Talk:0.999.../Arguments

MfD Result Notice

This page was the subject of an MfD discussion closed on 12 May 2007, with a keep result. Xoloz 18:36, 12 May 2007 (UTC)

Then... there're only 3 numbers?

OK... I learned about 0.(9) being the same as 1 in high school too...

But now I have this question.

1 = 0.99999... right?

2 = 1.99999... every ok so far.

does this mean that

0.99999... = 0.999999999...888... ?

or, to put it into words, is 0.(9) the same as 0. followed by infinite 9s AND infinite 8s? is that even possible?

or, even without knowing what number it would be, one could argue the following:

Since a number, like 1, is equal to the infinitely slightly "previous" number, there are only 3 numbers: zero, an infinite positive number and an infinite negative number.

Think about it... every number has an infinitely small 'equal'. At the same time, this 'equal' should have its own infinitely small equal, and so on...

Comments? —Preceding unsigned comment added by 201.246.25.69 (talk) 05:00, 18 January 2010 (UTC)

Within the real numbers, a number with infnitely many 9s followed by infinitely many 8s is impossible. Every digit only has finitely many other digits before itself.

Your "only three numbers" line of reasoning is flawed for two reasons. On the one hand, 0.999... is not "previous" to 1, not even "infinitely slightly" (or infinitesimally). They're exactly equal. (Again, within the real numbers - there are other number systems where a number denoted as "0.999..." is less than 1, but those number systems have significant drawbacks compared to the reals and are not in general use.) On the other hand, even if 0.999... managed to be both equal to and less than 1 (which is by definition impossible), not every real number has more than one decimal representation. For example, 0.333...=1/3 has just one, and there's no "infinitely slightly previous" number for 0.333... (actually, there isn't even one for 0.999... itself). Huon (talk) 06:23, 18 January 2010 (UTC)

Infinitesimals were thought to have numerous advantages by Leibniz, Euler, Cauchy, and others. Which drawbacks are you referring to? Tkuvho (talk) 11:09, 18 January 2010 (UTC)

Leibniz, Euler and to a degree even Cauchy predate the formalization of analysis (I'd set that at the age of Dedekind and Weierstraß). The specific drawbacks of the possible number systems depend on the specific number system, of course. For example, the surreal numbers and the closely related Hackenstrings have problems with multiplication: 3 * 0.333... = 1 > 0.999..., which is counterintuitive (and if you consider just the Hackenstrings, which are more closely related to decimal (or binary) representations than the full surreals, they won't form a ring). Often there will also exist numbers which cannot be represented by decimals any more, making decimal representations a bad choice to model those number systems in the first place. The article also mentions Richman's "decimal numbers" which don't form an additive group. Among the hyperreals there's no reason to denote any specific number (except, ironically, 1) as "0.999...". In any case the existence of infinitesimals will mean that you cannot define a meaningful metric on your number set (because the definition of a metric depends on real numbers and non-infinitesimal distances - of course that could be modified), and while you probably still can define a topology, it will behave very strangely. I can't think of a case where you have a topological group, limits are unique and the sequence (0.9, 0.99, 0.999, ...) converges. Huon (talk) 15:52, 18 January 2010 (UTC)

Your comment is thoughtful but not fully informed. What complicates the situation is that there are two separate issues involved: the mathematical/foundational issue, and the educational/pedagogical issue. I am not sure which is the one you would be interested in sorting out. At any rate, mathematically speaking, keep in mind at least the following two points: (1) the surreals cannot be used as foundation for analysis, since they lack the transfer principle; (2) the distance function on the reals has a natural extension to the hyperreals by the transfer principle, and in fact it can be constructed concretely in the ultrapower construction. As far as the students are concerned, at this level they are not interested in technical details of analysis and especially not topology. The real question is, whether their conception is an erroneous one, or rather a nonstandard one. Here the latter term is not referring to a particular mathematical theory, but to the possibility that their intuitions are coherent in the sense that they can be vindicated in a suitable mathematical framework. Tkuvho (talk) 12:15, 20 January 2010 (UTC)

I am no expert in non-standard analysis and may thus err in my opinions thereof. But I don't believe the hyperreals really are an improvement over the reals for educational purposes. While the hyperreals allow us to consider numbers infinitesimally less than one and may in this regard conform to the students' intuition, I dare predict that teaching hyperreals in high school would be complicated and not more thorough than the current high school education concerning decimal representations. In effect, we'd probably just have moved the regions where student intuition and mathematical model diverge.

I'm also still not convinced that the hyperreals are a good mathematical model for the 0.999... issue. As I noted before, there's no canonical hyperreal number (except 1) to be denoted "0.999...". I'm avare of the Katz&Katz paper you added, but I believe that firstly, the choice of $0.999...=0.{\underset {[\mathbb {N} ]}{\underbrace {999\ldots } }}=1-1/10^{[\mathbb {N} ]}$ is arbitrary - there are lots of other hyperreals which might just as well be denoted "0.999...". Secondly, it's an abuse of notation because the "0.999..." notation suggests nines all the way down; it would be better to denote Katz' number as $0.{\underset {[\mathbb {N} ]}{\underbrace {999\ldots 9} }}$ to show that it does have a last nine. Thirdly, I may be mistaken but I believe $[\mathbb {N} ]$ isn't even a well-defined hyperreal but depends on the choice of ultrafilter - and giving a specific ultrafilter is a non-trivial task. For example, is $[\mathbb {N} ]$ greater or less than the hyperreal given by the sequence a_n=n+(-1)ⁿ/n? Greater if the set of odd integers is contained in the ultrafilter, less otherwise. Huon (talk) 16:23, 20 January 2010 (UTC)

Thanks for your thoughtful comments. I was hoping we would narrow down the discussion to make it more manageable, to either the educational issues, or the mathematical/foundational issues, but you seem to be interested in both, which is just as well. I would like to make several points. (1) On the educational side, note that a couple of weeks ago user 67.161.232.156 spoke about the possible existence of a number of the form .000...1 (with infinitely many zeros), at the "User:ConMan/Proof that 0.999... does not equal 1" page. The suggestion was rebuffed by Tango who pointed out that ".999...8 is just as nonsensical as .000...1." Now user 67.161.232.156 might be wondering why his .000...1 is necessarily nonsensical, but Huon's .999...9 (with infinitely many 9s) makes sense. Why is his intuition erroneous, while yours not erroneous but nonstandard? (2) On the foundational side, in your comment above, you used the definite article in describing Weierstrass's formalisation of analysis. This is a possible viewpoint, but many historians view it as "a formalisation", not "the formalisation". Postulating Weierstassian formalisation as the unique foundation for analysis, of course, pre-determines the outcome of any discussion of infinitesimals, which were eliminated from the continuum by Weierstrass. As you may be aware, there are alternative foundations analysis that do not eliminate infinitesimals. One is by Robinson as discussed; another by Lawvere in smooth infinitesimal analysis. (3) .999... as a real number is well defined, and nobody claims it equals 1-(.1)^[N]. I don't think there is any abuse of notation here. (4) your question in the context of [N] boils to down to asking whether the infinitesimal defined by the sequence < (-1)ⁿ/n >, is positive or negative (the question has nothing to do with [N] itself). You are right, this depends on the choice of an ultrafilter. I am not sure why the sign matters so much. Note that the sign of this infinitesimal determines the parity of the hypernatural [N]. You are perhaps aware that the existence of a free ultrafilter is a consequence of the axiom of choice (the same goes for existence of maximal ideals, Hahn-Banach theorem, ...) Note that in the presence of the continuum hypothesis, all hyperreal continua are isomorphic. Furthermore, there are alternative constructions not relying on choice. Tkuvho (talk) 09:26, 21 January 2010 (UTC)

Concerning formalization of analysis I meant "the age of formalization of analysis", not specifically the Weierstraß approach. As I said, I'm no expert on non-standard analysis, and I wasn't aware of the specifics of Lawvere's approach; thanks for the reading suggestion.

The trouble with 0.000...1 and 0.999...9 is, of course, context. One needs to clarify what that last digit represents. I specified that it should mean a digit given by a (more or less) specific infinite hypernatural number, all within the realm of the hyperreals. 67.161.232.156 doesn't specify context, but his basic intuitive assumption that there should be a greatest number less than 1 is as wrong in the hyperreals as in the reals. He also seems to assume that there's some end to 0.999... (just like Katz) - something I'd say is the wrong kind of intuition (and 67.161.232.156 is a good example for common intuitive assumptions that can be wrong beyond repair, so please don't tell me that we should try and make our mathematics conform to intuition unless your number system also gives a greatest number less than 1). We wouldn't denote 1-10^-Googol by 0.999...; neither should we denote 1-10^-[N] by 0.999...

Finally, concerning the structure of the hyperreals and ambiguity of [N]: I am aware that free ultrafilters exist, and that the models of the hyperreals are isomorphic no matter what ultrafilter we use. But do these isomorphisms really map the hyperreals represented by the sequence (1, 2, 3, ...) onto each other? Can you prove that? I am willing to believe that the hyperreals represented by the sequences which eventually become constant are mapped onto each other, but anything beyond that seems dubious to me. Thus, the hyperreal represented by (1, 2, 3, ...) using one ultrafilter may not be equal to the one represented by the same sequence using another ultrafilter. In that case [N] would be ill-defined unless you also specified the ultrafilter - mere existence isn't enough. Or you end up with saying that [N] is just the class of hyperreals represented by that sequence, and not a specific hyperreal at all. Huon (talk) 12:06, 21 January 2010 (UTC)

Thanks for your thoughtful comments. I think perhaps we should narrow down the discussion to the educational issues. The foundational issues are only tangentially related to the present page. The full power of the hyperreals is not needed to handle the .999... issue. A common kernel for infinitesimal theories that can account for the ".999..."<1 phenomenon resides in primitive recursive arithmetic, in the context of the fraction field of a non-standard model of arithmetic. Such models were already constructed by Skolem thirty years before Robinson. You correctly point out that [N] will not necessarily be preserved by an automorphism. At any rate, on the educational front relevant here, I have to reserve judgment concerning your statement that user 67.161.232.156 "seems to assume that there's some end to 0.999... (just like Katz) - something I'd say is the wrong kind of intuition". We should try to listen to what education people say about this. Published work by Robert Ely shows that intuitions such as those of 67.161.232.156 are not erroneous, but rather nonstandard. If I thought they were erroneous I would not have pursued this matter on this page. Tkuvho (talk) 09:23, 22 January 2010 (UTC)

Sure

You have to be sure though when claiming that it is one, to point out that it is really (1 - .0000...1)
While the difference between .999... and 1 is infinitely nothing, it cannot be dismissed because it is everywhere.
.000...1 is the first thing there is greater than zero, and it's between every change between every two numbers all the way up to one. It's everywhere, but it's nothing. I believe that it is the graviton number, but I cannot prove it. --Neptunerover (talk) 16:31, 21 January 2010 (UTC)

I have never heard of a graviton number, but I don't think there is any useful number system (except the integers) where a "first thing greater than zero" exists. We usually want our numbers to form a ring, that is, to allow multiplication. What's 0.1 times the "first thing greater than zero"? Even worse, we also want our numbers to allow for cancellation, that is, for a≠0 we'll require that a*x=a*y implies x=y. I believe these pretty straightforward requirements alone imply that if there's a first thing greater than zero, it's one.

Consequently, the difference between 0.999... and 1 is not the smallest positive number (in all number systems I'm aware of). In the real numbers, the most widely used, it's exactly zero. In more exotic number systems, such as the hyperreals, it's possible to redefine 0.999... so that there is a non-zero infinitesimal distance between 0.999... and 1, but there still is no smallest positive number, and smaller infinitesimals exist. Huon (talk) 17:14, 21 January 2010 (UTC)

This is from the infinitesimal article: "In the 20th century, it was found that infinitesimals could also be treated rigorously. Neither formulation is wrong, and both give the same results if used correctly." My suggestion is that if both formulations work, maybe one formulation might work better for certain calculations. Using a number set that contains all infinities is better than one with no upper limit to contain anything, at least when you have to deal with infinities. In figuring out gravity, they might be better off using a set with limits, although still infinite. --Neptunerover (talk) 06:05, 22 January 2010 (UTC)

That's referring to using infinitesimals for calculus, considering things like dx to be infinitesimals rather than just shorthand for a limit. The sets used in that wouldn't have 1-0.999... being a non-zero infinitesimal. I don't know what you mean by "a set with limits", but the real numbers work perfectly well to figure out gravity, infinitesimals wouldn't help. The gravitational constant (in SI units) is really small, but it's still a perfectly normal, finite number. The only time infinities come up with gravity is with black holes and I don't think using a different number system would help there, not on its own, at least. --Tango (talk) 06:15, 22 January 2010 (UTC)

The infinities you mention with gravity are the exact ones I mean. They are the reason Quantum Mechanics cannot currently be meshed with gravity. Infinitesimals are perfectly valid for use in mathematical figuring, but the practice was mostly abandoned in the second half of the nineteenth century because somebody formalized calculus using the system of numbers it now uses. But that doesn't mean there's not another way. --Neptunerover (talk) 07:03, 22 January 2010 (UTC)

A graviton number would represent the force of gravity, which is very small. By reasoning with this number set, .999... would be the opposite of the graviton number, aka. the speed of light, which is the upper limit. If we do all our calculations within the decimals, all products should stay within the set. --Neptunerover (talk) 17:49, 21 January 2010 (UTC)

As I said, I've never heard of graviton numbers before, and I believe you made up the notion. If not, can you point me to a peer-reviewed article or a text book where I might read up on graviton numbers? Also it's not really an improvement to introduce too much physics into a pure math discussion. For example, I don't see why anything representing the force of gravity should have the dimension of a velocity, which would be necessary to compare it to the speed of light. You could also answer my original question: What's 0.1 * (1-0.999...)? Those are all decimals, so the result should still be a decimal, right? What decimal? Huon (talk) 18:08, 21 January 2010 (UTC)

I suppose it would be a very very small one. I don't think that's any weirder than any of the other weirdness that comes out the number sets that are generally used. I don't really know how the math would work within the set. Each number itself represents a velocity, from zero to max. I don't really know; can C be 1 in the E=MC(squared) equation? --Neptunerover (talk) 18:31, 21 January 2010 (UTC)

What very very small one? Write it down. "Each number itself represents a velocity, from zero to max." That makes no sense. The speed of light cannot be 1, since 1 is a dimensionless quantity and the speed of light is a speed. It can be 1 light year/year (and we often do use units where the speed of light is 1 unit since it makes the maths much easier), but that is still a speed. When you talk about graviton numbers, do you mean the gravitational constant? That is a very small number in SI units, but not infinitely small - it's about a ten-billionth. --Tango (talk) 18:37, 21 January 2010 (UTC)

I think 1 might be available for the speed of light, since time stops and space looses its meaning, so dimensionlessness may fit. --Neptunerover (talk) 19:24, 21 January 2010 (UTC)

I'm not certain how it would work. Something might get canceled. For instance, say the product of .1*(1-.999...) could remain .1 having some sort of a remainder representing the gravitational force against it at that moment. And yes, the gravitational constant is what I was referring to (thanks). The Graviton is a hypothetical particle that transmits the force. --Neptunerover (talk) 18:59, 21 January 2010 (UTC)

Neptunerover is correct regarding c: It is possible to use a system of natural units where c=1. Unfortunately for him, it is then just a small step to using units where the gravitational constant also equals 1, ie c=1=G, and G is not small compared to c: Planck units. But back to 0.999... and 1: Is 0.1 * (1-0.999...) less than 1-0.999...? If so, then 1-0.999... obviously is not the smallest positive number.

By the way, physics almost exclusively uses the real numbers (or the complex numbers) where 0.999...=1 exactly. Mathematicians can construct other number systems, but they are rarely used (thus the article claims by default that 0.999...=1), and their use gets ever rarer the closer the subject is to physics. Might I suggest reading up on decimal representations? Huon (talk) 19:36, 21 January 2010 (UTC)

Hey thanks. I always hated math. I think it's very one-sided as far as the brain is concerned. --Neptunerover (talk) 19:51, 21 January 2010 (UTC)

How perfect is this? (from the planck units article)
"Planck units are often semi-humorously referred to by physicists as God's units. They eliminate anthropocentric arbitrariness from the system of units."
And then what do these silly guys do next? They apply arbitrary anthropocentric concepts of measurement to the units, and then they get confused (or confounded, rather; I'm not saying it's anyone's 'fault', as my intent is not to point out a fault, but rather to suggest something else that might work. I've done plenty of things only to slap myself in the head in retrospect).
What is 'God's straightedge', and what is 'God's clock'? I'm guessing not ours.
Consider if we, instead of arbitrarily saying that the speed of light is X units of an arbitrary distance covered per arbitrary time segment, why not just say C is one, (or .999..., since time would have to stop completely at the speed of light making it dimensionless and possibly unreachable.) What if we started at the upper limit, the speed of light, call it 1 (or 100 or 1000? Whatever it is, it's the max, and the idea is to keep it an easy number and not just random), and then we figured out what the other values are going down from there? --Neptunerover (talk) 20:31, 21 January 2010 (UTC)

Also from the Planck units article, "Referring to G = c = 1, Paul Wesson wrote that, 'Mathematically it is an acceptable trick which saves labour. Physically it represents a loss of information and can lead to confusion.'". Note specifically the loss of information bit -- if you want it to have any meaning, you've got to put the units back eventually. But we're far afield from the original discussion of .999.... — Lomn 20:41, 21 January 2010 (UTC)

That actually is what we do. For historical reasons, it's not a nice round number, though. We define the speed of light to be exactly 299,792,458 m/s. We then define the metre from that (the second is defined in terms of caesium atoms). If we had known about special relativity thousands of years ago, then it would be a nice round number, but we didn't, so we have to put up with an un-round number to maintain consistency with old units that everyone is used to using. I don't know what you mean by "arbitrary anthropocentric concepts of measurement", though... --Tango (talk) 20:44, 21 January 2010 (UTC)

I mean like a meter or an inch. These are distances arbitrarily created by man, though not without use, but I believe they lead to confusion. I do admit that it might sound radical to suggest that perhaps the numbers we generally use might be confusing us (at least in Physics, which is very much math), and so I'm sorry for making the suggestion. --Neptunerover (talk) 21:02, 21 January 2010 (UTC)

What do metres and inches have to do with Planck units? Of course, one can convert from Planck units to other units of the same dimension and those conversions are usually mentioned when the Planck units are defined (since the people reading the definition will be familiar with the other units so can get a better feel for what these Planck units are if they know the conversion), but that's all. --Tango (talk) 01:08, 22 January 2010 (UTC)

While this is far off-topic for 0.999..., just imagine a speed limit of 5*10^-8 on the roads. Does that sound simple? Isn't it confusing if I tell you I live less than a heartbeat away from you? Shall we meet in a hundred trillion trillion trillion planck times? Oh, and please bring a few dozen million Planck masses of cookies (actually, since the Planck mass is dimensionless, just bring a few dozen million cookies...); I'll supply a googol of beer (Planck masses or cubic Planck lenghs? No matter, both are equally dimensionless...), ok? Our usual dimensions are appropriate for everyday tasks - and while one might redefine lengths and it wouldn't matter much, it certainly is an advantage that our units of time are (more or less) related to day length. There are other advantages of not using Planck units. Some final remarks: Saying c=0.999... is, for mathematical reasons unrelated to physics, the same as saying c=1 (that's the actual point of this article). It's also not true that "time stops at the speed of light". What would that even mean? Consider that 1 Planck time is much less than a second - are you saying that time stopped a tiny fraction of a second after it began? It took me more than 1 Planck time to write this, I can assure you, and time didn't stop in between. Finally, calling the number 1000 "easy" is again anthropocentric. We happen to have ten fingers and thus use base ten; if we had twelve, we might use a base 12 system and would consider 1728 "easy". Or how about 1024? Huon (talk) 00:18, 22 January 2010 (UTC)

They don't know what a second is. They might think they know how to measure one, but they don't know what it is. (At least not as far as I can tell). All the cells in our bodies do stuff lightning fast in a second. What's a second to them? How long does a second for us last for them? Is it because of our great mass in comparison to them that our time should be experienced differently? The human measurement of a second should not be universally applied. How many moments are in a moment depends upon how small (or big) you are. The Bible says a day for God is 1000 years for a man. God's a big guy. Not meaning to bring up religion, that's just my example that is documented (although with all the translations... who knows what it's supposed to be saying, and who relies on the Bible as a valid factual reference anyway? Anyone?) --Neptunerover (talk) 00:29, 20 January 2010 (UTC)

And I don't think 1 could ever actually be reached to where time would "stop." I think the change into a one could be like a supernova. It would be the fusion in the center of a star where there is no empty space between anything. --Neptunerover (talk) 06:35, 22 January 2010 (UTC)

You've gone past some simple misconceptions about mathematics and are now well into the realms of pseudoscience, so I'm going to bid you farewell. --Tango (talk) 06:42, 22 January 2010 (UTC)

Adios, amigo. --Neptunerover (talk) 07:14, 22 January 2010 (UTC)

Powers of ten, I consider easy, like the metric system as opposed to the foot/yard/etc. system. The idea of a meter is arbitrarily made, so that when we figure what the planck length is, based on a meter, we come up some fraction of a meter that is really only related to an arbitrary length. But if we started with the planck length and then went up from there, longer distances would be measured in gigaplancks (or I don't even know if that would be very long). I'm not really suggesting changing all the numbers we use for everyday purposes, which are generally served pretty well by the number systems we currently use for them. But to figure out the universe, I think we should start by using relevant units. --Neptunerover (talk) 03:59, 22 January 2010 (UTC)

Scientists do use relevant units. That's why Planck units exist - a scientist (called Planck, surprisingly enough) created them so he could use them to figure out the universe. They aren't used exclusively since it is often more convenient to use more conventional units, or units created for a particular purpose (eg. electron volts and parsecs), but they are used when it makes sense to use them. I don't see how any of this is relevant to 0.999..., though... --Tango (talk) 04:21, 22 January 2010 (UTC)

I just started out saying that .999... is different from 1 by the exact same difference there is between any two consecutive decimals. In order to get a 1, something, no matter how infinitely insignificant and ultimately unidentifiable, still something gets added to .999... --Neptunerover (talk) 05:20, 22 January 2010 (UTC)

"Consecutive decimals" do not exist. Either two numbers (1 and 2, 0 and 1345, 0.999999999 and 1) are distinct, in which case there are infinitely many other distinct numbers between them, or they are not distinct (0.999... and 1, 2/4 and 3/6, x+0 and x) and are therefore exactly the same in value. Not only are there infinitely many numbers between any two distinct numbers, but there are uncountably many, an infinity itself infinitely larger than your garden-variety infinity. So please, to close out, don't try to introduce your theory into the encyclopedia at large. — Lomn 14:11, 22 January 2010 (UTC)

You are splitting hairs with words, and your assumptive challenge is taken for what it is. I'm not interested in having an argument with know-it-alls; there's no point since they already know everything. Of course those people also want to control pages like this that are merely discussion pages about a topic. If you wish to 'frame' a debate, I suggest you go into politics. To close out, keep your assumptions to yourself, and if you don't like a discussion, stay away from it. --Neptunerover (talk) 07:35, 23 January 2010 (UTC)

Okay, forget about decimals for a second; they're misleading (which is kind of the reason for the existence of this page). Look at a line, and let points on the line represent real numbers. We have a point representing 0, a point representing 1, and half-way between those two points is a point representing 0.5. Now, remember that this is a platonic exercise in visualizing an 'ideal' line, and should be unaffected by the strange principles that we encounter when we get to small enough intervals that quantum mechanical effects become significant. Lomn's assertion above says that these points are dense, which is the name we give to the property that between every pair of distinct numbers, there is another number. We define real numbers this way because we want them to behave in this manner. It is analogous to the assertion that between any 2 points on a line there is another point. This is true in our platonic realization of a line, and it is true in Euclidean geometry.

Returning to decimals, they are a representation of real numbers, not the embodiment thereof. This representation consists of a list of digits between 0 and 9. A list of anything is characterized by a function from the natural numbers to 'anything'. So the symbol "0.999..." makes sense, because it is the mapping that associates every natural number with 9. Because there is no last natural number, the symbol "0.999...9998" makes no sense unless the 8 is in a finite place value (as Huon has pointed out below), in which case it is definitely not the 'next smallest' real number, since you could produce a number in between both by simply adding more 9's. So suggesting that 1 has a next smallest real number, which we denote "0.999...", leads us to the inevitable conclusion that 1 is somehow special in this respect, since "0.999..." certainly has no next smallest real number, and neither does any irrational number, or, for that matter, any number which cannot be represented as a rational number whose denominator is a power of 10. Why would you give such a limited class of numbers such a special class of significance, not to mention in the process destroy the property of density of the real numbers? --COVIZAPIBETEFOKY (talk) 16:31, 23 January 2010 (UTC)

I don't believe it is a limited class of numbers, at least not for the specific use I have in mind. I think of numbers as representing actual things which could be different than your average mathematician's approach. The way I'm looking at it, with the set (0,1{including all 'numbers' in between}), 1 does have a special place, because it is the top. I view the set as a ratio between emptiness and solidness. There can be nothing more solid than 1, because 1 is the absence of any 'holes' or spaces between anything. For instance, an atom is mostly empty space with a little thing in the center. That nucleus, I would consider a 1 because it consists of protons, etc. that are fused together without there being gaps between them. The atoms that make up a table each have 1 as their nucleus, but a table is not a 1 because it contains mostly empty space. The center of a star would be a 1 because even the atomic nuclei are fused together, making there be no empty space in the center of a star. Ones come in all different sizes because size is not what's important; what's important is solidness (in the number set that I use, in the way that I use it). Ones and zeros; that's computers, and I say it's the world. (Even if I can't explain it very well) --Neptunerover (talk) 18:12, 23 January 2010 (UTC)

But why should 1/10 have such special significance, where 1/3 does not? After all, 0.0999... is 'just below' 0.1, whereas no such number exists for 0.333... --COVIZAPIBETEFOKY (talk) 18:35, 23 January 2010 (UTC)

I'm not sure what you're asking me. .1 is different from 1/3 by lack of a repeating decimal. Can you expand on what you're asking me? --Neptunerover (talk) 18:55, 23 January 2010 (UTC)

Every finite decimal representation has a partner, which, according to us, is equal, and according to you, is 'just less'. No other decimal has such a partner. Why should finite decimal representations be so special in this respect? These are exactly those numbers which can be represented as a ratio of two integers where the denominator is a power of 10. Why should 10 seem to have such a mystical property that determines which numbers have infinitely close partners? After all, the particular choice of 10 is just an arbitrary one based on the number of fingers on our hands. --COVIZAPIBETEFOKY (talk) 19:36, 23 January 2010 (UTC)

I think 10 is more than just arbitrary, but you probably wouldn't agree with my reasoning (In fact I find it highly doubtful). I think it really all depends on how you want to divide up your 1. 10 'equal' parts, 12 'equal' parts... 10 is what I'm used to, and I think it is natural for us. As for generalizations, from my point of view, every decimal representation has two partners infinitely indistinct from itself, and each of those has the same. Following this line of reasoning, all numbers are ultimately indistinct from one another, making them all 1. This is that infinite density. (and if this paragraph makes sense, I'd be pretty surprised). --Neptunerover (talk) 20:04, 23 January 2010 (UTC)

Don't worry, I won't surprise you. But can you tell me which numbers are 'infinitely indistinct' from 0.333..., or 1/3, knowing that "0.333...332" is a meaningless set of symbols? And can you explain why 10 is so natural and non-arbitrary? --COVIZAPIBETEFOKY (talk) 20:14, 23 January 2010 (UTC)

/dedent

My definition of density is the opposite of yours, and exactly the same. You say between any two points there is one point; I say beside any one point there are two points. I'm talking about a peak, you're talking about a valley. It's the same difference, just opposites of each other. Can you see how arguing is futile here? We each are looking from a different direction at the same thing. (That is of course my opinion, while yours is your own and likely to be different than mine). Thank you for talking to me anyway. Sorry if it can be frustrating. --Neptunerover (talk) 07:54, 24 January 2010 (UTC)

WRONG. You got my definition wrong, and your definition makes no sense. How is a set such that every point has a first point above it and a first point below it 'dense'? Aren't the integers 'dense' under your definition? Every integer has a next integer (n+1) and a previous integer (n-1).

As for my definition, I did not say there is only one point between any two distinct points, and if you thought I meant that, you are deeply mistaken. All the definition says is that there is at least one point between any two distinct points, which implies that there are, in fact, infinitely many points between any two distinct points.

Let me explain: let a and b be distinct numbers in a system which is dense under my definition. Let c be a number between them, guaranteed to exist by density. Well, c is distinct from b (it is also distinct from a), so there is another number between those two, which we'll call d, which is, in turn, distinct from c, producing another number e, and so on. Thus there are infinitely many points between a and b. --COVIZAPIBETEFOKY (talk) 16:09, 24 January 2010 (UTC)

I understand your point of view, and I agree fully with your description of density. You're saying that there cannot be two numbers next to each other because in between them there is always room for more. My peak/valley comparison might not be very accurate now that you get me to look at it another way. I'll have to consider then what I meant.

Okay, what I'm thinking about has to do with the amount of change between any two 'adjacent' numbers, the difference between them will be some power of 10, meaning a 1 following however many zeros past the decimal point. In your example of a and b, if we take their separation as being .1, then the difference between either of them and c will be .01 or a lower power of ten. Each time the valley between two 'adjacent' numbers spreads for another number, the new difference represents a drop in magnitude. I'll see what you think of this, as I'm certain it's not complete. I don't have the right words to use here. --Neptunerover (talk) 17:00, 24 January 2010 (UTC)

I don't quite understand what you're trying to define. What is given, where do you intend to get? Do you want to start with the rational numbers (or the numbers of the form a/10ⁿ, where a and n are integers, n non-negative?) and then construct a number system where 0.999... differs from 1 (and where every number has a progenitor and a successor)? What other properties shall your desired number system possess? For example, shall it include all rational numbers? All reals? Shall it be a ring, or even a field? Or are you trying to do something completely different? Huon (talk) 18:00, 24 January 2010 (UTC)

I suppose I'm trying to figure out 'God's numbers', meaning a set of numbers that realistically represents the range of the universe (realistically to me, I cannot stress enough, because, as I've said, I find math difficult). Through the holographic principle, the 'inside' of a black hole can be considered a zero while the center of a star would be a 1. All the rest of the universe consists of the border between those two polar opposites. --Neptunerover (talk) 18:44, 24 January 2010 (UTC)

In your question though, you assert something that is seen as true by you and false by me, so I'll not be able to satisfy your question as posed. I could give you an example about 10 being natural for us, but my example would likely be viewed as nothing more than silly coincidence, and since my bet is that you consider coincidence irrelevant to reality, I think I should keep it to myself. --Neptunerover (talk) 20:26, 23 January 2010 (UTC)

/dedent

If you're not willing to share your views, how can we make sense of them to properly explain why they're incorrect? You'll notice everyone else has given up on you; that is because, as far as we can tell, you are spewing utter nonsense. Would you care to define a decimal representation? To us, it is a mapping from natural numbers to digits between 0 to 9. That is obviously not capable of representing "0.333...332", so you must have something different in mind when you use the words "decimal representation" than we do. Care to explain what your version of a decimal representation is?

For any countably infinite set, there exists a bijective function which maps the countably infinite set to the set of natural numbers, even if the countably infinite set contains the natural numbers. For example, the set of rational numbers - those numbers which can be written as a quotient of integers - contains the natural numbers as a subset, but is no bigger than the set of natural numbers since the rationals are countable: There is a bijection from the naturals to the rationals. --Neptunerover (talk) 17:30, 25 January 2010 (UTC)

You suggest that people abandon this discussion out of frustration, but I suggest another possibility that perhaps each in their own turn went "ah-ha!" and subsequently went off to write their own paper in hopes of recognition and perhaps a Nobel Prize. Neptunerover (talk) 08:40, 25 January 2010 (UTC)--

Don't flatter yourself. --COVIZAPIBETEFOKY (talk) 13:38, 25 January 2010 (UTC)

I think I've already stated that I consider them to represent a ratio between 0 and 1. --Neptunerover (talk) 22:34, 23 January 2010 (UTC)

I presume you mean a ratio of two integers, in which case you refer not to real numbers, but rational numbers. Those are incomplete; for example, they miss the square root of 1/2 (note that I have not defined the word 'incomplete' for you, and will not attempt to do so as it is rather complex, but that does not mean that you can magically insert your own definition; see here for a definition, if you wish). Also, the rational numbers are dense like the reals, so there is no greatest rational number less than 1, or any other rational number for that matter. --COVIZAPIBETEFOKY (talk) 23:20, 23 January 2010 (UTC)

Do you have a definition for unknown? Rather than .000...1 having no meaning (for me), I consider its meaning to be unknown. --Neptunerover (talk) 05:28, 24 January 2010 (UTC)

I'm not talking about a complete set. This set is filled with holes, and the only number without any holes is 1. --Neptunerover (talk) 07:30, 24 January 2010 (UTC)

First you say the set is filled with holes, and then you refer to a number not having any holes. Do sets have holes or do numbers have holes? I assume that for a set to have holes in it, you must mean that there is some sense in which something is 'missing' from it. How can this same vague property be applied to a number? --COVIZAPIBETEFOKY (talk) 16:09, 24 January 2010 (UTC)

Every other number besides one consists of a ratio of zero to one, making any number less than one include some amount of a hole with it. 1 is also a ratio, but there's no zero involved, it's just 1/1. --Neptunerover (talk) 18:44, 24 January 2010 (UTC)

And if you don't know what a mapping or a function is, I don't think I'm stretching the truth when I say you have gotten yourself in way over your head here. If that is the case, you should probably let this rest for a few years until you are properly introduced to these concepts. --COVIZAPIBETEFOKY (talk) 20:37, 23 January 2010 (UTC)

Your very first sentence (after the dedent) exemplifies the reason I could have for an unwillingness to share (and I have tried). We speak different languages as far as this topic is concerned, and framing the whole discussion from one point of view doesn't help two sides to understand one another, and especially when one side views its part in the discussion as being the corrector of the other side. In my personal opinion, Math textbooks are some of the most horribly written things I've ever experienced, and it takes a good teacher to translate one to a class. I'm certain you'd feel right at home reading one though. However, just because we speak different languages, that shouldn't mean we can't attempt to understand one another. All the experiences in my life have led me to this point with my current point of view, and there's no way I could ever expect anyone else to immediately get what I'm trying to say. But just because someone may be unable to explain something adequately to one, two, or however many people, that doesn't mean they are incorrect. I had no idea this concept would be so foreign to some people. I've gotten much out of this lengthy discussion, even if no one else has. --Neptunerover (talk) 22:18, 23 January 2010 (UTC)

Then perhaps my attitude is too pessimistic; obviously you must have come up with a perfectly acceptable set of numbers, along with well-behaved definitions of addition and multiplication on those numbers, such that every decimal (whatever your definition of a decimal may be) is associated with exactly one number in your system, and every number has a 'next' number and a 'previous' number.

I'd like to see that. Or some other evidence that you have good reason to believe that all mathematicians are wrong, beyond an emotionally charged rejection of a counter-intuitive notion. --COVIZAPIBETEFOKY (talk) 23:20, 23 January 2010 (UTC)

I'm not here to argue with you, especially when you add meaning to my sentences that was never there. I've read there is a big void between mathematicians and physicists. Everyone has their own angle, and it's the right one. I'm afraid I don't find your comments congenial. --Neptunerover (talk) 04:58, 24 January 2010 (UTC)

Okay, I'm failing at making myself clear. I'll try to clarify my point once again: you came here with the suggestion that the difference between 0.999... and 1 is ".0000...1", and we disagreed with you. In the system of real numbers, not only is this patently false, ".0000...1" is a completely meaningless set of symbols. However, that on its own does not mean that you were just spewing nonsense, because you may not be talking about the real numbers.

I am giving credence to your claims. I want to understand your point. As far as I can tell, you are making wild claims with no evidence. You are refusing to define your terms, because doing so will tie you down to those definitions. However, you are willing to wave your hands, and say "there's a number just less than 0.999..., which I'll call 0.999...98", without defining what those symbols mean. If you're trying to present an alternative viewpoint to the real numbers, then for gosh sakes, present it already! If you're not trying to present an alternative viewpoint, then I have sorely misunderstood your purpose in posting here. Perhaps a clarification on your part is in order. Unless you're just trolling?

Anyway, back to asking hand-wavey questions to try and pin down your point of view, since you so jealously refuse to share it openly. Since you have said that your number system, whatever it is, includes rational numbers, let me ask you this: what set of symbols, in decimal, would you denote as just less than 2/11, the repeating decimal 0.181818...? Is it 0.1818...17, or is it 0.1818...80? What if I add .5 to that number, giving 15/22 or 0.681818...? Does that make a difference to what the 'last' decimal should be? What's the last digit of 0.4673823123123123...? Since there is more than one repeating digit, is it not arbitrary what the last digit should be, and therefore how we should represent its adjacent decimal? --COVIZAPIBETEFOKY (talk) 16:09, 24 January 2010 (UTC)

And I would ask you "who's trolling who?" --Neptunerover (talk) 08:29, 25 January 2010 (UTC)

And... that's the end of my trying to converse with you. --COVIZAPIBETEFOKY (talk) 13:38, 25 January 2010 (UTC)

Here's one for you, how do you reckon .666...? --Neptunerover (talk) 14:00, 25 January 2010 (UTC)

I'll take a wild guess that he reckons .666... is 2/3. Do you reckon it differently? (Or have a different question in mind?) Phiwum (talk) 14:23, 26 January 2010 (UTC)

'C' seemed perhaps angry, and that got me thinking of an old Iron Maiden song. --Neptunerover (talk) 14:46, 26 January 2010 (UTC)

In one way, it might mean this red guy with pointy ears and a pointy tail and a pitchfork, but it must be considered that the ones who put forth that claim are also the ones who put forth plenty of other garbage concepts on the evolution talk page and so forth. The accuracy of how fanatics portray their fantasies should be questioned rather than arguing against their delusions. --Neptunerover (talk) 15:02, 26 January 2010 (UTC)

In a word, no. Your claim is wrong, and "0.000...1" has no meaning. A careless juxtaposition of the finite and the infinite does not a theory make. — Lomn 17:17, 21 January 2010 (UTC)

I don't believe I'm juxtaposing anything. I'm just saying that everything there is can be represented in the set between zero and one. I'm only referring to this one set, where anything above one is out of the question, while zero is the absolute bottom. No numbers outside of the set are needed or used. Does that clarify my statement? I'm only referring to decimals--all of them--between 0 and 1.--Neptunerover (talk) 17:38, 21 January 2010 (UTC)

"0.000...1" is such a juxtaposition. "..." describes an infinite sequence of 0s, to which you have appended a finite 1. You can have either infinite 0s, or finite 0s and a 1. Not both. I'm quite clear that you're referring to the zero-to-one range (which is infinite), but "0.000...1" is still just a string of characters with no mathematical meaning. Anyway, I see this has been properly ported to the Arguments page, where it may safely languish. — Lomn 18:16, 21 January 2010 (UTC)

Consider if what our speedometer read was, 0 to 1, with one being light speed, and zero being a dead stop. When we start rolling, what is our very first speed? It's going to be the very next thing up from zero, which will essentially be zero, but not quite zero. Every increase in speed is separated from the next lower speed by what is basically nothing, but still something. Like the separation between the iterations in a fractal. --Neptunerover (talk) 19:16, 21 January 2010 (UTC)

Trying to use physics to understand maths is backwards. It doesn't work like that. If you actually want to look at real world velocities then you will find that speed has no meaning on really small scales. Quantum mechanics gets in the way. Any time period smaller than the Planck time is meaningless (or, at least, not something we currently understand), so asking how far we travel in that amount of time is also meaningless. I'm far from an expert on these things, but if there is a quantization of velocity, then the quanta are finite, not infinitesimal. --Tango (talk) 20:05, 21 January 2010 (UTC)

I would think that Physics and Math should be helpful in elucidating each other. --Neptunerover (talk) 05:29, 22 January 2010 (UTC)

Not really. Maths helps you do Physics, but not the other way around. Physics helps you decide which axioms are worth investigating the consequences of, but that's it. --Tango (talk) 05:41, 22 January 2010 (UTC)

That's fine (it's just a restatement of Zeno's paradox), but it's unrelated to the original topic. — Lomn 20:03, 21 January 2010 (UTC)

New thread

Okay, so let me try and see this as best as I can...so you're saying that $0.(9)$ is the same as $1-0.000\ldots 001$. Now, upon first look of the conjecture, it seems to have valid meaning, I agree, because of its basis on the known that, for $n\in \mathbb {Z} ^{+}$, $\sum _{m=1}^{n}{\frac {9}{10^{m}}}=1-10^{-m}$. But, the error comes about from the consideration of $\lim _{m\longrightarrow \infty }10^{-m}$, which, as you can guess from examination of the identity, converges towards 0, thus making the limit itself zero. Saying that, therefore, would properly allow us to believe that $10^{-\infty }=0$, and thus that $0.(9)=1-10^{-\infty }=1$. (Is that understandable to everyone? I'm not sure; tell me if not.) --JB Adder | Talk 04:18, 23 January 2011 (UTC)

Since this is not really a continuation of the previous thread from a year ago, I made it into a new section. To answer your question, if $\infty$ is interpreted as an infinite hypernatural, then you can very well have a nonzero infinitesimal $10^{-\infty }$ or ${\frac {1}{\infty }}$, as John Wallis envisioned. Tkuvho (talk) 08:08, 23 January 2011 (UTC)

.999...9998

How does (.999...9998) differ from either (.999...) or (.999...9997)? --Neptunerover (talk) 09:22, 23 January 2010 (UTC)

What is .999...9998? The answer to your question depends on that. If the "8" (and the "7" in .999...9997) is the n-th digit for some natural number n, then the difference between 0.999...=0.999...9999999... and 0.999...9998 is 0.000...0001999...

If you've read the article's infinitesimals section (or certain parts of this page), we also discuss the (much less commonly used) hyperreal numbers, which allow a positional system where digits are numbered not only by natural numbers, but more generally by (possibly infinitely large) hypernatural numbers. But in that number system, the answer is still analogous: If the "8" is the h-th digit for some hypernatural number h, then the difference between 0.999... and 0.999...9998 (and between 0.999...9998 and 0.999...9997) is 0.000...0001999..., where the "1" is now the h-th digit and 0.000...0001999... may be an infinitesimal. But 0.999... = 0.999...999999... still has nines farther down the line and equals 1. Huon (talk) 12:46, 23 January 2010 (UTC)

I had a terrible algebra teacher one time; screwed me up bad. --Neptunerover (talk) 09:30, 25 January 2010 (UTC)

I think what we have here is a problem of different understandings of infinity. You're saying an infinite, never ending string of nines, while what I'm referring to is an infinite, never ending string of zeros which has a 1 at the end of it. Now that I consider it, I'm sure that it's the paradox that causes such difficulty in this being comprehended, but I see paradox as a very important factor in this universe of ours, and denying any possibility of its existence is comparable to an ostrich sticking it's head in the sand. In fact I think the very idea of infinity might be paradoxical.--Neptunerover (talk) 09:30, 25 January 2010 (UTC)

And here's the crux of the problem. "never ending string of zeros which has a 1 at the end". If it's never ending, where is the end? You can't have it both ways. If it's never-ending, than for ever nth digit, there's an n+1th digit that has the same value. If it has a 1 at the end, that means that there's a value for n that doesn't have an n+1, so it's not never-ending. --Maelwys (talk) 14:35, 25 January 2010 (UTC)

That's what makes it a paradox. Infinity here is between the decimal and the one. --Neptunerover (talk) 14:41, 25 January 2010 (UTC)

Yes, but the problem is that the paradox has no meaning on the common understanding of mathematics. And that's why people are having a hard time discussing this with you, because the rules of math are built to prevent this kind of paradox, so arbitrarily introducing one into your logic conflicts with common understanding, and makes it hard to discuss anything. For example: Which is a bigger number: 0.999...999 or 0.999...998? Well, obviously it would appear that the first one was bigger, because they both have the same "infinite" number of 9s, but it has one extra nine at the end where the other only has an 8. But then which of these is bigger: 0.999...999 or 0.999...9999? Again, they both have the same "infinite" number of 9s, but the second one has one EXTRA nine where the first one has no value. Does that mean it's bigger? Because it has "infinity+1" 9s? --Maelwys (talk) 15:14, 25 January 2010 (UTC)

Ah-ha, so is there any field of mathematics that deals with paradox? What about wave functions? --Neptunerover (talk) 16:14, 25 January 2010 (UTC)

The paradox doesn't need dealing with, it just needs you not to try and define inherently meaningless expressions. Why do you think wave functions would help? They have absolutely nothing to do with this... --Tango (talk) 12:37, 26 January 2010 (UTC)

My question was if paradox has meaning within the study of wave functions. The last person up told me I'm in the wrong area if I want to take paradox into account. --Neptunerover (talk) 15:07, 26 January 2010 (UTC)

Short answer: No. The mathematics of wave functions is neither ambiguous nor paradoxical. Huon (talk) 15:42, 26 January 2010 (UTC)

Lemme ask this, if I'm talking about a ratio between 1 and zero, then is what I'm talking about dividing by zero, which is perhaps the central paradox of my conundrum? (Not to mention, why it doesn't fly here) --Neptunerover (talk) 09:02, 27 January 2010 (UTC)

If you're talking of 1/0 you're indeed talking of division by zero, which is undefined in most contexts (excepting, say, certain projective spaces which neither contain infinitesimals nor are relevant to 0.999...). I don't think it's possible to construct a useful number system which allows division by zero and contains infinitesimals, and even if it were, it'd probably still be unrelated to 0.999... Huon (talk) 11:56, 27 January 2010 (UTC)

half way v. 9/10

There's the old story of "How long would it take to get to X if each day you traveled 1/2 of the remaining distance between you and X?" According to the question itself though, each day when you are through traveling, there will still be a remaining distance to the 'goal' equal to that which you just traversed that day. I don't see how this could be any different if you went 9/10 of the distance each day rather than only half way, other than getting closer faster, because the entire distance to the goal is still never spanned. I suppose that's very different than going 9.999.../10 of the distance each day. In that case the second day would be a very puzzling day. --Neptunerover (talk) 12:19, 26 January 2010 (UTC)

9.999.../10=1, so the 2nd day isn't confusing, just very relaxing since you have already arrived. --Tango (talk) 12:36, 26 January 2010 (UTC)

Neptunerover - there is no difference really. But I'm not sure what conclusion you're trying to draw? The halfway example presents a geometric series: 1/2 + 1/4 + 1/8th + 1/16th etc...and is equal to 1. Do you contest that? Because the same is true for 9/10: 9/10 + 9/100 + 9/1000.... = 1. Your specific application with walking I think is related to http://en.wikipedia.org/wiki/Zeno%27s_paradoxes but not really related to any equality or inequality with the given series.76.103.47.66 (talk) 09:11, 24 February 2010 (UTC)

Isn't this also valid?

Why is the following not valid? .9 does not equal 1; .99 does not equal 1; .999 does not equal 1; etc.

I think there is a paradox here. Proofs to the contrary do not make the proof above invalid. Why is this intuition not acceptable?Tristan Tondino (talk) 02:01, 5 February 2010 (UTC)

It is indeed a paradox. No number in the sequence is equal to 1, yet the limit is 1. Likewise, in the sequence 1, 1/2, 1/3, 1/4, ..., no number is equal to zero; yet the limit is zero. But you don't have a proof: Your argument has no bearing on the limit, only on the terms. Limits exist on the continuous real number line, not for discrete truth values.--Nø (talk) 08:24, 5 February 2010 (UTC)

If that line of reasoning were valid and implied that 0.999... didn't equal 1, we could also argue:

.9 does not equal 0.999...;

.99 does not equal 0.999...;

.999 does not equal 0.999...;

etc., and we'd have shown that 0.999... does not equal itself. Thus, the intuitive approach leads to self-contradiction. Huon (talk) 10:47, 5 February 2010 (UTC)

I confess that I don't understand why Nø claims that this is "indeed a paradox". There is simply no paradox to be found here. A simple reflection on the definition of limit will confirm that, for some sequences x_n, we have

for all n, x_n < lim x_n.

Tristan's observation is no deeper than that. Phiwum (talk) 15:49, 5 February 2010 (UTC)

Insertion: A paradox is when two lines of reasoning collide. Here, a "common sense" argument that might convince some non-mathemaricians collides with the strict logic of mathematics.--Nø (talk) 09:08, 6 February 2010 (UTC)

Reply: that is a much weaker notion of paradox than I've ever seen. Especially since one of the lines of reasoning here is simply invalid. The fact that a correctly understood conclusion differs from an invalid bit of reasoning to the contrary does not make a paradox in my book! (But our difference here is merely semantic, of course.) Phiwum (talk) 10:07, 6 February 2010 (UTC)

I think there is a mistake in your argument Huon: .9 does not equal 0.999... or maybe I have misread it. -- Phiwum, my sense is that these are different kinds of proofs -- the expression in question is "as the limit approaches zero." Is my "pseudo-proof"... logical and the other proof mathematical? Is this an acceptable distinction? Can we be certain there is no paradox without being committed to disunification? Or, in other words, do you mean (in layman's terms... i.e. mine) the language of the first pseudo-proof is meaningless (illformed) in mathematics?Tristan Tondino (talk) 22:53, 5 February 2010 (UTC)

Insertion: Your argument is not logic; it is maybe common sense. The mathematical argument is strictly logical.--Nø (talk) 09:08, 6 February 2010 (UTC)

I believe you misread my argument. We agree that 0.9 does not equal 0.999..., just as 0.9 does not equal 1. I repeated your argument and substituted "0.999..." for every occurence of "1". I believe you intended to argue that since all of 0.9, 0.99, 0.999 and so on are less than 1, so is 0.999... If that were valid I'd argue that since all of 0.9, 0.99, 0.999 and so on are less than 0.999..., so is again 0.999... - which cannot be true. Thus, just because something holds for all of 0.9, 0.99, 0.999 and so on, it needn't hold for 0.999... - and "not being equal to 1" is such a property just as well as "not being equal to 0.999...". Thus, your pseudo-proof is indeed wrong mathematically (and logically too, I'd say). Huon (talk) 23:32, 5 February 2010 (UTC)

Wow, so if I understand correctly, my non-proof is based on one concept "is less than" which cannot hold as we approach infinity... otherwise I would have to accept that 0.999... is less than 0.999... But... is there a proof for 0.999... equals 0.999... or is this an intuition?

Since, 0.9 does not equal 0.99; and 0.99 does not equal 0.999 etc.Tristan Tondino (talk) 00:05, 6 February 2010 (UTC)

Wow, I just read the talk page!?Tristan Tondino (talk) 00:31, 6 February 2010 (UTC)

"is there a proof for 0.999... equals 0.999... or is this an intuition?" It's not much of a proof, but since equality is supposed to be a reflection of 'sameness', we expect it to be reflexive (actually, we expect it to satisfy all the properties of an equivalence relation, which includes reflexivity). That is, x=x because x is certainly the same as x. --COVIZAPIBETEFOKY (talk) 01:06, 6 February 2010 (UTC)

But, "identity" is a complicated issue. There are many cases where x may not equal x - though this seems counter-intuitive and illogical -- at present the proof for 0.999.... equaling 1 seems to rely on an intuition as did the intuitive proofs above -- that mathematics cannot contain contradictions, but this is not established... At least on one reading of Godel's incompleteness theorem. Any thoughts? Tristan Tondino (talk) 01:57, 6 February 2010 (UTC)

Can you provide an example of x not equaling x (besides the obvious x being defined in one context to be 1 and in another context to be 2, and the x's in different contexts being unequal)? I'm really not sure what you have in mind there. --COVIZAPIBETEFOKY (talk) 02:23, 6 February 2010 (UTC)

a =√4 a=2 a=-2 2a=-2a a=-a This may be contextual of course, or just wrong.Tristan Tondino (talk) 02:35, 6 February 2010 (UTC)

Contextual? No. Just wrong? Bingo!

√(x) is defined for non-negative real x to be the unique non-negative real number y satisfying y*y=x. Of course, -y also satisfies the property (-y)*(-y)=x, but √(x) is the non-negative solution. So a=2, and a≠ -2. --COVIZAPIBETEFOKY (talk) 02:42, 6 February 2010 (UTC)

This is being explained to Tristan also at Talk:Square root. --jpgordon^{::==( o )} 07:05, 6 February 2010 (UTC)

Getting back to the problem. There is a fallacious argument in each of the article's proofs. If we cannot assume 0.999... = 1, since it is what we are trying to prove... we also cannot assume 1/3 = 0.333... since it begs the same question as the first proposition. In other words, the pseudo-proofs in the article do not succeed in anything more than hiding the problem. Tristan Tondino (talk) 15:09, 8 February 2010 (UTC)

First, there isn't a problem. But - you are right - some of the proofs are "pseudo-proofs", and the article does not try to hide that. Most students know and accept that 1/3 = 0.333..., and therefore they find arguments like "3 x 0.333... = 0.999... = 1" convincing, but in order to PROVE either of the statements "1/3 = 0.333..." and "1 = 0.999..." formally, you need to understand limits first.--Nø (talk) 15:25, 8 February 2010 (UTC)

I did do some calculus; but what we may be discussing are definitions of 1. 1 is not as simple as it looks. It is as unbounded as 0.999...; 1 could mean the infinite set of all calculations equaling 1. e.g. (2-1,3-2,0.999...) Tristan Tondino (talk) 15:59, 8 February 2010 (UTC)

Those are indeed all representations of 1. Unfortunately many people have trouble seeing that 0.999... is another representation of 1; hence the article. Concerning the 3*0.333... proof: The idea is that one can "show" 1/3=0.333... by long division (where one effectively hides all the limits out of sight); then one can conclude that 0.999... = 3*0.333... = 3*(1/3) = 1 without circular reasoning. Huon (talk) 16:11, 8 February 2010 (UTC)

The article makes it clear (or at least did so last time I read it) that in this context, "1" is a real number, and the definition is pretty unambiguous. I think your problem is not about the definition of "1", but about equality - but the answer to that is that two notations are equal if they represent the same real number.--Nø (talk) 16:16, 8 February 2010 (UTC)

"Sameness"... "equality"... "identity" are curious and generally very ambiguous to me. Even in a Mathematical context. So Nø, your observation is fair.Tristan Tondino (talk) 17:46, 8 February 2010 (UTC)

The terms are ambiguous on their own. What "equal" means depends on the objects you are talking about. You have to define it. Maths is most often phrased in terms of set theory, so all objects are sets and we define two sets to be equal if the each contain the other. We can define the real number, 1, as an equivalence class of Cauchy sequences of rational numbers. For that, we need to define an equivalence relation on those sequences, which we do (roughly speaking) by saying two sequences are equivalent if they get closer and closer to each other as you go further and further along the sequences. Decimal expressions don't actually correspond to real numbers, they correspond to Cauchy sequences of rational numbers (not all such sequences correspond to a decimal expression, though). We say two decimal expressions represent the same number if their corresponding decimal expressions are equivalent. 0.999... corresponds to (0.9,0.99,0.999,0.9999,...) and 1 corresponds to (1,1,1,1,1,1,1,1,1,...). If you examine those sequences you will see that they get closer and closer to each other (ie. for any positive rational number the difference between the nth terms of the sequences is less than that number for a large enough n), so they are equivalent and thus are part of the same real number. (All of this assumes you already have a working definition of rational numbers, which are the field of fractions of the integers, which are defined in terms of the Peano axioms.) --Tango (talk) 12:53, 11 February 2010 (UTC)

No! Equality is not ambiguous and its meaning does not depend on context. a = b iff they are (or denote) the same thing. The axiom of extensionality does not define equality on sets, but rather spells out an important feature of sets: sets with the same elements are equal.Phiwum (talk) 13:01, 11 February 2010 (UTC)

But 0.999... and 1 clearly aren't the same thing - they look different. In the context of real numbers, they are equal, in the context of strings, they aren't. Whether the axiom of extensionality defines equality on sets of just describes it is a question for mathematical philosophers, which I am not and have no desire to be. --Tango (talk) 13:24, 11 February 2010 (UTC)

Of course the syntactic objects 0.999... and 1 are not the same thing. But when we write 0.999... = 1, what we mean is that the object denoted by the two syntactic objects are identical. They are, of course, since both denote the real number one. The issue, then, is not that equality is context dependent, but rather that the interpretation of terms (what each non-logical syntactic object denotes) is context dependent. I suppose my complaint is a bit pedantic, but it seems important to me. Phiwum (talk) 14:36, 11 February 2010 (UTC)

Floor Function?

I'm sorry if this has already been discussed, but can't we just say that 0.999... does not equal one because of this function? I'm not arguing anything, just asking a question.

Floor m.n where m and n are strings of digits, is always m. So doesn't that mean floor (0.999...) = 0? Floor (1) = 1. —Preceding unsigned comment added by Goldkingtut5 (talk • contribs) 06:49, 11 February 2010 (UTC)

Your floor function is defined on decimal representations (or on pairs of strings of digits), not on (real) numbers. Of course "0.999..." is a different decimal representation of a number than "1", but they both represent the same number. For an analogy consider the map E on fractions given by E(p/q) := p. Then we have E(1/2)=1, E(2/4)=2, but still 1/2 = 2/4.

The article already discusses the "decimal numbers" which are in a 1-to-1 correspondence with decimal representations; they suffer from a lack of subtraction because 1-0.999... can't be defined in a satisfactory way. Huon (talk) 12:21, 11 February 2010 (UTC)

The floor function isn't usually defined like that. It is defined as the largest integer than is not larger than the number. The largest integer not larger than 0.999... is 1, since 1 is an integer and 0.999...=1. --Tango (talk) 12:38, 11 February 2010 (UTC)

I see. Thanks for clearing this up. Me, GKT5 15:10, 11 February 2010 (UTC)

And, it's me again, but consider real space. Pick a random point in space. The probability of choosing any one point that is not this picked point would be 1, then, right?Me, GKT5 04:59, 6 March 2010 (UTC)

You don't quite make sense, but I think you are talking about the concept of almost surely. That article should help you out. --Tango (talk) 05:54, 6 March 2010 (UTC)

Ahh, thanks!Me, GKT5 05:46, 7 March 2010 (UTC) —Preceding unsigned comment added by Goldkingtut5 (talk • contribs)

Another way

Here's another way. Since 1/0 equals infinity, then 1 = 0*infinity, and 1/infinity = 0. 0.999... plus 1/infinity = 1, then 0.999... plus 0 equals 1, and 0.999... = 1. 24.1.201.172 (talk) 01:56, 21 May 2010 (UTC)

The reason why it is like this is because i don't know how to format with actual math symbols since they introduced this new setup. 24.1.201.172 (talk) 01:59, 21 May 2010 (UTC)

Yeah, but 1/infinity = 0, that just proves that 1=0 (multiply each side by infinity, duh), so it won't work. 68071 (talk, contributations, something random)

Firstly, 1/0 is not infinity but is undefined. Secondly 0*infinity is indeterminante. 1/x as x->inf is 0. So 0.999...+ 1/x as x->inf = 0.999... = 1. This is just saying x + 0 = x because 0 is the additive identity. 86.172.185.252 (talk) 16:15, 26 August 2010 (UTC)

ipart formula from calculator

I have a TI-34 II calculator at home and it has the ipart function and the fpart function. The ipart stands for integral part, and if 0.999... was inputed into the calculator with the ipart function thingamajig (like this: ipart(0.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999...)), it will still be 0. If you do 1, it will come up as 1. Therefore, they aren't the same. 24.1.201.172 (talk) 02:07, 21 May 2010 (UTC)

I doubt that you really entered an infinite number of 9s. If not, then the calculator was, of course, right. The integral part (i.e., the floor) of 0.999...9 is 0 for any finite number of 9s.

It does not follow that 0.999... < 1 (where 0.999... has infinitely many 9s).

Tell you what: try it again, but this time, enter an infinite number of 9s. We'll wait for it. Phiwum (talk) 02:35, 21 May 2010 (UTC)

First of all, I agree with you. The person (my brother) who wrote the thing up there didn't punch in infinity digits of 9s, because it is impossible. The Sinebot thingy didn't know that was my brother (duh, it is a bot). Sixeightyseventyone (talk) 21:16, 21 May 2010 (UTC)

I give reason against my brother that 0.999... is equal to 1. The reason why is because:

0.999... = the infinite series 9/10 + 9/100 + 9/1000 + 9/10000 + ... , which converges to 1. Sixeightyseventyone (talk) 01:52, 22 May 2010 (UTC)

Understanding why 0.999... = 1 in simple terms

I see that many people are struggling to understand why 0.999... is equal to 1

I see many arguments that 0.9 is NOT 1, neither is 0.99 or 0.999 or 0.99999999999 ect... so why is 0.999... (repeating 9's) the same as 1?

It is because many people have a hard time truly understanding what infinity is. It is not a REAL number, there is no way in a string of digits like 489238902 to express infinity. What is happening with the above question is in your mind your are assigning and end to the string of 9's at some REALLY far away point. You are picturing (whether you know it or not) infinite 9's as being really really long, but subconsciously you are picking a distant point and stopping the string to say well if 0.99999 is not 1, why is 0.9999999(billions and billions of digits later)999 equal to 1? But it is not - the moment the infinite string of 9's ceases, it is NOT equal to 1.

A very important aspect of Decimal Notation (which is the number system we use) that you should understand is what numbers really mean. For example, the number 257 - we dont have a digit for that number, instead we use the single digit numbers 0-9 to express all other numbers in the system, and the way we used those digits to represent numbers was by creating "place values" - the one's place, the ten's place, the hundreds place and so on. Also, each place value is measured in magnitudes of the number 10. The ones place is 1, the tens place is 10, the hundreds place is 100 and so on

So what does 257 really mean? 2*100 + 5*10 + 7*1 which is 200 + 50 + 7 = 257 What about a number with non-0 digits after the decimal point? For example the number 624.7 - 6*100 + 2*10 + 4*1 + 7/10. For any non single digit value, you are REALLY expressing a sum.

The decimal place values are the tenths, hundredths, thousandths, and so on. They are expressed like this - 0.256 is 2/10 + 5/100 + 6/1000. So 0.999... is expressed - 9/10 + 9/100 + 9/1000 + 9/10000...

Ok? Now here is the part where you ignore your intuition. If you are absolutely stuck that all the mathematicians are wrong - then there is no point in trying to understand. All I am trying to do is explain it to people who want a way to understand it without using insane proofs that people with a limited math skill can't understand. I am going on the principle that people reading this at least a decent arithmatic skill and understand math to a very limited degree.

Ok - so the next thing about our number system you want to notice is exactly how we express decimals as fractions. Most people understand that 0.3 is the same thing as 3/10. How about 0.47? It is 47/100 right? You can go as far as you like - 0.12398 is 12398/100000 , 0.119 is 119/1000 and so on.

You can also express the value of a REPEATING fraction in a similar way. Before we used the place value to find the denominator (as we saw 0.47 was the fraction 47/100 because the "7" is the end of the fraction and it sits in the hundredths place). In dealing with repeating decimals, we use the maximum value of those place-values. An easier way to say that, is we use the place value minus 1. For example - we all know what 1/3 is. We also know it is the same as 3/9 - right? well 1/3 is expressed in decimals as 0.333... So you see the repeating digit there is "3" and it is only one digit. so it is not 3/10, that is wrong. It is 3/9 (which can be simplified to 1/3).

Take time, make sure you understand this part, convince yourself it is true with some examples. how would we write the following as a fraction? 0.454545.... Well we have two repeating digits this time - "45" is repeating, and that takes up 2 place values, but its not 45/100, that is wrong - The correct fraction is 45/99. (just a quick note, that is why before I said the max value of the place holder, because in the hundredths place, the max value is 0.99... it is also 100-1 if that is easier for you to remember). How about 0.111... Well we know its only one digit repeating, and so we use the one's place to determine our denominator because there are no other repeating digits after the first "1". So the proper fraction is 1/9. Go ahead and type 1/9 (1 divided by 9) in a caclutor and see what you get. It will be 0.111... One more because this is essential - what about 0.888... how do you think we write that as a fraction? Well you probably should know by now otherwise you really wont understand the rest. The answer of course is 8/9. Go ahead and check this out for yourself.

So what does all this mess have to do with 0.999... = 1? Well I want to make sure you understand what these digits REALLY mean.

Because you will see that what I have told you is true in ALL cases. Period.

So... what is 0.333... It is 3/9 right, we learned that. that simplifies to 1/3 what about 0.666... It is 6/9 right - and it simplifies to 2/3 finally how about 0.999... It can only be... 9/9 - right? and that simplifies to 1.

Does it make sense at first? No. It is super easy to understand? Of course not.

Sometimes you have to simply put your "common sense" notions aside. Because common sense only applies to what YOU are used to. It does not obey the laws of nature. Lots of things do not SEEM like the make sense, but when a proof is put down, it must be considered true - unless someone finds a way to completely disprove it. But really, once you get used to it, this idea is not that hard to see.

Try to understand that Mathematics was not invented in one quick swoop. No one just threw all this into existence overnight. We needed a number system, just a way to number things, count stuff, you know, basic easy ways to help keep track. Then we thought, well what if we want to use these numbers to figure something out? for example - multiplication didnt just exist out of no where. Numbering was invented, and people figured out rules for adding those numbers, because adding is helpful to our daily life. Then someone noticed when you have to add the same number over and over and over again it was tedious. 2 + 2 + 2 + 2 + 2 Who feels like doing that over and over and over again. So a shorthand was invented called multiplication.

Over time, we began to explore these numbers more and more. Finally we ran into strange numbers like pi and e. We ran into things that we couldn't make sense of, like dividing by 0. And we found unusual things that challange our common sense notions, like 0.999... = 1.

You see? It doesn't have to make sense. It's just a product of our invention of mathematics. Perhaps if some other system was invented we would not encounter things like this. It only exists because of OUR construction of the decimal notation system. It's not some official rule of the universe that would be here even if humans weren't - its just some funny little outcome that happened to exist due to our methods and design of the decimal notation numbering system.

I hope at least SOME people can understand better. I tried my best to explain it very slowly and simple - If you couldn't understand than I apologize, I am very advanced in Mathematics and sometimes going back to the basics of the basics is harder than the ridiculous stuff I am researching now. Alex DeLarge 10:03, 25 May 2010 (UTC) —Preceding unsigned comment added by A.DeLarge23 (talk • contribs)

Would you care to summarize your main points? Tkuvho (talk) 08:56, 26 May 2010 (UTC)

No. Because this isn't so easy for people to understand, and I feel it should be perfectly clear for anyone who is having a hard time. If you think it is too long to read, then you can surely find many many people who explain it very quickly, using precise mathematical proofs. If you do not understand those definitions and proofs, then you need this laid out in this fashion. I wrote this for people who have virtually no background in math other than maybe Elem and Middle School. In that sense, all the points are "main points" and if you just want a summary because you are too lazy to read it or you dont feel like trying then I can't help you. I am NOT trying to be rude, it is just true - and if you found this too complicated to understand than you will have to find someone else because I don't think I could explain it any simpler but I can almost guarantee you that someone could offer an even more simplified version. But If you still don't understand I would be happy to offer you an alternative point of view on the matter, however, if you want less reading, you will get more math, if you want less math, then you will get more reading. 108.2.103.208 (talk) 16:57, 26 May 2010 (UTC)

0.aaaaaaa = a/9, 9/9 = 1 --70.29.12.46 (talk) 07:41, 23 July 2010 (UTC)

What's the other point of view on the matter? Tkuvho (talk) 09:28, 27 May 2010 (UTC)

I suspect that you have missed Tkuvho's point. While you're mulling it over, you might want to review WP:SOAP. --Trovatore (talk) 21:27, 26 May 2010 (UTC)

What are you talking about? That wasnt any self-promotion, or opinion piece, it was an explanation of why 0.999...= 1 Without complicated proofs so that people who are not math experts can understand. I can offer any number of proofs for you, but unless you have a background in Mathematics you will not have a clue what I am talking about. I noticed a lot of people were confused and bothered by this fact, so I took time out of my day to help all those people who never went far beyond arithmetic or basic algebra to at least have some sense of understanding. How are the facts that 0.3 is 3/10 and 0.333... is 3/9 ect.. opinions or self promotion? that is ridiculus and uncalled for to accuse me of this. I simply stated FACTS. And then I followed up with a brief explanation of why it seems like it violates common sense and why these things pop up to begin with, by explaining that the number system was our invention that was originally needed for counting, and we kept taking more and more steps.

Those are not opinions. This is an argument page anyways, so isn't this a place to make your arguements based on 0.999... being equal to 1? Well that is what I did. And Tkuvho didnt have any point, how could I miss it?? He said "can you summarize" and I said no, not without adding proofs that people won't understand. However Trovatore if you think I am violating Wikipedias terms than I am removing this entry - now no one will get an easy explaination, they can go take 5 years of math and understand the proofs. And from now, I won't add anything that is 100% factual based, which is .... everything I would write. 173.62.181.145 (talk) 04:32, 1 June 2010 (UTC)

Mathematical facts really aren't up for debate, so when people who don't understand math come to 0.999...'s talk page to argue that, Wikipedia policy states that we should just delete those comments outright. However, this usually isn't very constructive, so we usually just move such comments to the Arguments page where we can explain to them why they're wrong, without cluttering up the main talk page.

Your comment here doesn't really do anything to that end, which is why people are saying that your explanation does not belong here. --Zarel (talk⋅c) 09:08, 1 June 2010 (UTC)

I think we are allowed to be more welcoming than that :) What I find interesting about the mathematical comments by the IP is the idea that the .9..=1 phenomenon is an artifice of the particular formal construction we chose as the basis of the common number system, and may in fact work out differently in other systems. This is certainly consistent with published work that's already reflected in the article. Tkuvho (talk) 09:42, 1 June 2010 (UTC)

Zarel explained the matter rather more gently than I would have. When editors begin their interaction with WP with that sort of 'tude, someone needs to make them aware what will and will not result in profitable interaction. --Trovatore (talk) 18:43, 1 June 2010 (UTC)

But this IS the arguement page - and I wasnt debating any mathematical facts. everything I said IS a fact, there was no debate. It was just an argument of why this is true, using simple terms. How is what I said ANY different than the people who pose the other proofs? I just did it in a more simple way because I saw many people here asking WHY because they dotn get it, because they cant read the proofs. I was offering my help to those people by stating the FACTS of mathematics, but in simple terms. NOTHING I said is not a fact when it came to the math.

0.3 IS 3/10, 0.333... is 3/9 and 0.999... is 9/9 or 3/3 or 1. These are not my interpratation of the facts. These are FACTS. These are the basic facts of our number system. It is all true, and it is a simple way for people to see it, as opposed to the algebra proofs and calculus proofs. Not everyone has this knowledge, so there are NO Facts I am debating, that is ridiculous. I have been studying Mathematics for my whole life. I Can offer you all kinds of complex proofs that are also facts - how is that different than offering proofs in a simple form, which is already on the Argument Page, this is NOT The talk page, that is why I posted here - it is another argument of why 0.999... is 1 - using the most basic of facts dealing with our number system. Not everyone took calculus or algebra to any real extent. What about those people? Are the screwed? Thats bullshit. I believe EVERYONE has the right to understand this. Thats what math should be about, it should be for everyone, not just someone who has a extended background. Go look in a 3rd grade text book and you will see these facts, that 241 is REALLY a statement of place value. Its how our number system works. And look up how we express decimals and fractions - .333 is 3/9 and .45454545454545.... is 45/99. It was just a way to help people. If anyone is going to argue that I am debating facts, or that I am biased, well thats ridiculous and wrong, this is an arguement page so thats what I did, i proposed another argument to the truth of 0.999...=1 using simple terms. If Im going to be accused of non-sense like trying to put up mathematical facts for debate then I want my entry deleted, and I Will keep deleting it over and over until it stays deleted. I took time out of my day because I wanted to help those who truly are bothered by this notion and want it to make sense. If wikipedia is not the place for people to find and understand information, then I will take my time elsewhere. 173.62.181.145 (talk) 15:50, 3 June 2010 (UTC)

A simple word of advice: repeatedly saying that you are an expert in mathematics and have been studying it all your life is not as likely to impress folks around here as you think. I'd wager that a number of regular contributors to this page are just as well-educated on the topic as you are. Phiwum (talk) 13:20, 4 June 2010 (UTC)

From my experience on this page, those doubting the equality are unlikely to study the arguments page for an explanation, even more unlikely to look at the archives. Thus, I doubt this explanation will actually do much good. Huon (talk) 13:57, 4 June 2010 (UTC)

Note that if you use binary, 0.1111... equals 1, and for octal: 0.777... = 1, ternary: 0.222... = 1, hexadecimal: 0.FFF... = 1, etc. 24.1.201.172 (talk) —Preceding undated comment added 18:50, 6 June 2010 (UTC).

"repeatedly saying that you are an expert in mathematics and have been studying it all your life is not as likely to impress folks around here as you think" I really hate people like you - shut up with that nonsense already. I wasnt boasting or acting like I have some superior knowledge of the subject. I was just letting it be known that Im not just some guy who is guessing. I have a background in mathematics, would you like to give me your email so I may send you some photocopies of credentials? No? Then dont waste time with a useless comment. I have sources for everything I Said, I just didnt believe they belong here on an arguments page because Im new to this. I though sources only belong on the articles, and if you notice, I didnt even try putting this there. I simply wanted to help all the obviously confused people understand this without complex proofs. What is so wrong with that? And so they don't think they are getting this info from just some guy who has zero background in math, I discussed a little of it, and then I was attacked and told I was inventing facts, so I had to get into it more. But I am trying to impress no one, so keep your stupid, useless comments about me to yourself. I'd wager that many contributers here are as or more educated than, but I bet a lot more arent. And I think if someone is going to be commenting on an article about how automobiles work, they should have at least SOME background in the subject. 173.62.181.145 (talk) 20:21, 16 June 2010 (UTC)

If you're so highly educated, you could at least try to practice some good spelling and grammar once in a while. --128.62.180.255 (talk) 17:43, 17 June 2010 (UTC)

Sure, by all means, I'd love to see your credentials. What makes you think I wouldn't? Just take my name and append @gmail.com. --COVIZAPIBETEFOKY (talk) 11:31, 18 June 2010 (UTC)

Me too! Send to sl@gmail.com.

Anyway, the current objection to your writing isn't that it's incorrect, but that it's irrelevant, as I've already said so many times before. --Zarel (talk⋅c) 14:40, 18 June 2010 (UTC)

0.999...#p-adic numbers

That section just proved that infinity equals -1. 24.1.201.172 (talk) 19:05, 6 June 2010 (UTC)

No, it just proves that with the 10-adic metric, the sequence (9, 99, 999, ...) isn't divergent but converges to -1. Huon (talk) 21:14, 6 June 2010 (UTC)

.999... doesn't mean anything

Huon's argument [above] that ends with ".999 does not equal 0.999...;" attempts to prove that the argument form is invalid. Instead it seems to say to me that .999... actually has no meaning at all. That seems sensible to me: How can you use infinity to define a real number when real numbers do not include infinity? Algr (talk) 09:49, 9 August 2010 (UTC)

Tristnan Tondino's argument there can usefully be interpreted in terms of the transfer principle. We have .9<.999..., .99<.999..., .999<.999..., etc. Hence by transfer, also the string containing an infinite hypernatural's worth of 9s still falls strictly short of the real .999...=1. The shortcoming of the real number system is that it contains no infinitesimals, which makes it impossible to capture this kind of intuition. This has been pointed out in published articles in leading education journals. Tkuvho (talk) 09:56, 9 August 2010 (UTC)

Meanings for symbols aren't handed down by God, you have to define things. The decimal representation in question is defined to represent the limit of the sequence .9, .99, .999, .... Further, based on the definitions of limits, it can be shown that 1 is the limit of the sequence.71.116.66.31 (talk) 17:31, 25 September 2010 (UTC)

It's a falacy

It's simply a falacy. 1 = 0.9999... is not true. If you assume that false axiom as true then anything can be demonstrated. In fact, 1 is aproximately equal to 0.999... If you substitutes "=" with "~=" then axiom is true and maths match again. Even clearer:

1 > 0.999...

And then every affirmation is true and everything recovers sense again. —Preceding unsigned comment added by 88.0.131.223 (talk) 10:38, 15 October 2010 (UTC)

You are incorrect. The article contains many proofs and arguments that they are equal, you have given no reason for why they are not equal. And if as you say 1 > 0.999..., then what is x = (1 - 0.999...) ? It is the difference of two real numbers, so must be a number, and cannot be zero if the numbers are not equal. --JohnBlackburne^words_deeds 11:34, 15 October 2010 (UTC)

Why isn't that example in the article? It's simple and it gives quite good insight about the concepts involved. Diego Moya (talk) 12:30, 15 October 2010 (UTC)

It is not in the lead because it begs the question: why is the difference zero? If it is the difference between two real numbers, it has no choice but be zero, but students struggling with the issue typically have not seen the development of real analysis required to understand it properly. The shortcoming of the reals is the absence of infinitesimals, and student intuitions about an infinitesimal gap between .999... and 1 are better captured in an infinitesimal-enriched number system. Thus defining .999... as a real number is merely changing the subject and is not likely to satisfy an inquisitive student, nor to help him understand calculus where such infinitesimals are extremely useful, as documented in recent educational literature. Tkuvho (talk) 13:39, 15 October 2010 (UTC)

"It is not in the lead because it begs the question: why is the difference zero?" - which is precisely why it should be included somewhere in the article (not necessarily the lead). It raises a very relevant question with just a few words. Such economy of language is invaluable when trying to explain (not define) the issue. Proofs can give a more logical and thorough understanding, but they require a lot of mental work. Insights give a "hook" to direct readers into the right mindset to understand the proof.

If you have to explain first what an infinitesimal is to get to the point where you explain why 0.999...=1, readers will be lost at the first 'let's epsilon be...' On the other hand, the "1 - 0.999..." makes them instantly say: "There's something funny at the end of this substraction... There's a missing concept I have to learn to understand it". This is the hook that explains how the knowledge they are learning will help them. Diego Moya (talk) 14:20, 15 October 2010 (UTC)

I don't think it's needed. It's not really a proof of anything, more a challenge to an incorrect assertion. It could be used as a basis for longer and more formal a proof by contradiction, but such are usually used only needed if a more straightforward proof is unavailable, and there are plenty of more straightforward proofs in the article. And it's OR, or at least I've not seen it written down anywhere.--JohnBlackburne^words_deeds 14:37, 15 October 2010 (UTC)

Ok, being OR is a good reason to not include it. But not being a proof of anything is an advantage, which is the point I was trying to make. Do you have a mathematical mind? I was trying to write content useful to people who don't have it (such as students), for whom proofs actually have a low explaining power and aren't really that much valuable. Diego Moya (talk) 14:48, 15 October 2010 (UTC)

Proof asked for on the talk page

On the talk page, 210.4.96.72 asked if someone could prove the following statement:

${\begin{aligned}summation\,of\,the\,sequence(0.1+0.01-0.1+0.001-0.01+0.0001-0.001...)=0\\\end{aligned}}$

There's one caveat: I'm not quite sure what he means by "summation of the sequence". Let's first introduce some notation: Let a₀=0.1, $a_{2k}=-10^{-k}$ for k>0 and $a_{2k+1}=10^{-(k+2)}$ for k>=0. Then the sequence (a₀, a₁, a₂, a₃, ...) is the sequence (0.1, 0.01, -0.1, 0.001, -0.01, ...). I'll prove that $\sum _{i=0}^{\infty }a_{i}=0,$ where I use the standard definitions for "sum of a series", which may not be what 210.4.96.72 had in mind. Namely, by definition, $\sum _{i=0}^{\infty }a_{i}:=\lim _{n\to \infty }\sum _{i=0}^{n}a_{i},$ where finite sums can be formed simply by adding the terms. Despite the similar notation, infinite series actually aren't formed by adding the terms, since addition of infinitely many terms isn't defined. Furthermore, by definition of the limit, we say that for a sequence (s_n), we have $\lim _{n\to \infty }s_{n}=s$ if for every ε>0 there exists an n₀ (depending on ε) such that for all n>n₀, we have |s_n-s|<ε.

So with these definitions and the notation from above, what I'll have to prove is that for any ε>0 there exists an n₀ such that for all n>n₀, we have $|\sum _{i=0}^{n}a_{i}-0|<\epsilon .$

• Fact 1: Let n₀ be any natural number greater than (4/ε)+2. Then for k>(n₀-2)/2, we have $10^{-(k+1)}+10^{-(k+2)}<2*10^{-(k+1)}<2/k<{\frac {2}{n_{0}/2}}={\frac {4}{n_{0}}}<{\frac {4}{4/\epsilon +2}}<\epsilon .$ This is what I aimed for; I chose the "4/ε" to arrive at a nice, round ε in the end.

Next, note that a₀+a₂ = 0.1-0.1 = 0, while for any i>1 we have a_2i-1+a_2(i+1) = 10^-(i+1)-10^-(i+1)=0. Thus, quite a few terms of the finite sums I'm about to look at will cancel out:

• Fact 2: For k>0 and n=2k+2 an even number greater than 2, we have $\sum _{i=0}^{2k+2}a_{i}=a_{2k+1}+a_{0}+a_{2}+\sum _{i=1}^{k}(a_{2i-1}+a_{2(i+1)})=a_{2k+1}.$ Every a_i for i from 1 to 2k+2 appears once on both sides of the equation.
• Fact 3: for k>0 and n=2k+1 and odd number, we have $\sum _{i=0}^{2k+1}a_{i}=a_{2k+1}+a_{2k-1}+a_{0}+a_{2}+\sum _{i=1}^{k-1}(a_{2i-1}+a_{2(i+1)})=a_{2k+1}+a_{2k-1}.$

Now we can combine the parts: Let n>n₀ be any (natural) number.

• Case 1: If n=2k+2 is even, we have $|\sum _{i=0}^{n}a_{i}-0|=|\sum _{i=0}^{2k+2}a_{i}|=|a_{2k+1}|=10^{-(k+2)}<\epsilon .$
• Case 2: If n=2k+1 is odd, we have $|\sum _{i=0}^{n}a_{i}-0|=|\sum _{i=0}^{2k+1}a_{i}|=|a_{2k+1}+a_{2k-1}|=10^{-(k+2)}+10^{-(k+1)}<\epsilon .$

Both cases need fact 1 for the very last equation (though case 1 could do with a weaker version), and each case also needs one of the other two facts for the second equation. QED. Huon (talk) 20:32, 20 October 2010 (UTC)

Worst article on wikipedia (moved from talk page)

I would nominate this article for "worst article on wikipedia". This article shows how the ignorant rabble can outvote the truth. No matter hown many people say the earth is flat, it's a sphere. No matter if Zeno says Achilles can't beat the tortoise, Achilles will win. No matter how many people continue to assert that 0.999 ... = 1, they are not equal equal. The *** limit *** of 0.999 ... is 1. That's it. Unfortunately, most people don't understand limits. It's not easy for the non-mathematician to understand. Most are so numerically challenged, they cannot even understand simple math concepts. Limits are far beyond them.

The article makes silly, pseudo-intellectual statements such as "the question of how two different decimals can be said to be equal at all". The article supposedly addresses "alternative number systems", but fails to mention other base systems. The reason many "initially question or reject it" is because it is not true. The business about "math educators" trying to figure out why students see through the fallacy is hilarious. The "math educators" are just showing their math deficiencies. "Has long been accepted by mathematicians" has no citation. The "proofs" shown on the page are ludicrous, and prime examples of logical fallacies, on the order of Zeno's paradox. The "proofs" remind me of the proofs of God's existence. Only one proof should be needed. Instead, many proofs are offered, because each one is flawed.

This article is a classic, and would be extremely humorous for any mathematician to view if it were satire. Since it's presented as fact, it's very disheartening to view. For those who would say "then edit it!", I tried that long ago. It's a total waste of my time. wikipedia is not peer-reviewed. It's rabble-reviewed. 174.31.157.82 (talk) 17:24, 8 December 2010 (UTC)

0.999... doesn't have a limit, it is the limit of a certain sequence. The article mentions other base systems in the second paragraph, noting that a similar effect appears there as well. Concerning peer review: Please go ahead, present peer-reviewed articles that support your claim that "the limit of 0.999... is 1". Huon (talk) 18:00, 8 December 2010 (UTC)

"The limit of 0.999..." is a nonsensical construct, which means as much as the phrase "the limit of 5". Numbers do not, in and of themselves, possess limits, although they may be the limit of a particular function or sequence as a variable approaches some other quantity. 0.999... is one particularly interesting expression of one such limit; that's all it is. (ESkog)^(Talk) 18:26, 8 December 2010 (UTC)

1) "The limit of 0.999..." is not a nonsensical construct. "0.999..." is the infinite series 9/10 + 9/100 + 9/1000 etc. The limit of the series is 1. That's all there is to it. 2) "0.999..." is not a limit. Whoever said that is showing their total ignorance of the concept of a limit. The number being approached (1) is the limit. 3) It is up to the editors of the article to show the peer-reviewed articles (in scholarly mathematical journals) asserting that "0.999... = 1". I saw none. 4) Yes, there is a sentence or two about other base systems. But in other bases, the phenomenon occurs in different ways, showing the fallacy of the simpler "proofs". Look at base 3, where the silly proof about 1/3 + 1/3 + 1/3 breaks down. 5) My comments being deleted by vandalism relates to my point about it being a waste of time trying to edit this "worst article on wikipedia". The whole thing about "0.999... = 1" is just a misconception, just like many other misconceptions of the numerically challenged. 174.31.157.82 (talk) 20:17, 8 December 2010 (UTC)

As far as (3) is concerned: Euler stated in his peer-reviewed 1777 article that 10=9.999... which is closely related to the claim contained in this page. Tkuvho (talk) 20:26, 8 December 2010 (UTC)

Why don't you actually ask a mathematician whether "0.999..." represents a sequence or a limit (that is, a number)? And of course the 1/3 + 1/3 + 1/3 proof breaks down in other bases! It is constructed for base ten. If you want a parallel proof for base 3, try 1/2 + 1/2. --COVIZAPIBETEFOKY (talk) 23:05, 8 December 2010 (UTC)

Well, 174.31.157.82, since you think we are numerically challenged, allow me to pose a numeric challenge to you. Here's a table.

Numerator	Denominator	Decimal representation
1	9	0.111...
2	9	0.222...
3	9	0.333...
4	9	0.444...
5	9	0.555...
6	9	0.666...
7	9	0.777...
8	9	0.888...
x	9	0.999...

Solve for x. Presumably you think x is not 9, so I would be curious to know what you think it is. 28bytes (talk) 20:55, 8 December 2010 (UTC)

0.999... is an example of a decimal representation of a real number. So are 0.333..., 3.14159... and 1.000..., though for 3.14159... the meaning of "..." subtly changes. The simple line of reasoning is that 0.999... should not be a different kind of object than, say, 1.000..., and that hardly anybody would argue that 1.000... should be anything but the number 1. I'm aware that this is, ultimately, convention, and I believe Richman argues that point in his 1999 paper, but that seems to be due to the abuse of notation by which we often do not clearly distinguish between a series and the sum of that series. Following Richman's line of reasoning would imply that the infinite decimals do not represent real numbers, but rather specific series whose sums are real numbers. That's mostly a cosmetic difference; if we take it seriously, it means that we can no longer compare infinite and finite decimals (because the former aren't numbers any more), but the system of such series is so boring that hardly anybody is actually interested in studying it. The numbers are the interesting objects, and the sequences and series are just tools. Thus, while I can't give a reference for that, I'd say basically every mathematician interprets infinite decimals as representing numbers, not series. If you insist, I can probably dig up various textbooks that do so. For example, Hans von Mangoldt's 1911 classic Einführung in die höhere Mathematik states on p. 147f that 0.090909... is a certain number, not a series, and the 0.999... case would be a direct analogy.

In summary 0.999... is not the series with sequence of partial sums (0, 0.9, 0.99, 0.999, ...), but the sum of that series, the limit of the sequence of partial sums. Huon (talk) 21:39, 8 December 2010 (UTC)

Simple: Given that every one of those equalities is wrong, x can mean anything you want it too. I can't believe that you guys are still trotting out that same tired circular 1/3=.333... "proof" that has already been discredited on these very pages. It is like those creationists who, when presented with real evidence, respond by finding a new audience to precent the same fallacies.

If .333... is a limit, and 1/3 is a number, then how can they be equal? Algr (talk) 04:27, 4 January 2011 (UTC)

A simple question:
y>1
y+x≥1
Solve for x

This question does not involve any infinities or "strange objects" or even any repeating decimals, and yet the real set falls apart over the idea. Algr (talk) 04:27, 4 January 2011 (UTC)

Where did you get the idea that .333... is a limit? 28bytes (talk) 04:51, 4 January 2011 (UTC)

Of course it's a limit. The right question is, where did Algr get the idea that a limit can't be a number? --Trovatore (talk) 04:57, 4 January 2011 (UTC)

In reply to the "simple question": The set of solutions is the set of pairs (x,y) such that x≥1-y and y>1. That's a well-defined set of real numbers. Obviously there is not a single set of x that will be the full set of solutions independent of y, though any non-negative x will solve the equation for every admissible y. So what?
Concerning limits and numbers: By definition the number a is the limit of the sequence (a_n) if for every real number ε>0 there exists a natural number n_ε such that for every n>n_ε we have |a-a_n|<ε. Thus by definition limits are numbers.
Concerning the "wrong" equations: Algr, what do you think is the decimal representation of 1/3? Do you think that 1/3 doesn't have any decimal representation? Wouldn't a system of representations that does not represent all your numbers be quite useless?
Finally, the 1/3=.333... proof isn't circular. It assumes that you have independently verified that equation, but doing so will hardly involve 0.999... - rather, one can use long division to divide one by three. The technical details may be hidden, but there is no circular reasoning. Huon (talk) 11:56, 4 January 2011 (UTC)

"the set of solutions is the set of pairs (x,y) such that x≥1-y and y>1." Restating the question is not an answer. x is not a range, but a single value for the entire range of y. You say "Obviously there is not a single set of x that will be the full set of solutions independent of y" but offer no explanation of this. You are just using the most obtuse language possible to avoid my point.

"Wouldn't a system of representations that does not represent all your numbers be quite useless? " HA HA HA! Almost all real numbers are impossible to describe using any known system. (Even counting psudo-descriptions like pi, and tricks like .1_{base 3}.)

"rather, one can use long division to divide one by three." Go on then, let me know when you are done. Long division is finished when you don't have a remainder. "1/3" is a division problem so restating the question is not an answer. Algr (talk) 08:24, 7 January 2011 (UTC)

Hi Algr. Let's try to adhere to polite language. It's only a math problem, after all. As far as .999... is concerned, could you let us know how you feel about it: is it (a) a process, (b) the infinitieth term in the sequence .9, .99, .999, etc., or (c) falls infinitesimally short of 1, or something else? If it is less than 1, how much does it fall short of 1 by, in your opinion? Tkuvho (talk) 11:27, 7 January 2011 (UTC)

Concerning the "simple question": For y=3, x=-1.5 is a solution. For y=2, x=-1.5 is not a solution. If there were a set of x such that it is the full set of all solutions independent of y, it would have to include -1.5 and not include -1.5 at the same time. That's impossible; therefore the set of x that solve the set of equations depends on y. If you want a single x that solves the equation for all y>1, try x=1. _{I feel you are trying to bog things down up to this point. One would not normally make those assumptions about an equation. Algr (talk) 08:54, 31 January 2011 (UTC)}If you want a single x that solves the equation for all y, that again does not exist because for y≤1, no x will do. I suppose you want to argue that there should be a smallest x that solves the equation for all y > 1 - again, that does not exist in the set of real numbers, but that's not a problem with the real number system. No matter what number system you choose, I'm pretty sure I can give you an equation that's not solvable within that number system. _{[Do you even realize that you are agreeing with me here?Algr (talk)]}(By the way, I also suppose you meant y>0 and not y>1.)_{[It doesn't matter, the equation works the same with any constant.Algr (talk)]}

Concerning representations: Every real number has a decimal representation, though not necessarily unique or finite. Of course we cannot write down an infinite representation in finite time, but it exists, and we can use it to compare real numbers. For example, I can write down enough of the decimal representation of pi to decide whether pi² is greater or less than 10.

Finally, long division. Let x be a real number between 0 and 1. Then x has a decimal representation 0.a₁a₂a₃... such that $x=\lim _{n\to \infty }\sum _{i=1}^{n}{\frac {a_{i}}{10^{i}}}.$ Then by the calculus of limits ${\frac {x}{10}}={\frac {1}{10}}\lim _{n\to \infty }\sum _{i=1}^{n}{\frac {a_{i}}{10^{i}}}=\lim _{n\to \infty }{\frac {1}{10}}\sum _{i=1}^{n}{\frac {a_{i}}{10^{i}}}=\lim _{n\to \infty }\sum _{i=1}^{n}{\frac {a_{i}}{10^{i+1}}},$ and x/10 is represented by 0.0a₁a₂a₃... Now on to 1/3, first step of the long division: 1:3 = 0.3 with a remainder of 0.1=1/10. Thus, 1:3 = 0.3 + 0.1:3 = 0.3 + 1/10*(1:3). Let 1/3 be represented by 0.a₁a₂a₃... and 1/10*(1:3) by 0.0a₁a₂a₃... as above. Then 0.3 + 0.0a₁a₂a₃... = 0.a₁a₂a₃... There is no carry in that addition; we can add all digits independently. Thus, we see that a₁=3, a₂=a₁, a₃=a₂ and so on, with a_n+1=a_n in general. Thus, a proof by induction shows that for all n, a_n=3, and 1/3=1:3=0.333... No circular reasoning involved, and done in finite time. Note also that I need limits only to prove that division by 10 shifts the digits one place to the right; if you accept that, the proof does not need limits at all. (There's another technical detail hidden in that proof, but it's still not circular reasoning, and I'd assume most people would accept it outright if I pointed it out.) Huon (talk) 12:57, 7 January 2011 (UTC)

"Finally, the 1/3=.333... proof isn't circular. It assumes that you have independently verified that equation -Huon" But that is the very definition of circular. When the meaning of infinitely recurring decimals is questioned, it is insulting to simply point to another infinitely recurring decimal and assume that that one shouldn't be questioned as well. Algr (talk) 08:54, 31 January 2011 (UTC)

No. The definition of a circular proof is that you presuppose what you intend to prove. The 1/3=.333... proof doesn't do that; it presupposes something else. And you wouldn't try to prove 1/3=.333... by arguing about 0.999...; thus it may be incomplete (though I did give a proof of 1/3=.333... above), but it's definitely not circular. Besides, the proof assumes that some people are willing to accept that 0.333...=1/3 without a full proof (because if 0.333... were not a representation of 1/3, what would be?); it is not meant as a complete formal proof in its own right. That still does not make it circular, though.

Concerning "bogging things down" and "not making such assumptions about an equation", could you please be a little more explicit? I don't understand what you mean there. Huon (talk) 13:26, 31 January 2011 (UTC)

I am baffled that you don't see how that argument is circular. I don't see what else I can do but repeat myself. Your proof about an infinitely repeating decimal involves the same assumption about another infinitely repeating decimal. Algr (talk) 17:11, 23 February 2011 (UTC)

Algr, I am quite baffled at what the problem your inequalities pose for the real numbers is supposed to be. Could you please explain it? Eric119 (talk) 16:24, 31 January 2011 (UTC)

The problem is that with the reals as you define them, you can create ranges that are equal, but not identical. 0<X<1; 0≤X≤1 and 0≤X<1 all have range 1. But if I say "a range of one centered on 1/2" it is not possible to tell if 1 or 0 fall inside or outside that range. So "Range 1" has three possible meanings. Algr (talk) 17:11, 23 February 2011 (UTC)

Actually it has four; you forgot 0≤X<1. But how does making 0.999... unequal to 1 solve that? What set would "a range of 0.999... centered on 1/2" be? And is "a range of 0.999... centered on 0.999.../2" different from that (and if so, how)? How would you distinguish the sets 0<X<0.999... and 0<X≤0.999...? Huon (talk) 17:27, 23 February 2011 (UTC)

Anthony's Arguments

As I said three years ago:

Removed. There is no reason for Wikipedia to be providing a platform for Anthony; it's time for him to go elsewhere. I'll be removing all his nonsense as it appears. --jpgordon^∇∆∇∆ 15:40, 23 October 2007 (UTC)[1]

--jpgordon^{::==( o )} 17:52, 30 January 2011 (UTC)

Sorry Anthony, while many of us disagree about .999...=1, I really can't help you. Some of what you wrote sounds like ideas we have discussed, but too much is sloppy terminology and misunderstandings. If a wrong process yields the right answer through luck, you still don't really know anything. Algr (talk) 08:36, 31 January 2011 (UTC)

I've semi-protected the page again for a while. --jpgordon^{::==( o )} 03:48, 1 February 2011 (UTC)

And, in his usual charming fashion, he's proceeded to vandalize the main article pages. Perhaps If We Used His Typographical Conventions! He Would Understand! That We Don't Want Him! Here At All! --jpgordon^{::==( o )} 17:03, 1 February 2011 (UTC)

Anthony, go away, forever. Protected again. --jpgordon^{::==( o )} 20:54, 4 February 2011 (UTC)

Non-explanation

So explain this,

We take it as given that .9999... has to continue for infinity in order to equal 1? Why is that? Isn't it because no ammount of 9s after the decimal point could represent 1 exactly? Therefore it must continue for infinity in order to be considered an exact representation. But isn't inability to physically stop and the need to continue appending 9s the very definition of an approximation? Not unlike we consider 3.14159... to be an approximation of pi. It doesn't matter that .999... follows a pattern when .314159 does not. Both could be theoretically calculated to an infinite precision. This suggests that there is a difference between the representation .999... and actual endless .999999999999999999999999999999999999999. But both .999... and the longform both remain with the exact same problem: while .999... conveniently adapts non-practical idea of infinity even at infinity still remeains but an approximation because there is no ends for infinity. (This, by the way suggests that infinity itself isn't really a number but a process but I won't go into that.).

Also:

1/3 = .333333333... therefore .3333333333...*3 =.999999999... = 1 is an example of an something that doesn't really explain anything. At least not to everyone. Why should a person who doesn't take .999999999... to be a number that represents 1 feel any differently about .33333333333... accurately representing a third of 1? In other words, this proof starts out with an assamption that is just as questionable as what it is trying to prove.

Why should 1/0 = x therefore x*0 = 1 be considered to be inaccurate whereas the example above is taken to hold true? I know the result of 1/0 is undefined but it is a precedent for laws of division and mathematical operators being not 100% reliant. From the point of view of mathematical operations we have a very similar set-up.

Another thing, to people insisting that .999... is a number and not a process. I have no problem grasping the concept of infinity and infinite repetition but answer me this then, if the fact that the full form version of that number (i.e. not the shortcut .999... form) with endless nines cannot actually be be finished being written down (or said, or, what that, matter USED) than how can you say it is a number in this when you cannot even finish it in this form?

Again, the obvious conclusion is that something that cannot be finished is an approximation by definition (again, I'm not talking about .999... form here but the full meaning behind the form). —Preceding unsigned comment added by 198.30.78.254 (talk) 16:03, 9 February 2011 (UTC)

Part of the confusion may be from misinterpreting the "..." to mean "approximately." The formulation "1/3 = 0.3333333..." is very different from "1/3 ≈ 0.3333333", which may be causing you trouble. Another thing that may help is to consider pi, which can't be "finished" (i.e. all its digits written down), but is a defined number and not an approximation. 3.14 is an approximation, but as with the example of 1/3, "pi = 3.14..." is different from "pi ≈ 3.14". Does this help? 28bytes (talk) 17:05, 9 February 2011 (UTC)

I'll address some of your points in greater length; if my explanations are insufficient we should probably break this up into various subsections for the different points you raise.

Concerning 0.333...: Firstly, there is no other candidate for a decimal representation of 1/3 but 0.333...; therefore it might be easier to believe that 0.333...=1/3. Secondly, you can do long division to prove that, by actually dividing one by three. (No, you will not finish in finite time, but mathematicians invented proofs by induction to solve that problem.) In effect, the 0.333... proof is not as mathematically rigorous as others; it's meant for people who believe 0.333...=1/3 in the first place.

Concerning "number vs. process": The set of infinite decimals such as 0.999... is meant to represent the real numbers. That's why we're interested in them (and that's why number systems such as Richman's decimals, where 0.999... no longer equals 1 but we lose subtraction in the process, are little more than nice toys). We know there are uncountably many real numbers, and we also know that there are only countably many decimals of finite length. Thus, representing the real numbers requires decimals of infinite length. I don't need to be able to write down all the digits to be able to work with that number. Actually, the countability line of reasoning above implies that you cannot find a method to write every real number in finite time - whatever you try, there will always be some real numbers which require infinite time to write down. You can call that a major drawback of the real numbers, but that's the price we have to pay for the properties that make the reals more useful than the rationals for many purposes.

And concerning approximations: If 0.999... were an approximation of 1, what's the error? By how much did we miss the mark? It's easy to show that the absolute value of the error is less than any positive rational number, and the only real number satisfying that condition is zero. Again, that we cannot write down 0.999...'s long form in finite time does not imply that 0.999... is an approximation, imprecise, not a number (unlike 1/0) or whatever - it's just an effect due to a built-in property of uncountably large sets. Huon (talk) 17:15, 9 February 2011 (UTC)

Explanation erroneous proving digit manipulation (moved from talk)

${\begin{aligned}x&=0.999\ldots \\10x&=9.999\ldots \\10x-x&=9.999\ldots -0.999\ldots \\9x&=9\\x&=1\end{aligned}}$

The point is this. Such important feature as invariability of significant number length for changes of this number in result of some mathematical operations is not taking into account in these calculations. Multiplication of decimal fractions on 10 is execute as well as for integers. This implyies what besides transposition of the decimal point on one numeral to the end of number the zero is added. Distinction is that non-significant zero in decimal fraction can be cutting. Interestingly that the infinite circulating decimal can be written in the form when direct mathematical operations with low-order digit are possible. The example of manipulation with numbers must be written as follows:

x=0,999...
x=0,(9)9 //Recording of infinite circulating decimal in the fit form for mathematical operations with the low-order digit//
10x=9,(9) //Number shift in the register, the low-order digit accepts zero value and cutting//
10x-x=9,(9)-0,(9)9
9x=9-0,(0)9
x=1-0,(0)1
0,999...=0,999...

Consequently "a" was equal 0,9999... and has remained equal 0,9999... because such important value as infinitesimal decimal unit was missing in the previous calculations. It was infinitesimal blunder. — Preceding unsigned comment added by Leonid 2 (talk • contribs) 09:00, 4 March 2011 (UTC)

Unfortunately I find it quite impossible to fathom what this is supposed to mean. Can you clarify it? JamesBWatson (talk) 09:39, 4 March 2011 (UTC)

I believe Leonid 2 says that when we multipy 0.999... by 10, the result should still have the same number of nines and that we should therefore have to add a zero at the end. This is wrong. 0.999... has countably infinitely many nines, one for every natural number. 10*0.999... = 9.999... still has countably infinitely many nines, even though it looks as if it had one more. Adding or removing finitely many elements from an infinite set does not change its cardinality. Besides, since the digits of 0.999... correspond to the natural numbers, there is no "last" digit after which the nines end. Huon (talk) 12:00, 4 March 2011 (UTC)

I have read about my explanations - "I find it quite impossible to fathom what this is supposed to mean". "I belive this is wrong". Probable cause is my bad English or the explanation was too short in spite of clearness. There are Hemming’s words - "The purpose of computing is insight, not numbers".
The program of information processing called Digit manipulation contain very gross bug connected with infringement of the order of performance of mathematical operations. Mathematical operations of addition, subtraction and multiplication of decimal fractions execute in the certain order and it is indifferent when these operations execute by a digit machine or on a sheet of paper. It is the order of processing from younger meaning bit to the senior. If addition of infinite periodic decimal fraction with an integer still can be written down in the form of 9,(0)+0,(9) having added to the integer an infinite chain of zeroes then in case of multiplication the problem gets fatal sense.
It seems that multiplication of decimal fraction on 10 is simple mathematical operation connected with carry of a decimal point on one digit. Actually for calculation it is necessary to execute some the consecutive strictly certain instructions. It is necessary to write down number, necessarily to add 0 by the end of record (initialization of operation of multiplication on 10, but at calculations on a paper usually it is not considered) and to add a sign on executed operation (for example, ^).
But number 0,999... hasn't the end. Correct operation of multiplication is prohibitive in this case.
If number was written 0,999...=0,(9) or such as of dimensionless linear array 0,(9)=0,(999...999) correct operation of multiplication is prohibitive too.

10*0,(9)=>0,(9)0^=>0,(9) //empty operation because shift on step =0 is impossible act//

Explanation: Linear array (9) is dimensionless, addition non-significant 0 in lower order digit don't change it. Even adding or removing finitely many elements from an infinite set (or cleaning of the lower significant digits) does not change its cardinality. Infinite set remains infinite.

For reception of an opportunity of shift of all array you must create control element (buffer entry). You must receive rights for changes most lower digit. Number 0,999... is infinitely, but if it had the end, it would come to the end on 9. Therefore number 0,999... write such as 0,(9)9 or execute the operation of global unconditional assignment.

x=0,(9)9

Explanation: Adding or removing finitely many elements from an infinite set does not change its cardinality.
But if ∞-1=∞, whereas ∞+1=∞+1.

Now indifferent if quantity of nines in the array (9) is endless, this array will be in computing such as unit and addition even only one 9 activate of the array overflow.

10*0,(9)9=>0,(9)90^=>9,(9)0=>9,(9) //execution of instruction//

But after operation of global unconditional assignment it is impossible to get rid from buffer digit for operations of addition and subtraction.

10х-х=9,(9)-0,(9)9=9,(9)0-0,(9)9

After the further calculations we receive very important volume - infinitesimal decimal unit - 0,(0)1.

x=1-0,(0)1=1,(0)0-0,(0)1

It is interesting element with original features. It is the mathematical tool which allows to formulate precisely many the existing mathematical proofs and it will be useful in exact numerical calculations.

Really 9*0,(9)=8,(9)1. 
The operation 9x=9 in programm called Digit manipulation is the operation of local un/conditional assignment x=1.

I found the bug in programm called Digit manipulation. Corrected result has the dead level 0,999...999==0,999...999 and verification can be. Probably I made a mistake. People said in Ancient Rome - "Errare humanum est". I ask argue away computing without words "I believe" and "I trust what 0,(9)=1".
My English isn't Shakespeare's language. Probably some considerations is no correct in English. I left text of this message on Russian on my talk page in English Wiki.Leonid 2 (talk) 11:10, 7 March 2011 (UTC)

You seem to do mathematics from the perspective of computer science. This is not helpful when dealing with infinite sets (such as the set of digits of 0.999...). For example, when multiplying 0.999... by 10 it is not necessary to shift the digits of 0.999... consecutively; we also do not need to add a 0 (or a 9) at the end of 0.999... - as you correctly said, 0.999... doesn't have an end. While you cannot build an automaton that will multiply 0.999... by 10 one digit at a time or do the multiplication on paper, that does not mean that 0.999... cannot be multiplied by 10.

The way you try to do the multiplication leads to infinitesimals, but the real numbers do not contain non-zero infinitesimals; that's the Archimedean property. Huon (talk) 21:32, 7 March 2011 (UTC)

"You seem to do mathematics from the perspective of computer science." Yes, of course. Computer science is the part of mathematics. Properties of part can't differ from properties of whole.

9*0,(9)=? //Question. How many will be, if multiply 0,(9) by 9?//

9*0,9=8,1
9*0,99=8,91
9*0,999=8,991
9*0,9999=8,9991
9*0,99999=8,99991
...............
9*0,(9)=8,(9)1
9*0,(9) no equivalent 9 //Verification//

Do you have more questions? Leonid 2 (talk) 12:25, 11 March 2011 (UTC)

Yes. What number system are you using? It's not the real numbers nor any other system of numbers I recognize. Huon (talk) 17:32, 11 March 2011 (UTC)

This is the decimal system. So you can multiply on a sheet of paper on a calculator or in your mind.

You said "While you cannot build an automaton that will multiply 0.999... by 10 one digit". It is quite possible. I showed how can do it and this result isn't differ from result on paper. //10*0,(9)9=>0,(9)90^=>9,(9)0=>9,(9)//

You said "real numbers do not contain non-zero infinitesimals". Why not do it? Such infinitesimals can be added or taken away from the real number if such number was written down definitely.

I wrote few lines of calculations and any schoolboy can tell what 9*0,(9) isn't equivalent 9. Infinitesimal decimal unit is displayed in the buffer zone in the calculations. The program of information processing called Digit manipulation contains fatal error. The Archimedean property can not correct direct errors in computing. Sorry. Sooner or later this error will be corrected.

We think in different languages. English very good language but it not contain feedbacks typical for Russian (declensional endings). User who is more inclined to understand an alien language said on my talk page "Your point about infinitesimals is a perfectly valid one".

Any discussion it is effort try to find the compromises. For example, you consider what an endless array inside of a number in parentheses it is unacceptable. Very good! Let's write down such array only in braces 0,(9)=0,(999...999)=0,(999...999)9=0,{9}9. Let's talk about special features and fields of application of the infinitesimal recurring decimals. Let's talk about how to make any calculations with recurring decimal infinitely correct. Do you want try to find the compromises? Leonid 2 (talk) 06:11, 13 March 2011 (UTC)

I don't think your number system is the same decimal number system the article attributes to Richman in its section on alternate number systems. Richman's system does not contain non-zero infinitesimals; it avoids those by disallowing certain subtractions (such as 1-0.999...). Your number system seems to contain some infinitesimals such as 0.(0)1. What operations on these infinitesimals are allowed? For example, is 1/0.(0)1 well-defined, and if so, what is it?

Regarding infinitesimals, the reals are simply defined in a way that disallows non-zero infinitesimals. If you add infinitesimals to the number system, it's no longer the real numbers. Why not use another number system with infinitesimals? Because adding them breaks other things which are more useful than the infinitesimals. Take, for example, topology. For a non-zero infinitesimal ε, is the set (-ε, ε) an open set? If it is not, the limit of the sequence (0.9, 0.99, 0.999, ...) is not unique any more. If it is, the sequence no longer converges at all. And while historically the process may have been different, nowadays the importance of decimal representations to mathematics is that they describe the real numbers, not the other way around.

Finally, I'm not sure I understand your calculations. You say you can multiply 0.999... by 10, but what you actually write down is 0.(9)9 * 10. Is 0.999...=0.(9)9 or 0.999...=0.(9)? You also seem to suggest that 0.(9)=0.{9}9; is the latter equal to 0.(9)9? If not, what's the difference? If it is, why is there a difference between 9+0.999...=9.(9)9 and 10*0.999...=9.(9)? Huon (talk) 11:03, 13 March 2011 (UTC)

As to Richman's "decimal numbers": If I understand correctly, it's not quite as ad-hoc as disallowing 1-0.999... specifically. There is simply no subtraction operation at all.

Infinitesimals, I think, are largely a red herring in this context. In the most common way of dealing with infinitesimals (à la Robinson) it is still not the case that 0.999... is less than 1. That is, if your ... means that you have 9s in all (standard or nonstandard) locations, then you still get 1. You might suggest that it means you have 9s in all standard locations but 0s in the nonstandard ones — but that expansion simply does not exist at all (in the nonstandard model), because if it did the nonstandard model could identify the standard integers, which is not allowed.

If you have 9s up to some nonstandard location, and 0s after that, then you do indeed get a value infinitesimally less than 1, but it's now not clear what justification you have to refer to this number as 0.999.... --Trovatore (talk) 19:26, 13 March 2011 (UTC)

I'm not want the discuss about the full article 0.999... . I do not care 0,(9) is equivalent to 1 or not. But the section of the article entitled Digital manipulation contains an obvious error which must be corrected. This error can be detected by direct calculations on a sheet of paper. In the multiplication 0.999 ... by 9 you will never get 9, no matter how many nines after the decimal point will be, and this should be displayed on the main page of the article. Leonid 2 (talk) 08:08, 19 March 2011 (UTC)

Within the set of real numbers, which the article specifies in its very first sentence, that calculation is correct. This is basically a variation of Hilbert's hotel which shows that addinng an element to a countably infinite set gives a set of the same cardinality as before - which means that 9+0.999... and 10*0.999... have the same number of nines. Huon (talk) 11:32, 19 March 2011 (UTC)

There is no need to discuss the multiplication recurring decimals by 10. In addition to the mathematical operation of multiplication there are additions and subtractions too. I wrote about this in detail above. Try to get 9 in direct multiplication 0.999... by 9. Leonid 2 (talk) 08:39, 20 March 2011 (UTC)

Why would I need to do that? The proof does not use direct multiplication by 9, but by 10. Multiplication of 0.999... by 9 would a priori give a result of 8.999..., leading to 9*0.999...=8+0.999... and by extension 8*0.999...=8. Huon (talk) 11:54, 20 March 2011 (UTC)

There is very much a need to discuss multiplication by 10, if you are claiming that 9*0.999... = 8.999 ≠ 9, because the proof at hand uses multiplication by 10 to prove that 9*0.999... = 9. You have to exhibit a flaw in the proof. --COVIZAPIBETEFOKY (talk) 14:28, 20 March 2011 (UTC)

I asked to refute my computations. I got the misty arguments about the properties of infinity about the properties of a real numbers and so on, instead of other computations. But I'm not want the discuss about the full article 0.999... and catch the wind in a net. Therefore once more.

9*0,999...=? //Question. How many will be, if multiply 0,999... by 9?//

9*0,9=8,1 //Start of the computing//
9*0,99=8,91 //Continuation of the computing//
9*0,999=8,991
9*0,9999=8,9991
9*0,99999=8,99991
................... //Continuation of the computing//
.....................
....................... //Stop when you will be tired or are finished these computations// Шутка 

9*0,999...=8,999...9991 //Answer. Any can enter the infinite number of nines instead of the dots by oneself//

Such decision is possible even for a scholar from an elementary school. In the multiplication 0.999 ... by 9, you will never get 9, no matter how many nines after the decimal point will be. The section of the article entitled Digital manipulation contains an obvious error. 9x=9, if x=0.999... it is the blunder. I wrote about this in detail above how this blunder arises.
Don't you agree with me? It may be. Which means that the listing must be of the your step-by-step computations 9*0,999...=9.

Ah, yes. I almost forgot. You asked how many it will 1/0.(0)1.

 1/0.(0)1=1(0).0 

Leonid 2 (talk) 09:11, 22 March 2011 (UTC)

The refutation of your computation is that numbers like 8.999...9991 simply do not exist within the real numbers and their decimal representations. Each digit after the decimal separator corresponds to a natural number (the 1st digit, the 2nd digit and so on). To what natural number should the "1" in 8.999...9991 correspond? Since for each natural number n, the n-th digit after the decimal separator of 0.999...*9 can be shown to be a nine (by induction), 9.999...*9 equals 8.999... Similarly, 0.999...*10 equals 9.999... and not 9.999...9990, which again is not well-defined.

Your notation suggests that you indeed want to use some non-standard number system that contains infinitesimals and infinitely large numbers. The real numbers, which this article (along with most of mathematics and physics) is concerned with, don't behave in that way. You might be interested in the hyperreal numbers, but even there, where 0.999...999000... can be considered a well-defined number (as Trovatore points out), we still get 9.999...999000... = 10 * 0.999...999900... ≠ 10 * 0.999..., and the canonical number to be called "0.999..." still equals 1. Huon (talk) 14:25, 22 March 2011 (UTC)

The empty argument instead of the actual calculations again. Statement x=0.999... is first line in the section Digit manipulation. Statement 9x=9 is fourth line. It is false statment if x=0.999... It is direct assignment x=1. Leonid 2 (talk) 10:54, 26 March 2011 (UTC)

That fourth line is not a direct assignment, but a consequence of the third line because 10x-x=9x and 9.999...-0.999...=9. You can get the latter result by a direct subtraction. Every digit of 9.999... after the decimal separator is a 9, so is every digit of 0.999... after the decimal separator. Thus, for every digit after the decimal separator, subtraction will yield a 0, and 9.999...-0.999...=9.000...=9. Huon (talk) 11:21, 26 March 2011 (UTC)

No need to abuse by the mathematical analysis if there are problems with arithmetic - direct calculation errors. Leonid 2 (talk) 13:15, 26 March 2011 (UTC)

There are no direct calculation errors. A different direct calculation may arrive at a result of 8.999..., but that's not an error, but rather an indication that 8.999...=9. Huon (talk) 14:16, 26 March 2011 (UTC)

Leonid 2, what is 10*0.999... - 0.999..., by direct calculation? --COVIZAPIBETEFOKY (talk) 23:19, 26 March 2011 (UTC)

If the number is written as 0.999 ..., then it can not be handled in the mathematical operation multiplication by 10 from the low-order digit. It is indeterminateness of the mathematical operations of multiplication by 10. Leonid 2 (talk) 10:08, 30 March 2011 (UTC)

Thank you. You have talked so much about the properties of rational numbers for me. Finally, I understood everything. The number 0.999... is not rational. It is the recurrent decimal fraction. It can not be represented as simple fraction. In converting simple fractions to decimals you may lose some information irreversibly. The numbers of multiples of 10 can not be divided by 9 or 3 evenly. 10/9=1+1, 100/9=11+1, 1000/9=111+1, 10000/9=1111+1, ... (10/3=3+1, 100/3=33+1, 1000/3=333+1, 10000/3=3333+1, ...).
The infinitesimal decimal unit 0,(0)1 proved to be useful once again.

1/9=0,(111...111)1+0,(000...000)1
1/3=0,(333...333)3+0,(000...000)1
0,111...=1/9-0,(0)1
0,333...=1/3-0,(0)1
0,999...=1/1-0,(0)1

The numbers 0.111... and 0.333... is not rational too. Addition and subtraction with decimal infinitesimals for the such numbers is admissible mathematical operation. Special form of recording is compulsory regulation in this case. Some features of the infinitesimal decimal unit 0,(0)1 on my talk page - Russian and English text.
But I do not care 0.999... is equivalent to 1 or not as before. Leonid 2 (talk) 10:08, 30 March 2011 (UTC)

Such intuitions cannot be implemented in the real number system, but A. H. Lightstone implemented them in the context of an ordered extension of R, see there for some details connecting up with research in modern mathematics education. Tkuvho (talk) 17:48, 30 March 2011 (UTC)

The intuition that "[t]he numbers of multiples of 10 can not be divided by 9 or 3 evenly" was implemented by Lightstone? News to me. While I agree that such intuitions cannot be implemented in the real numbers, they cannot be implemented in any extension of the reals, either. Your point would be more convincing if the various intuitive notions of how 0.999... differed from 1 agreed with each other, but they don't. For example, Algr seemed to want to have a greatest number less than 1, another idea that cannot be realized in the hyperreals and which probably cannot be realized in Leonid 2's number system either. Huon (talk) 18:10, 30 March 2011 (UTC)

In general I try to encourage students when their intuitions can be made mathematical sense of. In this case, I was referring to his intuition that the "infinitesimal" 0.(0)1 can be useful, and also the intuition that 0.111... with an infinity of 1s can be seen as different of 1/9. He is right in his sentiment that ordinary arithmetic operations on infinite decimals are ill-defined, as was also pointed out by Richman in his article cited here. Thus, to multiply two decimals, we need to start at the right-most digits because of possible carry-over. But real infinite decimals do not have a right-most digit. Therefore there are subtle foundational issues involved here having to do with the details of the construction of R. That's certainly one useful number system. So are its extensions, as envisioned by Leibniz already. Tkuvho (talk) 19:43, 30 March 2011 (UTC)

I calculated quantity of solutions for distinction 0,999... and 1 on my talk page "Quantity of solutions for distinction 0,999... and 1" for two cases (if quantity of nines after decimal separator is infinity and if quantity of nines after decimal separator is the more of infinity). General quantity of solutions for distinction 0,999... and 1 is equal ∞+2 solutions including solution in Lightstone's extended decimal notation. Probably 0,999... is not equal 1. Confidence is more 100%. I ask argue away these calculations only by other calculations. Leonid 2 (talk) 07:10, 10 April 2011 (UTC)

Define "more than infinity". Nongendered (talk) 10:46, 11 April 2011 (UTC)

"More than infinity" means that the quantity of digits in an infinite sequence to the limit can be equal to ∞+N where N can take values from 1 to ∞. Further increase in the length of it makes no sense. Section the article entitled Digit manipulation contains fatal error for any precision of calculations. Leonid 2 (talk) 05:39, 17 April 2011 (UTC)

There is no error in the article. A number of competent mathematicians have worked on it, and any possible error would have been eliminated by now. You should realize that the proofs given in the article are operating in the context of the real number system. Here arithmetic operations are not defined on the infinite strings of digits themselves. Rather, the operations are defined by taking finite n-strings of digits, performing the arithmetic operations on them, and then taking the limit of the resulting values as n tends to infinity. Before you can meaningfully challenge this approach, you must specify the number system you are working in, as an alternative to the real numbers. Note that in order to affect the article, such a number system would have to be sourced in the literature. Tkuvho (talk) 06:36, 17 April 2011 (UTC)

I have read "such a number system would have to be sourced in the literature". Wikipedia rules tolerate for deviation from this statement for mathematical calculations.

Define "real number system". Numbers containing infinite sequences it is already the special numbers which impose the certain restrictions on performance of mathematical operations of multiplication or division. In multiplication or division such decimal fractions by 10 shift not only decimal separator. It is compulsory for the shoolchilds as well as for the competent mathematicians.

If the quantity of nines after the decimal separator is infinite, in this case for such numbers exists high end system of recording - Lightstone's decimal notation. Such number system allows to solve tasks provided that the length of number after decimal separator is strictly and invariably equal to infinity. For example, 1=0,;9;+0,;0;;1. Mathematical operations of division will be performed in form of the irreducible infinitesimal remainders which are Lightstone's numbers too. For example, 1/9=0,;1;+0,;0;;1 or 1/6=0,1;;6;+0,;0;;4 or 1/7=0,;142857;+0,;0;;1. The rational fractions 1/3 and 1/9 can be converted in decimal fractions only as polynomials. Form of recording is indifferent combination.

Lightstone's numbers can be multiplied by any finite number without restrictions. Mathematical operations of division by numbers multiple 10 for such numbers can be written down only in the form of quotient 0,;9;/10=0,;9;/10, 0,;9;/100=0,;9;/100, ... or 9,;9;/90=1,;1;/10. Leonid 2 (talk) 06:22, 27 April 2011 (UTC)

Defining the real numbers is rather easy, provided you are familiar with the rational numbers. The real numbers are the set of equivalence classes of Cauchy sequences of rational number, where two Cauchy sequences (a_n) and (b_n) are equivalent if and only if the sequence (a_n-b_n) converges to zero in the rational numbers. The correlation to decimal representations is then quite obvious: The decimal representation a₀.a₁a₂a₃... represents the number given by the equivalence class containing the sequence (a₀, a₀.a₁, a₀.a₁a₂, ...), which is indeed a Cauchy sequence. Addition, subtraction and multiplication can be defined by doing the corresponding calculations on representatives of the equivalence classes; division requires us to choose a representative of the denominator (b_n) such b_n≠0 for all n (such a representative can always be chosen, and the quotient is always a sequence, but not a Cauchy sequence if (b_n) converges to zero). That's all it takes to define the real number system based on the rationals, which I assume we are all sufficiently familiar with.

Lightstone's decimal notation (which is not what you actually use) has several features you probably would not like. First of all, Lightstone's hyperreal decimals still contain the number 0.999...;...999... where every digit is a 9, and that number is still equal to 1. So you still have the effect that a number with nines all the way is equal to 1, and you haven't really gained anything with respect to 0.999... Secondly, there are some decimals which do not represent any number. Take for example 0.333...;...000... where all digits corresponding to a natural number are 3, and all digits corresponding to an infinite hyperinteger are 0. That's a well-defined decimal representation - but as Lightstone himself points out, it does not represent any hyperreal. I think I'll leave the explanation of the finer points where whatever you do differs from Lightstone's approach to Tkuvho, who is more of a friend of the hyperreal numbers than I am. I'll just note that Lightstone's system of representing the hyperreals does not require any infinitesimal remainders. Huon (talk) 10:59, 27 April 2011 (UTC)

from the other side

ok, and apologies if this has been covered before. An informal 'handwaving' argument but one that might help overcome some intuition blocks:

It should be obvious to all that 1.1 > 1. Similarly, 1.01, 1.001, 1.0...01 etc are all >1 for all such numbers with a (finite) string of '0's.

So what happens with an infinite string of 0's? It should be obvious that 1.0...=1 (there being no 'last place' for the digit 1)

Note that the difference of 1&1.1 is the same as the difference of 1&0.9

Similarly for 1.01 & 0.99; 1.001 & 0.999 and so on.

It (informally) follows that the difference of 1 & 1.000...(equality) is the same as the difference of 1 & 0.999... Yossarian68 (talk) 23:10, 23 April 2011 (UTC)

Nope. What you say "should be obvious" is the very statement you are claiming to prove. There is no situation in finite mathematics where you can shift a digit until it becomes irrelevant, so it seems very sloppy to assume that this ALWAYS works with infinite decimals. These various proofs all involve infinity doing things that are contrary to how the rest of mathematics works, and there is never any explanation as to why infinity would do X instead of Y. Multiplying by ten gains the power to alter the number of significant digits in a number, approximations become exact due to the LACK of any point where this can happen. At one point dividing both sides of an equation by zero became an acceptable way to 'prove' equality. What am I to make of these strange deviations from mathematical norms? Algr (talk) 23:35, 23 April 2011 (UTC)

You don't find it obvious that 1.0... (1 with a never-ending string of 0's) =1? Then you're right. I can't help. 'Shifting the digit' certainly does not make it irrelevant. However, when there's nowhere to put a 'last digit' then the last digit is irrelevant as by definition it doesn't exist! Yossarian68 (talk) 23:50, 23 April 2011 (UTC)

Placing a "1" at finite rank n to obtain 0.000...1 is paralleled by placing it at infinite hypernatural rank H, so as to obtain an infinitesimal 0.000...;...01. This clearly can't happen in the real numbers. Tkuvho (talk) 04:39, 24 April 2011 (UTC)

but we're talking about the Reals, so that's fine Yossarian68 (talk) 11:26, 24 April 2011 (UTC)

That's precisely my point. You are talking about the Reals, but many editors here and students in classrooms around the globe are told that 0.999...=1 before they have learned about the reals. It is one thing to say that one of the shortcomings of an otherwise extremely useful number system is the absence of infinitesimals. It is another thing altogether to insist that such an equality is built into the nature of the universe and physical reality itself, as it were. Tkuvho (talk) 12:56, 24 April 2011 (UTC)

And we are told that 1+1=2 before we are taught about the integers let alone the Peano Axioms so I'm not sure what your point is? (And the first line of the article specifies that 0.9... is a real number.) As for the ontological status of any mathematical model, I'll leave that to another article and/or a discussion over several beers Yossarian68 (talk) 18:40, 24 April 2011 (UTC)

It is obnoxious to try to deflect a question about .999... and 0.000...;...01 by locking the discussion into the Real set. It is no different then 'proving' that 1/2 does not exist by insisting that only Natural numbers be used in the discussion. I've said this before, BTW Algr (talk) 05:13, 25 April 2011 (UTC)

"obnoxious"? 1/2 isn't defined in the Naturals, so need Rational nos at least to discuss it. Root 2 isn't defined in Rationals so need Reals. Root -1 not defined in Reals so need Imaginary... However, 0.999... is defined in the Reals(as 1) so don't need hyperreals which would make no difference anyway as Reals are a subset?Yossarian68 (talk) 20:53, 25 April 2011 (UTC)

One divided by two IS defined in the Naturals. It is Zero remander One. If I were to insist that someone who was just learning fractions should be told that Zero remander One is the ONLY answer, wouldn't you find that to be obnoxious of me? Algr (talk) 04:05, 26 April 2011 (UTC)

It is best to avoid terms such as "obnoxious" in this type of discussion. As far as the reals are concerned, they are an extremely useful number system, but between 70 and 80 percent of the students reject the limit definition of .999... Namely, they agree that the sequence .9, .99, .999, ... gets closer and closer to 1, but they still think .999... falls short of 1. This does not merely have to do with uniqueness or non-uniqueness of decimal representation, since the students similarly reject the identification of .333... with 1/3 once it is pointed out to them that this appears to imply .999...=1. So declaring that zero, point, followed by an infinity of 9s is necessarily a real number amounts to changing the subject, as far as the students are concerned. Tkuvho (talk) 04:50, 26 April 2011 (UTC)

Tkuvho, do you have a source for those student percentages? It's pretty obvious that doubt is widespread, but I don't think I've ever seen it quantified, and the article doesn't seem to contain such a number, which would obviously be highly relevant. Huon (talk) 10:00, 26 April 2011 (UTC)

The source is given at list of common misconceptions. You are right, we should add it here as well. Tkuvho (talk) 10:12, 26 April 2011 (UTC)

Very interesting discussion enhanced by its illicit nature as sure we shouldn’t be having it on Wikipedia. I’m interested Tkuvho and Algr what if anything you infer from these students’ intuitions about the truth (whatever that means) of mathematical statements. I’m not clear if your concern driven by ontological, epistemological or pedagogical considerations (or all three)? And what if anything is the relation of your concern with such intuition to Intuitionism as a philosophy of mathematics? (I of course started this chain with an idea I thought might help overcome the intuition blocks of accepting 0.9...=1 albeit within the structure of the reals and so on).Yossarian68 (talk) 21:27, 26 April 2011 (UTC)

The "illicit", as you put it, infinitesimal "1-.999..." is an intuitive aid in learning the calculus, e.g., understanding the notion of instantaneous slope. I can try to provide a figure if you are curious. Tkuvho (talk) 04:15, 27 April 2011 (UTC)

0.999_ does not equal 1

http://uncyclopedia.wikia.com/wiki/0.999...

In mathematical theory .9 is a specific number. It is not half or a fraction of a number. It is .9 and it remains .9 in the equation. Unless an equation calls for rounding of the numbers, then .9 remains solely .9 in the equation. Therefore 0.999_ is a infinitely repeating number.

In mathematical practice there are no exact numbers. As reality is incapable of producing an exactly equal number in relation to something. So in mathematical practice 0.999_ cannot equal 1 regardless.

In mathematical theory where rounding occurs then 0.999_ equals 1.

If rounding is not established or .9 is specifically established as being an exact number, then no amount of crying and bitching changes it to 1.

You've been Uncyclopediaed, yo. 58.7.214.181 (talk) 03:50, 19 May 2011 (UTC) Harlequin

The "rounding" you referred to can be mathematically implemented either by Fermat's adequality or the standard part function, you may find those pages helpful. Tkuvho (talk) 03:59, 19 May 2011 (UTC)

Perhaps you should go back and read where I specifically, in detail, referred to equations where it wasn't being "rounded". But if all you can come up with is "no, no...if they are not infinitely repeating, then it totally equals 1", it's just further proving my point. 58.7.214.181 (talk) 04:03, 19 May 2011 (UTC) Harlequin

And yet you're the one trying to use Uncyclopedia like it's a reliable reference or something. Do you realise that most of that article was written by people who wrote this article, who needed to blow off some steam after arguing with "equality deniers" on this very talk page? Confusing Manifestation(Say hi!) 04:11, 19 May 2011 (UTC)

I'm using Uncyclopedia as a reference? Haha, well done at further proving my point and again failing to read. Strange, seems I explained exactly how 0.999_ does not equal 1 without anything relating to the Uncyclopedia article on the subject. I merely pointed out another site that makes fun of the bitching and crying those like yourself are doing.

I gave the equation separate from any other site. Which you should have known had you, you know, actually read anything. Hahaha.

0.999_ repeating in mathematical theory can only be 1 if it is not infinitely repeating. Your claim here is that somehow an infinitely repeated number will magically have an end and that it will magically round up once it has reached that end, even when the equation it is in specifically states it doesn't. Like ive stated, no amount of bitching and crying changes that fact that an infinitely repeating number does not have an end.

It's pretty hilarious really. I even explain that where it isn't infinitely repeating and is in an equation that "rounds" the numbers that it would equal 1 (again, had you actually read anything) and yet you seem to think that an equation where it can't even reach an end to "round" would somehow still do so even when it's specifically mentioned that none can occur?

Try again. This time without trying to take credit for other sites material. Especially when the best "argument" (or lack of) you can come up with is that "some guys" created the other article (despite it's refutation of "their" supposed claim that 0.999_ ALWAYS equals 1) and because "they did" it is therefore magically wrong. 58.7.214.181 (talk) 04:39, 19 May 2011 (UTC) Harlequin

Yo Har, is there a last "9" in your 0.999_ after infinitely many 9s? Uncyclopedia says there is. Do you agree? Tkuvho (talk) 04:15, 19 May 2011 (UTC)

I notice you don't know what infinite means. There is no "end", so there can't be a "last 9". Try again. Especially since I didn't state anything about Uncyclopedia being correct and specifically provided a detailed explanation of why the claim "hurr 0.999_ always equals 1 no matter what" is a false statement. 58.7.214.181 (talk) 04:39, 19 May 2011 (UTC) Harlequin

OK. Would you care to comment on the nature of the number 1-0.999_  ? Tkuvho (talk) 11:26, 19 May 2011 (UTC)

This is such a fascinating argument to me. As someone who's knowledge of mathematics goes about as far as long division I'm perfectly willing to accept the fact that 0.9 recurring equals 1, simply because I defer to the wisdom of people much smarter than me in this subject. If mathematics professors say it's so, who am I to judge? But the moment someone get's a little bit of knowledge about this subject they think they have what it takes to prove the experts wrong. Why not just accept it and move on? --86.182.1.71 (talk) 18:39, 15 June 2011 (UTC)

.9 repeating is just another way of writing 1 but with all its lesser place values being presented. Usually one is simply given the maximum place value: 1/1, but obviously the number can contain any combination of lesser place values. 1/10ths, 1/100ths etc. For example 100/100 is 1 too but for the 1/100ths. 900/1000 + 10/100 = 1 representing place values 1/100ths and 1/1000ths. Of course there is obviously an infinite amount of lesser place value delimitations intrinsic to the number 1 and these will add up to 1 just like any smaller finite combination. And when we list out these additions we get 9/10 + 9/100 + 9/1000... etc. Unless you think it is mathematically impossible to represent every possible lesser decimal place value you basically have to admit .9 repeating is exactly 1.76.103.47.66 (talk) 00:30, 6 August 2011 (UTC)

Aren't there two conflicting definitions ?

Hello. I am not terribly knowledgeable in mathematics. Some years ago I posted here a comment, which used non-conventional notation, trying to reason that 0.(9) couldn't be equal to one, because then all numbers would be equal to all other numbers (if you nullify the difference, even infinitesimal, between two numbers, then what's to stop you from generalizing that to the whole real axis?). I received then two comments. One simply said that I was using non-standard notation and so my argument didn't follow. Another one said that it was possible I used it, in so far as I was actually able to use it and convey my meaning, and that there might be some reason to my assertion. He then directed me to some higher mathematics I couldn't understand (p-adics).

But enough with introductions. I have revised my thoughts in various ways, and I am also now able to present my simple argument using conventional notation. So here it is:

Consider the sequence 0.9, 0.99, 0.999, 0.9999, 0.99999, etc. It is obvious to anyone that the limit of this succession is one. But the definition of a limit tells us that this limit can never be actually reached. Therefore there is never actually an equality between 0.(9) and 1, except in the limit itself.

The same holds true to the sequence 0.1, 0.01, 0.001, etc. It's limit is zero. Therefore the difference between 1 and 0.(9) is only zero in the limit.

I think that a parallel could be drawn to the point in the exact center of a wheel. I think it can both be said that it rolls or that it doesn't roll, depending on the exact context. There may be a confusion regarding two different "levels" of reality, that is, a confusion between an assertion and a meta-assertion.

Joao.g.madureira (talk) 16:48, 17 October 2011 (UTC)

0.(9) is not the sequence 0.9, 0.99, 0.999, 0.9999, 0.99999, etc., but a real number. (And by the way, it's not the definition of the limit which tells us that the limit can never be actually reached; there are other sequences where the limit is reached. As a trivial example consider the sequence 1, 1, 1, etc.) The only real number which could be represented by the sequence (0.9, 0.99, 0.999, 0.9999, ...) in this context is its limit.

There are ways to develop number systems where the basic "numbers" are themselves (equivalence classes of) sequences of real numbers, but those do not have a canonical element "0.(9)" - see for example the hyperreal numbers; there is a hyperreal number represented by the sequence (0.9, 0.99, 0.999, 0.9999, ...) (up to the choice of an ultrafilter, but let's ignore that for now), but it's not the same as the hyperreal numbers represented by the sequences (0, 0.9, 0.99, 0.999, ...), (0.99, 0.999, 0.9999, 0.99999, ...) or (0.9, 0.99, 0.9999, 0.99999999, ...), and which of them should be called "0.(9)"? Huon (talk) 18:08, 17 October 2011 (UTC)

Here's a different reply to João, which I hope will help convince him. João, there's an incorrect assumption in what you say, namely that a "limit can never be actually reached". That is 'often' true of limits, but not always. As a counterexample, consider the sequence a(n) = (1+(-1)^n)/n = 0, 1, 0, 2/4, 0, 2/6... Its limit is clearly 0, and a(n) does take on this value, infinitely often in fact. I suspect that this misconception about limits is part of the trouble many people have with the equality discussed in this article.

Another point, already made by Huon, is that by "0.(9)" we really mean the limit of the sequence (0.9, 0.99, 0.999, 0.9999, ...), rather than the sequence itself. So in a sense you're right that there are two distinct concepts at play here, the sequence on one hand and its limit on the other. What you (and many others) appear to misunderstand is that "0.(9)" by definition denotes the latter, not the former. Let me repeat that: by definition, 0.(9) = lim(0.9, 0.99, 0.999, 0.9999, ...) -- not 0.(9) = (0.9, 0.99, 0.999, 0.9999, ...)! Though this is a technical point, and so perhaps not as persuasive as my previous one. Regards. FilipeS (talk) 14:49, 6 February 2012 (UTC)

Rounding decimals

I'm sorry, I do not normally log in to Wikipedia because I do not edit information, mostly because I only come here to learn about things I do not know about (and how can I add to a topic I don't know about?) but also because I do not know how all of this actually works. So if I am doing something completely and horribly wrong, then I apologize, I just needed to say this. I also do not really have more than a high school education in mathematics, so I am not any sort of mathematician expert, nor even proficiently learnéd in math. Although, I have tended to have an unusually extraordinary understanding of mathematics, especially such basic mathematics as converting fractions to decimals.

However, I cannot accept the fact that 0.99999... = 1 and after reading the proofs, I understand why. The proofs all make sense, but I have the reason/problem which causes this to happen, and it is because infinite decimals are not perfectly accurate representations of fractions, which is why they are infinite. They go on forever because those fractions cannot be created into decimals. 1/3 become 0.33333.. and so on forever, because this is the -closest- representation of it that we can create with a decimal, but it isn't perfect. Therefor, 0.9999... is the closest representation of a fractional 1 that we can produce with a decimal, however it does not truly equal 1 because it is more like an estimation or rounded number.. it is like rounding 4.9 and 4.9 both to 5 and adding them together to make 10, and saying that 9.8 = 10 because 4.9 + 4.9 actually equals 9.8. That is the same as "rounding" 1/3 into 0.333... and so on forever. Therefor the actual number 0.999... itself is NOT truly equal to 1, it only seems that way if you multiply 1/3 * 3 as 0.333... * 3 equaling 0.999... because 0.333.. is "rounded" and therefor like the 4.9's I rounded to 5. I don't know if that makes any sense to anyone but it makes perfect sense to me, and I can try to explain it further if no one is understanding me. I am also unsure if there is any way to make a proof for this because there isn't much of a way that I can show that infinite decimals are rounded or estimations, other than by using the proofs already here.

Obviously if 0.333... * 3 =/= 1, then 0.333... is not a perfect representation of 1/3 in decimal form. It is just the closest. — Preceding unsigned comment added by Sweetnaivety (talk • contribs) 06:43, 27 October 2011 (UTC)

... and thus movement is impossible, and Achilles could never reach the tortoise. I sympathize with your reasoning; being a computer programmer, I see mathematics as a series of processes executed one step at a time. But you should understand that intuition doesn't mix well with infinite processes, that's why we use mathematical theorems instead. In this case, the decimal representation is not a rounding approximation, it's the exact representation of the fraction. It's exact because it's defined as a limit of an infinite series, and limits are by definition the exact value of the series.

One advice: when your intuition disagrees with the results of a solid mathematical proof, you should usually give a vote of confidence to the proof instead of trusting your intuition; the formal math is more likely to be correct. But at the same time try to understand the proof to see where it deviates from common sense. You'll usually find that the proof is using some definition that you were not taking into account. Diego (talk) 12:57, 27 October 2011 (UTC)

Let me add: What's the difference 1/3 - 0.333...? It's non-negative (because surely 1/3 is not less than 0.333...), but less than any number of the type 10^-n (because 0.333...+10^-n > 0.333...334 > 1/3). That shows it needs to be an infinitesimal. But by the Archimedean property the real numbers do not contain infinitesimals except 0. Since we operate within the set of real numbers, the difference must be 0; there was no rounding going on.

One might argue that the real numbers are the wrong number system; that instead we should use a number system which does include non-zero infinitesimals. But those number systems are all less connected to decimal representations than the reals, and they suffer a wide selection of drawbacks (depending on the specific number system). The real numbers are the system most widely used in mathematics and almost exclusively used in physics. Huon (talk) 13:17, 27 October 2011 (UTC)

I mean, it isn't -really- rounding, it is -like- rounding. It is not "exact" because it cannot be, or else it would not go on forever. Decimals are basically fractions of 10, and you are trying to conform fractions of thirds to tens.

I see mathematics as a basis of fact, where everything always means what it means. 1 + 1 will always equal 2, no ifs ands or buts, plain and simple. Math is simple, and my mind is simple, so my intuitions about math are pretty much always correct. I mean, just from learning how to calculate area and volume, by myself I was able to formulate a calculation/theory for the 4th dimension in elementary school, only to find out in middle school that my theory was already known. I never had to pay much attention in my math classes, because I instinctively, intuitively, already knew how to do everything.

Now because my mind is simple like math, it cannot accept that 0.999... is exactly equal to 1. They must be different numbers, because only 1 can equal 1. When you take 1 and divide it by 1, you do not get 0.999... so if these infinites were "exact" representations, then shouldn't you be able to get 0.999... if you divide 1 by 1, the same way you get 0.333... by dividing 1 by 3? But you cannot.

The simple proof which shows that 1/3 = 0.3333... , 1/3 * 3 = 0.3333... * 3 , 1 = 0.9999... assumes that 0.333... is an exact representation of 1/3, but if it is not, then the proof is wrong. In fact, I see it more as proof that 0.3333... is not exactly 1/3 or else when multiplied by 3, it would equal 1.

Interestingly, when I calculate 1/3 on my cell phone calculator to get 0.333333, and then multiply it by 3, I get 0.999999. However, when I put in (1/3)*3 and calculate it, I get 1.

Now, I would write more, but I am a half hour late for class right now, and I need to go. But I will be back to add more later, so please do not think this is all I have to say. I also am not sure what signing is or how to do it on these articles/posts. ---- — Preceding unsigned comment added by Sweetnaivety (talk • contribs) 16:39, 27 October 2011 (UTC)

Let's see it another way, with an example that is related to powers of ten (a different intuition than the one you're using). Think of this equation: 0.9 = 1.0 This is equivalent to the equation 0.9999.... = 1.0000..., or saying that the substraction 1.0000... - 0.9999... is equal to 0.

Why can we think that it should be true? Look at the first decimal position, the one inmediately after the dot. It is a 0 in the 1.0 case, and a 9 in the 0.9 case. Does that mean that 0.9999... and 1.000... are at a distance of one tenth? Of course not! Their distance must be smaller, since you now that 0.9999... is bigger than 0.9. The distance within both numbers is not within the 1/10 range; we could say that you get different digits because the distance, if any, has been "carried" from a smaller decimal. So we get that the first decimal position, even though they have a different value, are not enough to say that 1.0 and 0.9 are at diferent places in the real line with a granularity of one tenth. A different representation has placed the two points within the same interval of length 1/10, somewhere between 0.9 and 1.

By the same reasoning, you can say that the second decimal position (the 0.09 added as the second element in the 0.9999... series, as compared to the 0.00 that you add to the 1.0000...) is not enough distance to separate 1.0 and 0.9 within a range of one hundreth; both numbers must be closer than 1/100.

So far we have a series of repeated comparisons betweens 0's and 9's decimals, and none of them have been enough to define a distance between the two numbers - that represented by 1.0 and the other represented by 0.9. Now the funny thing is that this comparison process can go on forever, because the series is infinite, and every time you won't be able to find any distance between both numbers! For you to find what is their distance, you'd have to finish the series to get the final '9' that would generate a distance. But because the process is in-finite (no finish), all you get is a process generating more and more digits, and none of them is able to set any distance apart within the two numbers. If you're not able to find any distance between two points in the real line, you must conclude that both of them are the same point, no matter that you followed two different paths to arrive to that point.

How good is this simple intuition for you? This is how I understand it intuitively, I hope it helps. Diego (talk) 18:32, 27 October 2011 (UTC)

Then what is 1.0 - 0.11111...? — Preceding unsigned comment added by Sweetnaivety (talk • contribs) 20:55, 27 October 2011 (UTC)

Short answer: 1.0 - 0.111... = 0.999... - 0.111... = 0.888... = 8/9. Some more comments: Firstly, your calculator is no proof either way because it uses a finite amount of digits; it simply cannot show anything whatsoever about 0.999... You also say 1 cannot equal 0.999... because "only 1 can equal 1". For example, 2/2 equals 1. 2-1 equals 1. cos(0) equals 1. There are a lot of terms that look differently from 1 but equal 1, and 0.999... is one of them. Finally, you say that long division of 1/1 never gives 0.999... That's true, but unsurprising and irrelevant. Long division is built in a way so that every calculation gives a single unique result. So no matter whether 0.999... equals 1 or not, we cannot expect long division of 1/1 to give one result part of the time, another at other times. Let me instead present you with this line of reasoning based on long division:

You will surely agree that long division yields that 0.999.../1 = 0.999... Now try the long division 1.999.../2 - and you will see that you still get a result of 0.999... Thus (1+0.999...)/2 = 0.999... = (0.999...+0.999...)/2, and multiplying by 2 and subtracting 0.999... yields 1 = 0.999... How is that?

As an off-topic remark, you can sign your posts by adding four tildes (~~~~) at the end. The wiki software will automatically turn that into your user name, a link to your talk page and a timestamp. That's how I sign, too. Huon (talk) 21:54, 27 October 2011 (UTC)

I think you've just destroyed his illusion that math is simple :-) Diego (talk) 07:17, 28 October 2011 (UTC)

To finish off that illusion, I should link to the Banach–Tarski paradox... Huon (talk) 10:49, 28 October 2011 (UTC)

Yet you are assuming that 0.999... = 1. What then, is 0.999... - 0.09? Without converting the 0.999... to 1 first?

I think the problem mostly is that we think that 0.333... x 3 = 0.999... when that is not true. 0.333... x 3 = 1, NOT 0.999... but there is no way to represent that 0.333... = 1/3 because 1/3 is not representable in decimal form, which is the whole reason the decimal goes on forever. If 1/3 could be properly represented in decimal form, then it would not become infinite, it becomes infinite because there is no end in trying to convert it to decimal form, which means it never actually becomes equal to 1/3, or else it would end. Sweetnaivety (talk) 16:10, 10 November 2011 (UTC)

Short answer first: 0.999... - 0.09 = 0.90999... You are correct that I assumed the equality 0.999...=1 in order to calculate 1 - 0.111..., but I need not do so. Consider the following:

1 - 0.1 = 0.9

1 - 0.11 = 0.89

1 - 0.111 = 0.889

..., therefore:

1 - 0.111... = 0.888...

There is no missing "9" at the end because there is no end to the eights. For a more formal proof, you could employ limits and use that 0.111... is the limit of the sequence (0.1, 0.11, 0.111, ...) while 0.888... is the limit of the sequence (0.9, 0.89, 0.889, ...).

Next, you claim simultaneously that 0.333... x 3 = 1 and that "1/3 is not representable in decimal form". That obviously cannot both be true. If 0.333... is the number which when multiplied by 3 equals 1, it is by definition the decimal representation of 1/3. I assume you meant that 0.333... x 3 equals 0.999..., but not 1.

The remainder of your argument says that an infinite decimal cannot properly represent anything. We could argue whether the number represented by 0.333... is indeed 1/3 or whether it is a little less than 1/3 (and it's the former, by the Archimedean property line of reasoning I gave above). But now you seem to say that 0.333... simply does not represent any number. Why not? Why should we disallow an infinite decimal to represent something? After all, the entire point of having decimals in the first place is to have them represent certain real numbers - and the claim that just by being infinite they can no longer do so seems more of a philosophical than a mathematical objection. And it leaves you with lots of representations which do not represent anything, and conversely with lots of numbers without representations. That strikes me as rather unhelpful. Huon (talk) 17:17, 10 November 2011 (UTC)

I did not disagree that 1 - 0.111... = 0.888...

I also realize that my statement was contradictory, that 0.333... x 3 = 1 yet that 0.333... is not a perfect representation of 1/3. But if 0.333... is actually a perfect representation of 1/3 then that just means that our brains cannot comprehend the true value of an infinite decimal and we do not realize that 0.333... x 3 = 1. Though I still think that an infinite decimal is really not perfect representation of the fractions they represent. But either way, I still believe 0.999... =/= 1

However, if 0.999... - 0.09 = 0.90999... then how can 0.999... = 1 since 1 - 0.09 = 0.91?99.153.244.196 (talk) 08:40, 13 November 2011 (UTC)

Because 0.90999... = 0.91, as your own symbol manipulation shows.

BTW, a finite decimal is not a perfect representation of the fraction, but an infinite recurring decimal is. Your problem seems to be that you have no intuition about the properties of an infinite recurring decimal, and you keep treating it as if it was a finite-but-very-long one. Always remember that infinite objects can have different properties than all the finite objects that you find along the way.

Also, don't think too hard about the true value of a number. Mathematical objects have a meaning according to the rules each particular mathematician wants to use. If you change the rules (axioms), you also change the meaning of the resulting objects. Diego (talk) 09:52, 13 November 2011 (UTC)

"Equivalent" does not mean "Equal"

0.999... is equivalent to 1 in real number equivalence. They are NOT truly equivalence in mathmatical sense. We MUST use the curly equal sign to denote their equivalence, not the normal equal sign. If we mix up the two, then we can come up with a whole lot of bullshit.

Ex: If 0.999... = 1. Then 0.999... + 0.0...1 = 1 + 0.0...1 -> 1 = 1.0...1.

Since we already know 0.999... = 1, replace 1 in the above equaltion and you get 0.999... = 1.0...1.

Repeat the step infinitely, then you prove that 1 = infinity. — Preceding unsigned comment added by Ssh83 (talk • contribs) 21:10, 1 November 2011 (UTC)

This line of reasoning fails because 0.0...1 is not a well-formed decimal representation. A decimal has one digit after the decimal separator for every natural number - a first digit, a second digit, a third digit and so on. To what natural number is that last "1" in 0.0...1 supposed to correspond? Huon (talk) 21:35, 1 November 2011 (UTC)

As a matter of fact, Ssh83 has it backwards. According to standard mathematical terminology, 0.999... and 1, just like 1/2 and 0.5, or 3^2+4^2 and 5^2, are, in each case, both equal and equivalent. FilipeS (talk) 15:03, 6 February 2012 (UTC)

But you missed one small detail...

0.9999... is not equal to 1. Remember when you were in primary school? Well, in my primary school, they taught that a number can only be represented in one and only one way as a decimal. Anything they teach in schools is true. Sure 1 can be represented in many different ways, but as a decimal, it can only be written only one way. I am aware that 1, 01, 1.0000..., and 1.0 are the same number, but that is still the same way of representing a number in decimal form, becuase (if you remember in primary school again) if you have a number with any amount of digits in it, then there will be an infinite amount of zeroes both behind it, and after the decimal point. — Preceding unsigned comment added by Nominalthesecond (talk • contribs) 19:16, 25 November 2011 (UTC)

If they taught you in primary school that decimal representations are unique, they were wrong. If 0.999... does not represent 1, what else should it represent? What did they teach about 0.999... in your primary school? Somehow I doubt they mentioned infinitesimals or non-standard analysis. Huon (talk) 20:25, 25 November 2011 (UTC)

Please do not deify schools. They are often woefully wrong. If you are interested in knowing things that are true, you should spend more time researching the things that you "know", and less time parroting the "facts" you learned in school. --COVIZAPIBETEFOKY (talk) 21:09, 26 November 2011 (UTC)

I think it's possible that you two need to take your irony detectors in and have them adjusted.... --Trovatore (talk) 21:11, 26 November 2011 (UTC)

Clear example why 0.999... != 1

Add 0.1 to 0.999...

0.1 + 0.999... = 1.099...

Now

1.099... mod 1.1 = 1.099...

Try the same with 1

Add 0.1 to 1

1 + 0.1 = 1.1

Now

1.1 mod 1.1 = 0

What now.... :)

Identity:9C7C71F43DD9DC1E604F3F21BB760A402630D9B8404B33587193D16CE90439F25C1757C2FE6C10C612BC0BF0A3CBD1FBEB5FF20B2C4F2350C85C0CBBA523464C — Preceding unsigned comment added by 129.97.236.182 (talk) 05:39, 2 February 2012 (UTC)

1.099... mod 1.1 = 0. Care to show how you got a different result? Gustave the Steel (talk) 06:49, 2 February 2012 (UTC)

Sure, you assumed 1.1 == 1.099... and then got the result of 0, don't assume that and actually do the division using any division algorithm, You'll get 1.099...

I didn't assume anything. The equality of 0.999... to 1 has been proven, and the equality of 1.099... to 1.1 is a logical consequence of that. Gustave the Steel (talk) 15:55, 1 March 2012 (UTC)

Its true that 1.099... will act in much the same way as 1.1 in any conceivable operation other than modulus(only one I can think of at the moment).

Saying that 0.999... = 1 has with it much the same hazards as assuming 1/infinity = 0,

You're assuming that a real number can be created by dividing by something that isn't a number, and that an infinitesimally small value, regardless of its composition, is the same as the lack of a value.

By that logic x = 0.00...1 = 0.00...2 = 0.00...3 = 0.00...9 = 0.0...10 and so on, all of which equal 0

Or equivalently

That

x = 1/infinity = k/infinity,

Take the limit as K approaches infinity and you end up with indeterminate, but we've already said that 1/infinity = 0

Thus

x = 1/infinity = k/infinity = k * 1/infinity = k*0 (Which applies because you assumed that 1/infinity = 0)

Take the limit as K approaches infinity and you end up with 0, not indeterminate,

This problem arises because

0 != indeterminate != infinity

0 is a concept itself, granted, but you cannot simply equate 1/infinity to 0, and likewise 0.999... to 1 It works for all practical purposes I can think of other than modulus but it's not theoretically correct.

It is however correct to say that: The limit as n -> infinity 1/n = 0 or The limit of n/1 as n approaches 1 is 1 — Preceding unsigned comment added by 129.97.236.182 (talk) 10:05, 2 February 2012 (UTC)

Precisely. 1/infinity is not zero, but infinitesimal. See 0.999...#infinitesimals. Tkuvho (talk) 13:56, 2 February 2012 (UTC)

Not quite. In the context of the real numbers, 1/∞ is undefined (because ∞ is indeed not a real number), not infinitesimal and not zero. There are enriched number systems which contain a type of "infinity" for which 1/∞ = 0, and there are other enriched number systems for which 1/∞ is some infinitesimal (though in the latter case you usually would not denote an infinite number by "∞" because there tend to be many different infinite numbers). Anyway, I don't see what this is supposed to tell us about 0.999..., which clearly is a real number. Huon (talk) 14:29, 2 February 2012 (UTC)

Hi Huon. Nice to hear from you. I have the feeling that all I have to do to bring you back to wikipedia is... mention infinitesimals. Note that John Wallis, who first introduced the "∞" symbol, used it as an infinite integer. He also denoted a typical infinitesimal by... precisely 1/∞. Tkuvho (talk) 15:12, 2 February 2012 (UTC)

What do you mean by "1.099... mod 1.1"? Usually both arguments of the modulo operation are integers. FilipeS (talk) 16:24, 24 February 2012 (UTC)

From the article you linked to: "Although typically performed with a and n both being integers, many computing systems allow other types of numeric operands." Gustave the Steel (talk) 15:55, 1 March 2012 (UTC)

Thanks. Overall I agree with you, but I'm a little confused. You wrote that 1.099... mod 1.1 = 0. I can accept that, if we assume that the remainder must be an integer. But if we're using a fractional base then we should probably allow fractional remainders... FilipeS (talk) 17:07, 7 April 2012 (UTC)

We do allow fractional remainders, but since 1.1=1.0999..., the remainder in this case is still zero. Calculating modulo 1.1 gives the quotient group ℝ/1.1ℤ. Huon (talk) 17:29, 7 April 2012 (UTC)

Why .999... does NOT equal 1

This argument will go on for a long time...all because we use base 10. Anyway...

.9999999999....=1-.0000000...1 and .9999999999...+.0000000...1=1 — Preceding unsigned comment added by Phillies9513 (talk • contribs) 00:53, 24 February 2012 (UTC)

Am I right? You cannot deny that this isn't correct.

Plus, the common proof that since 1/9=.11111... and so on until 9/9=.99999999=1 is false because the only reason that 1/9=.11111... is because that is the closest number you can get to a number multiplied by 9 that will equal 1, in base 10. So i'm saying that 9/9 does equal 1 (duh) but 9/9 does not equal .999999.... and so therefore, 1 does not equal .999999999... — Preceding unsigned comment added by Phillies9513 (talk • contribs) 00:52, 24 February 2012 (UTC)

Hi! I'm afraid you're treading well-worn territory here, so much that Wikipedia has made a page that is specifically for discussing counter-arguments to the proposition 0.999... = 1. Please see Talk:0.999.../Arguments. Please could you the FAQ at the top of that page as well? There are also ten previous archives of the arguments talk page, where you can see how similar counter-arguments fared. -- The Anome (talk) 00:59, 24 February 2012 (UTC)

Sigh, if only both of you could have just read the article and seen that this is already provided as a count point. It's not however because we base our number system on powers of ten, the same would be true with a number system based off powers of 12 or anything else, it is because we base our number system on real numbers which don't include non-zero infinitesimals.AerobicFox (talk) 04:19, 24 February 2012 (UTC)

In kindergarten I was told that 2 comes just after 1.Now I know that's not true,because there is 1.5,but 1 does come just after 0.999999999999............ There's just no other number between them!Think of 1/6. It equals 0.1666............6666666667 — Preceding unsigned comment added by 86.104.51.117 (talk) 15:59, 21 May 2012 (UTC)

It is evident that you really don't understand repeating decimals. There is no last decimal in the expansion of 1/6. The exact decimal is 0.16666..., with a '6' in every finite position after the first. Phiwum (talk) 01:24, 22 May 2012 (UTC)

Every number minus itself =0 1-0.999...=0.000.....000....01 so 1 doesn't equal 0.999... — Preceding unsigned comment added by 188.25.224.87 (talk) 08:40, 23 May 2012 (UTC)

Wouldn't the equation 1-0.999...=0.000.....000....01 imply that 0.999...=0.999...999...99000...? Since 0.999... has nines all the way, there is no place in 1-0.999... where anything but a "0" could occur. Huon (talk) 09:53, 23 May 2012 (UTC)

So you mean that 0.000.....01=0 You say that 1=1.000....01.If 0.999...=1 then 0.999...=1=1.000...01 so 0.999...=1.000...01,but this is false because there is a number between them.So 0.000...01 NOT = 1,then 1-0.999... NOT = 0 and so 0.999... NOT = 1 — Preceding unsigned comment added by 188.25.220.81 (talk) 18:59, 23 May 2012 (UTC)

What I mean is that 1-0.999... does not equal 0.000...01. If 0.000...01 is supposed to have finitely many zeroes, it is larger than 1-0.999..., and if it's supposed to have infinitely many zeroes, it's not well-defined; for example it's not the decimal representation of a real number. (To pre-empt Tkuvho, it's not a representation of a unique hyperreal number either; at best it's an entire class of hyperreals, and it's still not the difference between 1 and the one hyperreal 0.999... whose decimal representation has nines all the way. If you're no fan of the hyperreals, you can safely ignore this side note.) Huon (talk) 19:18, 23 May 2012 (UTC)

Take x=0.999... and y=1. 2x=1.999..98 (that is 1.999... - 1/infinity) and 2y=2 Because there is a number between them those two are not equal so 2x≠2y. Therefore x≠y and 0.999...≠1