or any of that, but as the results of long division. If we take the correctness of long division as an axiom, along with the correctness of decimal addition, then the rest follows without circularity. It so happens that mathematicians have constructed several theories that meet these axioms, including sums, limits, and suprema. But if we work with the axioms directly, we can profess agnosticism over what exactly decimals are, and focus instead on how they work -- a viewpoint much closer to the popular one. Melchoir 01:30, 28 October 2006 (UTC)
--"(try 0.999... + 0.999... and see what happens)" - ok. Working left to right (ick, but pretty hard to go right to left like we normally would with addition, neh?) ... Clearly, in the units digit, we have a 1 carried from the 9+9 in the tenths. Then, in the tenths, we have 9+9 gives 8, plus a carried 1 from the hundredths gives 9. This clearly continues and gives 0.9999...+0.9999...=1.9999... But wait! If 0.9999...<1, then 1-0.9999... must have some positive value t. Then, clearly, 2-1.9999... would also have value t. But 0.9999...+0.9999... would equal (1-t)+(1-t)=2-2t. So 2-t = 2-2t?! Clearly, this is false for any positive t, so the assumption that 0.9999...<1 is false. -BlueSoxSWJ 11:38, 25 October 2006 (UTC)
You proved my point. You said "ick". You said: "Pretty hard to go right to left like we normally would with addition." In fact, impossible. You cannot carry out "normal addition" on "0.999... numbers". The only way you can add 0.999... and 0.999... is to assume that 0.999... is equal to 1, and that's circular reasoning again. If you could get past the first step (adding the numbers), you would have gotten somewhere, but the first step doesn't work. Similarly, you cannot carry out normal multiplication and subtraction in most cases with "0.999... numbers". So the "algebra proof" is bogus, because you cannot do operations "0.999... numbers" for the most part. The best you can say is that the limit of the geometric series represented by 0.999... is 1. The number 0.999... is infinitesimally close to 1, but never quite gets there. Dagoldman 12:07, 25 October 2006 (UTC)
The name of this article is "0.999...", not "A complete development of all of mathematics starting from the ZFC axiomatization of set theory" (or any other one of your favorite foundations of mathematics). The article must take certain mathematical truths as known to the readers (and preferrably, explained in other articles), and build up on them. Why don't you question the statement "3 * (1/3) = 1"? Because you know that is true, and you do not expect us to provide a proof of that. If you happened not to know that, you would probably want to read our articles on multiplication and fraction. Likewise, if you do not know several basic facts about decimal expansions (such as that 1/3 = 0.333...), you should read articles such as decimal representation and long division (and I am by no means saying those articles are necessarily very good - but that is a completely irrelevant issue here). All of the proofs you criticize leave a lot as preliminary knowledge, and if you do not have the necessary sources to establish this knowledge, you should move on to the more advanced proofs, which assume less preliminary knowledge (pretty much, just of rigorous constructions of real numbers, on which I do believe we have good articles). By the way, did you read those proofs before presenting to us the challenge of providing a proof which is more to your liking (there's no point in repeating a proof which is already in the article, right)?
Well, where's the proof that 0.333... is exactly equal to 1/3, which is the issue at hand? It's a totally different question than proving "3 * (1/3) = 1", so don't hide behind that. You say long division is a basic fact that proves 0.333... = 1. I read the article on long division and decimal representation. They have no proofs of 0.333... = 1, so why mention them? Do the division yourself and see. It clearly shows just the opposite. As you move to the right, there is always a remainder! That's why the division continues infinitely. At each step, it's always 10 / 3. At each step it never gets to 4. 0.333... is infinitesimally smaller than 1/3, and that's what keeps the long division repeating endlessly. The defenders of 0.999... = 1 keeps saying "Trust us. The proof is somewhere else". I must admit I have not yet gotten through proof five (Cauchy sequences), but the first four "proofs" are bogus, as I have clearly explained. Dagoldman 18:44, 25 October 2006 (UTC)
As for your other questions... Well, there really is a limit (pun unintended) to how much we can repeat the same things over and over again, so please take some time to review the different replies on this page (like the ones where the article is about real numbers, and there are no positive infinitesimal real numbers, and the definition of the number represented by a decimal expansion - and thus of 0.999... - is a limit of a certain sequence, etc., etc...) -- Meni Rosenfeld (talk) 13:43, 25 October 2006 (UTC)
Yes, you can. 0.999... is a real number. The real numbers, being, as they are, a field, have the operations of addition and multiplication that are defined for every element of the field. You can add and multiply 0.999... with itself or any other real number as much as you like. Maelin 13:30, 25 October 2006 (UTC)
Yes, you can add them, but you do not know the result. The "algebra proof" blithely produces results that cannot be verified or shown to be valid. Like I said, try adding 0.999... and 0.999... by hand and see what happens. You can't produce a known result. Dagoldman 18:50, 25 October 2006 (UTC)
0.99999999
0.99999999
----------
1.8
  18
   18
    18
     18
      18
       18
        18
         ...
----------
1.9999999...
Looks fairly obvious to me what's gonna happen. Within the confines of the logic, the "0.999... == 1" result is undeniable. The issue at hand are the boundaries, jargon, specialization, etc. Perhaps the biologists laugh themselves sick because mathematicians have no idea about the difference between complete and incomplete metamorphosis? mdf 20:34, 25 October 2006 (UTC)
Your addition example reveals another good proof that 1 - .999... = 0
since the difference must double, but tries to remain the same.
 
0.999999..           + t = 1  (lets call the difference t)
0.999999..           + t = 1
----------
1.8
  18
   18
    18
     18
      18
       18
        18
         ...
----------
1.9999999...          + 2t     =  2
1.9999999...          + 2t - 1 =  2 -1 (subtract 1 from each side)
 .9999999...          + 2t     =  1    (Nuts, now we have 2t)

-.9999999...          -  t     = -1   (subtract the original 0.99999999 + t = 1)
                         t     =  0
QED Jeff Carr 03:06, 26 October 2006 (UTC)
How does the difference double? If I say the difference between 0.9 and 1.0 is 0.1, does that mean the difference between 1.9 and 2.0 is 0.2? Of course not! --Wooty Woot? | contribs 01:01, 28 October 2006 (UTC)
Using your comparison, the difference would be between 1.8 and 2.0 (x2 on both sides), not 1.9 and 2.0 (+1 on both sides). - KingRaptor 01:54, 28 October 2006 (UTC)
Ah, I was making the mistake of rounding off the example. Still, though, who is to say 0.999.. x 2 is 1.999....? Doesn't that assume the conclusion? --Wooty Woot? | contribs 18:44, 28 October 2006 (UTC)
It is consistent with the conclusion, but that doesn't mean it assumes the conclusion. That's a pretty common error. That 0.999... x 2 = 1.999... is demonstrated in the formatted boxes above. Melchoir 18:50, 28 October 2006 (UTC)

Can infinity be reached ?

Moved from main talk page. --Brad Beattie (talk) 12:27, 25 October 2006 (UTC)

According to my mathematical knowledge (which isn't necessarily perfect), infinity cannot be reached, we can only tend to it. As a result, the solution to this issue should actually be written as:

lim(1 − 0.1x) = 1
x->+∞

As a result, though I don't question the result which is that 0.999...=1, I wonder if this cannot be considered as a graphical oversimplification. Metropolitan 10:50, 25 October 2006 (UTC).

It is not clear what you mean by "reaching" infinity, but yes, we can discuss infinite quantities as long as we properly define our terms. And no, 0.999... = 1 follows rigourously from the definitions - nothing "graphical" about that. -- Meni Rosenfeld (talk) 13:49, 25 October 2006 (UTC)
Saying that 0.999... = 1 sounds to me to be exactly the same as writing 1 - 0.1+∞ = 1, or if you like better 0.1+∞ = 0. My only problem in this is simply the use of +∞ as a real number, which it isn't. That's all. That's the reason why I was insisting on the fact that infinity cannot be reached, we can only tend towards it. Metropolitan 15:43, 25 October 2006 (UTC).


+∞ is definitely not a real number. If that's what you mean by "unreacheable", then we agree. I don't see why you would call it that way, though - not being real does not make it somehow "illegal". The important thing is to pay extra attention to what we wish to say about it. -- Meni Rosenfeld (talk) 16:27, 25 October 2006 (UTC)
If infinity and infinitesimals aren't real numbers then isn't an infinite string of nines simply an impossibility? So saying .999.... equals 1 is essentially a way of saying .99... doesn't exist, you should simply say 1.--T. Anthony 21:48, 25 October 2006 (UTC)
No. Note that "real number" is a technical term with a specific meaning, and just because infinity isn't a real number doesn't mean it's outside of mathematical reasoning. Melchoir 21:55, 25 October 2006 (UTC)

Real Numbers v Maths

Moved from main talk page. --Brad Beattie (talk) 12:29, 25 October 2006 (UTC)

I feal the reason that people hate this artical so much is the author has faild (at least at the start) to state that 0.999... = 1 is ONLY true when dealing with the branch of pure maths interested in real numbers. NOT in all maths.

If this were true then Pi =3.1415 and the squair root of -1 would not exists. This artical needs an introduction stating that "0.999... = 1 only holds true for maths which is only interested in real numbers".

Can someone please make this change. I do not want to do this because my english is very poor, and I do not wish to make a mess of the artical (mostly because some people on here will probably use it as an excuse to remove my changes.)

BTW, for all the people who are going to argue with me, Real number maths is only a small part of maths, and as an Electronic enginere I concider myself to be at least a passable mathmatiton working in a field which REQUIERS you to work beyond real numbers.

The situation doesn't change if you go to the complex numbers, either. In any case, the article is already explicit about where it assumes the real numbers and about the alternatives. To go any further by drawing attention away from the real numbers and toward the more obscure systems would be a violation of WP:NPOV#Undue weight. Melchoir 11:54, 25 October 2006 (UTC)
This is not undue weight. The whole article is about the status of 0.999..., and that status depends on the system in which it is defined. Your view that real numbers are the only numbers about which we should care, is something that I would say is not NPOV, and evidently something not shared by the many people who intuit numbers that are not real, and the many many mathematicians working in numerous different number systems. I know you know more maths than me, so I'm surprised to see you shoving so much that is interesting and useful under the carpet.
The part of maths where 0.9999... is not equal to 1 is negligible and of no applicability that I know of. Admittedly there are further parts of maths where people don't need the real numbers, but since reals are an important part of the complex numbers, both in their definition and as a subset, I doubt much of physics or electrical engineering are among them. --Huon 14:36, 25 October 2006 (UTC)
This isn't the case. The intuitive notion of 0.999... not equalling 1 is due to an intuitive notion of infinitesimals, which are the basis for the original formulation of calculus, and also for more modern formulations. If you really mean that calculus has no applicability, then I think we disagree most emphatically.
It's been a long time since I did maths and I never did get in to complex numbers but I'd have to say, arguing Pi=3.1415 because 0.999...=1 sounds just plain silly to me. Nil Einne 15:52, 25 October 2006 (UTC)
You misunderstand. If we deny that 0.999... = 1, we have to fundamentally change how we do mathematics in the real numbers. 0.999... = 1 is a result of the basic properties that the set of real numbers has. If we don't like that result, we can change those basic properties. But those basic properties are responsible for ALL of the results, including all the ones we like and use. If we change them to eliminate a mildly counterintuitive but mathematically insignificant result like 0.999... = 1, we have to completely rebuild an entire number system. Maelin 01:16, 26 October 2006 (UTC)
That really depends what you mean. When a mathematician is using a real number system, he can be happy in the knowledge that 0.999... = 1, when someone is using their intuitive system, then they don't need to completely rebuild an entire number system to say that 0.999.. != 1 because people knew that 2 + 2 = 4 long before Peano or any mathematician invented a formal system in which this was the case.
Well, there may be some place for number systems based on first intuition instead of an axiomatic approach, with neither the requirement of rigour nor of consistency, but that place sure as hell isn't in modern mathematics. Maelin 14:47, 26 October 2006 (UTC)
I'm not sure if you're replying to me. But I don't deny that 0.999... = 1. Indeed it does. However it seems pretty silly to me to claim that Pi = 3.1415 because 0.999... = 1. | 0.999 is an infinite series which ends up being 1. Pi is just well Pi. If we claimed 0.9999=1 then I guess you could claim 3.1415 = Pi but we're not... They're completely difference concepts and mixing the two seems silly to me Nil Einne 18:55, 26 October 2006 (UTC)
Causal relationships don't really exist in mathematics. It isn't very sensible to say that one theorem is true -because- another theorem is true. One might say that one theorem can be proven using another theorem, or that understanding the truth of one theorem will help one understand the truth of another theorem, or that two theorems are conceptually related, but the truth of one theorem can't really be considered the be the "cause" of another theorem. What you could say is that all true theorems we chose our axioms in such a way. In this situation, all mathematical results stem from the same cause - the axioms. If we don't like one of our results, we can change our axioms to fix it, but that will then affect all our other results, too. Maelin 23:32, 26 October 2006 (UTC)

Reals vs. Surreals

0.9r is 1 for real numbers, because they lack infintessimals -- since 0.9r and 1 are infintessimally different, if you have no infintessimals, then they are indistinguishable, and so are the same values.

0.9r is 0.9r for surreal numbers, because they have infintessimals.

I think that main reason that this article has so many disagreements hinges on this basis.

Some people intuit infintessimals, and some do not. —The preceding unsigned comment was added by 12:42, 25 October 2006 (talk • contribs) 61.84.255.151.

Actually, just because the surreals have infinitesimals doesn't mean that 0.999… < 1 in the surreal numbers. It depends on what meaning you give to the notation. For example, the sequence (0, 0.9, 0.99, 0.999, …) does not converge in the order topology of the surreals, so you can't define 0.999… to be its limit. So what alternative definition would be appropriate?
Oh, and while some people intuit infinitesimals, it's a huge stretch to say they intuit the surreals. That would require some kind of evidence. Melchoir 12:59, 25 October 2006 (UTC)
People generally don't intuit a formal system, they intuit the rough outlines of a system that has all kinds of nonorthogonal behaviour. Mathematical formal systems tend to be orthogonal. The intuitive system of many people on this page contains reals and infinitesimals. That they haven't created strict rules on what operations are allowed on some of these notions that they intuit does not mean that these notions are wrong. It also doesn't mean that they aren't useful in specific circumstances (eg. calculus). Mistaking the intuitive system for a formal system, eg, the reals, or the surreals is what is wrong.

well it isnt equal but frightening near

Moved from main talk page. --Brad Beattie (talk) 13:43, 25 October 2006 (UTC)

0.999... isn't 1, it's just very close. It is similar if we try to think about space and how big it is.

But here is a comparison: I have been cloned and now there are two identical people (1=1), but then the clone loses 1 atom (so he becomes 0.999..., as 1-n=0.999... where n is an infinitely small number). Now he isn't identical to me anymore, he's just similar. So 0.999... is 1-n where n is an infinitely small number.

Anyway interpret this as you like because the difference between 0.999... and 1 is smaller than the difference between the two clones. ;) In conclusion, I remember some great man said, "The difference between 1 and 1,000,000 is smaller than the difference between 1 and 0."

Please correct typos. :D

Okay, your cloning argument has two major flaws. Firstly, you are misusing the concept of equality. When we express a fact about two numbers in the form a = b, we mean that a and b are the same numbers. Twins are not the same thing, they are different things in every respect. Mathematical equality cannot be used in this way to express facts about the similarities between two different objects, since those objects are not mathematical objects but physical real world things.
Secondly, all humans are made of finite numbers of atoms. If you consider the number of atoms in a person at any given time to be x, then you take one away, they have x - 1. If you want to express x - 1 as a multiple of x, as in x - 1 = ax, then the coefficient is NOT 0.999..., it is a = x / (x-1), which is a finite number. It is small, but larger than zero, and not infinitesimal. Maelin 13:48, 25 October 2006 (UTC)
If 0.999... is less than 1, than there must be a number x such that 0.999... < x < 1. As a matter of fact, there must be an infinite amount of possible values for x. I challenge you to find even one possible value for x. Stebbins 13:57, 25 October 2006 (UTC)


Why should anyone accept that if a < b then there must be some some c such that a < c < b. Why could we not say that 0.999... and 1 are infinitesimally close where, by defintion, infinitesimally close means that a is infinitesimally close to b iff a < b and there is no c such that a < c < b. We could even "prove" it by saying that since 0.9 followed by any number of 9s can never get you to 1 and since there is no number between 0.999... and 1, 0.999... and 1 are infinitesimally close (as are 611.999... and 612 and so on). Davkal 14:10, 25 October 2006 (UTC)

The answer is simple: We expect to be able to calculate a + b, and then to calculate c = (a + b) / 2. Since a < b, we have a+a < a+b and a+b < b+b, or 2a < a+b < 2b, or a < c < b. Of course, we could conjure up some weird structure where something of the above doesn't hold, but if we can't even do simple arithmetic, what have we gained? -- Meni Rosenfeld (talk) 14:17, 25 October 2006 (UTC)
Because if you deny the Archimedean property then you are no longer talking about the real numbers. One might as well go to an article that says that, in the integers, 3 * 5 = 5 * 3, and say, "Why should anyone accept that a * b is equal to b * a?" It's one of the properties of the integers that multiplication is commutative. Similarly, it's one of the properties of the reals that infinitesimals do not exist. If you want to define a number system where they do exist (such as the surreal numbers), where you can have 0.999... < 1 as you so desire, then go for it, but it WILL NOT be the real numbers. Maelin 14:24, 25 October 2006 (UTC)


It seems a pretty weird structure already where something that is obviously less than 1 needs to be taken to be equal to 1 in order to make it possible for me to say that 3 + 5 = 8. I don't think that's true anyway. What would happen if we agreed that 0.999... was less than 1. It seems to me that the only thing we could not do is certain particular calculations involving just those recurring numbers, but since we can't do them anyway (we just pretend the recurring numbers are something else and do the calculations with those numbers instead) not a lot would appear to have been lost. Davkal 14:32, 25 October 2006 (UTC)

Okay, that is a good question. What would happen if we just declared that 0.999... < 1? Well, firstly, we have to decide how exactly we're going to just declare it. Mathematics is based on the idea of axiomatic systems. You start with a small number of axioms, which are your basic inherent truths that you will allow without proving. You need some axioms to give you something to start from. Then you work with them and derive theorems, which are statements that you know are true, because your axioms are (by definition) true and your reasoning is (presumably) valid. Note that when I say something is "true" here, I really mean "it follows from the axioms", since the connection between a "true" mathematical theorem and what is a "truth" in the real world is extremely vague and very philosophical and it doesn't have much to do with maths itself. Anyway, this is how mathematics works - first you choose some agreeable axioms, and then you explore all the results that follow from them.
So, that's all good. Now, what happens if we decide we don't 'like' one of our theorems? We really only have two choices here. We can try a different set of axioms, thus defining a new mathematical theory, and see if that makes things work more to our liking (this is like saying, "The real numbers are no good, let's define our own number system, where we can have 0.999... < 1"). Or, we can just put up with it, and accept that sometimes, agreeable axioms can lead to surprising results. What we -can't- do is just "agree" that one of the theorems is wrong. The reason is that the only control we have over what theorems are true and what theorems are false is in our choice of axioms. If we just add a new axiom that says, "Theorem whatever is false", we have suddenly made our mathematical theory inconsistent - the axioms lead both to a theorem, X and its negation, ~X. This is disastrous. An inconsistent mathematical theory is useless, because you can prove anything in it. EVERYTHING can be proven true in an inconsistent system, once you've found the inconsistency.
So we're stuck with our two choices. We can either try other number systems, and see if we can make one that works how we want it, or we can accept the occasionally surprising but always interesting consequences of our original axioms. Maelin 14:58, 25 October 2006 (UTC)
Hmm... Well put, but this doesn't really add much to my comments on the "Approaching it differently" thread.
Anyway... so much to say, and so little stamina to say it. I'll be brief, and I hope I get the message through.
  1. We can't do arithmetic. That renders every current theorem about real numbers (and many mathematical theorems involve reals in some way) incorrect and useless. That's quite disastrous.
  2. We gain nothing. If we can't do anything with those numbers, why would we want them?
  3. We're not "pretending" anything. 0.999... and 1 are two different notations for the same number, just like φ and (1+√5)/2 are two different notations for the same number.
  4. Decimal is arbitrary. The new numbers' entire claim for existence is a certain feature of decimal expansions. The system you'd get for binary, say, or a non-radix notation, would be different. Basing a mathematical system (as opposed to a convenient notation) on the fact that we have 10 fingers is silly. Decimal expansions are not some "universal truth", in the sense that different expansions must mean different numbers, but only a convenient way to represent real numbers, which unfortunately has some quirks - such as, every number which is an integer divided by a power of 10 has 2 different representations. So 1 and 0.735 have 2 representaions each (do you see again how arbitrary the 10 thing is?).
  5. In short, arbitrarily declaring 0.999... < 1 means making most of mathematics crumble to satisfy some whim. Not a very good trade if you ask me. -- Meni Rosenfeld (talk) 16:58, 25 October 2006 (UTC)

Infinity

Moved from main talk page. --Brad Beattie (talk) 13:45, 25 October 2006 (UTC)

Just another point about the fractional proof. Question
What is 1+infinity?
Answer infinity

For the same reason any of the proofs that use arithmetic on a number with an infinite series of numbers cannot hold. I dont care how you do it With these proofs people just seem to pick and choose their representaion of numbers. Im supposed to accept that 0.999...x10=9.999...
and therefore 9.999...-0.9999 = 9
But follwing this logic
0.999...x10=10!!!

I am still of the opinioin that 0.999.. is equivalant to 1(Not equal, but equivalant) but I think that some of the proofs simply don't hold. Graemec2 13:37, 25 October 2006 (UTC)

We're not 'doing arithmetic' with infinite series. We are doing arithmetic with one single real number. It is the limit of an infinite series, and such things are perfectly well defined - in fact, in one construction of the real numbers, numbers are defined as the limits of infinite series. By the way, 0.999...*10 = 10 is true. In fact 0.999...*x = x for every x in the real numbers. Maelin 13:53, 25 October 2006 (UTC)


It may be best to look at it as an infinite sum.

The article defines 0.999... as \sum_{n=1}^\infty \frac{9}{10^n} though without the summation symbols (at least if I remember correctly). Let's take this out a little farther.

S=\sum_{n=1}^\infty \frac{9}{10^n}=9\cdot\sum_{n=1}^\infty 10^{-n}
so multiplying S by 10 you get 10\cdot S=10\cdot 9\sum_{n=1}^\infty 10^{-n}=9\sum_{n=1}^\infty 10\cdot 10^{-n}=9\sum_{n=1}^\infty 10^{-(n-1)}=9\sum_{m=0}^\infty 10^{-m}
where in the last step I substituted m=n-1.
Since an infite sum of the form \sum_{m=0}^\infty r^m=\frac{1}{1-r} (where |r|<1) we have 10\cdot\ S=9\cdot\frac{1}{1-\frac{1}{10}}=10. Thus S=1 which was already defined as 0.999... A similar proof was already in the article. Do you have a problem with infinite sums in general or just in this particular case? For example, I've never seen anyone argue the fact that \sum_{n=1}^\infty \frac{1}{2^n}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1 but this is merely the representation of 0.999... in binary (see Zeno's Paradox). Unmasked 17:56, 25 October 2006 (UTC)

Rational Number

Could someone provide me with a formal prof that 0.999... is a rational number. Graemec2 14:16, 25 October 2006 (UTC)

The proof you're looking for is in this section of the article. The proof that 0.999... is a rational number is exactly the same as the proof that it is equal to 1. Stebbins 14:21, 25 October 2006 (UTC)
Claim - .999... is a rational number.
Proof - .999... = 9 * .111... = 9 * 1/9 = 9/9 which is clearly rational. qed
-bobby 14:24, 25 October 2006 (UTC)

No not QED

Proofs of your kind are not acceptable. Claim 0.999...=1 0.999...*1 = 1 =>1 / 1 = 1 There is no proof there. Multiplying a number by constant,rounding it up and then dividing by a constant is not an acceptable proof.

If 0.999... is a rational number you should be able to give me a proof that doesnt involve you you starting with the assertion 0.999..=1

Graemec2 15:17, 25 October 2006 (UTC)

Okay, we start with any of the half-dozen equivalent definitions of 0.999…. Then we conclude, by the corresponding proof, that 0.999… = 1. Then we notice that 1 is a rational number. Happy? Melchoir 15:40, 25 October 2006 (UTC)
I'd like to see you provide a proof that \frac{\sqrt{2}}{\sqrt{2}} is rational, which doesn't go like "√2 / √2 = 1, and 1 is rational". Proving that a is equal to b, and that b has a property P, proves that a has property P. It doesn't get more basic than that. -- Meni Rosenfeld (talk) 17:02, 25 October 2006 (UTC)

All rational numbers have a repeating decimal representation (this includes numbers such as 1.000...) - actually it is better to say that a number has a repeating decimal representation if and only if it is rational. Since 0.999... is a repeating decimal it must be a rational number. I know this isn't a formal proof but you can start with the pages I referenced and hopefully convince yourself that my first sentence is true. You may even be able to find a formal proof of it that meets your standards. Unmasked 17:04, 25 October 2006 (UTC)

From the article on Rational number:

"In mathematics, a rational number (commonly called a fraction) is a ratio or quotient of two integers, usually written as the vulgar fraction a/b, where b is not zero."

Consider: There are no two integers, a and b, such that the decimal expansion of a/b would result in 0.999.... Therefore, by the above definition, 0.999... is not a rational number. 71.72.135.142 05:00, 27 October 2006 (UTC)

Take a = 1 and b = 1. Both are integers, and a/b = 1/1 = 1 = 0.999... . The long division algorithm happens to only give one of the decimal representations of the result (in this case, it gives the 1.000... representation rather than the 0.999... representation), but that doesn't change the fact that the result of the division is 0.999... -- Meni Rosenfeld (talk) 09:55, 27 October 2006 (UTC)

Untitled section

Moved from main talk page. --Brad Beattie (talk) 14:40, 25 October 2006 (UTC)

.999... does not equal 1. I'm not going to bloody vandalize the page, but it CAN'T be.

I just had to say that. I'll go read the article and reason it out. I'll either come back and delete this, or argue why .999... doesn't, in fact, equal 1. Liquid Entropy

I took out your expletive since I expect some kids who take to math might be interested in the discussions regarding this topic. I look forward to hearing from you after you peruse the arguments on the main page. -bobby 14:47, 25 October 2006 (UTC)
Experience shows that people without sufficient mathematical background have a difficulty understanding the article (we should probably do something about that - even if their criticism is wrong, the fact that they choose to criticize means something about the article). I recommend that you read the replies on this page - some of them are very insightful.
By the way, the real number \frac{1}{\sqrt{e^{\pi}}} "can't" be one of the values of ii, but it is. Truth is often incompatible with intuition. -- Meni Rosenfeld (talk) 17:16, 25 October 2006 (UTC)
And in case anyone is interested about the formula Meni mentioned, see the Euler's identity article. Interesting stuff. Dugwiki 17:24, 25 October 2006 (UTC)

A simple, but major, error caused by rounding

All of the proofs that have been presented, I have noticed, seem to work off a finite representation generated by rounding or truncation of an infinite decimal, and extended by assumption the original decimal. For example, the first proof assumes that the final digit (should there be one) in the recurring decimal for \frac{1}{3} is a 3; for all we know, it could, in fact, be a 4, which, when multiplied by 3, returns the number to 1 again.

The most common error I have noticed, and this is one which I have even caught myself on, although only once, is the decimal-to-fraction algorithm (which is taught in many schools in Australia and, I assume, around the world). Because of the repetition of the singular digit, it is assumed that there is a rational representation for it; the result generally ends up formulating the assumption that the two numbers are alike.

The only proof, at least that I have noticed, that disproves the assumption, is the Egyption fraction-esque representation (it uses a similar structure to Egyptian fractions); this form explicitly shows the inequality of the recurring decimal: \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{9}{10000} + \cdots With each new fraction of the form \frac{9}{10^n} \ (n \in \mathbb{N}, n \geq 1 \mbox{ to clarify}), the difference between it and 1 lowers to \frac{1}{10^n}. Because the decimal is recurring, i.e. it has no terminating digit, then it can be assumed that n extends to \infty, making the difference \frac{1}{10^{ \infty }} (as far as I know, it has not be either proven or disproven that \frac{1}{x^{ \infty }} = 0 (x \in \mathbb{Z} \backslash \{ 0 \}), an infinitesimally small number, and an irrational real, making 0. \dot{9} irrational as well.

Saying that, like I said, it is solely an irrational real that has been given a rational (let alone integral) 'synonym' through rounding, and it should be accepted that the equality is untrue.

(I have also come to notice that \frac{0. \dot{9}}{2} = 0. \dot{4} \dot{9} \neq 0.5. Can someone explain that?) --JB Adder | Talk 15:18, 25 October 2006 (UTC)

No. There does not exist a real number, rational or irrational, that has anything to do with 0.999…, except the real number 1. Melchoir 15:37, 25 October 2006 (UTC)
JB-adder-- your arithmetic is wrong-- \frac{0. \dot{9}}{2} = 0.4 \dot{9}=0.5. --Alecmconroy 16:33, 25 October 2006 (UTC)
There is no last digit. You seem to say "maybe there is, maybe there isn't" and that's wrong. Also, ∞ is not a real number, and in this context (as opposed to other contexts) we have not defined arithmetic operations on it, so we cannot say that the difference is 1/10∞. We're not rounding anything, we're following definitions and theorems rigorously. -- Meni Rosenfeld (talk) 17:08, 25 October 2006 (UTC)

0.999 ≠ 1, but whole fractions work

One way to think of it is that 1/3 does not equal 0.333. How is this possible? Let's say each decimal of 1/3 is a third of the way to 0.333...4 (infinite amount of 3s). You then get two-thirds of the way to 0.66666...7, and when you get to 3/3 the 0.999... result is then impossible if thought of in those terms, because the infinite amount of 9s would be cancelled out by a whole number fraction at the very end.

This also easily explains longer decimals such as 1/7; 0.142857 repeating. Using 0.142857142857...1/7. Using the same logic above, even this bundled-up mess works well.

Which is proof that not only can infinitely small numbers exist independent from their whole numbers, but they can disprove the theory that fractions lead to 0.999... results, if thought of in such a way that is. Both arguements disprove each other in this case.

TurkAlbert 15:15, 25 October 2006 (UTC) -- Happy Trails.

I don't even know where to begin.
first, there is no 4 in the decimal expansion of 1/3 and no seven in the decimal expansion of 2/3. Long division proves this readily enough. Also, when you start an infinite string, there can be nothing after it. Hope this helps. Argyrios 15:24, 25 October 2006 (UTC)
I should add it doesn't even make sense to add a 4 when rounding... Nil Einne 15:47, 25 October 2006 (UTC)
when you start an infinite string, there can be nothing after it. Why not? If an infinite series is achievable, then why can't you do anything afterwards? If an infinite series is NOT achievable, then .333... will never exactly equal 1/3. Algr 18:01, 25 October 2006 (UTC)
This is what I was confused about in the article. When you assume that 0.999.. = 1, by saying that 1/3 = 1.33..., you are engaging in circular logic. You're saying a repeating decimal is equivalent to a fraction (1/3), because a repeating decimal is equivalent to a fraction (3/3). When you divide 1 by 3, then multiply it again, your infinite series is losing an infinitesimally small digit. --Wooty Woot? | contribs 18:11, 25 October 2006 (UTC)
It's not circular logic because we can show that 1/3 = .3333... using basic long division. Feel free to take pen and paper in hand to verify that the integer 1 divided by the integer 3 gives an infinitely repeating series of 3's. As to your assertion that we lose an infinitesimally small digit is lost when we do the multiplication, I would counter that we are actually losing an infinitely small digit. Any number that is infinitely small is in fact 0 (think of the limit as n->infinity of 1/n which indeed equals to zero). Hence, the multiplication does not change the value at all and the proof remains valid. -bobby 18:56, 25 October 2006 (UTC)
Long division shows just the opposite. Do the division yourself and see. As you move to the right, there is always a remainder. That's why the division continues infinitely. At each step, it's always 10 / 3. At each step it never gets to 4. 0.333... is infinitesimally smaller than 1/3, and that's what keeps the long division repeating endlessly. Dagoldman 19:14, 25 October 2006 (UTC)

I would like to again suggest that anyone that thinks that .999... is not equivalent to 1 PROVIDE A SOURCE. If you can cite a single published work by someone with a degree in any math or science field that says that .999... doesn't equal one, perhaps you will gain some credibility. 192.80.211.11 20:45, 25 October 2006 (UTC)greg

Is this what is ultimately at the core of mathematics? When logic fails, you submit to authority? Jesus is infallible too, but most priests know better then to try to answer such a question by showing off their credentials. What is Wikipedia for? It is not for smart people to be idolized, but for helping others understand. You won't get far in life by telling people that without a degree, their questions just don't matter. Algr 23:11, 25 October 2006 (UTC)
This is the core of Wikipedia. We are running an encyclopedia, not a mathematics research journal, not a classroom, and not a practice arena for getting far in life. Demanding sources is never wrong. Melchoir 23:24, 25 October 2006 (UTC)
We're not debating the article, we're debating the concept. --Wooty Woot? | contribs 01:08, 26 October 2006 (UTC)
When someone asks what Wikipedia is for, I'm going to tell them. Melchoir 01:16, 26 October 2006 (UTC)
I'll provide the second half of the answer: No, that is obviously not what is at the core of mathematics - Exact definitions and rigorous proofs are. But it is not a trivial thing to rigorously construct all of the properties of decimal expansions, and trying to present it here will create more problems than it solves. That's why we are providing references, so that anyone who challenges our claims can check them for a more detailed development. Only if you find a reason why the sources do not, in fact, support our claims, or a reason why the development in the sources is flawed, do we have something real to discuss. Oh, and another thing I have mentioned in the past, but not in the current debate: The fact that every single source or proffesional mathematician supports our claims, does not in itself mean we our right - but it definitely means that the burden of walking the extra mile to truly understand what we are dealing with here is the opponents', not the propponents'. -- Meni Rosenfeld (talk) 15:25, 26 October 2006 (UTC)

Perhaps a stupid question

I probably should read the article but am I right that since 0.999...=1 anything else with X999... equals (X+1). For example. 3.9999... = 4 | 1.5334999...=1.5335 ? Of course, I'm not particularly sure if you'd ever come across these in reality but I guess the concept remains the same Nil Einne 16:02, 25 October 2006 (UTC)

Yes; see 0.999...#Generalizations. Melchoir 16:16, 25 October 2006 (UTC)

Random question...

So does that mean that:

  1. 1.111... is also equal to 1?
  2. 1.999... equal to 2? —The preceding unsigned comment was added by Ejh102 (talk • contribs) 15:57, 25 Oct 2006 (UTC)
1; No, 1.111... is equal to 1+\frac{1}{9}
2: Yes →AzaToth 16:07, 25 October 2006 (UTC)
First one, eh say what? Why will 1.111...=1? This doesn't make sense to me. Second one, well I have the same question see above Nil Einne 16:04, 25 October 2006 (UTC)
No, 1.111... doesn't equal 1, because base "1" isn't the lowest digit-- 0 is. Note, however, that 1.000... DOES in fact equal 1. Or another way to show you-- the difference between 1.111... and 1 is at LEAST .1 ( ie, 1.111...-1>.1). In contrast, if you try to find a positivenumber that is DEFINTELY smaller than the difference betweeen 1 and .999..., you won't be able to find one. --Alecmconroy 16:26, 25 October 2006 (UTC)

For doubters, note that 0.xxx... is defined as the limit of a sequence

I noticed in some of the "counter-arguments" above that try and show 0.999... does not equal 1 rely on some confusion over what 0.999... represents. I think it might be useful to explain what that symbol 0.999... means.

"0.xxxxx..." is actually a short-hand abbreviation for saying "the limit of the sequence {0.x, 0.xx, 0.xxx, ...}" That is, it is the limit of the sum of all numbers x*10^(-k) from k=1 to infinity.

I'll leave it to the article and the reader to prove for themselves that the limit of the sum of 9*10^(-k) for k=1 to infinity equals 1. But I think once you understand that the definition of 0.999... is that it is the limit of a sequence, it might help clear up some of the confusion leading to some of the "counter-proofs" mentioned on this talk page. Dugwiki 17:15, 25 October 2006 (UTC)

Unfortunately, this has been stated, one way or another, too many times, and yet does not seem to have any effect. Truth is often difficult to accept, and I'm afraid some people just choose to remain confused. -- Meni Rosenfeld (talk) 17:21, 25 October 2006 (UTC)
I agree 100% that the limit of the geometric sequence represented by 0.999... equals 1. I have stated that repeatedly. But that's not what the article says. The word "limit" does not even appear in the article until way down. Dagoldman 19:00, 25 October 2006 (UTC)
No, .999... doesn't represent a sequence, it represents the limit of one. 144.82.106.32 21:26, 25 October 2006 (UTC)
That's exactly what I said above. It represents the limit of a sequence, and that limit equals one. Dugwiki 22:10, 25 October 2006 (UTC)
This is probably because most people are going to look at the sequence {0.9, 0.99, 0.999, ... } and say "Gee, every member of this sequence is less than 1. So if the limit is a member, then the limit must also be less than 1. Or the limit is not a member of the sequence ..." and at this point the person's brain explodes, wondering how the limit of a sequence isn't actually in the sequence. mdf 21:18, 25 October 2006 (UTC)
YES YES YES YES YES. This is the DEFINITION of 0.999... This should be at the top of the article page, as this is the beginning, middle, and end of this so-called argument.

A Naive Question

What is x, such that x-1=0.99999...? --Rlwest 13:03, 25 October 2006 (UTC)

x = 1, since 0.9999... = 1 -- The Anome 13:05, 25 October 2006 (UTC)

That sounds to me like "Affirming the Consequent"... Can you come up with something that says it without presuming the equality? And does "equality" really apply to non-exact numbers? --Rlwest 13:41, 25 October 2006 (UTC)

There is no such thing as non-exact numbers (though there may be non-exact numerical data in applications). If asked, "what is x such that x-1 = 444 / 148", is there something wrong with replying, "444 / 148 = 3, therefore x = 1/3"? 0.999... is just equal to 1, there is no way around it - the same way 444 / 148 is just equal to 3. -- Meni Rosenfeld (talk) 14:02, 25 October 2006 (UTC)
Well put Meni. Just to be clear, there is no need perform mathematical operations on .999... without substituting 1 because we have proved that .999... = 1! That's the beauty of proofs! -bobby 14:07, 25 October 2006 (UTC)

Show your work

I've seen a lot of arguments back and forth on this subject over and over again, so let's do what our teachers told us to do: show your work. I.e. prove your answer in a format that even the lamest of morons could follow the numbers.

What we know

Most high school algebra teachers have taught their class the method to find the fractional form of a repeating decimal. And they show the proof through the most common method, taught to us in preschool: reverse the problem to check your answer. So with that in mind, a small lesson:

One of the most fun fractions to play with is 1/7, which equals 0.142857... Now the way to determine the fractional value of a repeating infinite decimal is to use 9's. Take the section of the sequence that repeats, and place it over a series of nines with an equal number of digits. Ex. 142857/999999 Notice there are six digits on top and bottom. This it the fraction used to determine the value of 0.142857... Repeating. Reduce the fraction (you all do remember how to reduce, right? If not, break out a calculator and divide the bottom by the top, I promise it will always come out to an integer) and you are left with 1/7, our original fraction. Now check your work by reversing the problem. Multiply the fraction 1/7 by 142857 to arrive at the fraction 142857/999999, and likewise divide one by seven (I advise a calculator, unless you like long division that continues into infinity) and there you have it, reversal-proof that 0.142857... repeating equals 1/7.

If you care to try it with a few other fractions with prime denominators to get used to the concept go ahead and do so.

Where we go from there

Now we come to the wonderful 0.9... Repeating. We repeat the process, taking the sequence, which is 9 by itself, and place it over a series of nines with an equal number of digits, i.e. 9/9. Do I need to go on?

Unfortunately, this is not really a proof, just an example. It's a nice thing to think about, but not really what we need here. -- Meni Rosenfeld (talk) 14:03, 25 October 2006 (UTC)

Cogent objection?

Do we want to answer this objection more explicitly?

(Dispute. If you cannot make the assumption that 1 = 0.9.... then you also cannot make the assumption that 1/3 = 0.3...., in both cases it is true that the limit is equal to 1 and 1/3 respectively, however the arguement that 0.3... is strictly less than 1/3 is as valid as the argument that 0.9... is strinctly less than 1)

It may be enough to observe than 0.9999... is not the result of long division, so the cases are distinct. Septentrionalis 16:37, 25 October 2006 (UTC)

If you wrote a computer program that prints real numbers in decimal, and, when given a "1", it starts to print a bunch of nines instead, virtually everyone reading this would conclude this program has a bug. If the mathematicians burst in and explain that, "well, you know, this program, as long as it runs forever, is actually quite correct" would be dismissed just as easily as someone who suggested that, by golly, you can indeed unload your groceries into the kitchen, one molecule at a time! This is ultimately the source of the difficulty with this business. People know what is "right", Occam's razor is a very deep computational element in the human brain. Probably the mathematicians know this, since this particular result has essentially no use beyond the novelty factor. mdf 18:14, 25 October 2006 (UTC)
Well actually computers don't and shouldn't deal with infinity and mathematicians don't deny this and I'm pretty sure nearly every single one would agree a computer program giving the result as 0.999.... instead of 1 is buggy. Just because they mean the same thing doesn't mean you should use them interchagably. Nil Einne 18:39, 26 October 2006 (UTC)

Open set proof?

Isn't it possible to prove that 0.999...=1 by use of open sets? In other words if S=(0,1) is an open set then 0.999... is not a member of S since (0.999...,1) is not an open set. However, 0.999... is a limit point of S so it is in the closure of S (since it is a limit point) which shows that 0.999...=1. Unmasked 19:02, 25 October 2006 (UTC)

Well, reasoning with (0.999...,1) before determining what 0.999... is seems dangerous. Melchoir 20:16, 25 October 2006 (UTC)
OK I suppose what I'm saying is that it should be possible to prove that 0.999... is not in S with this line of thinking. I'm not sure why you moved this to the Argument's section as I'm just wondering if this is another method to prove 0.999...=1. Unmasked 20:38, 25 October 2006 (UTC)
Well, this page isn't just for cranks; any mathematical discussion that doesn't concern the article can get dragged here. Anyway, sure, you can prove that 0.999... is not in S, but at that point, since 0.999... <= 1, you've proved 0.999... = 1 already! Melchoir 20:43, 25 October 2006 (UTC)

I don't understand the part that goes "since (0.999...,1) is not an open set". Actually, (0.999...,1) is an open set, though not one that usually goes by the name (0.999...,1). --Trovatore 02:16, 26 October 2006 (UTC)

Exactly. For real numbers a and b, the set (a,b) is defined as the set of real numbers x such that a<x<b. So the set (0.999...,1) = (1,1) is the empty set, which is indeed open. --Kprateek88(Talk | Contribs) 14:51, 29 October 2006 (UTC)

Proof that 1 does NOT equal 0.999...

lets assume 1 = 0.999...

if x = 1, and x is then multiplied by 2, and then 1 is subtracted, x will still equal 1 if this process is repeated infinite times, x will still equal 1.

now, if 1 = 0.999..., then the same result should be reached if x equaled 0.999... instead of 1

but if you multiply 0.999... by 2 and subtract 1, you get 0.999...8. this then becomes 0.999...6, then 0.999...2, and so on. Repeated infinitly, you would end up with negative infinity, rather than 1.

therefore, 1 cannot equal 0.999...

i hope that all made some sort of sense, let me know if you see anything wrong with it

Diabl0658 19:45, 25 October 2006 (UTC)

  • The problem is that there is NO "last decimal place" in this expression.Jerry picker 19:48, 25 October 2006 (UTC)
The problem with your "proof" is that we are talking about real numbers, not surreal numbers, so the notation 0.999...8 is not allowed. →AzaToth 19:49, 25 October 2006 (UTC)

in that case, what about 1 - (2 / infinity) instead of 0.999...8

Diabl0658 19:58, 25 October 2006 (UTC)

  • Division by infinity is not defined on the real numbers because infinity is not a number, so (2 / infinity) is meaningless. 38.112.22.178 20:01, 25 October 2006 (UTC)

See it like this:

f(x) = 1- \frac{2}{x} = \frac{x-2}{x}

\lim_{x \to \infty}f(x) = 1 →AzaToth 20:11, 25 October 2006 (UTC)

I dont remember how to read those kind of equations to be honest. What about [0.8 + 1.8(0.11) + 1.8(0.12) + ... + 1.8(0.1n)] would that be a legal equation for real numbers? Diabl0658 20:19, 25 October 2006 (UTC)

Yes, it's a legal equation. You can rewrite it as:

0.8 + \sum_{x=1}^n1.8(0.1^x)

→AzaToth 20:41, 25 October 2006 (UTC)

A new proof for the skeptics

I will provide here a new proof that 0.999... = 1 (in the real numbers, of course). You'll tell me which step you disagree with (or would like to see more justification for), and we'll move on from there.

  1. 0.999... is not greater than 1. In other words, 0.999... <= 1.
  2. So 1 - 0.999... >= 0. Denote c = 1 - 0.999...
  3. Let n be any natural number.
  4. There exists a natural number m such that 10^m > n.
  5. 10^(-m) < 1 / n.
  6. 0.999... is greater than 1 - 10^(-m).
  7. Therefore, c < 10^(-m).
  8. So c < 1 / n.
  9. From 3-8 it follows that c is less than any number of the form 1/n, where n is a natural number.
  10. According to the Archimedean property of the real numbers, there is no positive real number with the property that it is less than any number of the form 1/n, where n is a natural number.
  11. Therefore, c is not positive.
  12. Since c >= 0, and is not positive, c = 0.
  13. 1 = 0.999...

QED

My guess is that step 10 will be the lucky one! Before responding, please read the link about Archimedean property. -- Meni Rosenfeld (talk) 20:45, 25 October 2006 (UTC)

No, I am sorry Huon but it fails long before step 10. Step 7 is already untrue because you cannot prove this for all m. If you tried, then you would end up with 0 < 0 which is nonsense. 198.54.202.254 19:04, 28 October 2006 (UTC)
I do like this proof, because no matter what 0.9999... is defined to be, as long as we agree on the Archimedean property and on some properties any number with some right to call itself 0.9999... should have (i.e. not being greater than 1, but being greater than 0.9999...9 with some finite number of nines), we conclude it must be equal to 1. But I agree that step 10 will be doubted by the fans of infinitesimals. --Huon 20:56, 25 October 2006 (UTC)

(for clarity, I have removed an addition by →AzaToth 21:08, 25 October 2006 (UTC)) -- Meni Rosenfeld (talk) 09:09, 26 October 2006 (UTC)

I still do not think that your point that it is "not greater than 1" is any help because it assumes implicitly that both cases are available, and hence, it is not surprising that one of the cases pops out the bottom. Any real differentiation between 0.999... and 1 would have to start off from a point where they are treated differently. Ansell 23:02, 25 October 2006 (UTC)
First, I'll say that the first step is like any other - I have not provided all the details, but am willing to fill in the blanks, right down to the fundumentals, upon request. The only steps where I used any property of 0.999... are 1 and 6, and I doubt anyone would disagree with any of them (specifically, a number with a lexicographically smaller decimal expansion than another cannot be greater than it). Anyway, it is unfortunate that people choose to continue to ramble on and on below, instead of replying to my very clear "challenge".
AzaToth, I apologize for removing your reformatting, but I think people will be confused to see 2 versions of the proof, and I think it's less readable with the differently sized fonts & rows. -- Meni Rosenfeld (talk) 09:09, 26 October 2006 (UTC)
So where is your formal proof? The above is clearly in error. Rosenfeld has no idea what he is writing about. 198.54.202.254 05:31, 29 October 2006 (UTC)


Meni, you're right that step 10 is the faulty link in this reasoning. The Archimedian principle appeals to an elementary misunderstanding of infinity. Claiming that 2*INF>INF as a proof is like claiming "gotcha last + infinity." The conclusion of the Archimedian principle is that the infinitely smallest positive number is not positive. Quite a contradiction. --68.211.195.82 14:12, 1 November 2006 (UTC)

Finally, a direct assault on the Archimedean principle (198.54.202.254 would be embarrased - that principle was already known to Archimedes, I presume). But you do it too broadly. If there were a "infinitely smallest positive number" - can you divide it by two? Is half that number smaller than the smallest positive number? Then a half of a positive number would not be positive. Quite a contradiction.
The conclusion of the Archimedean principle is that there are no infinitely small positive real numbers - or, exuivalently, no infinitely large real numbers. --Huon 17:04, 1 November 2006 (UTC)

Eh?

From my point of view this seems just like an argument over definitions. The normal (intuitive) definition of 0.9999... is of the largest number smaller than 1. You say this doesn't exist, but I don't see why this is a problem. It doesn't exist in a similar way that infinity doesn't exist, yet people find it useful to create a symbolic representation of it. What's the problem with a symbolic representation of the largest number smaller than 1 that we represent as 0.99999....? So it's not a real, big deal, I never said it was a real. You say that you can multiply it by 10, I say that as it's not a real, you can't just assume that you can do all the same things you can do with reals. You say it's not useful then, I say well it's not useful to you anyway, since you already have another and better symbolic representation for 1. I think for most people who first come across this, the point is not that you can make it equal to 1 in *your* system, it's just that if your system cannot represent such a simple and intuitive concept, perhaps it's your system that's wrong. You know, the strict mathematical system of real numbers is not the same system of numbers that everyone uses in everyday life, it just has some similar properties and operations.

Do not confuse your real world philosophy with the totally constructed maths world. In any realistic world, the definition step would take infinite time, and hence never complete. However, in a pragmatic maths society, they ignore the slight bulge in time, and assume the solution within their definition, for the sake of getting real work done. That pretty much sums up why a mathematician can "prove" their constructed concepts are "actually" right, they help people do work? Isn't that a good enough way to think. Truth is afterall a subjective concept ;) Ansell 23:41, 25 October 2006 (UTC)
I'm not sure if you're refuting me or agreeing with me. I think that this whole argument is exactly the kind of confusion that you describe. "Maths" people say that 0.999... = 1 because it does in their constructed maths world. "Normal" people say it doesn't because it doesn't in their alternative world.
Is there any evidence that most people are even aware of the possibility of multiple number systems, let alone that they have made a conscious choice to work within any given system, and that they are aware of the consequences? I've seen none. Melchoir 23:47, 25 October 2006 (UTC)
I'm not sure that that is relevant. In fact I think it's precisely because people aren't aware that there are multiple number systems that they feel compelled to accept the arguments that 0.999.. = 1. As soon as people start saying that "you can't do that operation because it's infinite", it becomes obvious that they are talking about a different number system, even if they don't realise it. Normal people didn't invent these nonintuitive number systems, mathematicians did, so it seems a little unfair that they then act as if their number system is the only one worth talking about or that just because I didn't invent a bunch of number systems then the one that I'm using is wrong.
I'd say most people don't shive a git one way or another, re: number systems. I wish they did, I've tried talking about all kinds of things with people, but few are interested in math, physics and the like. Basically, our anonymous fellow is quite correct. If you walk up to two people on the street and say "the ratio of your heights is 1", they will think "we have the same height". If, however, you say "the ratio of your heights is 0.999... (and continuing to say "nine")" they will think "one of us is ever so slightly taller than the other". if they interrupt you and ask who is the taller, and you say, "bzzzt, wrong bucko, according to centuries of real analysis, your heights are the same", they will say something like this: "Then why don't you just use '1' and be done with the matter?", and probably conclude you are just a pedantic asshole, out to waste their time. mdf 13:23, 26 October 2006 (UTC)
:-) and how wrong they'd be eh?

Oh, I get it

If you let 0.999... = 1 - 0.0...1, then obviously this value 0.0...1 will tend to zero as the number of digits approaches infinity. At infinity, it actually is zero. And as everybody knows, 1 - 0 = 1. It's pretty much an iterative infinite series, non? Sockatume 22:24, 25 October 2006 (UTC)

Of course, that would mean that 0.0...1 would = 0 which it does not as all infitesimals have a non-zero value (otherwise there would be nothing that was slightly more than zero without being zero, or any reason to state 0.0...1). It's amazing to me the extent everyone gets worked up over a convention that is used to make up for the failings of a number system based on the number of fingers/toes an "normal" human being has. 194.75.129.200 22:42, 25 October 2006 (UTC)
Infinitesimals do not exist in the real number system. Sockatume is right: it is both meaningless and incorrect to claim that there is an ultimate digit in an infinite sequence of digits. Stebbins 23:32, 25 October 2006 (UTC)
Personally I have never got my head around people putting the word "infinity" next to "actually". Ansell 23:03, 25 October 2006 (UTC)
What I started thinking in this discussion, even if mathematically wrong, is this. Let's say .00...1 exists as in it is a way of saying 1/infinity. Now then .999... is presumably 1-(1/infinity). So does 1-(1/infinity)=1? To my surprise it seems it has to. For math purposes I'm going to use X instead of 1.
X-(X/infinity)=X moving things around we get
X=X+(X/infinity) multiply both sides by infinity
X*infinity=infinity, the other side becomes X+infinity
infinity=X+infinity. See article on infinity for confirmation that infinity does indeed equal X+infinity.

Granted this doesn't make mathematical sense, but for someone tempted to believe infinitesimals are real it kind of works.--T. Anthony 00:03, 26 October 2006 (UTC)

First of all, even if your formula is not making any mathematical sense, it's logically wrong, here I'll do an attemt to provide an alternate story:
\begin{matrix} X-\frac{X}{\infty} & = & X\\ X & = & X + \frac{X}{\infty}\\ X\cdot\infty & = & X\cdot\infty +\frac{X\cdot\infty}{\infty}\\ 1 & = & \frac{X\cdot\infty}{\infty}\\ 1 & = & X \end{matrix}→AzaToth 00:33, 26 October 2006 (UTC)

Even if we abandon mathematics, I still don't see why .000…1 should equal 1 - .999… and not, say, .999…1 - .999…. Melchoir 00:06, 26 October 2006 (UTC)

It could I suppose. It was just a way of showing why the infinitesimal makes no difference if one insists on believing in infinitesimals. I think you'd rather people just kind of "quit whining and accept mainstream math", but I was trying to show how the intuition I'd also had can be subverted. I'm conceding .999.. is 1. I figured the response would be "but that's not math and I don't like how you did it", but I wasn't really doing it for your benefit.--T. Anthony 00:12, 26 October 2006 (UTC)
I didn't get how that subverted my intuition. There isn't actually an intuition in maths IMO. The basics of maths are founded in what people think really happens, and everything else is constructed on top of that in a more of less sequential manner. Proof that the foundations exactly describe the real world because maths is successful doesn't lie well with me. Which is possibly why I do not inutitively believe (or disbelieve) that X+infinity \neq infinity. The concept of a "thing" which can be used in place of a number, but itself is bigger than any number is the first logical inequality for me, and the rest kind of pile on after that. The any bit is a self-contradiction of terms. Ansell 00:35, 26 October 2006 (UTC)
Oh well, it just made sense to me. I guess ultimately I should just go back to not buying the .999 thing, but I did try to buy it for awhile. Not that it matters, we've moved on to other FAs.--T. Anthony 00:38, 26 October 2006 (UTC)
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