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1 does not equal .999 ... (moved from article talk page)

The fallacy, clearly, is that the notation ‘...’ upon which the proof relies, is intended to merely represent an undefined infinite sequence. It is a notation, not a factor. But it is used in this proof as an integral part of the value of the factor being compared to one, as though it was a defined mathematical quantity.

But the quantity ‘infinite’ is not specifically defined, (nor, I would argue, is it mathmatical). The result is nonsense. Would you like to have a cent for every decimal position in which the .9 has not yet reached the equivalent of one?

Look: If 1 = .999..., then 1.000... = 0.999... . Then subtracting 1 from both sides, 0.000...0 = -.000...1. Since .000...0 does not equal .000...1, the proposition is false. Professor Krepotkin (talk) —Preceding comment was added at 20:09, 13 June 2008 (UTC)

You might be interested in reading about finitism, it's a philosophy of maths that rejects anything infinite (I don't know if finitists generally reject infinite decimal expansions, though - someone else here will probably know). If you're going to reject the idea of having an infinite number of 9's, then 0.999... simply doesn't exist, so indeed it doesn't equal 1. However, if you do that, decimal expansions become pretty limited. If you limit yourself to terminating decimals, you can only describe numbers of the form ${\frac {a}{10^{n}}}$ for some integers a and n, which almost all numbers aren't (you're restricting yourself to a very small subset of the rational numbers). The expression "0.000...1" is meaningless even if you do accept an infinite number of digits - you can't have a 1 at the end, since an infinite sequence doesn't have an end (at least, not without getting into transfinite ordinals, which we don't want to do!). 1 - 0.999... is simply zero. There isn't a non-zero number infinitely close to zero, which is what you are trying to describe. We call such numbers "infinitesimals", and the real numbers don't contain any (see Archimedean property for the technical details). --Tango (talk) 20:31, 13 June 2008 (UTC)

Tango, you are overestimating the importance of almost all numbers. Most numbers are just as interesting as average novel written by one of the monkeys on typewriter. Almost all numbers belong to Poincaré's teratologic museum. Finite decimal expansions are limited in a good way. Tlepp (talk) 09:09, 14 June 2008 (UTC)

Sure, almost all numbers are never going to be used by anyone, but the numbers which can be written as finite decimal expansions are an extremely small set of numbers, far smaller than is required by the term "almost all". They are a tiny subset of the rational numbers. Even if you don't care about irrational numbers, can you really argue that you don't need all rational numbers? --Tango (talk) 17:42, 14 June 2008 (UTC)

I don't need infinite decimal expansions for rational numbers. When I think about two thirds, I prefer 2/3 over '0.666...'. When I want to discuss roots of $x^{2}-2$, I prefer $\pm {\sqrt {2}}$ over '1.414...'. For the circumference of a unit circle it's better to write $2\pi$ than '6.283...'. Three dots notation is rarely the best notation. How many different ways are there to assign 2.999... pigeons into 4.999... pigeonholes? Tlepp (talk) 18:15, 14 June 2008 (UTC)

I believe you said that Tango was "overestimating the importance of almost all numbers", not the decimal expansion of numbers. You appeared to have suggested that such numbers as $\pm {\sqrt {2}}$ and 2/3 are actually unimportant, however you notate them. In other words, you have started, unintentionally, perhaps, a semantical battle. Perhaps we should put that to rest before it gets to be a flaming war? ;) I will attempt to address what appears to me to be the actual issue at stake.

It is unavoidable that, in order to describe the real number set, an infinite process must occur. Some such infinite representations of the real numbers are Cauchy sequences and Dedekind cuts. Decimal expansions are another representation of the real numbers (see Construction of real numbers). If you don't accept the existence or usefulness of infinite decimal expansions, then you can hardly suggest that any of these other representations are much better, and by extension, you can't accept the existence of the real number set. In short, you shouldn't throw away infinite processes so quickly. --70.124.85.24 (talk) 19:48, 14 June 2008 (UTC)

In other words, you want to just ditch decimal expansions altogether? The good thing about decimal expansions is that, while they are probably never the best way to write any given real number, they can be used to write *any* real number. It's useful to have one way to write everything, since it allows direct comparisons. For example, which is bigger, ${\sqrt {10}}$ or $\pi$? It's difficult to tell without giving it some thought. Now, which is bigger, 3.162278... or 3.141593...? Now, it's easy. --Tango (talk) 21:44, 14 June 2008 (UTC)

Which is bigger, ${\sqrt {10}}$ or $\pi$? ${\sqrt {10}}\approx 3.16>3.14\approx \pi$. No need for three dots. I agree equalities look prettier than approximations, but the added beauty is an illusion with little practical value.

Important numbers are those that can be described with a finite number of characters, and it's a countable set. The real number set is uncountable. Almost all numbers (99.999...%) in our favorite number system are anonymous and can't be described.

99.999...% = 100% $\neq$ all

This is one of the few situations were '99.999...' seems better that shorter '100'. Somehow '99.999...%' notation seems to say a little more than it really does. Since 99.999...% equals 100% exactly, there's is no hint whatsoever that some meager set of numbers is not included in the 99.999...%. Tlepp (talk) 07:53, 15 June 2008 (UTC)

You need a better notation - $\approx$ doesn't tell you how close it is. If it's +/- 10%, say, then the inequality doesn't follow. You need something to specify that it's +/- 5 in the next decimal place, that isn't a standard meaning of $\approx$, to my knowledge. You also need a better notation than 99.999...%, it's better to just say almost all. It has a precise meaning that can't really be expressed in terms of percentages. 99.999...% of real numbers are not between 0 and 1, but not almost all. --Tango (talk) 14:37, 15 June 2008 (UTC)

Careless, careless percentages. Firstly, 100% = Almost all =/= all, infinite repeating nines never entering into it. Secondly, what kind of probablility measure are you using to talk about percentages of real numbers? Thirdly, let's forget about the utility of various sets of measure zero since the 'Professor' doesn't discuss anything more than a misunderstanding of simple decimal notation. Endomorphic (talk) 13:35, 16 June 2008 (UTC)

Not quite. We are not discussing probability measures, but just measures (the Lebesgue measure, say). 100% (or just a proportion of 1) would mean, if anything (though I haven't seen it written), ${\frac {\mu (A^{c})}{\mu (X)}}=0$. Thus $[0,1]^{c}$ contains 100% of the real numbers. Almost all means $\mu (A^{c})=0$. Thus $[0,1]^{c}$ does not contain almost all real numbers. -- Meni Rosenfeld (talk) 13:55, 16 June 2008 (UTC)

I take that back - specifically in real numbers, the most natural meaning of 100% would be $\lim _{x\to \infty }{\frac {\mu ([-x,x]\cap A)}{\mu ([-x,x])}}=1$. The rest remains unchanged. -- Meni Rosenfeld (talk) 15:16, 16 June 2008 (UTC)

You're right, it is a careless use of percentages - I explicitly said the percentages can't really be used to describe these things. But, if you really want to use them, then any reasonable way of doing so wouldn't work the way Tlepp wants it to. --Tango (talk) 15:04, 16 June 2008 (UTC)

The discussions almost all seem to be centered around mathematical theorems and proofs, yet whether 0.999... is 1 or not could be argued to be fundamental, and, as such, could not be proven mathematically, but dictated by logic/philosophy. If logic/philosophy dictates that 0.999... is not equal to one, then any theorem that 'proves' it does is wrong, rather than vice-versa. In terms of logic and philosophy, 0.999... can be very strongly argued to be distinct from 1. However, it is important to remember that mathematics does not really exist; it is a human construction to be a useful tool. As such, it may be useful to define 0.999... as being equal to one, even if it is distinct in terms of logic/philosophy. The sum of an infinite series is defined as its limit; it is useful for that to be so, and it is, for practical purposes, entirely accurate, even if it is philosophical incorrect. Hence the argument over whether it is philosophically correct or not is entirely mute. Captain Griffen (talk) 12:37, 20 July 2008 (UTC)

What alternative definition would you give for 0.999..., then? Oli Filth^(talk) 12:57, 20 July 2008 (UTC)

It's not about definitions of 0.999..., but whether infinitely close is the same as identical. Logically, it isn't, but practically, it is. Maths goes with the practical definition here, but that's arbitary, and doesn't mean that it applies outside of maths, nor does it mean that 0.999... == 1, only that they are taken as being the same. My point is that this debate cannot be argued from a mathematical stand point, as it is to do with something fundamental (ie: is infinitely close the same as identical). Captain Griffen (talk) 13:25, 20 July 2008 (UTC)

The question of whether 0.999... = 1 depends entirely on how one defines the symbol 0.999.... Without a definition, the question is meaningless (in the same way that asking whether k?37##!8 = 1 is meaningless). If one chooses to define the symbol 0.999... as the limit of the infinite series (i.e. the standard definition), then by definition, it will equal 1. If one chooses to apply an alternative definition, then a different result may arise (albeit a non-standard one). However, without such an alternative definition, this direction of debate is all but moot. Oli Filth^(talk) 16:08, 20 July 2008 (UTC)

All philosophical questions are meaningless. /sarcasm

Gian-Carlo Rota has criticized analytic philosophy with following words:

Whereas mathematics starts with a definition, philosophy ends with a definition. A Clear statement of what it is we are talking about is not only missing in philosophy, such a statement would be the instant end of all philosophy. - - The prejudice that a concept must be precisely defined in order to be meaningful, or that an argument must be precisely stated in order to make sense, is one of the most insidious of the twentieth century.

Captain Griffen's point should be interpreted in philosophical context. Tlepp (talk) 20:41, 20 July 2008 (UTC)

That might be relevant if we didn't already have a clear definition! If we go back to philosophy, then according to you, we work toward a definition, but we already have one! Why go backward? And more pertinently, where would we go back to? Oli Filth^(talk) 20:57, 20 July 2008 (UTC)

Why do math students bother with (verified) proofs? Why don't we just learn the theorems and formulas? Tlepp (talk) 21:20, 20 July 2008 (UTC)

No one I've ever spoken to has ever taken '0.999...' as anything except 0.9 recurring, which even the wiki article agrees with. 0.999... may be taken in maths generally to be the limit, but that does not mean it is defined as that. Hence it can be broken down to whether the infinitely small gap between 0.9 recurring and 1 should be considered, given it is infinitely small; this is to do with the philosophy or fundamentals of maths, and cannot be resolved by mathematical theorems, since mathematical theorems must be based on those same principles or fundamentals. If a theorem disagrees with the underlying philosophy, then the theorem is falsified - not vice versa. Mathematics is a tool, however, and so assumptions are made where they are useful (eg: 0.9 recurring being equal to 1, even if it is a philosophically contentious arguement). The reasoning behind 0.9 recurring being one in much of this debate is flawed for these reasons. Captain Griffen (talk) 21:13, 20 July 2008 (UTC)

According to the Wikipedia article on this, 0.999... is the same as 0.9 recurring, ie: 0.999999 to infinity. That disagrees with your assertion of what the standard definition is (in my work with mathematics, I've never been told that is the definition of '...', and it would be in direct contrast with the use in normal English). It is taken as being the limit, and it is for all practical purposes, but it is not [i]defined[/i] as being the limit, in my view (if it is, then that wiki article needs updating, as it is alledging 0.999... is synonomous with 0.9 recurring, which you are saying is not the case; 0.9 recurring is not the same as the limit of 0.9 recurring). Captain Griffen (talk) 20:24, 20 July 2008 (UTC)

The article is fine. 0.999... is the same as 0.9 recurring, and 0.9 recurring means by definition the infinite sum 0.9 + 0.09 + 0.009 + ..., which by definition means the limit of the sequence of finite partial sums 0.9, 0.99, 0.999 ... . I don't know what you mean by "0.9 recurring is not the same as the limit of 0.9 recurring" - the "limit of 0.9 recurring" is a meaningless statement AFAICS. Mdwh (talk) 00:01, 22 July 2008 (UTC)

You're right, it is a meaningless statement. I believe the article discusses the misconception that 0.999... refers to a process ("It gets closer and closer to one but never gets there" and similar). It is not, it is the limit of a process. --Tango (talk) 00:51, 22 July 2008 (UTC)

It's not really a matter of defining 0.999... to equal one, it's about defining the real numbers. If you take the standard definition of the real numbers (which turns out to be an extremely useful definition for all kinds of things, hence it being standard), then any reasonable definition of 0.999... as a real number will be equal to one. You either have to not have 0.999... as a real number (so it's basically just a meaningless sequence of symbols), or you have to change the definition of a real number (which would make maths useless for describing much of the real world). --Tango (talk) 17:14, 20 July 2008 (UTC)

The definition of real numbers (a useful tool) used in maths has no relevence to the actuality of it. What I'm saying is that arguing from maths about the basis of maths doesn't make any sense. Captain Griffen (talk) 20:24, 20 July 2008 (UTC)

I will give Captain Griffen and Tlepp this much: Yes, it's true that there are nontrivial philosophical issues involved in saying that the real numbers are the "correct" realization of the intuition of continuity, and it's true that the resolution of these issues is not simply a matter of mathematical proof in the ordinary sense, nor about stipulating a formal definition and seeing where it leads. It's also true that you don't have to have a precise definition before you can discuss something.

However: These issues have been resolved; there is a broad consensus to take the reals as, by default, the structure that realizes the intuition of a continuous line. This is for good reason, though the reasons are not as easily explained as the proofs in the article. The resolution is not the of sort of once-and-for-all definitive type that you get from a mathematical proof; in principle the question could be revisited if some genuinely new way of looking at things came to light. There is, however, no evidence of that here. And while you don't need a precise definition before you start, the continued failure to find a relevant way of making the doubters' ideas precise (or even minimally clear) is pretty strong evidence that they are not likely to be fruitful. --Trovatore (talk) 23:43, 20 July 2008 (UTC)

(Aside) When I wrote the above, I was under the impression that Tlepp was arguing on Captain Griffen's side. A more careful reading shows this is not true. --Trovatore (talk) 00:12, 21 July 2008 (UTC)

You almost seem to be agreeing that the way this is dealt with is not down to mathematical proof, but convention based on what is useful (which is my point), based on a contentious logical point (mainly that infinitely small = 0). This is useful, as it allows stuff like digit manipulation without maths breaking apart. (For example, the digit manipulation 'proof' assumes a 1:1 infinite-digit mapping after the decimal point, which isn't necessarily valid; without such an assumption the result could be used to argue subtraction of infinite length don't have any real meaning.) I agree in mathematics we should take 0.999... to be equal to 1, but purely because it is useful and as part of other assumptions made in maths, rather than any underlying logical reason. The same digit manipulation 'proof' can be used as good evidence of techniques which become valid thanks to that assumption. Captain Griffen (talk) 10:04, 21 July 2008 (UTC)

Wrong. Whether there are infinitely small non-zero numbers (i.e., infinitesimals) or not is simply irrelevant. In at least the common theories of non-standard reals, we have

• There exist numbers x such that 0 < x < 1/n for any natural number n.
• 0.999... = 1.

You're off on a red herring. The fact that there are no infinitely small non-zero distances in the standard reals is irrelevant to this proof, since there are theories in which such infinitesimals exist and 0.999... = 1.

To be sure, the section on infinitesimals isn't so clearly written. A few non-standard systems are mentioned, but for the most part, the article does not say whether 0.999... = 1 in these systems. Perhaps someone who knows this stuff better than me can fix that section? Phiwum (talk) 12:08, 21 July 2008 (UTC)

I'm not arguing that 0.999... isn't = 1, but that the arguments used to 'prove' it are erroneous when based purely on mathematical theorems. The article is fine to start with, particularly clear with "As a result, in conventional mathematical usage, the value assigned to the notation "0.999…" is the real number which is the limit of the convergent sequence (0.9, 0.99, 0.999, 0.9999, …)." - it is clear it is [i]assigned[/i] the value one as the limit by convention. Other parts, particularly in the 'proofs' are more confused on this (many of which are fatally flawed as proofs, particularly the 1/3 and digit manipulation based 'proofs', but others as well, given that they rely on assumptions that may or may not be considered valid). Speaking of the 1/3 based one, why is it even there, given that it assumes 1/3 == 0.333..., which is an assumption denied by anyone who does not accept 0.999... == 1. Captain Griffen (talk) 15:15, 21 July 2008 (UTC)

That's not a problem with 0.999..., it's a problem with all infinite decimals. An infinite decimal is defined as the limit of the sequence, it's the only reasonable definition. You could define it some other way, but you wouldn't have a useful notation (it wouldn't represent a real number). It's all just notation, so you're right that there is no underlying logic to it. The notation is defined to mean what it's most useful for it to mean, in this case, that results in 0.999... being an alternative notation for one. --Tango (talk) 16:00, 21 July 2008 (UTC)

I'd agree with all of that, except rather than the only reasonable [i]definition[/i], I'd say the only reasonable way to handle it (the difference in that being 3! is defined as 3 times 2 times 1, but has a value of 6; six can replace 3!, but 3! is not defined as 6). The arguments have mostly tended to revolve around mathematical theorems rather than arguing from maths-as-a-tool, including the wiki article, which, I'd say, is less effective in informing, and less accurate in explaining and showing the reasons behind it. Captain Griffen (talk) 23:27, 21 July 2008 (UTC)

Captain Griffen, you may be surprised just how many mathematical novices accept that 0.333... = 1/3 and doubt that 0.999... = 1, at least until they are shown that the former implies the latter. Of course, one cannot consistently believe that 0.333... = 1/3 and 0.999... = 1 (along with other basic facts about addition), but the fact is that, for whatever reason, the former is often intuitively regarded as less controversial than the latter. Phiwum (talk) 17:40, 21 July 2008 (UTC)

I agree with Phiwum (a few posts above) that the article is sadly weak in its coverage of decimal expansions in non-Archimdean fields. Back in the day, I did some ad-hoc browsing in a couple research libraries and Internet searches, but I couldn't find anything more solid than what you see. I'm not an expert in that area, so it's entirely possible that I just didn't know where to look. You might want to bring this up on the main talk page, where it might get more positive attention. Melchoir (talk) 22:27, 21 July 2008 (UTC)

I don't know of any other number system that uses decimal expansions, at least not routinely. You can use a similar system to describe p-adic numbers, but I don't think 0.999... would have any meaning there, I believe you can only have infinite expansions to the left, not the right, with p-adic numbers (the opposite of with real numbers). --Tango (talk) 00:51, 22 July 2008 (UTC)

The need for infinitesimals in the real world

THis discussion has an irrational fear of infinitesimals. But it is easy to come up with real world situations where infinitesimals are necessary. Consider the simple example of a tire (circle) on a road (line) In the real world, two solid objects cannot occupy the same space. So if the surface of the road is point 0, then how would you describe the surface of the tire at it's bottom point? The tire's vertical volume would be x>0, but if there is no "lowest value" to that equation, then the tire has no bottom edge - clearly an absurd failure to describe the situation. Thus values like .000...1 and .999... must exist in the real world, if not in real numbers. Algr (talk) 21:30, 13 June 2008 (UTC)

Considering solid objects is not really helpful here. In the real world, at the sort of microscales relevant to this article, the atoms of both the tire and the road are not at all "solid", and are best described by wave functions, and not only is interpenetration possible but the whole idea of "bottom edge" is also meaningless. Indeed, as the scale reduces further below the Planck distance, even ideas like time and space become difficult to reason about, and the actual nature of physical reality at those scales -- including whether or not spacetime is actually a continuum -- is as yet unknown. And all of this may happen no more than 30 or 40 places to the right of the decimal place: only a tiny way along the digits of 0.999...

Fortunately, none of this matters when we are discussing the reals, which we can reason about regardless of whether or not they actually describe physical reality at those scales. -- The Anome (talk) 22:00, 13 June 2008 (UTC)

If the real numbers have no connection to the real world, then why are we talking about them? Why not debate Star Trek transporters or which is the first Pokimon? In your effort to define yourself into correctness you've turned the whole Real Set into a useless fantasy that can't even deal with something as simple as a tire on a road. Algr (talk) 23:28, 13 June 2008 (UTC)

(Pure) Mathematics does not attempt to describe the real world. Scientists use mathematics to describe the real world, and that's why mathematicians get funding, but it's not why they come up with the maths. They do it for the sake of knowledge and beauty - it's all abstract. We don't need a real world application in order to consider it worthwhile to talk about maths. That said, the real numbers are very useful for describing the distance between a tire and a road - that distance is a non-zero finite real number. The electrons in the tire and the electrons in the road repel each other, so they are always separated by at least a small amount. If you want to ignore that aspect of the real world and consider an idealised tire and road, then you always need to ignore the bit about two objects not being able to occupy the same space, and the distance is just zero. --Tango (talk) 23:53, 13 June 2008 (UTC)

When the tire's lowest point is 0, the intersection of circle and line is a single point, which has volume 0, so there's no volume shared by tire and road. If the tire's lowest point were some infinitesimal, which infinitesimal would it be? What's at the point in between 0 and the tire's lowest point? Would there be an infinitesimal air-filled space between the tire and the road? Would they actually touch? Huon (talk) 00:12, 14 June 2008 (UTC)

The tire in your example doesn't "touch" the road. There is an electromagnetic field separating them. The distance is measurable. Tparameter (talk) 00:22, 14 June 2008 (UTC)

If one is more interested in philosophy than mathematics, http://plato.stanford.edu/entries/boundary/ is a better starting place than this article. Melchoir (talk) 00:46, 14 June 2008 (UTC)

Absolutely. Physical objects are not describable as point sets, made of continuous smooth "stuff" with clear boundaries, and it's not clear how to define these boundaries even if you could do so, so all arguments relating infinitesimals to physical objects are unfortunately irrelevant. Arguments based on common-sense understandings of reality sadly evaporate in the face of the weirdness of actual physical reality. Different physical regimes apply at different scales and energies, and since there is no practical reason to develop intuitive understandings of the universe as it behaves at large or small energies or scales (they're either irrelevant to ordinary life, inaccessible, or you're dead from heat/cold/vacuum etc.) evolutionary selection processes didn't bother to equip us with the necessary intuitions. -- The Anome (talk) 13:29, 14 June 2008 (UTC)

Off-topic but somehow related: intuitive vs formal continuity http://www.maa.org/devlin/devlin_11_06.html Tlepp (talk) 16:46, 14 June 2008 (UTC)

Nice article - thanks for the link. Now I don't feel quite so bad about not getting epsilon-delta proofs of continuity the first time they were explained to me! --Tango (talk) 17:50, 14 June 2008 (UTC)

Odd. I don't recall having any trouble with this back in the day. I guess either I have suppressed the memories of the difficult struggle, or they were better explained to me. I do agree, though, that the idea that "a continuous function is one that can be drawn with a single stroke of a quill" is a blatant lie - the Weierstrass function cannot be so drawn. -- Meni Rosenfeld (talk) 20:11, 14 June 2008 (UTC)

Better explained may well be it - I didn't get it when the lecturer went through it, but once my tutor did an example explaining it slightly differently, it all clicked into place (on second thoughts... I wonder if I was entirely paying attention the first time...). It gets worse when you get into general topology, though - "the pre-image of an open set is open". What on Earth has that got to do with anything? It works, and does everything you want of it, but it doesn't bear any relation to drawing lines on a pieces of paper without lifting the pen... For a start, it's talking about the inverse of the function, rather than the function itself, which is just weird... I think you just have to accept it as an arbitrary definition and get on with it... --Tango (talk) 21:31, 14 June 2008 (UTC)

Somewhat amusing to see point-set topology applied to the PA/MD border. But in any case the article seems to have made one fairly elementary topological blunder in claiming that the second realist theory requires one of the touching bodies to be topologically open and one of them closed -- why shouldn't the boundary be partitioned between them in some much more complicated way? --Trovatore (talk) 03:42, 14 June 2008 (UTC)

In Prof. Varzi's defense, he does say "These theories ... need not be exhaustive and can be further articulated...". It may simply be that partitioning the boundary is an unpopular option. And I can see a possible reason why: to partition a line in a fair and uniform manner you'd need to introduce non-measurable sets. For me, this is not much better than saying that a given boundary point might be assigned to one body or to the other, but we have no way of deciding which! And if you're going to allow that kind of indeterminacy then you might as well do away with the partition and let the whole line lie in limbo, at which point we've arrived full circle back at option 2.1.2 and therefore throw our hands up in the air and go get drunk. Melchoir (talk) 05:07, 14 June 2008 (UTC)

Actually I just thought of another argument against partitions: they're no use if the boundary in question is a single point. Which state does the center of the Four Corners Monument belong to?

(Nothing personal; it's obviously an interesting question...) Melchoir (talk) 05:24, 14 June 2008 (UTC)

Well, one of these things is not like the other. I've lived in New Mexico, I've lived in Colorado, I've lived in Arizona, but I've never lived in Utah. So obviously it belongs to Utah. --Trovatore (talk) 08:13, 14 June 2008 (UTC)

Ah yes, Brentano's classic 1903 Utahn Non-Residential Symmetry Breaking Argument. Melchoir (talk) 19:08, 14 June 2008 (UTC)

But what seperates the interior from it's own boundry? Endomorphic (talk) 14:26, 16 June 2008 (UTC)

Nothing. You can get arbitrarily close to the boundary while remaining in the interior. There is, however, a difference between "arbitrarily close" and "infinitesimally close". Any point in the interior is a finite distance from the boundary. --Tango (talk) 15:01, 16 June 2008 (UTC)

That's the mathematics of it, yes. But the link discusses the philosophy, a boundary as an ontological neccessity coming between distinct objects. Boundaries are things that seperate things; but if a boundary is a thing then there must be some thing seperating the boundary from each thing it seperates. And so on. /end{Devil's Advocate} Endomorphic (talk) 15:46, 16 June 2008 (UTC)

THis discussion has an irrational fear of modulo arithmetic. But it is easy to come up with real world situations where modulo arithmetic is necessary. Consider the simple example of a tire (circle) on a road (line) In the real world, when the wheel has rotated one whole revolution, it's in exactly the same orientation as when it had not rotated at all. Thus equations like 1=0 must hold in the real world, if not in the real numbers. Endomorphic (talk) 14:26, 16 June 2008 (UTC)

What does modulo arithmetic have to do with anything? Different situations in the real world need to be modelled using different mathematical objects. Distances are generally modelled using normal real numbers, angles are modelled using real numbers modulo 2*pi. That different things need different models doesn't make the models wrong. --Tango (talk) 15:01, 16 June 2008 (UTC)

Endomorphic's post is a parody of Algr's, designed to deliver roughly this message. -- Meni Rosenfeld (talk) 15:10, 16 June 2008 (UTC)

We've had this discussion before. 1=0 is correct if your variable is exclusively measuring the angle of the tire; Like Tango says; different things need different models. Hyperreals are inevitable as soon as you have inclusive and exclusive ranges. Algr (talk) 01:40, 14 July 2008 (UTC)

What would you model using hyperreals? Please give a specific example, including explicitly what hyperreal you would use. --Tango (talk) 01:57, 14 July 2008 (UTC)

Gee, I dunno, how about a tire on a road? Has anyone brought that example up? Algr (talk) 03:50, 14 July 2008 (UTC)

I asked for an explicit hyperreal that you would use for that scenario. --Tango (talk) 16:29, 14 July 2008 (UTC)

Hyperreals won't really solve ontological or epistemological problems of border. Hyperreal numbers are points on the hyperreal line and point is completely characterized by it's location. No two points can occupy same location. Two distinct hyperreal numbers can be infinitely close to each other, yet there's always a positive distance between them. If the border of a road and a tire isn't allowed to intersect, we can make their distance smaller than any positive rational number, but we can't make it zero. Thus the tire is not on the road, it is flying above the road. If an infinitesimally small bug is sitting on the road, you can drive(=fly) over it, and it won't get crushed. Tlepp (talk) 07:32, 14 July 2008 (UTC)

Tlepp is right, of course. To be a little more explicit: At the top of this section, you claimed the reals were not sufficient to describe a tire on a road. Let me quote the relevant section:

In the real world, two solid objects cannot occupy the same space. So if the surface of the road is point 0, then how would you describe the surface of the tire at it's bottom point? The tire's vertical volume would be x>0, but if there is no "lowest value" to that equation, then the tire has no bottom edge - clearly an absurd failure to describe the situation.

So let's try to describe the situation with hyperreal numbers. Again the surface of the road is point 0, which happens to be a well-defined hyprerreal number. But as with the real numbers, the hyperreals don't offer a "lowest value" to the (in)equation x>0, and in the hyperreals, the tire still has no bottom edge if you want it to be the smallest solution to x>0. What has been gained by using the hyperreals? Nothing. Huon (talk) 12:35, 14 July 2008 (UTC)

Well, to be fair, his point wasn't that the tire be 'as close as possible' to the road, but just 'infinitely close'. But it's still a ridiculous example, since, in the real world, objects are never infinitely close (at least not in the everyday world. I believe it's been suggested that inside a black hole, all particles are literally occupying the same point in space, which constitutes being infinitely close together). --69.91.95.139 (talk) 14:02, 14 July 2008 (UTC)

This may be a misunderstanding. Tlepp talks about things being infinitely close, Algr explicitly speaks of the "lowest value" to x>0, and claims that hyperreals would somehow help. They don't. Huon (talk) 15:26, 14 July 2008 (UTC)

Sorry, you're right. And I was referring to Algr's example when I said 'his point'. Incorrectly so, I'll admit. --69.91.95.139 (talk) 15:42, 14 July 2008 (UTC)

2

"Any point in the interior is a finite distance from the boundary." - Does this mean that a point on the boundary is not within either area? In other words, any point on a state border is not within any state, even though it is within the interior of the USA? Algr (talk) 09:45, 18 August 2008 (UTC)

In mathematics, a point on the boundary is not in the interior of either the set or its complement. It may (or may not) be an element of the set, but the interior of a set is characterized by each element having an open neighborhood entirely within that set. I doubt that point-set topology is the best model of real-world geography and politics, though - where sets of measure zero (such as boundaries) belong to is rather irrelevant in those contexts. Huon (talk) 11:49, 18 August 2008 (UTC)

If you're feeling up to it, you're welcome to take a crack at my argument in regards to "actual infinite sum" vs "limit of a sum to infinity", found at Talk:0.999...#Alternative_proof, and reposted here. I really look forward to finding out what you have to say about this.

In regards to what the underlying definitions of real numbers are supposed to represent conceptually and intuitively:

Ok, let's start over. Your first sentence was:

Both the ${\sqrt {2}}$ and the .333... proofs fail because they both make unjustified assumptions about what happens when a process is repeated infinitely.

So, can you explain how exactly a justified assumption about infinite processes can be made? If so, then please do tell. Otherwise, you're just going to have to face the facts: in mathematics, infinity is a key concept, and sometimes, assumptions about infinity are necessary.

In answer to your entire paragraph, however, the definition of real numbers is not the result of "repeating a process infinitely" (and therefore your argument about repeating a process infinitely is irrelevant); common definitions follow one of two approaches to defining the reals: A) as a limit (Cauchy Sequences) or B) as a gap (Dedekind Cuts). If you considered each term in a Cauchy Sequence and repeated the process to infinity, of course we have no idea what you would ultimately get. However, that is not what Cauchy Sequences represent. It is not what we get after the infinite process (which is more a matter of philosophical speculation than anything else) that they represent, but what we get when we take the limit to infinity - a well defined mathematical concept, out of which ultimately pops your disputed Archimedean property. To see how this applies to your argument above, notice that 0.333... is actually a Cauchy Sequence.

The Dedekind cut is similar, though perhaps a little easier to wrap one's mind around; none of the numbers in the cut actually represents the real number; it is the gap left by those numbers that represent a real.

I'm sure you will have a well-prepared and thought through response to my answer. You don't have to worry about addressing my first question, as I have rendered it irrelevant. There; I've reduced your workload a bit, haven't I? Of course, ∞-1 is still ∞, so... --69.91.95.139 (talk) 20:19, 14 July 2008 (UTC)

I'm waiting (but don't worry; I'm not holding my breath). --69.91.95.139 (talk) 14:18, 24 August 2008 (UTC)

Sorry for the wait, calculus wasn't built in a day. (And I do have a life outside of .999...) Algr (talk) 20:30, 25 August 2008 (UTC)

No need to apologize; take your time. --69.91.95.139 (talk) 00:28, 26 August 2008 (UTC)

Inaccuracy of Decimals

Personally, I think that this entire article describes the inability of decimals to represent certain fractions. You can't "start" at ".999...". You have to reach it doing other maths, and the way that happens is when you are converting fractions to decimals. All these proofs, arguments, counter-arguments, etc., are moot when you consider that if you had just stuck with the fraction notation you would not have this situation. Basically what I am saying is: .999... = 1 not by these proofs, but by the fact that you only arrived at ".999..." because you did 1/3 + 1/3 + 1/3 in decimal form (i.e. .333... + .333... + .333...). I know I can't be the only person to think of this, especially when you consider all the brilliant minds involved. --MadDawg2552 (talk) 14:24, 26 July 2008 (UTC)

I see what you're saying. Fractions are usually a far better way of representing rational numbers precisely than decimals, however decimals can also be used for irrational numbers and for approximations of numbers to a desired degree of accuracy, which makes them very useful. So, that's why we define decimals the way we do. Once you do that, you then want to try and make sense of every possible decimal expansion, including 0.999... . That doesn't really come from 1/3+1/3+1/3, it comes from 9/10+9/100+9/1000+... . It doesn't generally come up naturally, so in most cases it doesn't really matter what it equals, but if you want to be able to talk about all decimal expansions at once you need to be able to interpret all of them, even the ones that never really come up. Does any of that make sense? --Tango (talk) 15:09, 26 July 2008 (UTC)

While I was at the employment agency just now, twiddling my thumbs, I came to the realization that 0.999... is the limit as i approaches infinity of the following equation:

$9\times {\sum _{n=1}^{i}{10^{-n}}}\equiv 1-10^{-i}$

i	value	equivalence
1	0.9	1 − 0.1
2	0.99	1 − 0.01
3	0.999	1 − 0.001
∞	0.999999...	1 − 10-∞ == 1?

It easily rounds to but is not exactly unity. Interestingly, this appears to also indicate that $0.111...=\sum _{n=1}^{\infty }{10^{-n}}\equiv {1 \over 9}-{10^{-\infty } \over 9}$, is infinitesimally less than 1/9. Has no one else bothered to state the problem in this way? D. F. Schmidt (talk) 21:57, 28 July 2008 (UTC)

Lots of people have looked at the problem that way. Your mistake is in assuming that 10^-∞ is greater than zero - it isn't, it's exactly zero. See Archimedean property. --Tango (talk) 22:13, 28 July 2008 (UTC)

More precisely, the limit of 10^-x as x goes to ∞ is exactly 0. 10^-∞ is undefined in the real numbers, because ∞ is not a real number.

Of course, that doesn't blow the issue out of the water; it only redirects the problem: should decimal expansions really refer to the limit of their increasingly accurate finite expansions? The answer to that is, naturally, a definitive yes. The reason for that has a long history, which can be summarized by saying that the real numbers have turned out to be the most useful number sets throughout the ages, in both applied and abstract mathematics. And, within the real numbers, decimal expansions are most meaningful as a limit. --69.91.95.139 (talk) 23:00, 28 July 2008 (UTC)

Also, there really isn't an alternative definition. How else could you interpret them that would be even the slightest bit meaningful? --Tango (talk) 00:34, 29 July 2008 (UTC)

Yes, that's pretty much the gist of it. --69.91.95.139 (talk) 00:59, 29 July 2008 (UTC)

Well call me proud, prejudiced, or what-have-you, but none of the proofs given in the article are as succinct and definitive as the one I posted yesterday. The Cauchy sequence in the article right now has an introduction that's difficult to follow (for me, at least) -- and in fact such an introduction wouldn't have to be given if this other proof was given in the article, along with the simple table I constructed. The digit manipulation proof is a little awkward too, in my opinion, and I wouldn't be surprised if anyone that exercises logic would reject it. But anyways, my limited understanding of math would indicate that:

1. Practitioners of mathematics usually tend to prefer infinite precision, such that they prefer symbols (π, e, etc.) – and I don't think that's only for ease of use. (Scientists and engineers – applying science – might prefer symbols for the sake of pragma, but not mathematicians. Likewise, object-oriented programmers understand that even though two or more object referents may have the same value, it doesn't immediately follow that the two referents are the same object.)
2. All scientists (though not necessarily engineers) including mathematicians (as one might suppose) prefer to use precise language to the point of saying (as I said above) that $9\sum _{n=1}^{i}{10^{-n}}+10^{-i}\equiv 1$. If this wasn't so, we wouldn't have "divide-by-zero" problems, right? -- We might just as easily say that it is generically and absolutely true that $0^{-1}\equiv \infty$. As with all other sciences, assumptions must be declared -- especially when a "proof" is in question.

Ultimately, (in of course my own opinion) the proof that "0.999... = 1" could use my quite simple approach, and the only assumption to be declared is that the limit as i approaches infinity, 10^-i = 0. No need for any other clunky explanations like the one found under "fractions" or "digit manipulation" (which looks to me like garbage anyways, to be perfectly honest stating -- as a matter of course -- my own opinion). D. F. Schmidt (talk) 23:27, 29 July 2008 (UTC)

For a mathematician, you're absolutely right, that's the based way to look at it, however for the layman, understanding that the limit is zero rather than some positive infinitesimal is difficult - that's basically the Archimedean property. Once you accept the Archimedean property, the whole problem basically disappears, there's nothing really left to prove, but it's a difficult property to prove because it requires technical details about how the real numbers are defined. The digit manipulations proofs, while not wholly convincing (they are valid, but it takes some effort to prove it), are possible to understand with only a knowledge of basic arithmetic. --Tango (talk) 01:21, 30 July 2008 (UTC)

I would like to add one more remark on this topic: The precision of the equation I rendered above is absolute, to infinity. When i is 1, we can say that -- to infinite precision (which the mathematician prefers) -- the number of significant 9's is 1. When i is infinity, we can say that -- to infinite precision -- the number of significant 9's is infinite. Even when the number of 9's is infinite, there is one small (infinitesimal) value by which it must be increased in order to equate it with 1. I contend that to say otherwise is to say that -- to infinite precision -- ∞^-1 = 0. Remember that (from what I can tell) the Archimedean property is based on heuristics (which see).

However, when the precision is dropped from infinity, it is immediately rounded to 1. But mathematicians don't like to round numbers where they can avoid it. Am I right? D. F. Schmidt (talk) 01:31, 1 August 2008 (UTC)

No, you're wrong. The Archimedean Property is not based on heuristics, it's an absolute theorem. It's implied by the definition of the real numbers. ∞^-1 really is zero (if it's defined at all), see Extended real numbers#Arithmetic operations. --Tango (talk) 01:55, 1 August 2008 (UTC)

Am I wrong? What part of Archimedean property#Archimedean property of the real numbers is not heuristic? Quoting from that section, "Is c itself an infinitesimal? If so..." And if not? D. F. Schmidt (talk) 03:37, 1 August 2008 (UTC)

The first paragraph of that section is a mathematical proof of the fact that the real numbers are Archimedean.

At one point in the proof, a certain number c is constructed. The goal is to prove that c = 0. The rest of the proof then proceeds by contradiction. It is assumed (by contradiction) that c is positive. First, in the case that c is infinitesimal, a contradiction is reached. Second, in the case that c is not infinitesimal, another contradiction is reached. This last case is the "if not" you refer to. Melchoir (talk) 04:28, 1 August 2008 (UTC)

I don't understand how the fact that 2c > c contradicts the fact that c is a least upper bound for Z. That seems to me like a counterpart statement is that 2 > 1 means my head will implode in 5 seconds. The rest of the "proof" is not at all edifying to me except to indicate to me that one's words are somehow heavier than another's. I can't comprehend how the sum of all infinitesimals could be any less than infinity. After all, what is differentiation but the study of infinitesimals?! What am I missing here? D. F. Schmidt (talk) 05:51, 1 August 2008 (UTC)

If c is a positive infinitesimal, then 2c>c is a positive infinitesimal, too. Then c is no upper bound for the set Z of positive infinitesimals, because there's an element of Z larger than c. Huon (talk) 10:35, 1 August 2008 (UTC)

Ok, I guess I can understand the explanation now, but for the sake of other people who might have my same problem comprehending this, I guess it'd be like saying since 2c is a positive infinitesimal then it too belongs in the set Z, which increases the size of Z and with it, the lowest upper bound. Still: what's the rationale that the least upper bound for all infinitesimals is any less than infinity? After all, thinking this way would lead me to conclude that you believe that the least upper bound for all real numbers is less than infinity. D. F. Schmidt (talk) 18:10, 1 August 2008 (UTC)

By definition, all infinitesimals are less than all finite numbers. That's what it means to be infinitesimal (the technical definition is phrased a little differently, but that's what it means). Therefore 1 (or 2, or 1/2 or any other finite real number) is an upper bound for the set of infinitesimals and 1 is less than infinity, so the least upper bound must be less than infinity. --Tango (talk) 20:07, 1 August 2008 (UTC)

By that definition, I guess my argument holds no more water. But do you think it might be possible, say, to define a number by its reciprocal? Such as, 2 = 2 because 1/2 = 1/2? And 3 = 3 because 1/3 = 1/3? If so, 0.333... = 0.333... because 1/0.333... = 1/0.333....

As trivial as it sounds, this form of definition might be worth something. (Even if it's not worth something, bear with me; what preceded this may be completely irrelevant to the rest of my comment, which may have merit.) If you've never thought about it, as you start from addition and subtraction and progress to algebra and then calculus, the steps taken to arrive at the higher level of math that you've attained look trivial and differential rather than immense as those steps may have looked before you took them. Well, at least that's how I remember it. Then again, the farthest I ever went in a meaningful way with math is calculus II. (I did take III, but I didn't retain any of that, and I took D.E. but I comprehended so little in the first place.)

Well consider that 0 is a legitimate value, and 0 = 0. There are many numbers that resemble 0. By way of relation, the ratio of Avogadro's number (6.022·10^23, which is far greater than 0) to a googol is rather close to 0: $6.22*10^{23}*10^{-100}=6.022*10^{-77}$. Certainly, (hypothetically speaking) if you were to take a googol number of objects minus Avogadro's number, the change would probably go unnoticed when observing a physical change except by observing the difference itself -- such as if the Avogadro's number of objects were found in another place within the arena of observation. Bear with me. Even if 100 of these clusters were taken from the googleplex, it would still go unnoticed unless you were specifically looking for it. $10^{100}-100*6.022*10^{23}=10^{100}-6.022*10^{25}=10^{100}$ (observably).

Given the vast differences between these very large numbers – not to mention the even more vast difference between a googol and the very small constants, such as Coulomb's constant – remember that just as the difference (using any value x ≥ 1) between 10^x and x is so large, would it be at all fair to say that 10^∞ is equal to ∞? Even with a value of x as low as 20, the ratio of 10²⁰ / 20 is great. In calculus I, I was taught (if memory serves) that if problems come up such as this – $10^{\infty }/\infty =?$ — you take the derivative of each, such as $10^{x}/x$ and determine whether the value then is infinity or 0.

But as I reckon it, it is a ratio between infinities. And speaking of them as such, they do have reciprocals. Those reciprocals are asserted to have infinitesimal values (that is, values equivalent to 0). What, then, is the ratio of 0:0? Well this is my assertion: That whether or not 10^-∞ is infinitesimal, it is not exactly 0, nor is it equivalent to 0 except by rounding or for purposes of practical usage. It is a near-zero value which is impractical to calculate or use, which when necessary rounds to (but is not equal to) 0. After all, if it (10^-∞) was zero, it would not have a reciprocal of 10^∞, right? Well, somehow I feel like I haven't conveyed my point very well, but one of you might get the idea. At least, I hope so. 68.60.8.26 (talk) 02:28, 2 August 2008 (UTC) -- D. F. Schmidt (talk) 02:31, 2 August 2008 (UTC)

All the stuff about various constants made no sense at all, I'm afraid. As for $10^{\infty }$, I guess it depends on what you mean by the expression. The most obvious way to interpret it is as $\lim _{n\rightarrow \infty }10^{n}$, in which case it is just infinity, however your sense that it should be bigger than infinity isn't entirely misplaced. Take a look at aleph number - the integers have a cardinality of $\aleph _{0}$, the real numbers have a cardinality of $2^{\aleph _{0}}$, which is indeed a bigger infinity (for information about which one it is, see continuum hypothesis). Changing that 2 to a 10 won't make any difference, as far as I can see. As for $10^{-\infty }$, if you're working in the real numbers, it really is exactly 0, not just approximately but precisely. It's clearly not finite, and there are no non-zero infinitesimals in the real numbers, so it must be 0. --Tango (talk) 03:48, 2 August 2008 (UTC)

As with most Wikipedia articles covering math (any math higher than someone might already have studied), those for the aleph number and continuum hypothesis are beyond my comprehension, and even my interest, for that matter. If I had studied it, or if I might later study it, I might enjoy it. So I guess this is as far as I go in this discussion.

I do understand sometimes the necessity of using higher math in order to evaluate singularities, such as x/0. The truth is, as is mentioned, subtracting x − 0 any number of times will not reduce x to a remainder of 0 or a number having the absolute value less than 0. But just as I mentioned that 10^x/x is >>> 1 (when x > 1), it would seem to me that that any similar evaluation would be similarly obvious. Likewise, 10^-x*x (which is the reciprocal) should indicate that it is indistinguishable from but non-zero. I suppose my point is that higher math is like higher theology (which is a topic close to me): the higher math builds upon the lower math, and sometimes it is necessary to use higher math in order to explain lower-math singularities. But when the lower math (probably at least one order above where the singularity occurs) does already explain a lower singularity, why do you need to appeal to even higher orders of math? If this comment doesn't ring any bells of truth, then I don't suppose I belong here discussing this matter. Thanks for your time and labor in your attempt to show me why 0.999... = 1. D. F. Schmidt (talk) 04:30, 3 August 2008 (UTC)

From elementary school up through high school and even some of college, mathematics is simplified to a low level for us, but requires higher level math to fully justify. It's perfectly normal. As for the aleph numbers: they aren't that complicated; let me give it a shot.

The aleph numbers are numbers which refer to the size of infinite sets ("cardinality" is just a fancy term for size). The natural numbers, integers and rational numbers are all the same size (a bit of a paradox, but fully justifiable; see Countable set if you're interested), represented by the aleph number $\aleph _{0}$. The Axiom of choice ensures the existence of a next cardinal number (the cardinal numbers could have been infinitely dense, like the rational and real numbers, but they're not if we accept the axiom of choice as true), which we denote $\aleph _{1}$. The Continuum hypothesis can be expressed in several different ways, as shown in the article. However, the key idea is that the cardinality of the real number set is the same as that of the power set of the natural numbers, that is, $2^{\aleph _{0}}$, a provable fact without the continuum hypothesis. The continuum hypothesis takes the next step by suggesting that the real number set represents the next infinity, that is, $2^{\aleph _{0}}=\aleph _{1}$.

HTH. --69.91.95.139 (talk) 12:02, 3 August 2008 (UTC)

See also article on well ordering. Why is the ordering well? I guess for same reason the real numbers are real and right-wing is right. Is cardinality and cardinal number a better term?

If we don't assume Continuum hypothesis, what ordinal number $\aleph _{1}$ is explicitly? (Compare: what (hyper)real number 0.999... is explicitly? Assume 0.999... hypothesis = Archimedean property) Tlepp (talk) 21:59, 3 August 2008 (UTC)

Of course a "well" ordering is not necessarily "better" than any other ordering; it's a technical usage. This is one of the reasons I prefer the modern spelling wellordering with no hyphen and no space—I think that makes it clearer that it's a technical term rather than a value judgment. --Trovatore (talk) 22:04, 3 August 2008 (UTC)

Were those questions intended for me? --69.91.95.139 (talk) 00:05, 4 August 2008 (UTC)

$\lim _{x\rightarrow \infty }{\frac {10^{x}}{x}}=\infty$ and $\lim _{x\rightarrow \infty }{\frac {x}{10^{x}}}=0$. That's something you would cover in a first course in real analysis (1st year of Uni, possibly even before that in a non-rigorous fashion). It's important to remember that ${\frac {\infty }{\infty }}$ isn't 1, it's indeterminate and can turn out to equal anything depending on where it came from. While 10^x and x both go to infinity as x increases, 10^x does so much faster, so the ratio of the two ends up being infinity. The reciprocal case works exactly the same way. --Tango (talk) 20:37, 3 August 2008 (UTC)

Yeah. Well, don't be surprised that your fairly short essay does not enlighten me, 69.91.95.139. It's probably not your fault. Tango, I completely understand that infinity/infinity is indeterminate without context, and as such, is a singularity (without such context, as I assert). And I understand that the reciprocal winds up, well, the reciprocal. And division by zero is a singularity. But 10^∞ and its reciprocal are neither simply "infinity" or "zero". Ratios or subtractions may cause them to evaluate (or be roughly equivalent) to zero or to infinity, but still are not identical with either. It's like you guys want to do mid-calculation rounding, which is a grave no-go in the math and science world. This is my assertion, and I think that's all I have to add. D. F. Schmidt (talk) 04:47, 4 August 2008 (UTC)

Well, it's not really defined at all, but if you want to define it, the only meaningful way I can see is to define it as the limit of 10^x as x goes to infinity, in which case it really is infinity. There is no approximation going on. (It's worth pointing out, we're implicitly working with the extended real numbers now, rather than the real numbers, but that's not really important except for precision.) --Tango (talk) 21:56, 4 August 2008 (UTC)

What seems to be missing from your consideration here is that you can't really evaluate ∞/∞; it is beyond evaluation. 10^∞/∞ on the other hand is--it evaluates to something that you would assert is equal to ∞. The reciprocal evaluates to something that you would assert is equal to 0. So now consider ${10^{\infty }/\infty } \over \infty$. If you had previously evaluated the numerator, the result of the overall equation would then be ∞/∞ which is again beyond evaluation. However, if you evaluated it properly like any algebra student knows to do, it would be obviously evaluated to ∞. I agree that when facing infinite or infinitesimal values, they are essentially infinity or 0; however, in order to continue to use these values practically (or at least more so), you need to not "round" (or whatever word you choose) them mid-calculation.

And if I'm right in saying the above, I suspect that perhaps no value which involves infinity is truly 0 or simple infinity. D. F. Schmidt (talk) 01:07, 5 August 2008 (UTC)

Yes, I see what you're saying. Evaluation of limits and basic arithmetic operations do not commute when infinity is involved (they do otherwise - I learnt it as the Calculus of Limits Theorem (COLT), although I suspect it has an alternative name, since a google search gives 5 results, 3 of which are on my uni's website and one is me on a Wikipedia talk page about 0.999...!). That's evidence in favour of simply not counting infinite limits as existing and just saying 10^∞ is undefined and leaving it at that. If you want to define it, though, you do need to be careful about when you take limits (which is what you're referring to as rounding - it's not a bad analogy, I suppose). --Tango (talk) 01:30, 5 August 2008 (UTC)

PS: It's not quite as bad as rounding early, though - with rounding, you'll get an answer at the end and it will be wrong, with this you just don't get an answer. --Tango (talk) 01:31, 5 August 2008 (UTC)

Lack of Knowledge

Moved from main talk page. --Tango (talk) 22:32, 20 September 2008 (UTC)

Is it just me, or is this discussion page just full of people talking non-mathematical non-sense? People saying that when one squares the square root of two one doesn't get two. There is too much non-technically, informal, non-rigorous non-sense on this page. It seems (from reading some of the user pages) that many people that have contributed are little more that, if at all, undergraduate students. People; please don't write anything unless you're sure... It's confusing, and a little annoying. For me 0.999... does not equal one. This is informal shorthand. The true statement is that

$\lim _{k\to \infty }\sum _{n=1}^{k}{\frac {9}{10^{k}}}=1.$

That does not say that the sum is ever equal to one for any k. It simply says that the sum tends to one as k becomes very large. To say that something is "equal to" and that something "tends to" are very different statements. For example, people have proved that almost all of the zeros of the Riemann Zeta function fall on the critical line. As the number of zeros tends to infinity the ratio of zeros on the line to those off it tends to zero. That does not say that all of the zeros fall on the critical line. Limits and equalities are very different things. So there! ;o) Declan Davis (talk) 22:08, 20 September 2008 (UTC)

So what you define "0.999..." to mean? --Tango (talk) 22:32, 20 September 2008 (UTC)

Sorry, I don't understand the question. Do you mean to ask "So what do you define "0.999..." to be?" Well, to me, it is a zero followed by a series of nines. You could write as many nines as you like but you will never end up with one. If, however, you want compute the limit, then the limit will tend to one. I suggest that you read "The Prime Number Theorem" by G. J. O. Jameson, London Mathematical Society, 2003, for some interesting discussion on the difference between limits and equality. It is a very subtle difference, I agree. Declan Davis (talk) 23:43, 20 September 2008 (UTC)

So you don't think 0.999... is well defined? In that case, it's a useless concept, so it's much better to stick with the standard definition which is that something ending with "..." refers to the limit. The limit is equal to 1, therefore 0.999... is equal to one. --Tango (talk) 00:00, 21 September 2008 (UTC)

Exactly! The limit is well defined and is equal to one. I just don't like people writing a zero followed by lots of ones. I'd mark that wrong in every exam. It's a little bit like root two: it doesn't have a decimal expansion, and writing ever more accurate decimal approximations to root two and then saying "oh look, it's root two" is all wrong. If the convention is to assume that

$0.999...:=\lim _{k\to \infty }\sum _{n=1}^{k}{\frac {9}{10^{k}}}$

then I guess the statement is correct, although I don't for one second condone the abuse of notation. I would still strongly recommend that you read the above book.

Declan Davis (talk) 00:08, 21 September 2008 (UTC)

That's the convention I've always known, and I think it's pretty universal. "..." means the obvious pattern continues to infinity - more rigorously, you take the limit as the length of the expansions tends to infinity. If you object to the notation entirely, then the question of what it equals doesn't even exist. However, the notation is standard, so if you're marking it wrong in exams I would expect students to appeal (assuming they get to see the scripts - my Uni won't show us them...). --Tango (talk) 00:19, 21 September 2008 (UTC)

No-one needs to appeal: I drum it into them not to write such things. Also, on second reading I'm getting a little worried about your authority on matters mathematical. From reading your posts you seem to have problems with εδ-proofs, not to mention the defintion of a continuous function. What did you say? "It's just plain wierd"? May I ask: what is your mathematical background, and what makes you feel that your are such an authority on all things mathematical? Declan Davis (talk) 00:54, 21 September 2008 (UTC)

Seeing as you like εδ-proofs so much, I thought I'd rewrite the limit it terms of an εδ-statement. Let

$a_{k}:=\sum _{n=1}^{k}9/10^{n}.$

Then the actual statement should not be that 0.999... = 1, but should be that for each ε > 0 there exists a natural number $K_{\varepsilon }$ such that for all $k>K_{\varepsilon }$ we have $|1-a_{k}|<\varepsilon .$ Declan Davis (talk) 01:30, 21 September 2008 (UTC)

Because 0.999... is usually defined as the limit of the sequence (a_k), that's the same statement as 0.999... = 1, except that you avoid writing 0.999..., which misses the article's point. As an aside, given that Tango has quite a history of contributions on these talk pages, and given that he was awarded reference desk barnstars for his help on math topics, I don't think one needs to worry about his credentials. Huon (talk) 01:39, 21 September 2008 (UTC)

Thanks for the advise Huon, but I was asking Tango and not you. Besides I was questioning his mathematical authority, and not his skill as a Wikipedia editor. He is obviously a very skilled editor. Thanks again for your thoughts. Declan Davis (talk) 01:54, 21 September 2008 (UTC)

While I agree that it is important to emphasize that the statement "0.999... = 1" is actually

$\lim _{k\to \infty }\sum _{n=1}^{k}{\frac {9}{10^{k}}}=1$

I do not agree that you should force your students to rewrite every single decimal expansion as its corresponding limit. That's ridiculously and unnecessarily dull repetitious work. And yes, infinite decimal representations are defined as the limit of their successively larger representations, by convention, in the vast majority of mathematical contexts (as are infinite series). --69.91.95.139 (talk) 01:47, 21 September 2008 (UTC)

I did not say that I force my students to rewrite every single decimal expansion as its corresponding limit. If I did then please leave a link. Please don't put words in my mouth (or should that be words in my hands seeing as I'm typing, I don't know). I was simply saying that if a student were to write, for example π = 3.142, then this would be incorrect. I shall repeat, once again, my objection: limits and equalities are different. Declan Davis (talk) 01:54, 21 September 2008 (UTC)

There is a big difference between π = 3.142 and π = 3.141..., the former is, of course, incorrect. The latter is entirely correct (although not very precise since being irrational there is no pattern to continue on). --Tango (talk) 11:02, 21 September 2008 (UTC)

I know there's a difference, but thanks for pointing it out just in case I didn't. I didn't ever say that there wasn't. People really need to stop trying to put words into my mouth. Instead, people need to read what I have written and stop infering things which may or may not be true. Once again: thank you for your thoughts, I found them most enlightening.  Declan Davis   (talk)  23:49, 21 September 2008 (UTC)

You have an extremely unpleasant argument style, Declan. Stop accusing people of "putting words in your mouth" every time they try to emphasise a point that you appear to be disregarding. Stop questioning other editors' "mathematical authority" (as if appeal to authority weren't even more invalid in mathematics than it is elsewhere). Stop mocking editors for typoes. Please read WP:CIVIL. This is not the place for you to smugly belittle others, it is a place to discuss mathematics, in particular the equality 0.999... = 1, in a civil manner. Maelin (Talk | Contribs) 00:10, 22 September 2008 (UTC)

The evidence suggests you don't know the difference. Why else would you compare 0.999... to 3.142? --Tango (talk) 13:42, 22 September 2008 (UTC)

Declan Davis has been perfectly civil. You guys are misrepresenting (or misunderstanding) what seems to be perfectly valid points of his. Algr (talk) 17:17, 22 September 2008 (UTC)

Of course he has! Every bit as much as you have! --69.91.95.139 (talk) 21:48, 22 September 2008 (UTC)

Thanks Algr, at least someone's reading what's actually been written. For the other editiors, please allow me a few words. As far "accusing people" of putting words into my mouth, let me present two cases in hand (both can be found above). The IP user 69.91.95.139 said "I do not agree that you should force your students to rewrite every single decimal expansion as its corresponding limit." I had never for one moment said that I force my students to rewrite every single decimal expansion as its corresponding limit. So that is putting words into my mouth. Tango said that "There is a big difference between π = 3.142 and π = 3.141...", which implies that I had made the point that that there wasn't a difference, which I hadn't. So that's coming close to putting words in my mouth. I have questioned people's mathematical authority because they have used phrases such as "it's pretty universal" when I don't think that they are in a position to comment on what is, or is not, universal. I find IP user 69.91.95.139's reaction quite funny: his way of saying that I haven't been civil is to be sarcastic and be uncivil towards Algr. I find that most amusing. Finally, Tango: it is clear that π ≠ 3.142. I made the point in an earlier post to tell IP user 69.91.95.139 that I didn't "force my students to rewrite every single decimal expansion as its corresponding limit" but that if they tried to use some truncated decimal expansion in place of a proper limit then their answer would be marked as incorrect.

 Declan Davis   (talk)  00:51, 23 September 2008 (UTC)

If people are misinterpreting you, it's because you are explaining yourself extremely poorly. If you're not getting students to rewrite recurring decimals, what did you mean by that comment? And, since no-one suggested a truncated decimal was correct, what was the relevance of your comment about 3.142? --Tango (talk) 00:58, 23 September 2008 (UTC)

I think that I have explained myself perfectly well: limits and equalities are different things. I have made this point several times. My point about π was that formally it does not have a decimal expansion. So to attempt to write a decimal expansion of π is incorrect, even though π is a well defined number. For me, I don't like the practise of writing 1/3 = 0.3(3). One third is 1/3 and it's best left as an object with the property that three times it is one. Real analysis can be a bit wishy-washy. Proving the completeness of the reals involves showing that a certain number's less than this, and more than that, without actually getting a grip on the number itself. This is much the same. Trying to put decimal expansions to limits, irrational numbers, or transcendental numbers is a bit sloppy. These numbers exist, and have formal definitions (using limits) but you can't put a string of numbers to them... or at least that's what I think.

 Declan Davis   (talk)  01:09, 23 September 2008 (UTC)

I disagree completely with your assertion that π does not have a decimal expansion. Every real number has a decimal expansion (indeed, many have two, as this article demonstrates). It may not have a finite decimal expansion, but that is another matter. I can even give you a closed form for each digit of π in any base of the form 2ⁿ, if you like. A third is indeed 0.333..., and 1/3, and also 3^-1 and all of these have the property that three times them is one. Maelin (Talk | Contribs) 01:57, 23 September 2008 (UTC)

Maelin: Well, that's your opinion, and you're allowed to have and express that. I disagree with you. We shall have to agree to disagree. I have been teaching in university for four years now, and one of my modules is "Fourier series and iteration" which includes a large amount of real analysis. I have been thinking about these things for almost ten years now. I suggest that you show this discussion thread to one of you professors at university, and see what they say. The whole idea behind a transcendental number is that it doesn't have a well defined decimal expansion. Please go and ask a teacher and let me know what they say. Until then: good luck in your studies, and have a nice week.  Declan Davis   (talk)  02:04, 24 September 2008 (UTC)

Declan, if you have been telling students that transcendental numbers do not have a well-defined decimal expansion, then you've been doing them a disservice. This is just not correct. The decimal expansion of pi, say, is perfectly well-defined, even in a constructivist sense (there is an algorithm for computing the digits of pi). You are welcome to your own private definitions, but it's a damn shame when you try to appeal to your own authority when others object. Phiwum (talk) 11:03, 24 September 2008 (UTC)

Declan, I realize you have a PhD in mathematics, so I don't have any idea why you think that these notions of yours are relevant to classical mathematics. It's a plain fact that every real number has a decimal representation and it's a plain fact that infinite decimal expansions are commonplace in classical analysis. You may have your own personal definitions and philosophy of mathematics, but they don't hold any particular weight for others. Phiwum (talk) 03:16, 23 September 2008 (UTC)

Once again, this is a case of people not reading what has gone before. I realise that it's tempting to skim-read a post and get straight down to writing your point. I have never said that "these notions of [mine] are relevant to classical mathematics". I've already made my point above. The article is too informal, it makes something out of nothing. The use of informal shorthand becomes so ingrained that people forget that it is shorthand. Then they start to believe that there is some kind of mathematical voodoo. For me, this whole article is vacuuous. If, instead of using the shorthand, we wrote out the limit then there'd be no problem. students don't understand the shorthand. A large chunk of the article is about why people don't understand or why they disagree. If we wrote out the truth first time then there'd be much less trouble. It's like an idiom in a foreign language. It make perfect sense to the native speaker, but to a student it seems crazy and needs to be explained.  Declan Davis   (talk)  01:56, 24 September 2008 (UTC)

Declan... You forgot to ask Algr about his (her?) mathematical background. Was that really because of Maelin's advice, or did you just decide give up on that argument technique when it started working against you? ;) --69.91.95.139 (talk) 11:10, 23 September 2008 (UTC)

IP user 69.91.95.139: Not at all friend. I didn't find any need to question Algr's mathematical authority since s/he didn't make any statements like "it is universal". They didn't profess to know the whole picture. When people without the requisite experience and/or knowledge make claims (without acceptable sources) then I will naturally question their knowledge and authority. I am a scientist. If I did not question the validity of the information presented to me then I would not be doing my job.  Declan Davis   (talk)  01:56, 24 September 2008 (UTC)

It's true that the "..." notation is not a rigorously defined mathematical symbol, as far as I know (and indeed, Ellipsis#In_mathematical_notation states this). However, the article clearly states the recurring decimal, and also lists the other notations such as using a dot or bar over the 9 (which I have always been taught as denoting a recurring decimal).

As it happens, I raised the possible ambiguity over using ellipses, way back in Talk:0.999.../Archive_6#Use_of_ellipses. But do you agree that $0.{\bar {9}}$ equals 1? Or if we were speaking it, that "0.9-recurring equals 1"?

It's not about what the majority of editors think, it's about verifiable sources and not personal opinions. And you may have been asking Tango, however, this is a public page addressing issues about a Wikipedia article, so it is fair for any editor with an interest to respond. Mdwh (talk) 02:47, 21 September 2008 (UTC)

You guys always resort to "verifiable sources" when you run out of arguments, but this page is not the article, it is for people who practice scientific skepticism of what they read. Such people will reject .999... = (1/3)x3 as circular logic, and you have defined limits and the Archimedean property in ways that make those arguments equally circular. Algr (talk) 07:45, 23 September 2008 (UTC)

I only brought up verifiable sources in response to him asking about people's authority. I'm happy to try to explain the mathematical proof for those who do not understand it. Mdwh (talk) 01:37, 24 September 2008 (UTC)

Not a Proof?

The 0.999... = 3(1/3) argument is not a proof. It is a heuristic argument designed to show how a potentially counterintuitive result can be seen as a simple consequence of a hopefully less counterintuitive result. Someone who rejects that 1/3 = 0.333... recurring is perfectly entitled to reject that it demonstrates 0.999... = 1, but this does not mean they are entitled to reject proper proofs. Limits are defined via epsilon delta proofs and are fully formal, fully rigourous and fully legitimate definitions. The Archimedean property of the real numbers is a direct result of its construction. You are simply practising intellectual dishonesty by refusing to try to understand the arguments raised here, and you should not pollute Declan's entirely unrelated discussion with your little sniping comments. Maelin (Talk | Contribs) 08:38, 23 September 2008 (UTC)

O'RLY? The article has 0.999... = 3(1/3) as the first example in the proofs section and presents it as a perfectly valid reason why people ought to believe that .999...=1. You don't see anything "intellectual dishonest" about that? I see that as a perfectly valid reason to distrust the premise and suspect that all the other proofs are equally flawed. If I submitted a research paper for peer review and had this kind of error at the top, would you expect people to take my other findings at face value? Algr (talk) 18:23, 23 September 2008 (UTC)

A research paper is aimed at an audience of experts. Wikipedia is aimed at an audience of laymen. The article isn't intended to be a rigorous mathematical paper, it's intended to be an encyclopaedia article. If we started with a detailed rigorous proof referencing the constructions of the real numbers and the Archimedean property and limits and convergence and everything else we would just scare away our readers. The first "proof" is intended to convince readers of the equality and it is a valid argument it just misses out a load of steps (it relies on the fact that those kind of arithmetic manipulations work on convergent series, a fact that needs proving, as does the fact that the series converges - neither of those are within the scope of that first argument, though). The article then moves on to more rigorous approaches for those readers that want them. --Tango (talk) 18:42, 23 September 2008 (UTC)

Laymen can still understand logic. .999...=1 and .333...=1/3 both invite the exact same question: What does it mean for a decimal to extend to infinity? To answer one by assuming the other is not "leaving out steps", it is circular logic and insulting to the reader who wants to understand the real question. Algr (talk) 19:13, 23 September 2008 (UTC)

Most laymen cannot, however, understand the answer to your stated question, because the answer invokes a bit of high-level mathematics. It is most certainly not insulting to say that most laymen would not come even close to understanding the underlying reason for 0.999 and 1 being the same number, as the reason is a tad bit complicated.

And no, the 'reason' I refer to is not the Archimedean property; I refer to the property of completeness. The former is a consequence of the later. --69.91.95.139 (talk) 22:29, 23 September 2008 (UTC)

Intuitively, it's very easy to see what it means for a decimal to extend to infinity - it means it keeps going on forever and never stops. Making that rigorous is a little more difficult (it requires some 1st year analysis), which is why we skip it. --Tango (talk) 22:38, 23 September 2008 (UTC)

Yes, the process is easy to understand, it is the result which is the problem, I've already explained why. Completeness is just more circular logic because the Real space is only complete if you assume that .999...=1. If non zero infinitesimals exist, then the real set is not complete. Algr (talk) 00:27, 24 September 2008 (UTC)

No. The real number set is complete because of its construction. 0.999... = 1 is a result of its completeness. Nobody sat down and said, "Okay, let's make sure 0.999... = 1 is true just to offend the sensibilities of people without formal maths education. Hum, that means we have to make the real set complete. Okay, complete it is." We don't construct the real numbers in the way we did simply to make 0.999... = 1, we construct them that way to make an interesting and useful number system. Completeness is part of that construction. And the equality arises from that. Maelin (Talk | Contribs) 00:38, 24 September 2008 (UTC)

If the real number set is defined as complete, then you can't say that any describable value is not a part of it. "The highest value of x<1" is just as valid a description of a number as "The square root of two". Both values have unambiguous positions on the number line, (actually √2 has two positions.)... Algr (talk)

Sure I can. I can describe a value, "a number x that squares to -1", and yet, that number is not a part of the real number set. Your confusion here arises because you have absolutely no idea what complete means, and are instead pretending that you do for the sake of continuing the argument. Maelin (Talk | Contribs) 05:42, 26 September 2008 (UTC)

Congratulations! You have at last recognized that the Real set is not the appropriate venue answer to every mathematical question. Had you failed to make this leap you would have had to insist that there is no √-1, just as there is no highest value for x<1. Algr (talk)

But you have never described an extension of the real numbers that would contain a "highest value for x<1". And please don't bring up the hyperreals: they do not hold that property, and wishing they did will not make it true. --69.91.95.139 (talk) 12:18, 28 September 2008 (UTC)

No-one has claimed that the real numbers are the appropriate venue for every mathematical question, but they are the appropriate venue for this one since decimals are a notation for real numbers. --Tango (talk) 14:18, 28 September 2008 (UTC)

... But that is beside the more serious point. I can't believe you are defending teaching laymen a clearly wrong proof (.333...x3) just because it makes them think what you want them to! That is no better then if I made a fake moon rock and and started using it to "prove" that the moon landings were real. Once you descend to that kind of tactic, you have no right to expect anyone to trust what you say. Algr (talk) 05:18, 26 September 2008 (UTC)

The proof isn't wrong, it just has a few gaps in it. It's not intended to be rigorous, it's just intended to convince laymen of the result. The rest of the article goes on to more rigorous proofs. --Tango (talk) 15:29, 26 September 2008 (UTC)

Gaps? Yes, you left out the part where you assume that .999... = 1. Algr (talk)

Where, exactly, is that assumption hidden? Please specify PRECISELY where we have assumed that 0.999... = 1. You need to point out exactly which particular inference we make, whcih requires 0.999... = 1 to be true, and which will fail to be true if the equality is untrue. Don't say "it's an underlying assumption of the whole proof," this will merely prove you have no idea what you are talking about, as usual, and that you are just trying to be inflammatory. Don't pick apart this post. Don't accuse me of bullying. Just answer the question. Where did we make that assumption? Maelin (Talk | Contribs) 06:50, 27 September 2008 (UTC)

It is by extension of assuming that .333=1/3. Also, both http://www.dpmms.cam.ac.uk/~wtg10/roottwo.html and Construction of the real numbers show that the equivalence is simply written into the definition of what a real number is, so any proof involving real numbers is circular logic. Here is the exact quote:

__

So when you say "the real number x" what you really mean is "the decimal expansion x"?

Well, it's not always what I think of when I talk about real numbers, but if you insist on a precise definition, then I can fall back on this one.

Does that mean that 0.999999.... and 1 are different numbers?

Oh yes, I forgot about that. Different decimal expansions correspond to different real numbers except in cases like 2.439999999.... equalling 2.44. So I suppose my definition is that real numbers are finite or infinite decimals except that a finite decimal can also be written as, and is considered equal to, the "previous" finite decimal with an infinite string of nines on the end. Happy now?

__

Firstly 0.333 does not equal 1/3, 0.333... does. You need to be more precise in your notation if you want to be taken seriously. Secondly, long division proves that 1/3 is 0.333..., you don't need 0.999...=1, you just need that "10 divided by 3 is 3 remainder 1", do you dispute that? I don't see what that quote has to do with anything, that's not a rigorous definition of the real numbers. The real numbers are defined either in terms of Cauchy sequences or Dedekind cuts, neither or which rely on 0.999...=1. --Tango (talk) 16:23, 27 September 2008 (UTC)

Precise? You and Maelin make all sorts of spelling errors, and I never comment on them. I may not always succeed, but I do my best to understand what you are really trying to say. Assuming that a remainder will just disappear if you push it far enough away doesn't sound very rigorous to me. Finally, the first line of Dedekind cut includes " and A contains no greatest element." Where does that come from? Algr (talk) 02:50, 28 September 2008 (UTC)

The "no greatest element" line is to avoid an ambiguity. If you omitted it, you could construct two "Dedekind cuts" for every rational number, say 1/3: One with A containing the rational number, the other with B containing it. I assume that you're about to say that this artificial construction is what forces 0.999...=1, but that's not the point: If you tried to construct a number system out of partitions of the rationals without this restriction, you'd double more than just those numbers with a terminating decimal expansion. There's not even a decimal representation for most of the numbers you'd get (say, for the doppelganger of 1/3). If you tried a straightforward generalization of addition, you'd also realize that some of those numbers don't have an additive inverse. Huon (talk) 11:16, 28 September 2008 (UTC)

Infinite decimal expansion?

Every real number has a decimal expansion (indeed, many have two, as this article demonstrates). It may not have a finite decimal expansion, but that is another matter.

Well this brings up an interesting point. If you start with the decimal expansion for π, changing any digit in that expansion will result in a value that is not π. Correct? So what happens if you change an infinitely distant digit? If π has no last digit, then do further digits become irrelevant at some point? And if the idea of an infinitely distant digit somehow has no meaning, then how could any decimal expansion be described as infinite? Algr (talk) 07:35, 23 September 2008 (UTC)

For exactly the same reason that the set of natural numbers is infinite, but no individual natural number is infinite. --Trovatore (talk) 07:43, 23 September 2008 (UTC)

So we are back to "infinity only works when we want it to." If infinite digits exist, then you have non-zero infinitesimals. If they don't, then .333... approaches 1/3, but can never equal it. Algr (talk) 07:51, 23 September 2008 (UTC)

0.333... can't "approach" anything, it's a fixed number. The main difference here is between "infinitely many" (as in "infinitely many digits") and "infinitely large". Obviously there can be sets containing infinitely many elements without any of those elements being infinitely large. The set of real numbers between 0 and 1 is one example, the set of natural numbers is another. The set of digits of 0.333... is bijective to the set of natural numbers, as there's a first digit, a second digit and so on. Since there are no infinitely large natural numbers, there's no "infinite" digit. Huon (talk) 10:35, 23 September 2008 (UTC)

But that's the point Huon: "0.333..." is not a fixed number, it's short hand! It's actually a limit. Let $a_{k}:=\sum _{k=1}^{n}{\frac {9}{10^{n}}}$, then when you write 0.333... you actually (should) mean $\lim _{k\to \infty }a_{k}$. In much the same way that when we write $\int _{0}^{\infty }f(x)\ dx$ we actually mean $\lim _{a\to \infty }\int _{0}^{a}f(x)\ dx$. Any good calculus text book will make the second point quite early on, but to save on paper and ink we write the upper limit as $\infty$ with the understanding that we really mean the limit.

You're right when you say that (in the limit) there is no infinite digit. In $\lim _{k\to \infty }a_{k}$ we do indeed get a bijection between the digits in the decimal expansion and the (non-zero) natural natural numbers, this is given by $n\mapsto {\frac {9}{10^{n}}}$.

 Declan Davis   (talk)  11:37, 23 September 2008 (UTC)

Quite right, Declan, $0.333...=\lim _{k\to \infty }\sum _{i=1}^{k}3^{-i}$ by definition. But you're quite wrong when you suggest that 0.333... is therefore not a number. It is trivial to prove (as you know) that

$\lim _{k\to \infty }\sum _{i=1}^{k}3^{-i}=1/3$

and hence 0.333... is a number. It's the same number as 1/3.

You seem to think that limits are different than numbers, but when a limit $\lim _{k\to \infty }a_{k}$ converges to, say, $b$, then we write $\lim _{k\to \infty }a_{k}=b$ for a reason. Namely, in such situations, $\lim _{k\to \infty }a_{k}$ means $b$ in exactly the same sense that $2+2$ means $4$. (Admittedly, such limits aren't usually terms in first-order logic, because function symbols are taken to represent total functions, but this is a technical fact of no essential consequence.) Phiwum (talk) 11:55, 23 September 2008 (UTC)

Sorry, I didn't explain myself properly. What I meant to say is that the shorthand "0.333..." itself is not a number. It's just that: shorthand. The limit that it represents is of course a real, well defined number. I still probably haven't explained myself totally, but I hope you can see what I'm trying to get at. And don't worry, I am a mathematician and I fully understand limits. That's why I have written that $\lim _{k\to \infty }\sum _{n=1}^{k}{\frac {9}{10^{n}}}=1$. For me, the whole point is that this quirk of notation has lead to people thinking there's some kind of mathematical voodoo going on. The use of such shorthand leads to interesting expressions such as "0.999... = 1".  Declan Davis   (talk)  12:16, 23 September 2008 (UTC)

The situation with 0.333... is no different than the situation with $\lim _{k\to \infty }\sum _{n=1}^{k}{\frac {3}{10^{n}}}$ or, indeed, with 1/3 or 2 + 2. In each case, we have a term ("0.333..." or "2 + 2", say). These are syntactic objects which represent certain numbers. In each case, the representation depends on relevant definitions. There is, perhaps, a sense in which the ellipsis is a bit informal (I don't know of any formal definition of the ellipsis), but in practice the meaning is clear.

In any case, I must admit this isn't what I thought you were saying. Seemed to me that you were saying 0.333... is "really" a limit and hence not a number and certainly not 1/3 (although it "approached" 1/3). That opinion is, of course, contrary to standard mathematical exposition. But now you seem to be saying merely that 0.333... is an informal (though nonetheless well-defined) shorthand for a particular limit and this fact may escape mathematical novices. Perhaps so, though I don't see it as a persuasive argument that we should drop the shorthand and write the limit out instead. Phiwum (talk) 13:40, 23 September 2008 (UTC)

I fail to see your point. We can agree that "0.333..." is a notation that denotes a limit and thus a number, but that's more semantics than mathematics. Still neither 0.333... nor any limits tend to anything, and for the sake of extreme precision, I'd also disagree with Phiwum's notation that "$\lim _{k\to \infty }a_{k}$ converges to, say, b": The sequence (a_k) may converge to b, the limit, if it exists, is a number and can't converge. Huon (talk) 12:34, 23 September 2008 (UTC)

Well, that's your opinion and you're entitled to it. We shall have to agree to disagree. Have a nice day Huon.  Declan Davis   (talk)  12:41, 23 September 2008 (UTC)

Declan, I'm still trying to understand exactly what your contention is here. Which, if any, of these statements do you find objectionable?

• "0.999..." is a shorthand for an infinitely long string, consisting of a zero, a decimal point, and an infinite string of nines
• That infinitely long string is a legitimate representation of a single real number
• The real number represented is $\textstyle 0.999...:=\lim _{k\to \infty }\sum _{n=1}^{k}{\frac {9}{10^{n}}}$
• This interpretation is, in some sense, the "correct" interpretation that we should promote on Wikipedia
• Since the infinitely long string represents this real number, then so too does the shorthand 0.999... for that string
• The real number represented, that is, the limit described above, is equal to the real number 1

I must confess I'm not sure exactly what it is that you dislike about the article. Please try to summarise your opinion in a self-contained form, and we will hopefully be able to avoid any further issues of people misinterpreting you as above. Maelin (Talk | Contribs) 13:58, 23 September 2008 (UTC)

Maelin, my objection was posted a long time ago way up top. It's that I don't like the lack of formality and the use of shorthand; well shorthand's okay provided people remember that it is just that. Out of your list I would object to

• That infinitely long strings are legitimate representations of real numbers.
• This interpretation is, in some sense, the "correct" interpretation that we should promote on Wikipedia.

I agree with the other points. For me it's a bit like power series. In Taylor's work he actally says that a smooth function can be written as a polynomial plus some smooth remainder. It's the same with numbers like π, people talk about the k-th digit of π but when they're saying is that we can write π as a finite desimal plus some other transcendental remainder. With irrational numbers, we can write root two as a finite desimal plus an irrational remainder. With numbers like 1/3 we can write it as a finite decimal plus a rational remainder. That's what we formally mean when we try to write down decimal expansions of π, root 2, 1/3, etc. In the case of 0.999... we write it as a finite decimal plus some remainder expressed by a limit. People are mixing up equalities and limits. It's a subtle difference but people were moving between one and the other as if they were the same. I am a pure mathematician and I prefer formal rigour to handwaving. Now I'm sure that some people will immedietly attack that comment, by saying "It's not handwaving, it's a fact you idiot", or something along those line. But it's not a fact, it's shorthand. The discussion has totally got out of hand, and is nothing to do with the article anymore. People started to put words into my mouth, so I replied to that. Then some people didn't think that people were putting words into my mouth, so I replied with evidence, and this has continued ad nauseum. Most of my replies that been to clear up things people have said about what I have said. I have tried to bow out of this discussion a few times, since it has degenerated into nonsense. I shall attempt to do the same thing again... although I am sure that someone will have something to say that prompts a response from me.  Declan Davis   (talk)  14:16, 23 September 2008 (UTC)

"1" is a shorthand for "the successor of the natural number postulated to exist by the fifth Peano axiom" (based on the version of the axioms in that article, I know not everyone formulates it exactly the same way). Are you suggesting we shouldn't use the notation "1"? Mathematics is full of shorthands, they are what make it feasible to actually use the concepts for anything. As long as the shorthand is well defined, as "0.999..." is, there is no reason not to use it. --Tango (talk) 14:42, 23 September 2008 (UTC)

What I understand Davis to be saying is that you should only use shorthand if you are certain that the reader will understand what said shorthand really means. It is unfair of you, Tango, to take this to the ludicrous extreme of suggesting that a wikipedia reader might not know what "1" means. There are reasonable and unreasonable suggestions for confusion. If you are roman you might think that IV=4. Others might say that IV=Intravenous therapy. Many writers here seem to have no regard at all for their readers. But an encyclopedia exists to inform people; It's pointless to be "right" in a way that does not inform.Algr (talk) 18:55, 23 September 2008 (UTC)

Tango: I've just read the article on the Peano axioms, and the choice of 0 or 1 is simply a choice of the base case for the inductive argument. The article also says that the axioms have been used "almost unchanged", i.e. they have been changed, and that mathematical logic was in "its infancy" when they were constructed. I reject this as a useful source. Algr: Please, call me Declan.  Declan Davis   (talk)  01:32, 24 September 2008 (UTC)

I have no idea what but the limit could be meant by that shorthand 0.999... And the article's first sentence and the "Introduction" section should make it crystal clear. Huon (talk) 19:23, 23 September 2008 (UTC)

Well this is what I meant by my earlier comment [1] about ellipsis - is it simply that he objects to the use of ellipses as "shorthand"?

Would he agree that $0.{\bar {9}}=1$?

Would he agree that $\sum _{n=1}^{\infty }{\frac {9}{10^{n}}}=1$?

I see that he agrees with $\lim _{k\to \infty }\sum _{n=1}^{k}{\frac {9}{10^{n}}}=1$, so it seems to be he's not disputing the mathematical facts here, he's just saying that people shouldn't use certain notation. Is this correct? (Although I'm not sure why one shouldn't use notation that is well defined - I can see a plausible argument regarding the use of ellipsis, but not with these other notations. If you are worried that the reader might not understand, then why do you assume that they do understand the notation that you are happy to use, such as summation signs, infinity, and limits?) Mdwh (talk) 02:17, 24 September 2008 (UTC)

Algr, as an aside, you seem to suggest in the above discussion that limits are somehow ill-defined. Could you elaborate? For you convenience, I will provide you with definitions. I will even use words instead of mathematical symbols, just for you.

$\lim _{X\to p}f(X)=L\Leftrightarrow \forall \epsilon \in \mathbb {Q} [\epsilon >0\Rightarrow \exists \delta \in \mathbb {Q} (\delta >0\land \forall X[0<|X-p|<\delta \Rightarrow |f(X)-L|<\epsilon ])]$.

$\lim _{n\to \infty }X_{n}=L\Leftrightarrow \forall \epsilon \in \mathbb {Q} [\epsilon >0\Rightarrow \exists N\in \mathbb {N} (N>0\land \forall n\in \mathbb {N} [n>N\Rightarrow |X_{n}-L|<\epsilon ])]$

--69.91.95.139 (talk) 11:14, 25 September 2008 (UTC)

You shouldn't use real numbers in the definition until everybody agrees what a real number is.

if and only if for every rational number ε > 0 there exists a rational number δ > 0

Although this leaves uniqueness unanswered. Tlepp (talk) 11:54, 23 September 2008 (UTC)

Good point. Changed accordingly. To that last sentence, I assume you mean the question of the uniqueness of a limit, as in, there is only one L which meets the definition? I would say that is implied by the equality, although, before a particular limit can be used, of course, it would have to be proved that that limit has exactly 1 solution L to the above definition. --69.91.95.139 (talk) 12:01, 23 September 2008 (UTC)

∏håπks ƒδr µs¡n∂ wδr∂s ¡πs†æd δƒ må†hεm冡çål s¥mßδls. Algr (talk)

I hope that's not sarcasm. I can always replace the definition that uses words with that using mathematical symbols if you want.

How long did you spend on that comment, anyway? --69.91.95.139 (talk) 10:08, 24 September 2008 (UTC)

There's probably a webpage somewhere that does it for you. --Tango (talk) 12:39, 24 September 2008 (UTC)

No web page. It's my personal spin on L33T speak. And it is sarcasm. How about words for ε, δ, and the intent for L and p?. As far as I can tell, it just looks like you are going on about completeness some more. Algr (talk)

I have addressed your needs above. However, in answer to your questions:

ε represents how close we want f(X) or X_n to get to the limit L.

δ represents how close we have to get X to p to get f(X) within ε of L.

L represents the value of the Limit.

p is the x-value approached to get the limit of f(X).

You didn't ask, but:

N represents how large n has to be in order for X_n to be within ε of L.

The definition works a bit like a really bad game: I give you an ε, and you give me an N or a δ to get within ε of the limit.

In a way, yes, I am going on about completeness. However, it really has more to do with you having a problem with the limit definition that you don't even know.

Happy bashing. :) --69.91.95.139 (talk) 11:09, 25 September 2008 (UTC)

Algr, it's time to stop this

"as far as I can tell". Algr, let's face facts. At this stage, your convictions here are stronger than your knowledge of maths is capable of rectifying. Someone to whom the 0.999... = 1 equality has little importance may be convinced by one of the "simple proofs" on the article. You, however, evidently believe the equality false with such conviction that it will be impossible, given your currently inadequate understanding of real analysis, for us to present a strong enough argument to convince you otherwise. You don't know enough maths for us to change your mind. So we have two reasonable options here to avoid this pointless conversation going around in circles forever.

• The first option is for you to leave, carry on believing whatever you want regardless of whether it's correct, and sit content in your ignorance-borne beliefs. This option will be easy but unsatisfying, and you will still be wrong.
• The second is for you to learn the maths. This will be more difficult for you as you have demonstrated a strong resistance toward learning any relevant maths in previous exchanges. Every time someone finally puts an unavoidable, unambiguous challenge to you to explain something or confess you don't know, every time someone finally lays out clearly exactly where your error in reasoning lies, you simply leave that thread and carry on elsewhere. So you will have to make a genuine attempt to learn a bit of maths BEFORE we can get to the question of 0.999... = 1. You will have to admit that you know less than other people regarding this issue. Don't worry, it won't be too hard. Nothing involved is beyond a first year university student level. Once you have learned the maths, if you still disagree, we will listen. But only after you have shown the commitment to actually learn what you're talking about.

So you have to make a choice now. I, and I'm sure many other editors here, are getting tired of you skimreading each post, trying to find a particular sentence or two, some little snippet that you can creatively interpret into something you can pull apart, while disregarding everything else. Ask yourself why you are still here, arguing this after all these months. Is it because you are after a bit of inflammatory internet argument with all its laughs, or do you really want to understand the mathematics involved and figure out the truth here? Maelin (Talk | Contribs) 14:22, 25 September 2008 (UTC)

Awww... You had to take the fun out of it!

Sorry, you're right. Of course you are. I absolutely agree that he should get some ecumicashun before he writes another word on this talk page.

But you still took the fun out of it. :( --69.91.95.139 (talk) 22:56, 25 September 2008 (UTC)

Declan Davis had plenty of "ecumicashun", and what good did it do him? You bullied him off of the forums by sheer volume, and pulled me back in because I had to tell you to be civil. So you are stuck with me. Forever. You guys think that the mathematical community is on your side, but your understanding is a parody of real mathematics, and you have all but admitted that this whole article is a con. BTW: When I abandon discussions, it is usually because they are going round in circles, or becoming distracted by the point, or I have a real life to attend to. Algr (talk) 04:58, 26 September 2008 (UTC)

We didn't bully Declan off, he left due to a disagreement of opinions. If you had any idea what he had been arguing, which you don't, you would understand that his issue is totally separate and distinct from yours. The mathematical community is on our side. Go ask any proper mathematician to look at our 0.999... article and then ask them whether 0.999... = 1 is true in the real numbers. I notice that you seem to conclude that discussions are going around in circles just when someone has finally refuted your point in a manner that you can't easily ignore. You are a troll, Algr. You aren't interested in getting to the truth, you are just interested in endless argument. You don't want to learn, you want to argue. Maelin (Talk | Contribs) 05:48, 26 September 2008 (UTC)

I want to learn, but not from people who directly advocate misinforming laymen and intentionally misinterpret what others have said. If you understood Declan Davis then why did you keep doing that? He was perfectly clear to me. The only reason you outnumber us is because every time someone who knows what they are talking about shows up, you guys overwhelm them and drive them away. You guys still have no idea what my objection is, nor do you care. It takes two to tango, and more then two to bully. Algr (talk)

We've watched you long enough to know that your objection is whatever lets you argue with us. For a while it was "we need the hyperreals so that we don't have 0.999... = 1." Before that it was "the real numbers are useless". You swing wildly between whatever claim you can use to float your arguments for a little while.

But maybe you've got a point. If you really think we haven't given you a fair go, prove that you're interested in one by making an effort to meet us halfway. You learn a bit of mathematics, we'll listen carefully and fairly to what you have to say. Or, if you prefer, we could just keep arguing pointlessly in this thread. Your call. Maelin (Talk | Contribs) 09:25, 26 September 2008 (UTC)

Just thought I'd drop by and see how all my old friends are getting on. I'm glad to see you're all getting on well. I shall not discuss any topic of the article since I have declared myself out of the debate, although I would like to say a couple of things. Maelin, you need to read WP:CIVIL, your last comments to Algr were just damn rude. I find you comment "You learn a bit of mathematics..." worst of all. As a 21 year old undergraduate you yourself have a lot of mathematics to learn. I had been tempted to mention this before, and thought it jolly bad form. There's a saying, how does it go? Oh yeah, don't throw stones if you live in a greenhouse. Also Algr was right, I left because I found it impossible to talk sense to anyone, it was like beating my head against a brick wall. Most people didn't seem to even read my posts fully, let alone try to sympathise with them, before getting on to their high horses and giving us all a speech. I have since been warned of the pitfalls to be encountered on these arguments pages.  Δεκλαν Δαφισ   (talk)  18:42, 3 October 2008 (UTC)

I should probably update that. I'm 22 years old now with a bachelors degree majoring in Pure Maths, completing an Honours year. But this is irrelevent because, as I said before, credentials are irrelevent in discussions of mathematics. Furthermore, Declan, you haven't been on this article for the two years that we've been going over the same old ground, over and over, with Algr. We've tried all manner of things. He ignores the rigourous proofs because he doesn't understand them, whilst claiming that the simpler, casual-layperson-targeted proofs are based on faulty reasoning. He latches onto ideas of which he has no understanding (the hyperreals and completeness, to name just two) and tries to use them to support his claims. Then he disappears once you've finally rebutted him in a way he can't weasel around, and turns up a while later jumping on somebody else's posts, claiming that he was busy. There are pages and pages in the archives of editors patiently trying to explain things to him while he employs standard troll tactics and ignores/misuses everything he doesn't understand. We have experienced his "discussion" style often enough to know that he is not interested in argument-as-a-means-of-establishing-truth, he is only here for argument-as-a-form-of-entertainment. Maelin (Talk | Contribs) 06:00, 4 October 2008 (UTC)

I see. Well you're right: I haven't been on this discussion page for two years. If what you say is the truth - which I'm not saying it isn't - then you are right to be annoyed. On a more touchy-feely note: we shouldn't allow ourselves to express these feeling of annoyance. If someone doesn't understand then we should try to be nice and help them understand. If they still disagree then it's often better to agree to disagree, move on, and focus on something more constructive.  Δεκλαν Δαφισ   (talk)  16:24, 5 October 2008 (UTC)

Unfortunately it is often not a case of people not understanding but rather of them not wanting to understand, which there is nothing much we can do about. Moving on is, indeed, often the best option. However, some of these people refuse to move on as well and will continue trying to argue their incorrect points of view in other sections. --Tango (talk) 16:28, 5 October 2008 (UTC)

2 (bis)

"not wanting to understand"? I've explored every explanation you've come up with, and went through the trouble of researching all sorts of new ideas just to be sure I understood what you were talking about. You on the other hand don't even seem to know what my position is. I'm not saying 0.999... = 1 is false, I am saying it is declaratory. The article is written to tell people that =1 is observational, like the speed of light, when it is really more like the speed limit on a highway Do you understand how important that distinction is? Algr (talk) 08:01, 11 October 2008 (UTC)

0.999... = 1 is declaratory in the same way as 1+2 = 3 is declaratory or as it is declaratory that the sum of a triangle's angles is 180°. There are lots of times we've been through this. Still, this is an encyclopedia, and we report on "declaratory" truths just as on observational ones. Huon (talk) 08:36, 11 October 2008 (UTC)

No, 1+2=3 is observational - any real world scenario would in some way reflect this result. (Unless you change the meaning of the symbols, but that would be arguing for arguing sake.) The sum of a triangle's angles could go either way - the measuring unit of a "degree" is declaratory, but the result of putting three angles from a flat triangle together is observational. It is impossible to understand the world if you can't distinguish observational from declaratory data. Algr (talk) 09:53, 11 October 2008 (UTC)

The definition of "3" is just as declaratory as that of "°" or "0.999...", and any real world scenario would reflect that 0.999...=1, too. But any real world scenario would reflect that we should use spherical (or, possibly, hyperbolical) geometry, not Euclidean geometry, where the sum of a triangle's angles is anything but 180°. Huon (talk) 09:59, 11 October 2008 (UTC)

I might agree that 1+2=3 is declaratory, since (ignoring real world considerations and models and just sticking with pure maths) 3 is defined to be the successor of 2, and "1+" is defined as taking you from a number to its successor. 0.999...=1, however, is not declaratory, it follows from the definition of the real numbers (which is declaratory, but makes no explicit mention of 0.999...). Neither is the fact that in flat space a triangle has 180 degrees, that follows from the axioms of Euclidean geometry and the definition of a degree. Both of those are declaratory, but neither explicitly mentions the sum of the angles of a triangle. In mathematics pretty much everything boils down to choosing an appropriate definition and then seeing what follows from it. Those definitions are declaratory (although usually chosen because they are useful for describing the real world) but what follows from them are not. --Tango (talk) 13:49, 11 October 2008 (UTC)

Just for the fun of it, I might define operations $\oplus$ and $\otimes$ on the real numbers, where a$\oplus$b := max(a,b) and a$\otimes$b := a+b. Let's add the element $-\infty$ to our number set, with the obvious rules that $-\infty \oplus$a=a, and $-\infty \otimes$a=$-\infty$ for all a. Then I have a number system that has more numbers than the reals, has two operations which are commutative, associative and distributive, has a neutral element of "addition", and the set without the neutral element of addition is a group under "multiplication". In effect, my attempt at a number system is at least as well-behaved as any system of almost-reals Algr might invent where 0.999... is not equal to 1. And 1$\oplus$2=2. So when 0.999...=1 is considered declaratory (which in effect means that the meaning of 0.999... is not observational; we could choose some other meaning than the limit of the sequence (0.9, 0.99, 0.999, ...), though I'd call that "arguing for arguing sake", too), so should 1+2=3 be considered declaratory, because I might choose some other meaning for the operation. Concerning the triangles, I agree that in Euclidean geometry, for flat triangles, the sum of angles can be proven to be equal to 180°. But to have a "triangle" denote a flat triangle is a special case, and if other sets of triangles (spherical or hyperbolic) were to become the "standard" triangles, we'd consider the statement "The sum of angles in a triangle is greater (or less) than 180°" true. Thus, while possibly not declaratory itself (actually the statement on triangle angles in full generality is simply wrong if I don't specify the geometry), the obvious interpretation is declaratory. This is just the same as interpreting "0.999...=1" to mean "The real number that's the limit of the sequence (0.9, 0.99, ...) equals 1". Huon (talk) 18:14, 11 October 2008 (UTC)

What did I just say about changing the meaning of the symbols? Are you sure I'm the one using "troll tactics?" Language and symbols are always declaratory, but the ideas that they represent may not be. I specified flat triangles because I knew you might pull out spherical & hyperbolical geometry, but you went through it all anyway. It changes nothing, because the geometry invariably dictates the sum of the angles, regardless of how they are measured. Maybe you two genuinely don't understand the difference between observational and declaratory facts. Algr (talk)

Huon said: "the statement on triangle angles in full generality is simply wrong if I don't specify the geometry" O'RLY? Then why is it proper to assume that .999... refers to the real set when that is not stated, and is in fact directly contradicted by the description of non-zero infinitesimals inherent in the question? Algr (talk) 22:23, 11 October 2008 (UTC)

Because there is no other reasonable interpretation of 0.999..., and I have no idea what you mean by the second part. --Tango (talk) 22:27, 11 October 2008 (UTC)

Algr, if you propose that there's some useful number system where 0.999... is not a real number and differs from 1 by a non-zero infinitesimal, you should give that number system. Unless you do that, your claims can't be discussed. If you're interested, I'll gladly help you construct such a number system, but I predict both a lack of usefulness and serious flaws that will be hard to reconcile with the "real world" (such as a lack of subtraction). Anyway, just as the triangle article mostly covers the standard flat case (stating that the sum of a triangle's angles is 180°), with just a short paragraph on non-flat triangles, so this article should (and does) mostly cover the standard case of real numbers, with short paragraphs for other number systems. Huon (talk) 22:59, 11 October 2008 (UTC)

Hello again everyone. I was just passing by and thought I'd pop in and say hello. This discussion thread seems to have got a little silly. It's nothing to do with the article, and is simply an argument. Let it go people. Take some deep breaths and reassure yourselves of your own intelligence. Come on, this is silly. There are thousands of maths articles that need working on. Focus your energies onto improving Wikipedia.

 Δεκλαν Δαφισ   (talk)  14:15, 13 October 2008 (UTC)

It's meant to be an argument, take a look at a title of the page! ;) --Tango (talk) 14:23, 13 October 2008 (UTC)

I have read the title of the page, and moreover the discription of this page: "This page is for mathematical arguments concerning 0.999..." There's a difference between "an argument" (i.e. a heated exchange of diverging or opposite views) and "a mathematical argument" (i.e. a set of reasons given in support of a mathematical idea). I think you reply is a perfect exemplification of my point. Thanks Tango.  Δεκλαν Δαφισ   (talk)  15:48, 14 October 2008 (UTC)

";)" represents a winking face and is used online to denote a joke. --Tango (talk) 15:59, 14 October 2008 (UTC)

I see. I thought for a moment that it was some strange new puntuation mark, or that you'd lost all ability to punctuate correctly ;o)  Δεκλαν Δαφισ   (talk)  19:54, 14 October 2008 (UTC)

Another number system is beside the point. (Although I have been working on one.) When has it ever been necessary to provide a correct answer in order to prove another idea wrong? I'm going to try another approach to this, and post in talk an alternative opening to the article. Algr (talk) 02:11, 22 October 2008 (UTC)

You're not trying to prove an idea wrong, remember? You don't dispute that 0.999... = 1 in the real numbers, which is the number system we explicitly mention we are using (and which is also, by far, the most widely-used number system of its type on Earth). Your dispute is that the real numbers are not the system everyone should be using. You claim that this number system is not good enough, and in order to make that claim, you need to provide a superior alternative. Maelin (Talk | Contribs) 05:45, 22 October 2008 (UTC)

Good enough for what? I don't even understand that thought, let alone support it. Algr (talk) 06:18, 22 October 2008 (UTC)

If you're not suggesting another system but the real numbers, and accept that in the real numbers 0.999...=1, why are we even having this discussion? Huon (talk) 10:35, 22 October 2008 (UTC)

The End (for me at least)

I am glad that so much conversation and debate has been generated about this article. I have learned a lot about myself, Wikipedia, and the human disposition over the last few days on this page. I think that we can all learn some lessons. Many of us (including myself) are too quick to judge, and too quick to dismiss. Many of us (including myself) have been rude and offensive. Many of us (including myself) have tried to put our opinion forward as fact. These are all natural things to do. I know that I shall try to learn from this discussion, and try to stop doing some of the bad things and try to do more of the good things. I guess one of the problems with mathematics is that it's too black or white, too right or wrong, and so our natural human competitiveness can come out in a bad way. I have seen this in every mathematical discussion page that I have read. All I can say is that some of us should agree to disagree. We should carry on working to improve Wikipedia. And finally, I hope that there are no hard feelings between any of us. After all: we're all on the same team.  Declan Davis   (talk)  02:39, 24 September 2008 (UTC)

I'm sorry to see you go, Declan. I think that a paragon shift is in the cards for mathematics soon. (Or at least the teaching of it.) It certainly won't be lead by me, but you may be a part of it. Algr (talk)

You right Declan. Lot of the people are fixed with their ideas and are not ready to change how they think. They like very much to say that they are correct and others are incorrect. I think you are wasting your effort trying to be nice. You have offered an 'olive branch' and no one (except Algr) have accepted it. Instead they ignore and they carry on to argue, almost like the wolfs that they eat the sheep. This is for why I do not write on talk pages anymore. Dharma6662000 (talk) 18:54, 24 September 2008 (UTC)

"This is for why I do not write on talk pages anymore.". Oops. 82.41.244.42 (talk) 20:38, 24 September 2008 (UTC)

Sorry. It is my bad English. What I wanted to say was that I do not write my ideas anymore. I had to write here because I have strong feelings about this and I need to say. Actually, you help to demonstrate my point. You rather laugh at what I write than understand what I mean. You understand the point I made, but you are happier to laugh and make yourself feel happy. This is exactly the point I was making, and you give example... thank you. Dharma6662000 (talk) 20:46, 24 September 2008 (UTC)

Declan's opinion simply does not reflect mathematical usage and terminology, despite his educational background. I can't imagine that we should pretend this is all a matter of opinion (rather than a matter of well-settled convention) for the sake of being nice. You may feel otherwise. Phiwum (talk) 21:27, 24 September 2008 (UTC)

Your profile it say that you study philosophy. For why you write mathematics page? Dharma6662000 (talk) 21:40, 24 September 2008 (UTC)

Because he/she is interest in mathematics, perhaps? The two subjects are actually quite closely related - they both depend heavily on logic. --Tango (talk) 23:30, 24 September 2008 (UTC)

As Tango says, it is possible to study both. You say something about my "profile", but I don't know what you mean. Perhaps you found a link to my home page, but if you did, you would learn that I'm not a mathematical novice (though I also do not have a PhD, as Declan has). In any case, I certainly don't ask that anyone rely on my authority. Phiwum (talk) 23:56, 24 September 2008 (UTC)

82.41.244.42 should take a look at WP:CIVIL, that comment was unnecessary. But since when has it become a custom to ask editors about their background? Editors should be judged by their contributions. Phiwum and Tango discuss mathematics, and to the best of my knowledge, their opinion is the mainstream one. If necessary, they could probably cite books to that effect. On the other hand, Declan's claims (for example, that irrational numbers shouldn't have decimal expansions) are anything but mainstream. For example, Cantor's diagonal argument for the uncountability of the real numbers exlicitly requires us to assign decimal expansions to (at least some) irrational numbers - in fact, it requires that any infinite sting of digits is indeed a representation of a real number. Appeals to authority won't help his case. Huon (talk) 23:29, 24 September 2008 (UTC)

It is the same. Declan have left the talk, you all say you are correct, so why you need to carry on talking? You are sure that you is write and he is wrong. Why you carry on to eat the sheep like a wolf? I do not see that you disagree with fact, but only with notations and conventions. This is a small thing. Why not move on? This is what he say earlier: "We should carry on working to improve Wikipedia." You say that you are right and he is wrong, so move on! Dharma6662000 (talk) 23:53, 24 September 2008 (UTC)

It's not really relevant to your point, but what a standard convention is is a fact. --Tango (talk) 23:57, 24 September 2008 (UTC)

No, Tango, this is not true. For some the natural numbers include zero, for others they do not include zero. A convention is a choice. You all agree on fact, but Declan does not like decimals and he likes limits. Others they say that decimals are OK. The fact is the same, the choice of view is different. I asked my professor about the limit question and he has just relied: "Yes, certainly, infinite decimals are limits. That is the only way to make sense of them. They say that, given ε > 0, there exists n(ε) such that the finite decimal with n places differs from a particular real number by less than ε. Of course this begs the question of the definition of a real number, but that is another story..." Dharma6662000 (talk) 00:09, 25 September 2008 (UTC)

No one disagrees that they are limits. I think you may have overinterpreted what your prof said. --Trovatore (talk) 00:13, 25 September 2008 (UTC)

Mike, you may missed something higher. Declan said (a long time ago) that infinite long decimals aren't defined and they are really limits. Most people said no. This is what I say. I know they are limits, you know, my prof know, Declan know. But people they like to argue, they like to be right, they like to be king. This is why they are how they are. Dharma6662000 (talk) 00:22, 25 September 2008 (UTC)

An infinitely long decimal, strictly speaking, denotes rather than is a limit, but most informed people asked whether it "is" a limit will elide that point and say yes. The limit that it denotes is a real number. There is no contradiction between something being a limit and being a real number. --Trovatore (talk) 00:28, 25 September 2008 (UTC)

I didn't say "convention", I said "standard convention". There is no standard convention on whether or not zero is a natural number, there is a standard convention on how to interpret a recurring decimal. Whether or not a standard convention exists is not a matter of opinion, it is a matter of fact. You can be of the opinion that the standard convention is a bad one, but you can't be of the opinion that it doesn't exist. --Tango (talk) 01:41, 25 September 2008 (UTC)

No I agree. The existance of a convention is a fact: it exist or no exist. But earlier you say "but what a standard convention is is a fact." You say that a standard convention is a fact. This is why I say that no is true. Existance and validness are different. Dharma6662000 (talk) 10:04, 25 September 2008 (UTC)

No, you're misinterpreting me. I think we agree completely. When I say "what a standard convention is is a fact" I mean the fact is "what the standard convention is", not "the standard convention". "It is a standard convention that recurring decimals denote a limit" is a fact. "Recurring decimals denote a limit" is not. --Tango (talk) 14:40, 25 September 2008 (UTC)

No, I no think that I misinterpreting. Please read what you has written in above. Maybe you talk too heavy and maybe you is changing mind, but that OK. I have dictionary and some time I understand what you say better that you do. What man write in moment is better shine on soul than what what man say after he thinks! I no like these English pages. For me it look like every English man he hate his brother. I never will understand the anglosaxon man. You was warriors in past, but time changes, why you no all be friends? Declan he try to make peace, but you all like to argue more. It very crazy. I hope you no insult by that what I say, but is what I say, is truth! Dharma6662000 (talk) 22:37, 25 September 2008 (UTC)

Where did your ability to speak English go? :-/ I'm pretty confident that I wrote what I meant, it's just a little confusing. Perhaps if I add some brackets it will be clearer: "(what a standard convention is) is a fact". That's distinct from "(that which is a standard convention) is a fact". --Tango (talk) 23:08, 25 September 2008 (UTC)

And where go your understanding of Civility? I may no be English, but I am not idiot. You write what you write. Maybe I should ask "Where your ability to write what you mean go?" Fact is fact friend. But actually no. You all seem to change fact when it please. Before you answer please read what you has wrote. I know you will not, that is why your next answer will be as silly as last. Dharma6662000 (talk) 02:59, 26 September 2008 (UTC)

Sorry, Dharma, I think it's pretty clear that you are the one misinterpreting Tango here. He wrote "what a standard convention is is a fact". The repeated 'is' makes his meaning clear: "the standard convention is x" is a fact. Your claims that you understand what he wrote better than he does are sheer arrogance on your part. Maelin (Talk | Contribs) 04:39, 26 September 2008 (UTC)

What a stardard convention is by wiktionary?

standard (adjective) 1. Falling within an accepted range of quality.

convention (noun) 1. A meeting or a gathering. (The convention was held in the arguments page.)

The arguments page convention has fallen outside an accepted range of quality. And that is a

fact (noun) 1. An honest observation.

Tlepp (talk) 06:43, 26 September 2008 (UTC)

I'm sorry if my comment about your English offended you, I'm just confused - you were writing in perfectly good English before and now suddenly your writing is full of mistakes. What changed? --Tango (talk) 15:33, 26 September 2008 (UTC)

Tango, a year ago I was trying to tell you what you just said right here, and you totally rejected it. Algr (talk) 09:38, 25 September 2008 (UTC)

Really? Could you provide a link? --Tango (talk) 14:40, 25 September 2008 (UTC)

It takes two to tango. For others may I suggest an article by Timothy Gowers.

A dialogue concerning the need for the real number system

or

The existence of the square root of two

Tlepp (talk) 10:23, 25 September 2008 (UTC)

These articles are quite good. Will respond later. Algr (talk)

In fact, you guys REALLY ought to read them yourselves... Algr (talk) 02:33, 26 September 2008 (UTC)

Done. Pretty interesting. --69.91.95.139 (talk) 11:00, 26 September 2008 (UTC)

Goodness, what the f*** was that all about? Formally, the statement 0.999...=1 is intended to mean the following: "the equivalence class [A_n] is equal to the equivalence class [B_n]", where A_n is the Cauchy sequence

$A(n)=\sum _{k=1}^{k=n}{\frac {9}{10^{k}}}\ ,$ and B_n is the Cauchy sequence B(n) = 1 for every n. Here the equivalence relation is on the Cauchy sequences of rationals, and is called "co-Cauchy-ness". That is, for two Cauchy sequences of rationals X_n and Y_n are co-Cauchy if for any rational ε > 0, for all n sufficiently large, |X_n - Y_n| < ε. The above equation *literally* means nothing else than the set of sequences of rationals co-Cauchy to A_n is equal to the set of sequences of rationals co-Cauchy to B_n.

Assuming you are willing to subscribe to the Cauchy sequences construction of the real numbers, this resolves all your disputes that make sense. —Preceding unsigned comment added by 212.183.136.192 (talk) 19:36, 23 October 2008 (UTC)

If people accepted any of the various standard constructions or properties of the Reals (or even just one, say, the Archimedian property) then all disputes would be resolved. People don't; that's why the argument exists. This page is like Cthulhu - if you're already happy that 1=0.999... then close your eyes and walk away :) Endomorphic (talk) 13:51, 27 October 2008 (UTC)

Fraction Issue

I've got a question involving the use of $1/3=0.333...$ It seems to me that 0.333... does not truely represent $1/3$. It is merely the closest we can get to $1/3$ using decimal numbers. It would be correct in saying that lim_x→3 x(0.333...) = 1, however 0.333... is not a limit.

0.333... is the limit of the sequence (0, 0.3, 0.33, 0.333, ...), which happens to be 1/3. Saying that lim_x→3 x(0.333...) = 1 is of course correct, but so is saying that 3 * 0.333... = 1. Huon (talk) 20:35, 18 November 2008 (UTC)

In fact since multiplication is continuous, they're equivalent. 82.41.43.170 (talk) 20:41, 18 November 2008 (UTC)

Number Lines

Well, not specifically about number lines, but since that is where the < and > symbols are used most, I'll use it as an example.

Suppose |x| < 1. I was told that this could include any number up to but not including 1, but it could include a number that was an infinitesimal amount smaller than 1, as long as it did not equal 1. Could someone please define said number, if it is not 0.999999...? What would it be? 0.99999...9999999...999999...9998? --58.161.211.14 (talk) 08:52, 6 February 2009 (UTC)

The real numbers don't have infinitesimals, whoever told you that was probably talking nonsense. (They might have been talking about some other obscure number system like hyperreals, but I doubt it - they aren't mentioned very often, except on this page!) --Tango (talk) 10:22, 6 February 2009 (UTC)

To 58.161.211.14: the difficulty you have found in defining a number just infinitesimally less than 1 -- or to get anyone else to do so -- is perfectly understandable. This is because -- if you want the generally understood definitions of arithemetic to apply consistently and real numbers to behave in various other reasonable ways -- there is no way one can exist. As Tango says above, you can create a system which can contain such a number, if you really want to, but only if you are willing to break one or more of the conventionally accepted laws of arithmetic.

The article goes into this in some detail, but the really solid proofs that there isn't such a number require the techniques of real analysis to understand completely. The article has references to other articles and external resources which explain these concepts in much more detail.

For more information, you might want to read Talk:0.999.../FAQ, which contains answers to the most frequently asked questions about this topic. -- The Anome (talk) 15:04, 6 February 2009 (UTC)

User 58.161.211.14, there isn't a real number just less than one. Think about making a journey, say a journey of one mile. One the first day you traveled half of the distance, then on the second day you traveled half of the remaining distance. Each day you travelled half of the remaining distance between where you were and your destination. Every day you would always move a little closer to your destination, but you would also never reach your destination (you're only ever allowed to move half of the way there). This example shows that there are arbitrarily many points between two given real numbers (starting point and destination), and that you can move as close as you like to a given real number without actually getting to it. The set of real x such that |x| < 1, i.e. -1 < |x| < 1 is an example of an open set. These were really mind-bending ideas to me back in the day, but after a bit of thought and a lot of examples you'll get it.  Δεκλαν Δαφισ   (talk)  23:39, 9 February 2009 (UTC)

"there is no way one can exist." - The Anome

Why is this even a problem? I can name three accepted mathematical concepts that are built on things that can't exist:

1) Infinity. Come back when you count to it.

2) The square root of negative one. This directly contradicts the fundamentals of basic mathematics. But once you learn to ignore that, everything is fine.

3) Two random points on a dart board. Since points have zero size, their are an infinite many points on a dart board. So the chance of two random darts hitting the same point are zero. Unless you define the second dart as hitting the same point as the first one. Then a zero probability event is guaranteed!

So given those examples, what is the problem with defining a maximum value for X<1? We would just assign it a variable ø and move on. Algr (talk)

You can define all kinds of things which, on the face of it, don't make sense in the real world. They do, however, need to be consistent with the rest of the mathematics you are using. If you define such a real number ø (and it would be a constant, not a variable), it would lead to contradictions. For example, if ø is maximal then what is 1-(1-ø)/2? If it's either 1 or ø, that contradicts ø being less than 1, if it's somewhere inbetween that contradicts ø being maximal. So, the only way you can introduce such a ø is to say that basic arithmetical operations don't work on it, which contradicts the real numbers being a field. --Tango (talk) 11:22, 10 February 2009 (UTC)

For #1 and #2, neither infinity nor the square root are real numbers, which is why Declan Davis specified "there isn't a real number just less than one". For #3, see almost.

Sure, you can define a, say, ø to be the largest number infinitesimally less than 1, but then you have to define what happens when you do arithmetic on ø - what happens when you divide by 2 and add 1/2? Is it greater than ø but less than 1? That would be what we mathematicians call a proof by contradiction that ø doesn't exist, something for which does not exist for infy or i.

Sure, you could probably define away the contradictions, but then you have a very messed up number system containing a very messed up number, and this is irrelevant to a discussion on 0.999... Are you trying to argue that 0.999... should be defined as this number "ø" instead of 1? The problem with that idea is:

1. Mathematicians have already defined 0.999... = 1. It's usually a bad idea to define something as two non-equivalent concepts.

2. The number system you'd need to construct with be so convoluted and inconsistent that it would not be worth studying, and there'd be no benefit of constructing it in the first place.

Am I missing something? --Zarel (talk) 20:18, 10 February 2009 (UTC)

Algr, I'm not quite sure that you have explained yourself properly. Could you please say more. Also, #3 is totally untrue. Two random points on a dart board make perfect sense, as do two random points on a line. I assume you have made some kind of attempt at sarcasm. That's not a very constructive way forward, and it doesn't seemed to have worked either. Please be more patient and logical when it comes to making posts.  Δεκλαν Δαφισ   (talk)  22:08, 19 February 2009 (UTC)

Algr's Number System

Ok. Let's actually get Algr to decide what properties his number system should satisfy.

(I'm talking to Algr here)

It won't be easy, but this is exactly what people who extended the Real numbers to include infinity and/or complex numbers had to worry about. It wasn't just a simple matter of deciding to give -1 a square root (although in that particular case, it turned out to work that way very nicely).

I suggest you take a look at this page on the construction of the real numbers for some help on deciding what kinds of properties should be satisfied. I have included the first couple properties which you seemed to suggest in the above discussion. You will need to consider whether you want to keep such properties as associativity, commutativity, and closure of addition and multiplication, existence of additive and multiplicative inverses, the properties associated with the ≤ operator, etc. Note that the additive and multiplicative identities are in the real numbers, so those properties are already satisfied.

(Now I'm talking to everyone)

Let's keep the list of properties at the end of this section (that is, the one titled "Algr's Number System"), so any further discussion should be placed between my comment and the list.

Here's looking forward to what Algr will come up with, --72.177.97.222 (talk) 12:50, 10 February 2009 (UTC)

I don't expect Algr to give detailed mathematical constructions. Here's my suggestion:

A=R(x) is a transcendental extension of degree one over reals. Elements of A are represented by P(x)/Q(x), where P and Q are polynomials with real coefficients. We may assume that P and Q have no common factors and the lowest degree term of Q has coefficient one. We define element $(a_{0}+a_{1}*x+...+a_{m}*x^{m})/(x^{k}*(1+b_{1}*x+...+b_{n}*x^{n}))$ to be greater than zero if either a₀ > 0 or ( a₀ = 0 and a₁ >= 1 ).

I leave it to you to figure out what properties are broken. Is 1-x a maximal element of the set defined by '<1'? Tlepp (talk) 18:47, 10 February 2009 (UTC)

I don't think it's even possible to say if it's maximal or not since one of the properties you've broken is ">" being a partial ordering. Consider 1 and 1+0.5x. 1-(1+0.5x)=-0.5x, -0.5<1 so -0.5x<=0, so 1<=1+0.5x, but (1+0.5x)-1=0.5x, 0.5<1, so 0.5x<=0, so 1+0.5x<=1. That means 1=1+0.5x, so unless you want to take equality as being broken (and I'd really advise against it!), your ordering doesn't work and there is no concept of maximality. --Tango (talk) 19:09, 10 February 2009 (UTC)

Yes I've broken total ordering. A<=B if and only if (A=B or B-A > 0) which is not same as (A is not greater than B). "0.5<1, so 0.5x<=0" is not true for the partial ordering I defined. Tlepp (talk) 18:52, 11 February 2009 (UTC)

Sorry, yes, you've broken total ordering, not partial ordering. So there is a concept of maximality, just not a particularly useful one. I don't know if 1-x is maximal in {a<1}, ignoring elements that aren't comparable to both 1 and 1-x, I'd have to think about it. --Tango (talk) 19:58, 11 February 2009 (UTC)

Why are we discussing something called "Algr's Number System"? What place does such a discussion have on Wikipedia? Is this original research? If so, it doesn't belong here. Is this published research? Why aren't we citing it? --Zarel (talk) 20:01, 10 February 2009 (UTC)

This page is something of an exception to the usual rules about what is appropriate discussion for Wikipedia talk pages. If this page wasn't here, we would be dealing with all these arguments on Talk:0.999..., whether we wanted to or not. This discussion in intended to help Algr, and others, understand why 0.999...=1 and why there isn't really any useful alternative. We're currently doing that by discussing the flaws in a possible alternative. --Tango (talk) 20:07, 10 February 2009 (UTC)

Ah, I understand. I thought this page was Talk:0.999.... --Zarel (talk) 20:24, 10 February 2009 (UTC)

It's certainly possible to construct a number system satisfying most properties Algr wants satisfied. I'm not sure how actually constructing such a number system will convince him of anything pertaining to 0.999..., though. It seems everyone is just arguing in the wrong direction. --Zarel (talk) 20:24, 10 February 2009 (UTC)

Algr is arguing that 0.999... ought to be defined to be something infinitesimally less than 1, we're trying to point out that doing so would be completely useless. --Tango (talk) 20:33, 10 February 2009 (UTC)

Isn't it simpler to point out that no one defines it that way? I mean, the Arguments section is for convincing people that 0.999... = 1. And if we wished to explain why it's not defined this way, shouldn't there be easier ways than the construction of a failed number system? --Zarel (talk) 20:48, 10 February 2009 (UTC)

Simpler, sure, but less convincing. "That's just the way it is" is hardly a strong argument. --Tango (talk) 20:54, 10 February 2009 (UTC)

To add to that, Algr has been here for a LONG time. I'm just trying to come up with more creative, more insightful, all-around better methods of attempting to convince him (or her? I really don't know) that he's simply wrong. I don't expect him to come close to being able to come up with a full-fledged number system (certainly not a useful one), but I have invited him to try and do so. A waste of time, perhaps, but I do occasionally have some spare time, so I'll try anyway. --72.177.97.222 (talk) 00:00, 11 February 2009 (UTC)

Well, here's one possible number system:

We shall call it: R ∪ ø

Arithmetic with ø shall be defined as the same as arithmetic with 1, with this exception: ø < 1

I think that's enough for the system to be internally consistent. --Zarel (talk) 20:48, 10 February 2009 (UTC)

Minus 1 from both sides, you get 0<0, that's a contradiction. --Tango (talk) 20:54, 10 February 2009 (UTC)

With two exceptions, then: ø < 1 and the properties of arithmetic on inequalities don't apply to ø. --Zarel (talk) 21:14, 10 February 2009 (UTC)

That's cheating! But seriously, you've just shown what we're trying to show - any attempt to do this results in you having to accept that some properties of the real numbers won't apply to this new set. However, I'm not sure your number system actually contains any (non-zero) infinitesimals since 1-ø=0, so you can't really say that ø is infinitesimally less than 1. --Tango (talk) 21:07, 10 February 2009 (UTC)

Well, of course some properties of the real numbers won't apply; that's true of any set. Not every square of a complex number is a positive number (I've lost so many points on math homework proofs when I forget I'm dealing with complex numbers and assume this).

And of course my system doesn't actually contain any nonzero infinitesimals - it doesn't need to.

The more important point is that it's perfectly possible to construct a set like R ∪ ø (or any other set satisfying most properties one wants), but it's relatively pointless to do so.

(And SineBot needs to be less zealous. It changed my signed comment into an "unsigned" comment!) --Zarel (talk) 21:14, 10 February 2009 (UTC)

Can you formalise what you mean by "Arithmetic with ø shall be defined as arithmetic with 1, with this exception: ø < 1", please? I'm not sure exactly what you mean. Maelin (Talk | Contribs) 01:21, 14 February 2009 (UTC)

I interpreted it as meaning "ø*x=1*x" for any standard arithmetic operation, *, and x in R ∪ {ø}. That actually introduces other problems, though - both addition and multiplication lose their identities (1ø=0+ø=1, rather than ø). And, without an identity, you don't have a meaningful definition of inverses, so subtraction and division cease to be well defined. --Tango (talk) 17:43, 14 February 2009 (UTC)

I saw two ways of interpreting it. One way, Tango described above. The other is to associate with every x an "infinitesimally smaller number" and an "infinitesimally larger number" (that is, x+ε and x-ε, where ε is a positive infinitesimal). This also presents its own problems. For example, what is ε*0.1? Is it ε, or is it yet another, smaller infinitesimal? If it is the former, what should (x+ε)*(y-ε) be? (perhaps it should depend on the relative magnitudes of x and y?) If it is the latter, then ø ≠ 1-ε. --72.177.97.222 (talk) 00:26, 15 February 2009 (UTC)

Well, we need not decide that introducing this ø has broken arithmetic. We need simply observe that if all arithmetic operations on ø give the same results as arithmetic operations on 1, then necessarily 1 = 1 × 1 = 1 × ø = ø, so we have only broken the strict inequality symbol (since ø < 1 but also ø = 1). The infinitesimal thing, I think, is going to cause more complicated problems. Maelin (Talk | Contribs) 03:37, 15 February 2009 (UTC)

Wait, why should 1 × ø = ø? 1 may just cease to be the neutral element of multiplication. I see this number system as "R with a double point at 1", with both points behaving just as 1 should, except that ø is, by definition, less than 1. A non-Hausdorff topological space which is no longer a ring or even an additive group, but otherwise almost well-behaved - addition and multiplication are still commutative, associative and distributive. We might even reintroduce a neutral element of multiplication by defining ø × ø := ø (instead of 1). Unfortunately, I don't see how we can have both a neutral element of addition and associativity.Huon (talk) 11:25, 15 February 2009 (UTC)

Huon has the right idea. I meant it the way Tango described it: For every function f, f(ø) := f(1) except the identity map. The problems 72 and Maelin describe are natural results of what happens when I introduce only a single number infinitesimally smaller than 1 to the reals. It's much easier than allowing multiple infinitesimals, upon which arithmetic must be defined - instead, I can just introduce ø define away inconsistencies.

(By the way, 1 × ø = ø is not necessarily true because ø is not a real number. This means that f(x) : x -> 1*x is no longer an identity map, but oh well, no one says the existence of a multiplicative identity is a requirement of a number system. And if it is, as Huon points out, it can be ø, although that would probably introduce unnecessary further inconsistencies.)

R ∪ ø, is, by the way, clearly not a group at all. You can't arbitrarily inject a single element into a group and expect it to retain its group axioms. An identity element cannot exist, since by definition no operation can produce ø. --Zarel (talk) 03:34, 16 February 2009 (UTC)

"<" is the one thing we can't break, since it is used in the requirements for the system (that there exist a maximal element less than 1). --Tango (talk) 12:54, 15 February 2009 (UTC)

Wow, what a great discussion! People are politely arguing their points, learning new things from others, accepting when another points out their mistakes, and so on! This section is certainly the epitome of an ideal exchange. There certainly isn't anything missing from this discussion, like, say, a guest of honor or anything.

In all seriousness, though: I can see you Algr. You are clearly active on Wikipedia about now. You don't have to contribute to the list of properties; we just want to find out more about the number system you were proposing earlier. That's what this discussion is for.

We're waiting. --72.177.97.222 (talk) 22:39, 13 February 2009 (UTC)

Whoo, I think you're being a bit impolite, yourself. I mean, WP:AGF; you could at least try asking on his talk page again, before implying all "Algr must be avoiding us!" --Zarel (talk) 00:43, 15 February 2009 (UTC)

Fair enough. I don't think I was really being mean about it, but I'll go ahead and ask on his talk page. --72.177.97.222 (talk) 01:18, 15 February 2009 (UTC)

---

Let's call Algr's number system A. These are the properties:

1. A includes the Real numbers. That is, $A\supset R$.
2. There is a maximal element of the set of all numbers X in A satisfying X < 1, which we will call ø.

Base Infinity

Hi all, I'm sorry I haven't participated in this, but real life is calling again. In any case, I didn't think that inventing a new number system was really central to what I have been trying to say here. (And I don't think it is good to let one article totally dominate what I do.) But since everyone seems to be waiting for me, here is the quick and dirty notion:

Base infinity starts like base ten or any other base, with a series of places, and digits to put in the places. But if the base is infinity, then the "ones" place contains the entire set of Reals. (Rationals might be all that is needed.) This has the minor inconvenience of requiring an infinite number of symbols to write, but this can be fixed with the notation below. The place that is "Tens" in decimal would be the countable infinities, and the place that would be the "tenths" becomes the first infinitesimal place.

To fix the infinite symbols problem, I'll use any representation of rational (real?) numbers as a digit. The underscore _ marks where each digit ends, and the next begins. Since Reals need a decimal point, I'll instead include "®" to mark the Real/Rational place. (Is everyone seeing that as an R in a circle? I know text changes on different OSes I'll have to use R if ® doesn't work for everyone.) So:

Note: I've improved how the notation works since this discussion took place. I'm leaving this as is for clarity. In the new system, the reals are to the left of the ®, and the ® behaves like a decimal point.

1 = ®1

1/2 = ®(1/2)

1/0 = 1_®0

.999... = ®1_(-1) (Note that each place has it's own sign.)

edit: ...999. = 1_®(-1) (Thus the Real place conforms to the Real number interpretation.)

®1 - 1_®0 = -1_®1

edit ®1_(-1) / ®2 = ®(1/2)_(-1/2)

Generally, each place sees the place to the left as infinity, and the place to the right as unity of whatever operation is active. So in the last case, the infinitesimal ®_-1 or ®_(1/2) does not effect the value of the real place when treated as a real. For the same reason, any form of infinity divided by 2 remains infinity.

All this is another case where I doubt that I am the first person to have thought of this. But I've searched Google and found nothing similar, so if it is out there, it probably is all described in very different terms. Algr (talk) 07:45, 16 February 2009 (UTC)

That's certainly an interesting number system, and I'd like to investigate its properties further if I get time, but I don't think it does what you want. If the ordering is as I expect (lexicographic ordering, essentially), then your definition of 0.999... isn't the biggest number less than 1. ®1_(-1/2) is bigger. --Tango (talk) 11:39, 16 February 2009 (UTC)

Each digit does not effect the next unless something infinite or zero happens within. Since -1 an - 1/2 are neither infinite nor zero, the perspective of the Real place is that they are identical. This means that the symbols <, =, and > need alternate versions to indicate if they are place specific or not. In traditional math, 1_®-1 = 1_®0. In Base Infinity you can proceed as if this were still true unless some operation were to create an infinite value that would then carry into the Inf place.Algr (talk)

That doesn't deal with the problem I mentioned of ®1_(-1) not being maximal, if you want a biggest number than is strictly less than 1 (which I thought was the goal of all this) then you need an ordering that gives you that. The natural ordering on your set is the lexicographic ordering, which doesn't give a maximal value less than 1. Your multiplication seems very counter intuitive, but I guess that isn't a reason to reject it. It certainly guarantees that the definitions of infinitesimal and infinity are met. I kind of see how you carry over infinities, but I'm not sure what you do with zeros. What is ®1_0 * ®0_1? If it's zero then you have zero divisors and they are horrible to work with. If it's a choice between 0.999...=1 and zero divisors, I would choose 0.999...=1 every time! --Tango (talk) 21:26, 16 February 2009 (UTC)

Furthermore, it's not distributive:

®1_1 * (®1 + ®1) = ®2_1 < ®2_2 = ®1_1 + ®1_1 = ®1_1 * ®1 + ®1_1 * ®1

And if I understand multiplication correctly, it contains zero divisors:

®1_0 * ®0_1 = ®0_0 = 0

Thus, while the additive structure forms a group, the multiplicative structure seems a little useless to me. Despite Algr's examples, division is very tricky, as many numbers don't have anything resembling a multiplicative inverse (what is ®1/®0_1 ? Probably 1_®0 by the rule for 1/0, but 1_®0 * ®0_1 does not equal ®1 again - if I understand multiplication correctly, it equals 1_®0_1.)

If I misunderstood multiplication (as I probably did), a more general definition might be helpful. What is a_n_..._a₁_®a₀_a_-1_...a_-m * b_k_..._b_-l in general? An even more basic question: Are only finitely many of the "base infinity" digits non-zero, or do we allow infinitely many non-zero digits? Huon (talk) 14:17, 16 February 2009 (UTC)

And if we're allowed infinitely many, is that in both directions or just one? If if just one, which one? I'd like a better definition of multiplication as well... --Tango (talk) 14:57, 16 February 2009 (UTC)

Multiplication is computed as convolution. Just like normal product but without overflow to next digit. 123 * 456 = 1*4*10^4 + (1*5 + 2*4)*10^3 + (1*6 + 2*5 + 3*4)*10^2 + (2*6 + 3*5)*10 + (3*6) = 4_13_28_27_18 = 56088 Tlepp (talk) 18:20, 16 February 2009 (UTC)

That might make sense, but is explicitly contradicted by Algr's "division by 2" example. It would also make the entire structure much better-behaved, by the way. Huon (talk) 20:35, 16 February 2009 (UTC)

Tlepp I'm not sure I've followed your intent here. Does my comment above help?Algr (talk)

Huon's question: Multiplying and dividing by ®_1 has the same effect that .1 would have in base ten. So yes, ®1/®0_1 = 1_®0, but 1_®0 * ®0_1=®1. In your second example, you added instead of multiplying. Note, I corrected the result of /®2 above due to this. Algr (talk)

Ok, that's better. Now I think you have the definition of multiplication that Tlepp was talking about, which is far better behaved than the one you had before. Now to work out exactly how well behaved it is... --Tango (talk) 21:37, 16 February 2009 (UTC)

I don't think you have multiplicative inverses, which is unfortunate. There doesn't seem to be an inverse for 1_®1_1, for example. Not having multiplicative inverses is a serious problem. Can anyone work out which elements do have inverses? --Tango (talk) 21:47, 16 February 2009 (UTC)

Doesn't this work? 1/(1_®1_1) = 1_®1_1 Algr (talk)

Doesn't (1_®1_1)*(1_®1_1)=1_2®3_2_1? If not then you'll have to explain your multiplication again... --Tango (talk) 22:07, 16 February 2009 (UTC)

X * ®_1 = ®_X  ; X * 1_® = X_®. So: X * ®_1 * 1_® = X Algr (talk) 22:19, 16 February 2009 (UTC)

Yes, but how does that help? --Tango (talk) 22:25, 16 February 2009 (UTC)

You're correct, Algr, similar ideas have been thought of before. Replace "infinity" (which is poorly defined) with ω, and I believe your system becomes the surreal numbers (or at least my understanding of them). The surreal numbers are, if I understand them correctly, one number system in which $\sum _{n=1}^{\infty }9({\tfrac {1}{10}})^{n}=1-\varepsilon \neq 1$, and this value can reasonably be written as "0.999...", but 1-ε is not the largest surreal strictly less than 1.

Oh, and, 1/(ω+1+ε) is definitely a surreal number, but I don't know which one it is.

(If I'm wrong about any of this, which is likely, feel free to correct me.) --Zarel (talk) 22:20, 16 February 2009 (UTC)

If Algr's number system were the surreal numbers, then I think his construction would be the standard one (it's far simpler than anything in our article on them, as evidenced by the fact that I almost understand his!). --Tango (talk) 22:29, 16 February 2009 (UTC)

Just to make sure I understand it correctly: "For the same reason, any form of infinity divided by 2 remains infinity." That doesn't mean it stays the same "form" of infinity, right? 1_®0 / ®2 = 1/2_®0 ?

In that case, the number system becomes distributive, and a ring, a commutative ring with multiplicative identity at that. If we restrict ourselves to finitely many non-zero digits, we won't get multiplicative inverses, and we have in effect constructed R[∞,∞^-1], the set of Laurent polynomials with real coefficients in one variable, which we call "infinity". Only homogenous polynomials (that is, numbers with just one non-zero digit) will have a multiplicative inverse. This number system is also a subring of the hyperreals, and it should be a proper subset, because for any infinite element "∞" the hyperreals contain things like ∞^½, while this system contains only integer exponents of infinity. It may contain the Hackenstrings as a proper subset, but I'm not sure whether that's the case in a meaningful way - the Hackenstrings don't contain infinite numbers, and I doubt they even form a ring.

Of course we still have not solved the first problem raised by Tango: This number system does not contain a largest number less than 1.

If we prefer to have a field (multiplicative inverses), we'd have to choose R(∞) instead, the field created from R by adding a transcendent element (which we again call infinity). I currently don't see an easy description for that field; it's definitely much larger than the ring of Laurent polynomials, but I doubt it's as large as the hyperreals - it might be the surreals Zarel mentioned, I'm not sure about those. Huon (talk) 22:32, 16 February 2009 (UTC)

R(∞) is isomorphic to the field of rational functions from R->R, isn't it? If that were the surreals, we would know it. --Tango (talk) 22:47, 16 February 2009 (UTC)

Well, the surreals also contain a bunch of numbers Algr's number system with infinity = ω doesn't contain, and division is ridiculously complicated in the surreals, but there definitely exists an injection between Algr's numbers (well, at least the rational subset) and the surreals.

And why is a maximal element less than 1 a requirement? I don't see why 0.999... has to be defined as a maximal element less than 1, just an element larger than any real less than 1. --Zarel (talk) 23:01, 16 February 2009 (UTC)

0.999... needs to be a specific element larger than any real less than 1 (and less than 1 itself). Taking the maximal such element makes a lot of sense, choosing any other element is likely to be rather arbitrary. If there is another reasonable way of choosing one, then great. --Tango (talk) 23:13, 16 February 2009 (UTC)

Well, there are much more reasonable ways than "the maximal element less than 1" - especially the difficulty in constructing a useful set containing such an element. The surreal number 1-ε definitely qualifies as a specific element larger than any real and smaller than 1, and it makes sense to equate that number with 0.999... As above:

$\sum _{n=1}^{\infty }9({\tfrac {1}{10}})^{n}=1-\varepsilon <1$

By the way, hackenstrings can also be injected into the surreals - every hackenstring represents a surreal number. A modified form of hackenstrings (specifically, Hackenbush and combinatorial game theory) is how surreal numbers are usually introduced by teachers. --Zarel (talk) 23:23, 16 February 2009 (UTC)

Oh, of course, if you actually have a concept of infinite sums of powers of 1/10 then you should take that approach. The most natural way of defining that sum for R[∞,∞^-1] (Algr's number system) gives you 0.999...=1, though (since its topology is induced from R - it's the cartesian product of countably many copies of R with the topology induced from the lexicographic ordering). I can't see any meaningful way of giving it a topology such that it doesn't contain R as a closed subspace. --Tango (talk) 11:43, 17 February 2009 (UTC)

Just one quick comment: extending the real to include infinity and the complex numbers wasn't really a problem. It was a very natural progression with many physical applications to support the idea; these include hydrodynamics and electrostatics (See Harvey Cohn's "Conformal Mappings on the Riemann Sphere", McGraw-Hill Book Company, 1967). I fail to see any real world applications in the perverted number systems proposed here except, of course, mental-masturbation.  Δεκλαν Δαφισ   (talk)  15:11, 22 February 2009 (UTC)

We're mathematicians. Finding real world uses for maths is the job of scientists and, somehow, they always seem to find something. There are even applications for p-adic numbers... --Tango (talk) 16:12, 22 February 2009 (UTC)

Yes, we are mathematicians. I am a differential geometer, and have been for many years, but my point is still valid: many mathematical (pure or applied) ideas with connexions to the real world are more natural and easier to develop, like the complex numbers or infinity. And tango, you seem to miss a key point: "Without pure mathematics, applied would be imputant. Without applied mathematics, pure would wither and die."  Δεκλαν Δαφισ   (talk)  18:53, 22 February 2009 (UTC)

Yes, we need the applications otherwise no-one would give us funding, but that doesn't mean we need to have any applications in mind when doing pure maths. They can come along later. --Tango (talk) 20:41, 22 February 2009 (UTC)

Well, we've already established on your talk page that you are not an experienced mathematician, and that you know very little of the professional world. So, as a student, please don't give the rest of us professional mathematicians a bad name. Those of us that receive money from various funding bodies try to reconcile that with our research.  Δεκλαν Δαφισ   (talk)  16:17, 24 February 2009 (UTC)

Nobody ever found a practical use for an idea that hadn't been thought of yet. For years, lasers were nothing but a pretty color show, and now they are everywhere. Algr (talk)

If you consider doing maths for maths sake as bad, then you're not my kind of mathematician. --Tango (talk) 22:32, 24 February 2009 (UTC)

2 (again)

I already told you guys the practical uses for 1 ≠ .999... and 1.000...1. They are notation for defining inclusive and exclusive ranges. Most ordinary people would find 1<X≤2 unintelligible, but could understand X= 1 to 1.999... It is true that Base Infinity does not truly define the highest element less then one, but it does have an unambiguous non-1 interpretation of .999..., which was the point, anyway. Algr (talk)

That's just notation, though, and it's a very bad one. X can't actually equal 1.999... because presumably X is a real number (if it is modelling some real life quantity) and 1.999... isn't a real number (with your interpretation). --Tango (talk) 21:55, 22 February 2009 (UTC)

Oh, an I'm assuming you meant 1≤X<2. --Tango (talk) 21:56, 22 February 2009 (UTC)

That's right, I wanted illustrate the problem with 1<X≤2 by showing how long it takes for people to notice that the two ranges are not the same. In the case of a dividing line between ranges, you specifically DON'T want .999... and 1.000...1 to be real numbers because that way any real number must be either above or below your dividing line, and thus clearly in only one range. Algr (talk) 05:52, 23 February 2009 (UTC)

But to say X is between 1 and 1.999... inclusive (which is what I would interpret by X=A to B - well, actually, I would interpret it as X ranging over the whole interval, rather than just being in the interval, I only know what you mean by context) is nonsense since it can't be 1.999... . I really don't see a problem with ≤ and <, they do the job and are perfectly easy to understand - I was taught them in primary school, as I recall, very early on in secondary school, at the latest. Oh, and nobody ever believes the "I was just testing" line! ;) --Tango (talk) 11:58, 23 February 2009 (UTC)

Tango, you and I have a good heads for mathematics, and so could be counted on to remember > vrs ≥. But a discussion about what is easy to understand leaves the field of mathematics and gets into Usability testing. I recall a radio show where the subject came up about when would be the 1000th hour that that station was on the air. They went through about 15 callers who didn't know how to figure this out, or didn't want to try, until I called myself and told them. Sadly, most people forget higher math the second they leave the classroom. If you need to communicate with regular people, you need to do the work of understanding what language they can work with. That is why I used board games and tax codes in my example. They don't use real numbers at the IRS. They use rationals, with a cut off at either $.01 or at whatever their computers round off to. To them, the Archimedean property is just as much 'mental masturbation' (I hate that term.) as my base infinity. Algr (talk) 22:03, 24 February 2009 (UTC)

So, the IRS don't use rationals, just integer multiples of cents? Endomorphic (talk) 09:24, 25 February 2009 (UTC)

Firstly, regarding the claim that "they don't use real numbers at the IRS"; I really doubt that they do everything in pennies; consider, for example, calculating daily-accrued interest on small sums; the roundoff losses would be very significant. If I was designing their systems, I'd work in some smaller unit like microcents to catch the low end -- at the same time, you'd need more than 32-bit precision in your integer counts to ensure that some of the very large real-world sums of money, measured in the billions of dollars, Secondly, all fixed-precision computer floating-point numbers are rationals anyway. -- The Anome (talk) 10:13, 25 February 2009 (UTC)

They probably use DECIMAL(32,6) or something equivalent. It doesn't matter, they're still irrelevant, they use discrete numbers. Endomorphic (talk) 15:33, 2 March 2009 (UTC)

'Citation needed?' I wasn't intending to put that into the article. But if common sense will do, it's simple: Ask any accounting program what 1/3 is. Most will tell you .33. Some may say .333333333, and none will give you any indication of an infinite decimal. Computers natively think in integers, and operating systems teach them finite decimals, which are a subset of the rationales. Trying to teach the Real set to a computer is probably possible, but would be slow as molasses. And what purpose would that serve? Are you going to write a check for $π any time soon? My tax accountant rounds everything to the nearest $1 Algr (talk) 07:29, 4 March 2009 (UTC)

Sorry, I'm completely missing Algr's point here.

He claims that some folk don't understand the difference between ≤ and <. Perhaps so, but so what? Is this an argument that we ought to define a new mathematical class that replaces the reals and satisfies 0.999... < 1? If so, it's a mighty weak and tenuous argument.

And then we go off onto a digression regarding computer representations of numbers, but once again, so what? Of course computers can not represent all the reals. Heck, they can't represent all the integers either. Should we also take this as an argument for redefining the integers as a finite set?

Sorry, Algr, I just don't understand what you're trying to prove here when you discuss the innumeracy of the common man and of machines. Phiwum (talk) 13:59, 4 March 2009 (UTC)

I'm proving that a non-one interpretation of ".999..." has practical real world uses for communication. I'll list this with my other points:

1) Outside of this discussion, no one ever types ".999..." when their intent is to describe a real number.

2) Mathematics has plenty of valid number systems in which 1 is not the first or most obvious interpretation of .999...

3) A non-one interpretation of ".999..." has practical real world uses for communication.

Put these together and it is clear that the opening sentence of the article is simply wrong. BTW, I should have said above that the ≤ vrs< is deeply compounded by the question of how you deal with multiple inequalities in a single equation. That is NOT everyday mathematics that people use in their lives, and given the state of the world right now I'd rather everyday people learn more about economics, history and logic then higher math. Algr (talk) 21:32, 4 March 2009 (UTC)

Okay, let's examine these.

1) Yes, and so? Very few people refer to Dr. John as Mac Rebbinack, but it is still a fact that Dr. John is Mac Rebbinack. And it is still the case that 0.999... denotes the same real number as that denoted by 1.

2) I have never encountered a number system in which 0.999... is not equal to 1, but I imagine they exist. So what? I could define a mathematical theory in which 1 + 1 does not equal 2. Nonetheless, in a (hypothetical) encyclopedia article about 1 + 1, it is natural to say that it is equal to 2.

But let's grant you a partial point and say that, in the real numbers, 0.999... = 1.

3) Let's suppose this is true, although I am dubious. So what? It does not change the point that in the real numbers 0.999... = 1 and that in almost every case in which one encounters 0.999..., it will denote the real number one.

But, let's take your suggestions to heart. We could change the opening this way: In mathematics, the term 0.999... could denote practically any imaginable mathematical object. Some people (one person?) say that it is better if it doesn't denote the real number one, but in practice, no one is much influenced by these people.

Better? Phiwum (talk) 21:59, 4 March 2009 (UTC)

For Phiwum:

1) And few people refer to New York, as the city in Lincolnshire, and yet it denotes that british city. Refusing to think outside of the Real set is the same as refusing to think outside of England to consider what "New York" means. Algr (talk)

2) Argument through ignorance. We were just talking about surreals and hackinstones, why didn't you encounter that? Algr (talk)

3) And in England, New York = Lincolnshire. Show me a single case where someone has some process that generates .999... (As opposed to introducing the text hypothetically for the purpose of saying that it equals one. Algr (talk)

Algr: "Show me a single case where someone has some process that generates .999... (As opposed to introducing the text hypothetically for the purpose of saying that it equals one."

Easy.

There are 3 doors. Behind 1 of them is a car, and behind each of the other 2 is a goat (like the Monty Hall problem). When trying to figure out which door the car is behind, we resort to probabilities:

P(car is behind door 1) = 1/3 = 0.333...

P(car is behind door 2) = 1/3 = 0.333...

P(car is behind door 3) = 1/3 = 0.333...

Theoretically, the probabilities should add up to one (because these events are both mutually exclusive and collectively exhaustive). However, you would tell us, the probabilities do not, in fact, add up to one. They add up to 0.999... This is disturbing; we can't guarantee that the sum of the probabilities will be 1 by satisfying mutual exclusiveness and collective exhaustiveness!

Back in the mathematical universe, however, this is not a problem, because 0.999... = 1, so the sum of the probabilities is 1.--72.177.97.222 (talk) 23:34, 5 March 2009 (UTC)

1/3 ≠ 0.333... That is just rephrasing the same assumption. (We've been through this before.) Also, why would you use Reals for this? Rationals are made for this kind of situation. Algr (talk) 00:02, 6 March 2009 (UTC)

The reals can do anything the rationals can do, since the rationals are a subset of the reals. How we represent numbers does not change what they are. Your assertion "1/3 ≠ 0.333..." is false. Your problem here is with positional notation: you don't like recurring decimals. Fine. Provide us with an alternative. Maelin (Talk | Contribs) 00:35, 6 March 2009 (UTC)

And the hyperreals can do anything that the Reals can do. But I don't think either of us want to do our taxes that way. Algr (talk)

Do "1 divided by 3" in long division and tell me what you get. --Tango (talk) 00:39, 6 March 2009 (UTC)

I get .1_{Base 3} :P Algr (talk)

But what is 1/3 in decimal representation? Surely 1/3 has a decimal representation, doesn't it? If not, your brand of decimal representation becomes pretty useless because it doesn't represent quite a lot of rather simple numbers appearing in daily life. Huon (talk) 02:22, 6 March 2009 (UTC)

π is pretty simple, but it doesn't have a decimal representation. In fact almost none of the Reals do. Yet we survive somehow. Algr (talk) 04:07, 6 March 2009 (UTC)

π has a decimal expansion. For any natural number you can say what digit in is that decimal place in the decimal expansion of π, that's all you need. You don't need there to be a simple pattern to the digits. --Tango (talk) 10:41, 6 March 2009 (UTC)

Algr: I used a very simple case. In that very simple case, the probabilities involved were rational numbers. This is not always so. For example, we can have a uniform distribution of over the real numbers between 0 and 1. Then it is pretty obvious that, eg, the probability of getting a real between 0 and 1/π is 1/π, an (GASP!) irrational real number.

There are, of course, much more clever ways of getting real numbers in statistics, but I can't think of any that refer directly to the probabilities themselves (without dealing with limits or purely theoretical objects, that is). What I can remember is that the expected value of the number of random variables in the uniform distribution between 0 and 1 it takes to add up to at least 1 is e (see e (mathematical constant)#Stochastic representations). --72.177.97.222 (talk) 12:13, 6 March 2009 (UTC)

3

Unfortunately there is no way to NOT interpret "0.999..." as 1 without breaking positional notation. Do you have a suitable alternative for positional notation in mind, Algr? Is it as simple and efficient as positional notation? Maelin (Talk | Contribs) 05:28, 5 March 2009 (UTC)

It seems to me that positional notation is broken as soon as you suggest two different ways of writing the same number. You must either except a 10^-inf place, or deny it, and either choice causes the Real set to break down. Algr (talk) 23:21, 5 March 2009 (UTC)

As long as each representation refers to a single, unambiguous real number, the notation works. Non-uniqueness is annoying, but not the end of the world. --Tango (talk) 23:22, 5 March 2009 (UTC)

In what way does non-uniqueness of representation in positional notation "cause the Real set to break down"? --jpgordon^∇∆∇∆ 00:01, 6 March 2009 (UTC)

If 10^-inf exists, then their ought to be digits beyond it, like .000...1. If 10^-inf does not exist, then .999... must always fall short of one. Algr (talk)

That's just not the case. Limits are a well defined mathematical concept and the limit of 0.999...9 as the number of nines goes to infinity is 1, there is no need for there to actually be an infinityth nine. --Tango (talk) 00:10, 6 March 2009 (UTC)

Algr, it is pretty obvious you still haven't studied limits. Just because sometimes you can evaluate a limit simply by substituting the limiting value into the expression, doesn't mean that this is always so nor that such an insertion should even produce a meaningful expression. This equality below may hold sometimes but it is NOT the definition, it is not necessarily true, and if it does hold, that is merely a happy coincidence.

$\lim _{x\to a}f(x)=f(a)$

Note that in some cases, the right hand side may not even be well-defined, whilst the limit is well-defined. For example,

$\lim _{x\to 2}{\frac {x-2}{x-2}}=1\quad {\mbox{whilst}}\quad {\frac {2-2}{2-2}}={\frac {0}{0}}={\mbox{?}}$

Limits to infinity are NOT defined by "sticking infinity in and seeing what you get", they are defined in more rigourous ways. For the most common definition, see (ε, δ)-definition of limit. Maelin (Talk | Contribs) 00:31, 6 March 2009 (UTC)

But aren't you precisely arguing my point here? You can't assume that .999... equals the limit of .9; .99; .999; ect. Algr (talk) 02:09, 6 March 2009 (UTC)

Actually the assumption that, if I go far enough, the sequence .9, .99, .999, ... will ultimately become arbitrarily close to .999... seems almost self-evident. Here we're talking real numbers, aren't we? If .999... were a real number other than limit of that sequence, it'd have to be either larger than the limit (making it larger than 1) or smaller than the limit (making it smaller than some .999...9 with finitely many nines). Both is rather absurd. Huon (talk) 02:41, 6 March 2009 (UTC)

You can if that's how it has been defined! Oli Filth^{(talk|contribs)} 09:34, 6 March 2009 (UTC)

Algr, how do you represent (0, 1] in your scheme? Eric119 (talk) 19:20, 5 March 2009 (UTC)

It is .000...1 to 1. Algr (talk) 23:03, 5 March 2009 (UTC)

Then where is that 1 in .000...1? I suppose .000...1 is more than just a string of characters but is supposed to actually represent a number in a meaningful way - which way? Huon (talk) 02:17, 6 March 2009 (UTC)

That 1 is in the same hypothetical place as the '9' that causes .999... to cary over and become 1. In Base Infinity it is ®0_1 Algr (talk)

Oh. There is no such hypothetical place; .999... does not "carry over and become 1"; .999... is 1. It's a number, not a process. --jpgordon^∇∆∇∆ 05:29, 6 March 2009 (UTC)

Limits are defined as the results of a process. If "..." denotes a limit, then nothing using that notation is a number. A process is not equal to it's limit, as the $\lim _{x\to 2}{\frac {x-2}{x-2}}=1\quad$ example shows. Algr (talk) 07:08, 6 March 2009 (UTC)

Yes, the process here is adding more and more 9's. 0.999... is the limit of that process, which is a number, the number 1. As you say, limits refer to the *result* of a process, which can be a number, they don't refer to the process itself. --Tango (talk) 10:40, 6 March 2009 (UTC)

Other number systems

Here is a sentence from the article Sexagesimal:

60 is the smallest number divisible by every number from 1 to 6.

Does this statement make sense? In the set of Reals, certainly not. Any Real is divisible by any other Real. Shall we go in and fix this obvious error? I don't think so. We know from context that if someone is talking about certain numbers being 'not divisible', by others, well that person must be talking about some other number system - in this case, the Natural numbers. So it seems that Reals aren't the only game in town, as far as numbers go. So why lock a discussion about .999... into the reals? As I've said above, no one ever types ".999... " when they want to convey a Real number. Algr (talk) 04:24, 6 March 2009 (UTC)

No one ever types ".999..." except for the purpose of this discussion. --jpgordon^∇∆∇∆ 05:30, 6 March 2009 (UTC)

Precisely, so how do you justify that the Real set is appropriate? In the natural numbers, 1.1 = 1 because the .1 is undefined and thus ignored. In Reals, infinitesimals are similarly ignored, so .999... = 1. But it is clearly not reasonable to discuss 1.1 or .999... within number spaces that are defined as being unable to accommodate their central features. Algr (talk) 07:16, 6 March 2009 (UTC)

Uh, no. There is no entity in the natural numbers denoted by the string "1.1" under conventional notation (unless you treat the . as multiplication). You don't get to ignore undefined things, if a notation is undefined then any expressions involving it are also undefined. 1 + (0/0) is not equal to 1, it is an undefined expression. Furthermore, the equality 0.999... = 1 has nothing to do with infinitesimals. It is a result of the limit definition of positional notation and nothing more. Maelin (Talk | Contribs) 09:20, 6 March 2009 (UTC)

Decimal notation (including a fractional part) always refers to real numbers, that's what it is a notation for. If the sentence you quote had said something like "every number from 1.0 to 6.0", it would have been wrong because that notation clearly refers to real numbers, not integers. Likewise, "every number from 1/1 to 6/1" would have been wrong because fractions always refer to rational numbers. The notation "1" is highly ambiguous, so you have to work it out from context, in this context it is clearly referring to the integer. --Tango (talk) 10:37, 6 March 2009 (UTC)

"" You don't get to ignore undefined things" - Well I'm glad to hear you say that, because what happens to the "carry the one" when you divide 1 by 3? Everyone seems happy to just say "limit" and ignore it. But that x-2 limit clearly shows that you can't assume that when a limit is reached the result will be at the position that the trend indicates. The trend for that example is fixed at 1, but the actual result of the trend is undefined. Tan(x) does something similar, where the trend is to + infinity, and then it suddenly becomes negative. There is simply no reason to assume that .999... = 1, just because the trend with finite numbers of 9s appears to be headed that way. Algr (talk) 00:32, 7 March 2009 (UTC)

One doesn't need to "assume" anything. Oli Filth^{(talk|contribs)} 00:42, 7 March 2009 (UTC)

Quite the contrary: First of all, the limit isn't "reached". It is not "what happens when I plug in infinity and evaluate". The limit is a number with specific properties. As I consider further and further elements of a (converging) sequence, they get ever closer to the limit - but unless the sequence becomes constant, I can't count on ever "reaching" the limit. Secondly, the limit is precisely what the "trend" indicates. In the x-2 example, as you say the "trend" is fixed at 1, and the limit is 1, too. Concerning the "carry the one": That obviously does not end, just as there's no end to the threes in 0.333... But at each step that remainder gets closer to 0. The limit of the remainder is equal to 0. Finally, I believe we begin to lose sight of the main advantage of decimal representations: The real numbers are important (even you agree when you state that the main objective of having 0.999...≠1 is to describe intervals of real numbers more clearly), and the decimal representations represent every real number. The usual way, some real numbers are represented in more than one way, but what you suggest would lead to real numbers without any representation, non-real numbers without any representation, and representations which do not belong to any real number. Are there any real numbers with infinite decimal representations in your system? Or are all infinite decimal representations an infinitesimal shy of their real number? Any interpretation of decimal representations in your base infinity system I can think of make them pretty useless as a tool do describe the reals. Huon (talk) 01:14, 7 March 2009 (UTC)

Jeez Louise

The main contention on this appears to be the wording. Infinity is a philosophy that is impossible to represent physically. It varies in definition from person to person, depending on what particular thing they happen to be talking about. Forget about people's opinions on the meaning of infinity in mathematics. Forget about the proofs. Just realize that .333... is the numerical representation for the fraction 1/3, and multiplying that by three gives you .999... (and a greater number of stops if you want to be whimsical). 0.999... is not necessary to describe anything, really. You can just as easily say 0.333... times 3 is 1. Some guy instead decided for some reason that it would be easier on young brains to call it 0.999... and was sadly mistaken. The only reason this article exists at all is because of all the controversy surrounding the term.Asperger, he'll know. (talk) 00:08, 19 March 2009 (UTC)

We've had people doubting that ".333... is the numerical representation for the fraction 1/3" before. .999 may not be necessary to describe anything, because 1.000... is much nicer for that task, but it's necessary for consistency. Every string of digits formed in the correct way should represent a real number, and if .999... didn't represent a number, the system of decimal representations would be flawed. Not critically so, but needlessly so. Huon (talk) 01:01, 19 March 2009 (UTC)

Well, if mathematicians call a third 0.333... I have no problem with calling three thirds 0.999... (.) Just do it in your own homes. I'm a fractor myself.

PS: I can't believe I missed that. "In physically". Asperger, he'll know. (talk) 01:18, 19 March 2009 (UTC)

Infinity is indeed impossible to represent physically, but calling it a "philosophy" is just an abuse of terminology. Infinity is a very well-defined concept, thankyouverymuch.

Nobody decided that 0.333... times 3 would be 0.999... instead of 1. It equals both, since they're the same thing. The whole misunderstanding comes up when someone tries to apply the 0.333... notation to the digit 9, resulting in the sequence 0.999..., and then due to a misunderstanding of mathematics declares it to be less than 1 since it "looks like it". --Zarel (talk) 20:26, 19 March 2009 (UTC)

By a philosophy, I mean anything that is a point of view, a mode of thought or a noun that is not used to denote an actual object. Anything that isn't. Somebody did indeed decide that 0.999... could be said to equal 1, or else it would not be said to be so. I never said that 0.999... and 1 were not the same thing and spent sentences explaining it. You may say that 0.333... times 3 is 1 or 0.999... (.) I only used that as an example because it is one of the few ways you can turn up a result like that. Also, a "well-defined concept" is a good definition of a philosophy, although not all "philosophies" are well-defined. In fact, "logic" is considered by Wikipedia to be a "branch" of philosophy, and "mathematic logic" a branch of logic, and therefore, (although it sounds cheesy to say so), philosophic by nature. If you would like me to call infinity a concept I can, but saying 'calling it a "...philosophy" is just an abuse of terminology. Infinity is a very well-defined concept...' sounds ironically close to "0.333... is 0.999..., not 1". Asperger, he'll know. (talk) 23:26, 19 March 2009 (UTC)

Nobody "decided" that 0.999...=1, it was deduced from the construction of the real numbers and the definition of decimal notation (those things were decided, but not arbitrarily). I think your definition of "philosophy" is rather too broad - by your definition "sound" is a philosophy, which seems absurd to me. --Tango (talk) 23:32, 19 March 2009 (UTC)

The term sound refers to the audible vibration produced in objects from an outside force. What humans describe as "sound" is the brain's interpretation of vibrations experienced in the inner ear - a "feeling". It describes a movement of particles in a direction that is not defined by the term. Think of a puzzle as an example. Is a puzzle a thing? Only if you decide to put it together. The pieces that make up the puzzle are real, and the term is used to describe an agglomeration of pieces placed in a way that is not explicitly specified by the term puzzle. In the same way, because our brain decides to interpret movement in the cochlea as sound, sound exists. While different pieces will change the appearance of the puzzle, it will still be called a puzzle. Really, whatever is convenient goes. Just as pretty much everything can be considered a wave, and everything can also be considered a particle, anything can be considered a concept or an object. I personally don't give a crap, and I hope to be completely finished with philosophy at the end of this sentence. My only real point is that someone, or a group of someones, decided that the mark "1" denotes a single thing. By the same point, someone decided that three threes is the mark that we call "nine". Someone put three dots (...) behind a series of commonly used numbers, and because the concept of one was not represented in the traditional fashion, they !assigned! it the value of 1. You pronounce oh and owe the same way, right? People decided that these would be pronounced the same way, despite their spellings. If not, people could always pronounce owe as ow, if they felt like it. Asperger, he'll know. (talk) 00:08, 20 March 2009 (UTC)

No, 0.999... was not assigned a value, its value was calculated. Decimal notation is defined in a particular, general way which results in 0.999... being interpreted as $\sum _{i=1}^{\infty }9\cdot \left({\frac {1}{10}}\right)^{i}$, which can be shown to equal 1 using the definitions of real numbers, limits, etc. (And I suspect "oh" and "owe" had their pronunciations before they had their spellings - spellings were not uniquely defined until fairly recently.) --Tango (talk) 00:34, 20 March 2009 (UTC)

Firstly, I, and a lot of other people, pronounce oh and owe the same way, (thus no plural for "pronounciations"), and secondly, I am not referring to people as a whole, but the people who actually pronounce oh and owe in the way I say it. Nothing in what I said made any suggestion that pronounciations were set before our time, and these people are deciding to pronounce oh and owe as o with a hat whenever they read aloud, not in any particular historical period. And while you can calculate stuff all you want, the truth is that 0.999... IS 0.999... in the same way that 9 can be calculated as 10 minus 1, but IS, simply, 9. Anything that holds any meaning to anyone has at least one general definition. Also... "particular, general" is an oxymoron. Asperger, he'll know. (talk) 00:58, 20 March 2009 (UTC)

You have an unhelpful and unproductive determination to latch onto words and semantics rather than see the point, this discussion of pronunciation and philosophy is completely irrelevent and serves only to distract from the main point. The real numbers are constructed to have certain properties that we find useful in a number set. We define a notation for them: positional notation so we have a way of discussing real numbers. Whilst there are real numbers we cannot perfectly denote in finite space, the notation at least lets us denote them to within any desired degree of accuracy. The notation system also makes certain arithmetic operations convenient, for example, digit-by-digit addition. This notation system is defined using the Real Analysis concept of limits, a well-defined, well-understood idea. This notation system delivers the mildly unintuitive result that the real number denoted by (the expanded form of) 0.999... is in fact the same real number as that denoted by (the expanded form of) 1.000..., and this is simply a side-effect of the definition. It has no significant consequences other than proving that the real numbers do not exist in one-to-one correspondence with decimal expansions, and causing a bunch of people with no understanding of limits to come and argue with us for days on a web based encyclopedia. Maelin (Talk | Contribs) 01:32, 20 March 2009 (UTC)

Saying point-blank that I'm rude in my writing is latching onto words and semantics rather than seeing the point :D. The problem with conversations like these that grow out of control is that it becomes difficult to pinpoint causes and effects. I started that terrible philosophy crap because he said he didn't like my version of "philosophy". No other reason. Secondly, the reason I "latched" onto his choice of wording is because I believed that he had misinterpreted me, or because I thought I saw some subtext... For instance, when he said "particular, general" I assumed that he wrote them for a reason, although the sentence still makes sense without them. I thought that maybe he had deleted part of a sentence accidentally. When he talked about "the pronounciations of oh and owe" I assumed he simply hadn't realised that I meant that they had the !same! pronounciation. Also, it is rather unnecessary to imply that this arguing is stupid, because arguing is not a one-man job. I should know, it's my religion's major sport. Asperger, he'll know. (talk) 02:18, 20 March 2009 (UTC)

You must realise I'm just doing this for fun. Asperger, he'll know. (talk) 03:05, 20 March 2009 (UTC)

::Eats popcorn and watches the show:: Algr (talk)

You misunderstand my point about owe and oh. You said "People decided that these would be pronounced the same way, despite their spellings." when, in fact, it is far more likely that people decided to spell them differently despite them having the same pronunciation. "0.999..." is a notation, it represents a real number, which we can deduce to be the same real number as is represented by the notation "1". It is important to distinguish notations from mathematical objects, I think that may be the cause of some of your confusion. --Tango (talk) 12:13, 20 March 2009 (UTC)

Sorry, on my talk page I have explained the reason for my lateness... Anyway, I would like to end my ridiculous arguments right now, and wish to apologise for any crossed words. I should really have written the oh and owe thing in the present tense, referring to the fact that, while different spellings are usually pronounced different, people ignore the general rule and nonetheless pronounce the two the same. I had an edit to the little section containing profanities that was quickly deleted below; you can check it on history. Goodbye, everyone.Asperger, he'll know. (talk) 04:01, 22 March 2009 (UTC)

And yet another dissident gets driven away. So it goes. Algr (talk) 18:21, 22 March 2009 (UTC)

I don't believe I ever voiced the opinion that 0.999... was not 1. I'm leaving due to the fact that I feel that this page, as a whole, is pretty silly, and any further comments on my part would serve no real purpose... My only problem with the main page is that the lead paragraph is, in my opinion, far too detailed. Asperger, he'll know. (talk) 19:31, 22 March 2009 (UTC)

I didn't think you did, but a LOT of people have given up on trying to fix this article. It is sad. Algr (talk)

Isn't it time to kill this page?

IMO, this page has long outlived its usefulness. It now seems to be stuck in an infinite loop with Algr (or occasionally a new face) dregging up the same old stuff periodically, and people who know what they're talking about telling him why he's wrong (usually another misunderstanding about what limits mean). A talk page for general discussion about the article topic isn't standard practice, and is generally frowned upon (WP:TALK). Surely by now this kind of circular chit-chat should be taken to user-talk, or off Wikipedia (i.e. an external forum)? Oli Filth^{(talk|contribs)} 18:45, 22 March 2009 (UTC)

This page exists to move such discussion away from the main talk page for the article so that that page can be used for its intended purpose. That hasn't changed, so I think this page should continue to exist. We just have to learn which arguments are worth responding to and which aren't. --Tango (talk) 19:38, 22 March 2009 (UTC)

I'd vote to delete this page, too. Any doubters may still visit User:ConMan/Proof_that_0.999..._does_not_equal_1 or various math forums to voice their concerns. If they want to argue on our talk page, let them be deleted. Gustave the Steel (talk) 03:30, 25 March 2009 (UTC)

Not a good idea. This page keeps the talk page much cleaner than it would be with all the more "imaginitive" folks that argue here. Tparameter (talk) 01:48, 26 March 2009 (UTC)

I'd vote to keep. It's a useful fool magnet. --jpgordon^∇∆∇∆ 16:17, 26 March 2009 (UTC)

I agree with Tango, Tparameter and jpgordon's arguments above for the continued existence of this page. "Argument" pages aren't usual practice, but this isn't a usual article, and special problems sometimes require special measures. Keep. -- The Anome (talk) 18:55, 26 March 2009 (UTC)

I take the point that this article gets a higher-than-average level of dispute, but as with any other talk page, surely the solution for off-topic material (as Gustave said above) is just to delete it? A page like this, IMO, just exacerbates the "problem"! (Don't worry, I'm not going to try arguing this into the ground! The consensus clearly seems to be "keep".) Oli Filth^{(talk|contribs)} 19:12, 26 March 2009 (UTC)

If you delete it, people put it back. If you give them somewhere else to go it allows people that are interested it correcting people's misconceptions to answer them here while not disturbing everyone else. --Tango (talk) 19:14, 26 March 2009 (UTC)

Keep, of course. This page exists because the main article is written in such a way that inspires distrust from people who truly want to understand the situation. In all this time I think I've learned enough to propose a true fix. I'm writing in in the talk page now. Algr (talk)

This page is more of a haven for trolls (see "Prove 6 is rational" below) than a forum for serious mathematical discussion. It doesn't keep vandals away from the article, nor does it prevent people from attempting to argue on the talk page (the histories of both pages show this). It's fun to come here and argue, but I'd still vote to delete this page. People with serious questions should be referred to actual math forums, and trolls should be punished according to standard policy. Gustave the Steel (talk) 15:56, 3 April 2009 (UTC)

You realize that thread was originally in Talk:0.999...? It's here because we don't want it there. Look, the trolls aren't going to magically go away if we remove this page; they'll go straight to Talk:0.999... and then what do we do? --Zarel (talk) 22:43, 3 April 2009 (UTC)

Yes, I do realize that. It's part of my point, actually; the existence of this page did NOT prevent people from arguing on the talk page, nor has it prevented people from adding falsehoods and original "research" to the article. If we delete this page, then we can handle trolls using the policies that work for every other Wikipedia page (delete stuff that doesn't belong, inform people of policies, ban repeat offenders). As it is, we are much more forgiving of trolls here than elsewhere, and I really don't see why our trolls deserve more consideration than other articles' trolls.

In fact, consider the intelligent design talk page. They have a simple warning: "This is not a forum for general discussion of Intelligent design. Any such messages will be deleted. Please limit discussion to improvement of this article." They don't have an Arguments page. If Wikipedia's default policies are good enough to protect intelligent design from trolls, can you really claim that 0.999... needs special policies above and beyond the norm? Gustave the Steel (talk) 03:44, 4 April 2009 (UTC)

I agree. We are much more forgiving of trolls here than elsewhere. And we shouldn't be. But I think the problem is that none of us know about general Wikipedia process. There aren't any administrators around to ban misbehaving users. There aren't even any users who know what Wikipedia policy the trolls are violating, and how to justify deleting the corresponding messages, and how to warn them. I'd greatly appreciate someone who did that. --Zarel (talk) 10:09, 5 April 2009 (UTC)

The guideline being breached is WP:TALK. Any editor can justifiably delete off-topic material from a talk page, and apply one of the {{uw-chat}} warnings to the user's talk page; if they still continue after a final warning has been given {{uw-chat4}}, then I believe that counts as vandalism, in which case they can be reported at WP:AIV. At that point, an admin will step in.

Of course, I'm not sure how that guideline applies to a page such as this, which technically shouldn't even exist! Oli Filth^{(talk|contribs)} 10:24, 5 April 2009 (UTC)

Cool, I think we can make the case. How do we start a vote? Gustave the Steel (talk) 02:12, 6 April 2009 (UTC)

WP:MFD. Oli Filth^{(talk|contribs)} 10:10, 6 April 2009 (UTC)

The last couple of new threads have only increased my conviction that this page has completely outlived its purpose, so I've listed it at MfD; see tag at top of page... Oli Filth^{(talk|contribs)}

No administrators? You have three in this very section of the page alone: Tango, The Anome, and jpgordon. Uncle G (talk) 00:15, 8 April 2009 (UTC)

True. What can administrators do, though? The only objectionable thing Algr's done is walk away from arguments he can't answer, which isn't against general policy. Even the 98 person generally posts things that resemble math (although the history shows some exceptions). Unlike ConMan's page or other math forums, this page's lack of guidelines provide trolls with freedom from administrative action. Gustave the Steel (talk) 01:12, 8 April 2009 (UTC)

Very simple proof that 0.999... does not equal 1

1 is a rational number. It can be expressed by comparing two integers. In other words, such a comparison which is a ratio is an average. There are infinitely many such averages that respresent 1. Now 0.999... is 'something' you cannot express by comparing ANY two integers, that is, there is no average that represents 0.999...

You can take any two lines of the same length and use their ratio to express 1. Now, are there any two lines whose ratio represents 0.999... ? Of course not. The only way you can justify your false logic and reasoning is by assuming that 0.999... is equal to 1. No, I am sorry, it is not. Now if you are going to talk about limits, then we are no longer discussing Euclid's rational numbers. We begin to delve into Cauchy sequences which are false because the very definition of a Cauchy sequence is false (it is circular for starters). Just be honest and delete this pathetic article. You will gain respect and perhaps people might start noticing what you write. In the meantime, you are nothing but a bad joke.

What really chokes me up is talk of infinitesimals, hyperreal and surreal numbers which don't exist. OMG, real numbers have never been well defined. In fact, we know nothing more than the ancient Greeks knew about real numbers and here you are discussing all this other poppycock! 98.201.123.22 (talk) 23:30, 2 April 2009 (UTC)

I'm sorry, you asserted "Now 0.999... is 'something' you cannot express by comparing ANY two integers, that is, there is no average that represents 0.999...", but it looks like you forgot to include your proof?

I am not the one who needs to prove anything. You are claiming that 0.999... is a rational number. So prove it.

We have many proofs in the article that 0.999... equals 1 (and thus is rational). If you claim otherwise, you must either refute our proofs, or prove that 0.999... is irrational. Feel free to read the FAQ, by the way; it may address some of your questions. --Zarel (talk) 06:33, 3 April 2009 (UTC)

My apologies, I was misled that with the heading "Very simple proof that 0.999... does not equal 1", there might in fact be a very simple proof that 0.999... does not equal 1. My mistake. For proofs that they are equal, see the article, or endless discussions on these talk pages. Mdwh (talk) 11:27, 3 April 2009 (UTC)

"Now, are there any two lines whose ratio represents 0.999... ?" Yes, 1/1.

No, 1/1=1. Read previous response.

You have done nothing to prove, or even suggest, that 0.999... and 1 are not equivalent. --Zarel (talk) 06:33, 3 April 2009 (UTC)

"Now if you are going to talk about limits, then we are no longer discussing Euclid's rational numbers." We are talking about the reals, as clearly stated in the article. Mdwh (talk) 00:01, 3 April 2009 (UTC)

The articles states a lot of things but unfortunately 'clarity' is not part of any of the statements.

I am not sure how the phrase "denotes a real number equal to one" could be misinterpreted. Mdwh (talk) 11:27, 3 April 2009 (UTC)

If 0.999... is not a limit of a sequence and does not differ from 1 by an infinitesimal (since you claim infinitesimals "don't exist"), then what is it? The archives contain a proof that if 0.999... is a real number (including the rationals), isn't smaller than any number of the form 0.999...9 with finitely many nines, and isn't larger than 1, then 0.999...=1. That proof doesn't use limits, Cauchy sequences or anything except that there are no non-zero infinitesimals within the reals.

Huon (talk) 01:05, 3 April 2009 (UTC)

Well, you claim that 0.999... is a well-defined real number. So you tell me what is the difference between 0.999... and 1. Also, tell me what is the number that differs from pi by an infinitesimal? You cannot use the word infinitesimal in your response as you are unable to define it and you are also unable to provide any example of an infinitesimal - whatever this nonsense is. Oh, and you still have not been able to provide two lines that express 0.999... 98.201.123.22 (talk) 01:29, 3 April 2009 (UTC)

This claim that we cannot express 0.999... as the ratio of two integers is a simple instance of begging the question. Of course we can, since we can show that 0.999... = 1, and 1 = 1/1. Thus, 0.999... = 1. Your rejection of this simple fact assumes that 0.999... is not equal to 1, and hence cannot serve as a proof that 0.999... is not equal to 1. Phiwum (talk) 03:41, 3 April 2009 (UTC)

Huon wasn't claiming that there are infinitesimals. As for two lines that express 0.999..., well, that's easy. 0.999... is obviously a repeating decimal; recognizing this, you can easily make it into a ratio. The result is 9/9, reducing to 1. Thus, for any non-zero real x, the ratio x/x equals 0.999... Gustave the Steel (talk) 04:19, 3 April 2009 (UTC)

That's what Huon indeed claimed. 0.999... is as irrational as pi or any other irrational number. If pi is defined as a Cauchy sequence, then pi is indeed rational which is of course nonsense. All cauchy sequences are rational. 70.197.200.143 (talk) 11:26, 3 April 2009 (UTC)

Actually, if you read the articles I specified earlier, you would not claim that 0.999... is irrational. It is a repeating decimal, and all repeating decimals are rational. The process to turn a repeating decimal into a ratio works on 0.999... and gives a result which is supported by other proofs.

Also, could you point out where Huon claimed the existence of infinitesimals? I'm looking for it, and I don't see it. Gustave the Steel (talk) 15:31, 3 April 2009 (UTC)

The difference between 0.999... and 1 is zero, because they're equal (see above for "You haven't proved that they're unequal, nor refuted any of our proofs that they are"). Infinitesimals do not exist in the real numbers, so stop bringing them up. --Zarel (talk) 06:33, 3 April 2009 (UTC)

The difference between pi and its predecessor is 0. Only problem is we don't know what pi or its predecessor is. 70.197.200.143 (talk) 11:26, 3 April 2009 (UTC)

To 98.201.123.22: one of your principal objections to 0.999... = 1 appears to me to be that, because "0.999..." and "1" are two visibly different notations, you believe that they must represent different numbers. However, this is not necessarily the case. Consider, for example, the notations "1/2" and "0.5000...". One is a simple fraction, the other an infinite decimal expansion. Nevertheless, they both denote the same real (and also rational) number. Similarly, "1/3" and "0.333..." both represent the same real (and also rational) number. Mathematically, there is no difference in kind between these cases and that of "1" and "0.999...". The (quite strong) intuition that "0.999..." must mean "1 minus a very tiny bit" is simply wrong. -- The Anome (talk) 09:36, 3 April 2009 (UTC)

Appearances are very deceptive Anome. Can you read? Go back and reread what I wrote before you write rubbish. 70.197.200.143 (talk) 11:26, 3 April 2009 (UTC)

To answer your questions: 1-0.999...=0, as others pointed out. A number x is an infinitesimal if for every natural number n, we have |nx|<1 (note that 0 does satisfy this definition, and I don't claim there are other real infinitesimals - that's the Archimedean property, by the way). Since there are no infinitesimals except 0 within the reals, a real number differing from pi by an infinitesimal is again pi. Concerning the ancient Greeks, I seriously doubt that Euclid expressly used rational numbers - the Pythagoreans knew about the irrationality of the square root of two, but Euclid was more concerned with geometry than algebra - though I have to admit that I've never read the Elements. Anyway, I don't think Greek math ever left the realm of the algebraic numbers, with the exception of pi - and they couldn't show that pi is transcendental. The claim that Cauchy sequences are always rational is almost ridiculously wrong - I can easily construct a Cauchy sequence converging to the square root of 2. And finally, what is the "predecessor" of pi supposed to be? Real numbers (and rationals, for that matter) don't have predecessors - between any two different real (or rational) numbers there are infinitely many others. Huon (talk) 14:21, 3 April 2009 (UTC)

Please do not call the writing of other users "rubbish".

That said, it does not appear that The Anome has misread anything. Please clarify. --Zarel (talk) 23:07, 3 April 2009 (UTC)

Prove 0.999... is a rational? Okay!

${\begin{array}{lrcl}{\mbox{Let}}&x&=&0.999...\\\Rightarrow &10x&=&9.999...\\&&=&9+0.999...\\&&=&9+x\\\Rightarrow &10x-x&=&9+x-x\\\Rightarrow &(10-1)x&=&9+0\\\Rightarrow &9x&=&9\\\Rightarrow &x&=&{9 \over 9}\end{array}}$ —Preceding unsigned comment added by Maelin (talk • contribs) 05:13, 3 April 2009 (UTC)

Proof 6 is a rational? Okay!

${\begin{array}{lrcl}{\mbox{Let}}&6&=&0.999...\\\Rightarrow &10*6&=&9.999...\\&&=&9+0.999...\\&&=&9+6\\\Rightarrow &10*6-6&=&9+6-6\\\Rightarrow &(10-1)*6&=&9+0\\\Rightarrow &9*6&=&9\\\Rightarrow &6&=&{9 \over 9}\end{array}}$ —Preceding unsigned comment added by 70.197.200.143 (talk) 11:19, 3 April 2009 (UTC)

Have you considered taking this to http://uncyclopedia.wikia.com/wiki/Talk:0.999... , where it might be more appreciated? If you're still unsure about what constitutes a proof and what doesn't, please try http://us.metamath.org/mpegif/0.999....html -- The Anome (talk) 12:32, 3 April 2009 (UTC)

I'll WP:AGF and assume you don't understand the flaw in the analogy.

Letting x = 0.999... is valid, because it is a definition of x, a way of saying "Okay, the next time I say 'x', I mean '0.999...'.". Letting 6 = 0.999... is invalid, since it is a comparison, so there first needs to be some proof of why 6 = 0.999... before it can be used in a proof. --Zarel (talk) 23:00, 3 April 2009 (UTC)

Amazing! A proof by contradiction that 6 ≠ 0.999...!

Erm... That was what the proof was supposed to be, right?

Reads title

Oh.

God.

It's worse than I thought! Faints. --72.177.97.222 (talk) 03:03, 4 April 2009 (UTC)

Actually the above proof is as invalid as all the other proofs you provide. Why? It is based on the assumption that 0.999... = 1. Also, it is invalid because cancellation does not apply to recurring decimals. To replace the mantissa with x is wrong. And yes, it leads to weird results - you can almost prove that anything is equal to 1. 98.201.123.22 (talk) 13:19, 4 April 2009 (UTC)

Yes, the above proof is invalid. The one above it is also arguably invalid, as it assumes the arithmetic properties hold for infinite decimal sequences. However, they do hold; it's just an assumption that requires proving, which leads us to the much higher topics of Analysis. And that proof most certainly cannot be used to prove that absolutely anything is equal to 1. I'd like to see a demonstration of that claim (and the proof that 6 is rational is not such a demonstration).

It might benefit you to take a look at the list of types of mathematical proofs and perhaps also Pseudomathematics to help you understand just how horribly screwed up the proof above that 6 is rational is.

If you still need help understanding why the latter proof is not at all like the former, I will be happy to help. However, you currently come across as a troll stirring up trouble by making the most obvious of mistakes. --72.177.97.222 (talk) 13:38, 4 April 2009 (UTC)

You contradicted yourself - first you agree it is invalid and then you state the proofs do hold. Hmmm, it seems to me you are the one who needs help. My major is in mathematics, so please, spare me your condescending attitude. I am not inferior to you. 98.201.123.22 (talk) 13:43, 4 April 2009 (UTC)

I agreed that it was invalid because it is incomplete. It is quite possible to complete the proof so that it will be perfectly valid. However, the completion of the proof requires some analysis, which you seem to prefer not to go into, so I have stayed away from there. --72.177.97.222 (talk) 13:49, 4 April 2009 (UTC)

Complete and valid proofs are given in the article, that do not require assumptions about the arithmetic properties of infinite sequences. Mdwh (talk) 13:26, 5 April 2009 (UTC)

Now answer the question: find two integers s.t. 0.9..= 1

It always happens on Wikipedia. When you cannot answer a question, your tactics are to stray into different topics. We are not going to discuss infinitesimals (no such thing) or the Arch. property. To say that a number is infinitesimal if it is not Arch, is not saying anything. This is much like saying that an irrational number is a number that is not rational - it really does not say much. We cannot define any characteristic of numbers by a non-attribute or an attribute that is lacking. Now let's just think about what 'rational' means: any rational number is one that can be measured completely. This distinction is important. Thus to state that 1/3 = 0.333.. in base 10 is incorrect. 3/10+3/100+3/1000+... Hmmm, looks like I will go on forever and still not know the measurement of this quantity. All I know is that I am thinking about 1/3 which is finite and measurable. Remember the ratio is a comparison of magnitudes. What two magnitudes are being compared if 0.999... is a rational number? Not 1/1 unless you assume these are already the same magnitude which these are obviously not. I do not care how close or far these numbers are or even if numbers exist between these. Please don't confuse the argument by asking if I can find a rational number between these - there are a lot of things one can't do. Besides the definition of reals according to Cauchy is that a 'limit' exists for the partial sums of the representative convergent sequence. But this limit is vague and ethereal and all we can say about it is that it is always rational. From this we realize also that irrational numbers and real numbers in general are not defined contrary to popular opinion. 98.201.123.22 (talk) 13:32, 4 April 2009 (UTC)

Dude... Stop pretending you know anything about the real numbers, rational numbers, and Cauchy sequences and try doing some research for a change. There are Wikipedia pages addressing each of these topics. No, it's not easy material, but if you want to argue about it, then you'll have to learn about it first.

The point of the Cauchy sequence construction of the reals is that there are 'holes' hidden in the rational numbers, and the Cauchy sequence is a technique for figuring out exactly where those holes are. Another technique for finding those holes is by Dedekind cuts, which are perhaps more intuitive. Look it up. --72.177.97.222 (talk) 13:47, 4 April 2009 (UTC)

Dude? You may want to think outside your pants for a change. what makes you think I am a dude? Continue adding posts - all doubt that you are indeed a fool is being removed.98.201.123.22 (talk) 13:57, 4 April 2009 (UTC)

Unless you mean for me to entertain the notion that you are a different species than me (that is, Homo Sapiens, in case you're not sure), you are most certainly a dude. "Dude", as it is used nowadays, is a non-gender-specific term. --72.177.97.222 (talk) 14:17, 4 April 2009 (UTC)

I thought that the feminine for "Dude" was "Dame". I can't recall "Dude" being used to refer to a woman anywhere. Algr (talk) 08:11, 6 April 2009 (UTC)

Before it started becoming a non-gender-specific term, the feminine was dudette. I have heard "dude" used to refer to women before. You may very well be correct in thinking that you have not, because trends vary in different locations across the world.

This is, of course, a meaningless argument over semantics, which has nothing to do with 0.999... --72.177.97.222 (talk) 11:42, 6 April 2009 (UTC)

Is it just me, or is the question rather empty? 5 and 6 are two integers such that 0.999... =1 - because that equality does not depend on a pair of integers, any pair of integers will do. Do you mean two integers such that their ratio is 0.999...? Then, as you've been told before, any pair (n, n) will suffice, since, as is proven in the article, 0.999... = 1. Note that I don't claim that 0.999...=1 because I can find those integers - it's the other way around, I can find those integers because 0.999...=1, which is proven differently.

To me it seems as if you start out with the assumption that 0.999...≠1 ("obviously not" the same magnitude) and reject anything that contradicts that assumption. Here is the old proof that I mentioned earlier - it doesn't use limits, it doesn't rely on manipulating infinitely long decimals, no tricks at all. I can use that proof to show the equality, and then 0.999... has all the properties of 1.

Oh, and to settle that once and for all: A real number is an equivalence class of Cauchy sequences of rational numbers, where two Cauchy sequences (a_n) and (b_n) are equivalent if and only if the sequence (a_n-b_n) converges to 0. There is a natural map from the rational numbers into the real numbers, where x is mapped to the equivalence class of the costant sequence (x, x, x, ...). Thus, we can consider the rational numbers as a subset of the reals; a real number is irrational if it is an element of the copmplement of this subset. Note that this definition does not rely on limits. If you want, I can of course also give definitions of Cauchy sequences and convergence. Huon (talk) 16:09, 4 April 2009 (UTC)

On that same page that you refer to, there is a section that refutes Rasmus' proof. See Rasmus' proof and its error. 98.201.123.22 (talk) 23:35, 7 April 2009 (UTC)

Have you actually read that section? The supposed "error" was that Rasmus started with a statement that is true for all natural numbers, and someone believed that then the statement would also have to hold for infinity, which it didn't. Of course the implication is flawed, and Rasmus' original statement, namely, that 0.999... is greater than any number of the form 0.999...9 with only finitely many nines, seems self-evidently true if 0.999... is in a meaningful way related to the rest of the number system. Huon (talk) 01:38, 8 April 2009 (UTC)

Actually this seems dodgy to me too. How can you discuss Cauchy sequences and convergence without already having the real numbers with which to define a metric space? Maelin (Talk | Contribs) 02:01, 5 April 2009 (UTC)

A sequence (a_n) of rational numbers is a Cauchy sequence if for any rational number ε>0 there is a natural number N such that for any pair of natural numbers n, m greater than N, we have |a_n-a_m|<ε. No reals involved. A sequence of rational numbers (a_n) converges to a rational number x if the complement of every open neighbourhood of x contains but finitely many elements of the sequence. Again, no reals involved. Huon (talk) 10:23, 5 April 2009 (UTC)

I have to reply to your comment "1/3 = 0.333.. in base 10 is incorrect." If you divide one by three, you get 0.333... (To prove, simply perform long division. You'll get 0.3, then 0.33, and then you apply induction to conclude that 0.333... is the answer). Thus, you must either admit that 1/3 equals 0.333... or that you think division does not work. (If you think that 1 divided by 3 produces some other result than 0.333..., please share what you think the result is.) Gustave the Steel (talk) 01:47, 5 April 2009 (UTC)

Augh, for the love of god... Stop using long division to justify claims on this article, that is only going to fuel the fire. The long division algorithm never terminates, don't claim that a non-terminating algorithm has an output. You can prove that 0.333... = 1/3 in other ways, but every time this long division thing comes up it is just more fodder for the trolls. Maelin (Talk | Contribs) 02:01, 5 April 2009 (UTC)

I was trying to explain the equivalence of repeating decimals to rational numbers using low-level mathematical concepts. Thanks for undermining the attempt. You really helped us all out there. Now we can go back to convincing him that repeating decimals are equal to rational numbers by citing evidence which he will ignore. Gustave the Steel (talk) 05:26, 5 April 2009 (UTC)

Claiming that the long division algorithm "proves" that 0.333... = 1/3 has never been a convincing argument, even to people who aren't questioning 0.999... = 1. I myself am highly dubious of it and I fully acknowledge that 0.999.. = 1. Just because a statement is true does not mean that every proof of it is valid. Every invalid argument should be undermined, whether it is yours or your opponent's. If you are going to treat this as though it is a war, some kind of us against them conflict in which every argument is a soldier and ought to be supported no matter how flimsy, please go elsewhere. You are only going to make it more difficult for us to conduct a reasonable discussion. Maelin (Talk | Contribs) 11:09, 5 April 2009 (UTC)

That's right. Bad arguments are the surest way to convince people that you are intentionally hiding or resisting the truth. Algr (talk) 20:09, 5 April 2009 (UTC)

Maelin: would you support removal of the digit manipulation proof that 0.999... = 1? It does, after all, suggest that 0.999... * 10 produces output, even though multiplication of a repeating decimal would be a non-terminating algorithm. If not, you must surely acknowledge that if a person cannot understand a formal proof of an equality, it is acceptable to attempt an explanation using simplified mathematics. 1/3 does equal 0.333..., and LD is a handy way to convince anyone past third grade of the truth, even if it's not precisely formal. Gustave the Steel (talk) 21:29, 5 April 2009 (UTC)

On 5 April 2009, at 21:29, Gustave the Steel said:

I have yet to cite long division as a proof.

On 5 April 2009, at 01:47, Gustave the Steel said:

If you divide one by three, you get 0.333... (To --->prove<---, simply perform long division. You'll get 0.3, then 0.33, and then you apply induction to conclude that 0.333... is the answer).

Just sayin', ya know? (But don't sweat it; we all make mistakes) --72.177.97.222 (talk) 22:40, 5 April 2009 (UTC)

Whoops. Yeah. OK, Maelin, I apologize for the nastier tone in my preceding post. I'll delete the offensive portions (and leave the second part). Gustave the Steel (talk) 22:44, 5 April 2009 (UTC)

Oh, and since you only want formal, verified evidence, I shall refer you (both) to this. Gustave the Steel (talk) 05:26, 5 April 2009 (UTC)

As 72.177.97.222 advises, please learn what you're talking about before you talk about it. Thanks.

Repeating decimals are defined to be limits. If you don't understand limits, you could try asking Wikipedia:Reference desk/Mathematics, but that is extremely out of scope of a discussion on 0.999... --Zarel (talk) 10:15, 5 April 2009 (UTC)

If repeating decimals are defined to be limits, then it should say so at the top of the article in order to demonstrate that .999...= 1 is by definition, and not observational. Algr (talk)

Back to the "by definition, not observational" argument, again??? My God, this dead horse has been beaten to hell and back!

It's observational. The limit is observed to evaluate to 1.

It's an incidental side-effect of the reals, not a definition. We've been trying to tell you that, and you never seem to get it. --72.177.97.222 (talk) 22:52, 5 April 2009 (UTC)

Well, if forever adding 9's to .999 eventually makes one, then maybe repeating a false statement like your "side-effect" an infinite number of times will make it true. Algr (talk) 08:05, 6 April 2009 (UTC)

If you showed an inkling of an understanding of what the mathematical concept of "limit" actually means, you might be taken more seriously.

So, you have a choice:

1. Do some research. Gain some understanding of what a limit is. Come back to us, and explain exactly the problem you have with that concept. Then we will discuss it with you.
2. Walk away and pretend nothing ever happened. A month or so later, come back to find us arguing with someone else, and jump right back in. Don't ever address any of the questions you were asked to address earlier. We will discuss new, fresh ideas with you, but ultimately, you will be stuck at a corner again, and you will have to walk away, again!

--72.177.97.222 (talk) 11:42, 6 April 2009 (UTC)

Algr, you are quite correct that forever adding 9's to 0.999 will never create 1. It will also never create 0.999...

Many people think, incorrectly, that 0.999... is a process or a function, rather than a number. They give it away in their language; they say that 0.999... "gets closer to" but "will never equal" 1 (as though 0.999... was moving). Actually, the process of adding 9's to a decimal over time will get closer to but never equal 1, but that's not what 0.999... is. 0.999... is a number with a static, fixed value, and proofs have shown it to be equal to 1. Your use of the phrases "forever" and "eventually" gives you away as someone who think of 0.999... as a process. Gustave the Steel (talk) 23:46, 7 April 2009 (UTC)

I have another neat proof

Let a = 0.999...

Then a^2 = (0.9 x 0.999...) + (0.09 x 0.999...) + (0.009 x 0.999...) etc
= 0.899999... + 0.089999... + 0.008999... etc
= 0.8 + 0.17 + 0.026 + 0.0035 etc
= 0.999...
= a

Since a^2 = a, a = 1 (or a = 0 but we know this is not the case). QED. —Preceding unsigned comment added by 91.105.25.46 (talk) 00:30, 6 April 2009 (UTC)

Nope. Line 1 should be:

Then a^2 = (0.9 x 0.9) + (0.09 x 0.09) + (0.009 x 0.009) etc

And I don't see how you get from line 2 to line 3, or line 3 to line 4. And finally, you need to demonstrate that 1 and 0 are the only values in which a^2 = a. I already know of another. Algr (talk)

91.105.25.46's line 1 is correct; Algr misses all of the mixed terms (0.9 x 0.09), (0.09 x 0.9), (0.9 x 0.009), and so on. Anon gets from line 2 to line 3 by adding digit-wise: 8/10 from the first summand, then 9/100+8/100 from the first two, then 9/1000+9/1000+8/1000 from the first three ...

The step from line 3 to line 4 could indeed do with an enhanced explanation; I'm too lazy do do that. The easiest way might be a proof by induction that $\sum _{i=1}^{n}{\frac {9i-1}{10^{i}}}=1-{\frac {n+1}{10^{n}}}$ - and then some sort of reason why the right hand side is indeed 0.999...

The last step is easily done; if a²=a, then a(a-1)=0, and since there are no zero divisors in our number set, a product is zero if and only if one of the factors is zero. So a=0 or a-1=0. Huon (talk) 10:55, 6 April 2009 (UTC)