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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://wiki.riteme.site/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)

Now, I cannot say I have a math degree, or any university degree. However I take great interest in mathematics and find this problem particularly interesting. I found this quite interesting as well. Say perhaps, the notion of 0.999... = 1 is incorrect. Would it not be correct to then say that 0.999... + 0.000...1 = 1? Removing that 0.000...1 would create a difference between the two. Rendering them unequal.

There are some strange number systems where 0.999... and 1 differ (those are not the standard real numbers but systems like the surreal numbers or Richman's decimal numbers where a number is, by definition, uniquely given by a decimal representation). In most of those number systems, 0.000...1 is not defined - either no number exists that, when added to 0.999..., gives 1, or that number does not have a decimal representation at all. That's because we cannot say where the "1" in 0.000...1 would be - since there's no "end" to decimal representations, we'd have 0.000...1=0.000...1000..., and conversely 0.999... would have to equal 0.999...9000... - that's not what 0.999... is usually taken to mean. There is the hyperreal approach of Katz where "0.999..." denotes any of a set of numbers of the type "0.999...9000..." with some fixed infinite amount of digits, and then "0.000...1" would indeed be a hyperreal number with the "1" in a certain place following infinitely many zeroes (or more precisely, a class of such numbers corresponding to the class of numbers Katz all calls "0.999..."), but that's pretty much a fringe view. The lone hyperreal number which has nines all the way is still equal to 1. So in short, in number systems where 0.999... does not equal 1 there is no need for the difference to even exist, much less be representable by a decimal representation. Huon (talk) 16:18, 28 January 2013 (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)

1 is not an "infinitely repeated number" and "Harlequin" obviously did not know anything about mathematics. — Preceding unsigned comment added by 91.185.1.174 (talk) 12:56, 6 September 2012 (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)

To what natural number does your last "9" in 0.999... correspond to? And why should it correspond to any natural number? What makes you think it should correspond to any particular natural number? What if he tells you that there is no last "1" because "..." means forever? I am surprised you think 0.999... is a well-formed decimal. I can't see it being any more well-formed than 0.0...1. 173.11.135.234 (talk) 20:31, 28 January 2013 (UTC)

There is no last "9" in 0.999..., they indeed go on forever. But 0.0...1 has a "1" (obviously), so where is it supposed to be? To which natural number does that digit correspond? Huon (talk) 21:51, 28 January 2013 (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)

You are right when you say I cannot deny that this isn't correct. --Trovatore (talk) 21:27, 4 November 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)

You're telling me that 1/6 = 0.1666...667? Ok, well if we assume that, then you can multiply your 0.166...667 (a finite decimal expansion, since otherwise that representation has no meaning, as Huon has already mentioned at length) by 6. Do you get 1? No, you get 1.000...0002, another finite decimal expansion. Hence, 0.1666...667 cannot equal 1/6, pretty much by the definition of what "1/6" is. 109.175.245.125 (talk) 19:48, 30 January 2013 (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 — Preceding unsigned comment added by 188.25.224.223 (talk) 18:22, 12 June 2012 (UTC)

The problem with that approach is that there's no last nine in 0.9999... which could produce the "8" in 1.999...98. If the decimal representation of 2x were indeed 1.999...98000..., then 0.999... would actually be 0.999...99000..., but it's nines all the way. Since the decimal representation of x=0.999... doesn't end with zeroes, neither does that of 2x, and there's no place where that "8" would appear: It's still nines all the way, and 2x=1.999... = 1+x. Huon (talk) 18:46, 12 June 2012 (UTC)

Decimals and incomplete division

YOU DO NOT UNDERSTAND DIVISION.

A REPEATING DECIMAL IS THE QUOTIENT PORTION OF AN INCOMPLETE DIVISION - THE REPEATING DECIMAL DOES NOT INCLUDE THE REMAINDER - EVEN AS IT BECOMES INFINITELY SMALL - THAT KEEPS THE QUOTIENT FROM FULLY EXPRESSING THE VALUE OF THE FRACTION.

1/3 IS NOT EQUAL TO .333... BECAUSE .333... IS ONLY THE AMOUNT THAT IS REPRESENTED IN THE QUOTIENT.

PLEASE DO NOT DISMISS THESE FACTS BECAUSE YOU ARE NOT YET ABLE TO UNDERSTAND THEM. — Preceding unsigned comment added by Mjs1138 (talk • contribs) 23:37, August 24, 2012 (UTC)

I have moved your comment from the middle of a two-year-old discussion to this section so it's easier to find the newest contributions without wading through the old comments and even fracturing others' comments.

I don't think your definition of repeating decimals is correct. Repeating decimals are a special case of decimal representations and as such are defined via sequences. The real numbers can be defined as the equivalence classes of Cauchy sequences of rational numbers (where two Cauchy sequences are defined to be equivalent if their difference converges to zero within the rationals), and the decimal 0.a₁a₂a₃... represents the real number given by the Cauchy sequence (0, 0.a₁, 0.a₁a₂, 0.a₁a₂a₃, ...) (for the sake of brevity I'll omit the proof that that's indeed a Cauchy sequence of rational numbers). The rational numbers can then be identified with the subset of real numbers given by constant sequences (x, x, x, ...). To show that 0.333... equals 1/3, I then have to show that the sequence (1/3-0, 1/3-0.3, 1/3-0.33, 1/3-0.333, ...) converges to zero. Again I'll omit the proof for the sake of brevity, but I can easily add it if required.

On the dispute resolution noticeboard I provided an outdated but still reliable source which gives a closely related definition: von Mangoldt, Hans (1911). Einführung in die höhere Mathematik. Vol. 1. Leipzig: Verlag von S. Hirzel. p. 148. I'm confident most other textbooks on analysis will give equivalent definitions; for example I'm pretty sure Jean Dieudonné's classic Foundations of Modern Analysis (1960) gives a rather similar definition, but I don't have that work at hand right now. If you want to give another definition via "incomplete division", please provide a reliable source so I can look up your definition. Huon (talk) 01:14, 25 August 2012 (UTC)

Wikipedia is wrong to say "the symbols "0.999..." and "1" represent the same number"

I know you guy are sick to death of people saying this but I do believe I'm speaking sense on this one.

The problem here is that people are visualising 0.999... as a finite concept and use finite equations to link it to a finite number (1). Although it is a finite number in essence, the representation uses a infinite concept (this is because each digit represents as value and it is the infinte chain of these values that give the number it's identity). Now this is were the equation 0.999...=1 provokes Galileo's paradox, because that we're essentially trying to do is force a infinite concept into a finite value, because taking infinity as a fixed concept always leads to contradictions somewhere along the line that just don't make snese.

Although 0.999... does appear to be exactly where 1 is what we don't take into accound is the concept that 0.999... is not the same type of number, it is construced using infinity, so as I've already said the two, following this logic, never be mixed. There are other paths of logic to follow of course that do show 0.999... to be equal to 1 but as to many things the desired 'answer' can be different depending on which logic you follow (see Exponentiation#Zero to the power of zero), which is the point I'm trying to get across. Wikipedia should never adapt one angle of view and use it as their dependance; doing as is no better than bias. All views should be taken into acount and the focus should be on the conflicts in arguments. Robo37 (talk) 22:52, 4 October 2012 (UTC)

First of all, Wikipedia should report what the sources have to say, and the vast majority of mathematical textbooks indeed agrees that 0.999...=1. There are some few exceptions where "0.999..." and "1" do not denote the standard real numbers (ie even "1" no longer denotes what you'd expect it to denote!), and the article does mention them in the section about alternante number systems, giving them at least as much weight as they deserve from their prominence in the literature. But unless you don't care about subtraction or believe that 0.999... should be not one number, but an entire class of different hyperreal numbers, we can safely ignore them. The authors of these papers for different number systems agree that within the real numbers 0.999...=1 still holds. And whereas zero to the power of zero can be defined differently depending on the context, within the real numbers there's no coherent way to define 0.999... as anything but 1.

Next, let me shortly discuss (and dismiss) Galileo's paradox. That paradox deals with different countably infinite sets and the question of whether those sets are of the same "size" or not. It does not deal with a comparison of finite and infinite sets, and I don't see where the question of whether two countably infinite sets are of the same "size" is relevant to 0.999... - I don't see any non-bijective map between countably infinite sets in this context.

I believe you're mistaken about how 0.999... is defined. Correct me if I'm wrong, but you seem to believe that 0.999... is obtained by adding 0.9 + 0.09 + 0.009 + ..., the result of an never-ending summation of ever smaller numbers. (If that's not what you believe, please explain what you meant by the "infinite concept" you refer to please be as precise as you can.) Admittedly that's what the notation suggests, but that's not how it's defined. If you're interested I can give a full definition based on Cauchy sequences of rational numbers, but that's rather tricky and proably not really helpful. The important point is that despite its looks, 0.999... is a number just as finite as 1, defined without any use of a number "infinity" or an endless process. In fact, both 0.999... and 1.000... (which of course is usually written as just "1") are the results of analogous definitions, and if you contend that one of them is an "infinite concept" that gets its value from infinitely many digits, the same is true for the other (only that the infinitely many digits for 1.000... are all zero instead of all 9, and that by convention we usually omit trailing zeroes after the decimal separator - that doesn't mean these digits don't exist).

My suggestion would be to read up on decimal representations in an introductory calculus textbook, or to have a look at our own decimal representation article. That article is a little shorter than I'd like because it assumes familiarity with the real numbers instead of introducing them via Cauchy sequences; the 0.999...#Cauchy sequences section actually gives a rather good introduction to that topic. Huon (talk) 02:02, 5 October 2012 (UTC)

What the.....?

I am a high school student and In class we had to change 0.9999........ in to a fraction. When I done this I got 9/9 which is 1 and I got confused and thought I might research about it. I simply can't grasp the fact that 2 numbers equal the same number, I'm no mathematician but It simply makes no sense. I also think that it can't possibly be an infinite decimal otherwise it could never be represented by a fraction but reoccuring decimals can never be accuratly written (just an idea). Danjel101 (talk) 12:02, 10 October 2012 (UTC)

Work harder on grasping the fact, since it is true. There are many representations for each number. Recurring decimals can always be accurately written, since all recurring decimals are representable as fractions -- probably by the method you just used in your class. --jpgordon^{::==( o )} 22:25, 10 October 2012 (UTC)

I'm no mathematician either, but using two representations to represent the same number is actually pretty common in the decimal number system. For example, we have no problem understanding that 1.000... is also the same as 1. --87.82.207.195 (talk) 11:49, 27 October 2012 (UTC)

But in your math class, didn't you also convert 9/9 into 1 without a problem? The simple fact is that there is always another name for any number, and 0.999... is one such name for 1. It's the same as 9/9 or 5-4 or 4.159234x10^0, and one simply has to learn how repeating decimals are simplified into rational numbers.

4.159234×10⁰ = 4.159234, not 1. But 4.159234⁰ = 1. Double sharp (talk) 03:37, 13 November 2012 (UTC)

0.999... is an ill-defined concept

Please DO NOT delete this section. It belongs here. I know you do not like it because it refutes all your silly arguments. But this is an arguments page,... https://www.filesanywhere.com/fs/v.aspx?v=8b6966895b6673aa6b6c — Preceding unsigned comment added by 60.1.42.128 (talk) 03:25, 23 January 2013 (UTC)

I'm no expert on the number-theoretical approach to the natural numbers (I prefer the Peano axioms myself), but I believe you haven't really understood set theory. For example, you seem to confuse the fact that every set has the empty set as a subset with the fact that the set $\{\{\}\}$ contains the empty set as an element: While for every set M we have $\{\}\subset M$, we also have $\{\}\in \{\{\}\}$. This doesn't hold in general: The empty set is a subset of the set {Knife, Spoon, Fork}, but it's not an element.

Seems to me you don't understand much at all. If you had read the article correctly, you might have noticed that I mentioned the confusion arising from the empty set being an element as well as a subset. So, perhaps a reread? 60.1.42.128 (talk) 08:12, 23 January 2013 (UTC)

Your proof on 0.999... and 1 suffers a very basic flaw: Not everything that's true for every element of a sequence must be true for the limit. Of course $S_{n}=\sum _{k=1}^{n}{\frac {9}{10^{k}}}<1$ for every n, but this property does not translate to $\lim _{n\to \infty }\sum _{k=1}^{n}{\frac {9}{10^{k}}}$. There's an easy counterexample that shows this line of reasoning is invalid: We also have $\sum _{k=1}^{n}{\frac {9}{10^{k}}}<\lim _{m\to \infty }\sum _{k=1}^{m}{\frac {9}{10^{k}}}$ for every n, but obviously we do not have $\lim _{n\to \infty }\sum _{k=1}^{n}{\frac {9}{10^{k}}}<\lim _{m\to \infty }\sum _{k=1}^{m}{\frac {9}{10^{k}}}$. Huon (talk) 04:46, 23 January 2013 (UTC)

No, no. Of course we do not have$\lim _{n\to \infty }\sum _{k=1}^{n}{\frac {9}{10^{k}}}<\lim _{m\to \infty }\sum _{k=1}^{m}{\frac {9}{10^{k}}}$. But that is not what the article says! May I suggest a reread? Your analogy is completely irrelevant. There is no comparison between your statement and conclusion and the article itself. 60.1.42.128 (talk) 08:12, 23 January 2013 (UTC)

(Edit conflict) I believe there are numerous flaws in the arguments just in the first few pages of the posted pdf, but that is not really relevant, because it appears to be arguing that 0.999≠1 if you change the basic definitions and number system. This is obvious, and the changes you propose (I assume you are posting here to propose changes) would have no more effect on this article than thousands of others which all use the widely accepted real number system. There is certainly no good reason this article should use a fringe definition when all the articles it links to such as repeating decimal do not.

What you "believe" is quite frankly not important.

Furthermore, I am not proposing ANY changes to the article. In case you had not noticed, this is the ARGUMENTS page. I do not for one moment believe that ignoramuses would change their wrong opinions. Let me ask you this - why don't you delete all the other sections that are no arguments at all? On a power trip here my little man? 60.1.42.128 (talk) 08:12, 23 January 2013 (UTC)

Actually, I will address a few obvious flaws:

1. In the argument on page 1 and 2 about the set definition of cardinal numbers (which shows how far from relevance to the 0.999... article it is) seems to be completely based on ignoring the distinction between an object and its contents, and incredulity that an abstract definition doesn't seem to be physically measurable. To put it in simple terms, if I have an empty bag inside a box, is the box empty?

An object and its contents you say? Who said anything about contents? Your analogy is obviously flawed: An empty bag is not the same as an empty box. To put it in simple terms for you - if I have an empty bag inside another empty bag, is the bag empty? Oh wait, which bag?! 60.1.42.128 (talk) 08:12, 23 January 2013 (UTC)

1. The real numbers are unnecessary to 0.999=1. The rationals are sufficient.
2. The top of page 8, it is assumed that a limit obeys inequalities members of the sequence obey. There is no basis given for this assumption, and in fact is false in the standard definition of the real numbers. I have tried to put into words what areas of maths this assumption breaks, but the simplest one is to say that it is not a useful (or standard) definition of limits.

You might try rewriting the paragraph. It makes little sense. A limit DOES apply to inequalities as much as it applies to equalities.

60.1.42.128 (talk) 08:12, 23 January 2013 (UTC)

1. The paragraph with red text on page 12 is again purely an argument on incredulity - if the algebra says two symbols are equal, but they don't FEEL equal, something must be wrong, so it must be modern academia. The exact same argument could just as easily be used with 2/2 = 1 showing that fractions are not really numbers.

Oh nonsense again. The red paragraph makes sense. What you have written does not make any sense. Again, your analogy is almost as poor as your understanding. 60.1.42.128 (talk) 08:12, 23 January 2013 (UTC)

1. An overall argument seems to be baseless claims that things are poorly defined (limits, 0.999..., sets, maths in general). A particularly circular argument continued across much of the paper but most obviously at the bottom of page 12 seems to be that 0.999... is not a number and therefore proofs that involve it are nonsensical and therefore nothing can be proven about it so it is not a number.

Do you even have any idea what a Circular argument is? 0.999... is not a number, and if it is not a number, then the algorithm would obviously produce weird results. Garbage in, garbage out. I can tell that this article has gotten under your collar quite a bit. I am sorry. The intention of posting the link here is so that others who are not trapped in group think like yourself, might have a chance from a different perspective - the correct one. The link is not annoying anyone else except you and all the other little Wiki puppets. Why don't you let others decide what they would like to read? Has anyone complained yet? Is there anything that is offensive or inflammatory? So, just leave this section as it is please. People do not need you to think or protect them. 60.1.42.128 (talk) 08:12, 23 January 2013 (UTC)

--Qetuth (talk) 04:52, 23 January 2013 (UTC)

Uh, I don't understand your response there. I have not deleted anything of yours that I am aware of, and the only one who has given any argument on where this belongs, or said anything offensive or inflammatory, is you.

• On point 1, I said bag and box to give two different names to avoid confusion. But feel free to consider bag A inside bag B. Bag B is clearly not empty. Although rereading the section I feel the far bigger issue is the constant irrelevant claims about magnitude and dimension in a defintion of cardinality.
• On point 3 about limits, I think Huon said it much clearer and all you say in response is "that is not what the article says." The article directly assumes, without any explanation, that if each element of a sequence is less than 1, then the limit, in effect the least upper bound, of that sequence must also be less than 1. Without that line there is no contradiction on page 8 and everything is fine. Since I have not seen a definition of limit which works that way and the article has not provided one, all I can do is say that this is not true by the standard defintion (specifically, in real, rational, and virtually all other commonly used continuous number systems - it would be true in eg cardinal numbers).
• On page 12: Okay, what happens if I put the 'object' 10/10, which I claim is not really a number, into the algorithm on page 11? I get 1 as output. But the input must by definition be the same as its output. So I should be suspicious that something is wrong with the input, and hence 10/10 is not always a number. My point: No part of the argument given on p11/p12 relies on any property of 0.999... besides the fact that if it IS a rational number it is equal to the rational number 1, and the claim that it is not always clearly defined. Both of these could be applied to other representations of 1. Using that as a basic of an argument to show that 0.999... is not clearly defined is the very defintion of a circular argument.

Okay, looking at your conclusion, I must quote: "But this is an arguments page,...". You posted an article to an arguments page with, to me, some obvious logical flaws, unexplained assumptions, and arguments from incredulity. I argued. Why? I suppose because I DON'T want people to be "trapped in group think" - I hope that more people, you or other readers of this, would learn to question what they read, and avoid logical fallacies. Anyone is free to read what you post, they are also free to read what I post, and I would hope they would be willing to decide what to think for themselves, but I know from experience many don't. --Qetuth (talk) 11:44, 23 January 2013 (UTC)

Magnitude and number (not dimension - it would help if you could read properly) have nothing to do with cardinality. Cardinality is a set-related concept. While cardinality requires that the prior concept of number already be established, number does not require cardinality.

Your analogy of bags is so hopelessly flawed that one can't even know where to begin explaining. If a set is empty and it can contain another empty set as an element, then it is no longer an empty set. It is an ill-defined object that contains a subset and an element that are identical.

10/10 is equal to 1, so you would get 1. But 0.999.. is not equal to 1. What part of this are you unable to comprehend? k/k is ALWAYS a well-defined number, provided k is measurable.

You are the one who needs to avoid logical fallacies. Your lack of reasoning is absolutely pathetic. 173.11.135.234 (talk) 09:21, 25 January 2013 (UTC)

I'll try to focus on the 0.999...≠1 proof because I believe the entire set theory discussion is rather irrelevant to it - no matter how we define them, we probably all agree on what the natural numbers should be.

On page 7 you argue (and I agree) that for all natural numbers k we have 1-1/10^k<1. So far, so good. You then state (and again I agree) that "academics" will point out that we need to consider the limit. It's rather obvious that there's no natural number k such that 1-1/10^k=0.999..., so we indeed need to do something more.

At the top of page 8 you write: $\left(\lim _{n\to \infty }\left(1-{\frac {1}{10^{n}}}\right)<1\right)\Rightarrow (1<1)$. I still agree with the implication: If we had $\lim _{n\to \infty }\left(1-{\frac {1}{10^{n}}}\right)<1$, then we would arrive at the obviously wrong conclusion that 1<1. So the premise must be wrong, and we arrive at $\lim _{n\to \infty }\left(1-{\frac {1}{10^{n}}}\right)\geq 1$ and thus 0.999...≥1.

I had assumed (maybe wrongly) that you intended to argue that since 1-1/10^k<1 for all k, we must also have $\lim _{n\to \infty }\left(1-{\frac {1}{10^{n}}}\right)<1$. This implication is wrong. If for some sequence (S_k) we have S_k<x for all k, all we can say about the limit is that $\lim _{n\to \infty }S_{n}\leq x$. I gave an explicit counterexample to the wrong implication that's a direct analogy to what I believed you intended to argue: For all k we have $1-{\frac {1}{10^{k}}}<\lim _{m\to \infty }\left(1-{\frac {1}{10^{m}}}\right)$, but the fact that all elements of our sequence are less than $\lim _{m\to \infty }\left(1-{\frac {1}{10^{m}}}\right)$ does not imply that the same is true for the limit of the sequence. So there's no reason to believe that the fact that all elements of our sequence are less than 1 should imply something about the limit being less than 1. Huon (talk) 17:35, 23 January 2013 (UTC)

Um, no. The premise is correct. The problem lies not with the premise, but the concept of limit. Your inference that somehow 0.999...>= 1 because of a wrong premise is quite shocking. I'll let you think about it.

"...but the fact that all elements of our sequence are less than $\lim _{m\to \infty }\left(1-{\frac {1}{10^{m}}}\right)$ does not imply that the same is true for the limit of the sequence."

In the same way, the lim (9/10+9/100+...) does not imply that the sum 9/10+9/100+... is the same as the limit of the sequence.

Look, I know that you find it absolutely intriguing that I chat with you on these pages. But really, I do not want to nurture a discussion with anyone. My suggestion is that you do not remove the link to the article, so that others can learn a view that is different to yours.

I do not care for your opinion or any of the Wiki sysop or admin opinions. So, please hold your two cents. I know your flawed arguments well. I also know that I am correct and have no need to change your opinion which is of no significance to me. Your opinion is all over the article. So take a deep breath, relax and learn that NO means NO. I do not want to have a discussion with any of you. 173.11.135.234 (talk) 09:29, 25 January 2013 (UTC)

.999... is just a decimal representation of the 10th division of a number being itself divided into 10 and so on recursively. It's not something to flip out over. (((1/10)10)-1/10)+1/10(((1/10)10)-1/10 +1/10) Which is taking 1 and dividing it into 10 equal parts then taking the 10th part and dividing that. Simplified 9/10 + 1/10(9/10 + 1/10) This is clearly one. But if we keep sub dividing the 10th part into the next 10th subdivision... 9/10 + 1/10(9/10 + 1/10(9/10 + 1/10)) = 9/10 + 9/100 + 9/1000 + 1/1000 and if we kept this up indefinitely we would get 9/10 + 9/100 + 9/1000 + 9/10000...So this is just taking 1 and dividing it up in a different way. Nevermind does it add up to 1, we never left 1, it was always 150.136.175.142 (talk) 18:57, 28 October 2015 (UTC)

Simple Visual Proof

For anyone who may benefit from this, here is a page that was built after a lengthy discussion over on the Talk page: Visual proof of 0.999... = 1

What a joke! Your page does not contain any proof - visual or otherwise. Do tell what "ad infinitum" means. As a homework assignment I would like you to imagine that you are immortal. Then try out the "ad infinitum" process you so ignorantly talk about. Tell me about your conclusions and especially when do you reach a stopping condition. Infinity is an ill-defined concept of your irrational mind. 173.228.7.80 (talk) 02:57, 5 February 2013 (UTC)

Keep the crazy talk to your topic and your site, Mr. Gabriel. Gustave the Steel (talk) 05:18, 5 February 2013 (UTC)

Imagining immortality is no problem. Just imagine that on every day, you know for sure that you will live to see the next one. You never actually reach infinity years old, but neither do you ever die. It's the same with 0.999... (well, assuming we're taking the standard definition of 0.999...): you never reach the infinitieth or last nine, but you never stop either. This infinity is more of a potential infinity than an actually realized infinity. Double sharp (talk) 13:50, 2 January 2015 (UTC)

Mr Gabriel (I assume), thank you for the feedback. Infinity is certainly a very difficult concept for the human mind to grasp. Some would say impossible. I expect that we can agree that there is no stopping condition for infinity. I was not ever intending to indicate that there was. But what does happen is mathematical convergence. So while the series never stops, the sum does converge to a definite number. You are very critical of the term "ad infinitum", but my purpose in using that term was specifically to indicate that the series never stops.

If you have any suggestions for improvement, I would be appreciative of that.

And if all anyone has to offer is negative feedback, I appreciate that as well, because there are always ways to use feedback constructively.

As for whether or not a proof is attained, it is clear to me that both images stand as proofs. I am totally open to the possibility that I am mistaken in this understanding, but so far no one has presented me with evidence that has persuaded me to change my view on that. The Square is offered as proof that this infinite series converges to 1/3. The circle is offered as proof that the 0.999..., when broken down into an infinite series, converges to 1. No one has to accept the validity of these proofs. And for anyone who wants to convince me that I am mistaken, then I would require a criticism that is more substantial than an appeal to emotion.--Tdadamemd (talk) 09:17, 5 February 2013 (UTC)

Sorry Tdadamemd, but you have just done the same thing visually that all these other 'proofs' do - you have your conclusion hidden as an assumption within your proof. In the circle example, every cut you make has an 'inside' and an 'outside', and only when these are added together does the circle equal one. Every cut is defined as excluding part of the circle. But then you propose that continued to infinity, there would be a cut that contradicts what you just defined and would exclude nothing. But you give no reason why this last cut would be any different from the others - you just declare the answer that you are supposedly proving. So the proof fails like all the others.Algr (talk) 07:05, 11 February 2013 (UTC)

Except that there is no "last cut", of course. I'd suggest a look at Zeno's paradoxes: You're suggesting Achilles will never overtake the tortoise. Huon (talk) 14:57, 11 February 2013 (UTC)

If there is no "Last cut" and every other cut divides the circle, then what does .999... refer to? Algr (talk) 06:31, 12 February 2013 (UTC)

0.999... does not refer to any cuts at all. It is the label (name) of a single specific point on the number line. The cuts only come into play when we draw an equivalence of that one number to an infinite summation of numbers. The cut circle is a representation of the infinite summation, not 0.999... directly. The direct representation of 0.999... directly is simply the whole circle of unit area without any cuts.--Tdadamemd (talk) 01:08, 13 February 2013 (UTC)

Your first sentence says that .999... is qualitatively different then .9 or .99. Then by the last sentence you just declare it to be 1, thus building your conclusion into your premise. This happens EVERY time. Algr (talk) 05:55, 13 February 2013 (UTC)

I do not see it that way at all. Yes, the circle has unit area before any cuts are made. But the equality is proven from the summation of the infinite series. So while unity was there a priori, it is also the logical conclusion from the proof. Instead of 'building from the premise', I see it as a premise and conclusion coinciding. (An alternative way of constructing a proof is to assume the opposite, and then logically arrive at a contradiction, therefore disproving the opposite.) You are certainly free to not believe the proof, but I don't see any criticism of this (to date) as invalidating it.

There is a deeper question that I see at the heart of the incredulity of this equality. That question is whether or not 0.999... and 1 are the same number. Two numbers can be different, representing different points, yet have equal value. Notice that:

1 - dx = 1

I would say that this is the big hangup that most people have regarding 0.999...=1, although many are at a loss on how to articulate precisely what they are thinking. The visual proof I've presented here in this section does not address that question at all. There are some who say that 0.999... and 1 are the very same point. There are others who say that 0.999... is the point that is one point to the left of 1.000... And there are others still who maintain that there are lots of points between the two. And in this category, there are those who say that there are lots - but a finite set of points in between, and there are those who say that there are an infinite set of points in between. Regardless of which of these cases is the truth, both 0.999... and 1.000... have the exact same value when "zoomed out" above the level of the infinitesimal. This is rigorously proven via many methods beside the one I've presented here.--Tdadamemd (talk) 12:38, 14 February 2013 (UTC)

Actually I'm with Algr for a change. For the last step in the proof - equality and not just equality up to a certain something - you need something more than just a pretty picture. You'll need the fact that we're talking of real numbers, and you'll need some analysis. Also, I strongly disagree with the statement that "1 - dx = 1" - you have not given any definitions, but by default I'd say 1-dx is an inhomogenous differential form, an element of $\Lambda \mathrm {T} ^{*}\mathbb {R}$, that is definitely not equal to 1. If that's not what you mean, you'll have to give definitions of "1" and of "dx" so that they can be added and so that 1-dx actually equals 1 - which would probably imply that either dx=0 or that 1 has no additive left inverse...

While technically there are other number systems which do contain infinitesimals and which might have a number 0.999... that's different from 1, I have yet to encounter an actually useful number system that does the deed - the closest we come is Katz' hyperreals where 0.999... is not a single number, but rather a set of numbers defined up to the choice of an ultrafilter - the hyperreals may be useful, Katz' definition of "0.999..." is mathematically useless and seems to be only of educational value when teaching (real) calculus. Huon (talk) 16:28, 14 February 2013 (UTC)

The analysis of the visual proof is written up in the text of that article. From memory, it says that the circle has unit area to start with. Then the series of slices corresponds to the infinite series that represents the equivalent of 0.999... A bonus of visual proofs is that they can be very straightforward, and this one fits that. (There's a visual proof of the Pythagorean Theorem that is a work of beauty.)

As for the dx equation, I was trying to state that if you take an infinitesimal away from 1, you still have a number that is equal to 1. It may not be the same number you started with, but both before as well as after have equivalent value in the realm of real numbers. ("Zoomed out" to the point where infinitesimals have no discernible value when taken individually.)--Tdadamemd (talk) 02:01, 16 February 2013 (UTC)

That seems an abuse of notation to me. Either you work within a number system that does have non-zero infinitesimals; in that case taking a non-zero infinitesimal away from 1 will leave you with a number unequal to 1. Or you work within the reals where there are no non-zero infinitesimals to begin with. What you denote as "equal" is actually "equal up to an infinitesimal", and while 1 minus an infinitesimal is indeed equal to 1 up to an infinitesimal (heh), that's pretty much irrelevant to 0.999... - I don't think anybody ever claimed it differed from 1 by more than an infinitesimal (except those cranks who believed we couldn't define it in the first place). Algr in particular believes there should be a non-zero, infinitesimal difference between 0.999... and 1, I think, but has yet to come up with a useful number system which will allow that. Huon (talk) 02:37, 16 February 2013 (UTC)

I am seeing you to be mixing concepts.

The value of infinitesimals is not an either/or proposition as you seem to be describing. All infinitesimals have value by definition. When that value is assessed from the perspective of the Reals, then their value is equivalent to zero - but it is not zero. It has infinitesimal value.

This is why the equivalence holds that you can subtract an non-zero infinitesimal away from 1 and the result is a number that is equivalent to 1.

From a philosophical standpoint, there are those who say that infinity and the infinitesimal form barriers that are hard limits to human knowledge. Beyond that on either side (hyperlarge or hypersmall) lies a realm that can only be grasped by the mind of God. For a human to enter that realm would require the merging with the mind of God, and this would appear as having gone utterly mad to all other humans in our normal realm of existence.

So advice to anyone trying to fully grasp the relationship between 0.999... and 1.000... beyond the simple equivalence that 0.999... equals 1.000..., be prepared to don the straightjacket! Ha. I expect that in physics there will also be pioneers who bridge the gap between the realm of the Newtonian and the realm of the Quantum. I had a professor in grad school who advised us to don't even try. He was a Nobel laureate. I expect that his basis for telling us that was because either he himself had tried and failed ...or that he tried and succeeded.--Tdadamemd (talk) 10:38, 17 February 2013 (UTC)

Time to remove this article from Wikipedia

See the following article but read especially pages 29 and 30. thenewcalculus.weebly.com/uploads/5/6/7/4/5674177/proof_that_0.999_not_equal_1.pdf You may NOT quote anything from this PDF because it has not been published in paper form. 173.228.7.80 (talk) 15:20, 7 February 2013 (UTC)

Wikipedia would never quote anything from a self-published PDF from an anonymous mathematician (the paper is signed only "The spirit of the Archimedes lives in me. I may very well be the last great mathematician."), let alone delete a well-sourced article on the basis of it. --McGeddon (talk) 15:28, 7 February 2013 (UTC)

And here I though you had no need to change others' opinions. If you want to discuss the math, you know where to find the arguments page. If you just want to claim you're right without engaging your critics and without any peer-reviewed publications on your side, that won't fly. As a completely unrelated aside, I'm pretty sure that under the doctrine of fair use we might quote parts of the PDF for purposes of commentary, criticism or research (among others) if there were anything worth quoting in that paper. Huon (talk) 16:46, 7 February 2013 (UTC)

Wikipedia is not in the business of accepting editorial demands from cranks. After excising from the document all of the insults and screeds against mathematicians and university departments and all of the logically unsound mathematics, there is nothing left worth quoting. — Loadmaster (talk) 19:15, 7 February 2013 (UTC)

I agree that .999... is not equal to one, but unfortunately that kind of writing is no way to go about proving anything. Algr (talk) 12:53, 9 February 2013 (UTC)

re: deleted comment - Anon, you have interesting ideas in your paper but you are killing yourself with your attitude. What is the point of being right if you drive everyone away from the truth? Algr (talk) 01:03, 10 February 2013 (UTC)

Time to face reality

There is no valid proof that 0.999... equals 1.

www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-171.html

www.filesanywhere.com/fs/v.aspx?v=8b6966895b6673aa6b6c

I think you ought to delete this article as it contains false information. 197.79.0.4 (talk) 16:19, 4 April 2013 (UTC)

I absolutely loved the part where Euler was wrong and where it is said: "mathematicians don't really understand limits that well. In fact all of them don't know what is a number." I'm glad we have you to enlighten us. 93.102.161.97 (talk) 08:06, 7 April 2013 (UTC)

197.79.0.4, you really should read Gödel, Escher, Bach, the chapter on formal systems and the infinite will be enlightening. It also backs you up, sort of. You'll feel vindicated as well as corrected. Diego (talk) 09:07, 7 April 2013 (UTC)

You really should learn to think for yourself. Have you ever been able to think for yourself? 197.64.80.75 (talk) 12:27, 11 April 2013 (UTC)

Mr. Gabriel, please do not continue plugging your theories here. They won't get any traction, and I think in the future we'll treat you exactly like Anthony R. Brown -- remove your postings immediately. --jpgordon^{::==( o )} 18:39, 11 April 2013 (UTC)

Debunking (0.99...)=1 by using successive multiplication

(1)*(x)= (x)

therefore we can deduce

1*(0.99...)= (0.99...)

If it is true that 1=0.99... that would mean that

2*(0.99...)= (1.99...)= 2

1*(0.99...)=(0.99....)=1

AND

(0.99...)*(0.99...)= (0.99...)

If we applied this logic consistently, then we would also have to say that (0.99...)*(0.99...)=(0.99...) which obviously poses a problem since

(0.99...)*(0.99...)=(0.99...)=1

[(0.99...)*(0.99...)]*(0.99...) ........... = 1

And so on.

If (0.99...) truly equals 1 then one could theoretically infinitally reduce a given number by multiplying it an infinite number of times by (0.99...), and that number would also equal 1 by this logic, since it would automatically have to equal (0.99...) which again equals 1. As an extension of this, 20 could equal 3, 5232 could equal 91, and so on. --88.73.3.251 (talk) —Preceding undated comment added 20:53, 27 October 2013 (UTC)

What do you mean by "infinitely reduce"? --jpgordon^{::==( o )} 05:31, 28 October 2013 (UTC)

You can no more "reduce" a number by multiplying it by 0.999... than you can "reduce" it by multiplying it by 1.

What I don't get is why (0.999...)ⁿ=1 should imply 20=3. Rather, 20*(0.999...)ⁿ=20*1=20. Multiplying infinitely often is a little more tricky because it needs to be properly defined, and while there are a few infinite-multiplication-related caveats, they don't actually apply here. We have:

(0.999...)^∞:=lim_n→∞(0.999...)ⁿ=lim_n→∞1ⁿ=lim_n→∞1=1.

So not much happens if you multiply 0.999... with itself. Huon (talk) 20:41, 28 October 2013 (UTC)

Consider a number between 0 and 1 and take for example a cube root of it infinite times. You'll have 0.999... as a result which you claim equals one. Or lim (m^1/3)^1/n = 1;(0<m<1);(n->∞). Note that some math operations could be reversed using opposing operations. 3+3=6 can be reversed with 6-3=3. 0.5^3=0.125 with cube root of 0.125=0.5. But acceppting 1=0.999... we are not able to get the original number anymore using the opposing operations. Thus will have lim 1^n = 1; (n->∞). What does all this mean? If we take a cube root of 0.3 for example 1 million times we can easy by opposing operations get 0.3 again. But if we do this an infinite times and if 0.999...=1. it is already impossible to get the original number (0.3 in this example). And this is a contradiction. At which point exactly we are not able to get 0.3 anymore? The answer is needed or we have to confess there is a flaw in the mathematical definitions we use concerning real numbers, 0.999... and limits. — Preceding unsigned comment added by 95.42.185.235 (talk) 10:12, 13 October 2014 (UTC)

The problem you're having here is with the idea of limit; taking a limit is not an operation that you can reverse (at least, not in the trivial way you mean here). You can't reach a limit by "applying infinite times the same operation"; by definition, "infinite" doesn't have an end. The (upper) limit is defined as "the smallest number that is higher than any other number you can reach by adding smaller and smaller digits". The limit itself is not inside the sequence of repeating numbers in the series, it's deliberately defined outside of the sequence; that's why 0.999... (defined as the limit, not the series) may be the same as 1: the limit is a number that is above 0.9, and above 0.99, and above 0.999, and above any other finite number that you can reach by repeating an operation a finite number of times, no matter how many; the smallest real number with that property is the number 1. You don't reach it by looking at the highest number in the sequence (as there's no such thing), but by finding the lowest number outside the sequence.

In terms of calculating and making mathematical demonstrations, there's no such thing as "repeating an operation infinite times", except as a shorthand for something else - instead, you find a different relation that always holds no matter how (arbitrary) many times you repeat the first one. The trick is that the limit is not defined by the '+' operation or the '^n' operation, it's defined by the '>' (greater than) operation. Diego (talk) 11:45, 13 October 2014 (UTC)

The real problem is that we do not know at which point exactly a finite number of repeating nines becomes infinite. And it is an ill definition. 0.999... should not be considered a real number, lust like 0.333... and so on. We do not need these numbers. 0.333... is approximation of 1/3 which we can always use. 0.999... is 1-1/∞ and if 1/∞ is not a real number, 0.999... should not be either. One can have 0.999... as a result only because of approximation and rounding.

The sum of geometric series is a limit too, and it was used as a proof so many times. Yes, lim (m^1/3)^1/n = 1;(0<m<1);(n->∞) does not mean that (m^1/3)^1/n = 1;(0<m<1);(n->∞). More, I think (m^1/3)^1/n can never be one. Still it is equal to 0.999... --95.42.185.235 (talk) 16:10, 13 October 2014 (UTC)

The real problem is that we do not know at which point exactly a finite number of repeating nines becomes infinite. It is known, and the answer is "at no point". You can never repeat a finite process and turn it into an infinite amount. Therefore, the infinite number has to be defined through a different set of rules; you can't use the rules defining numbers with a finite amount of digits to describe numbers with an infinite amount; the later are handled through alternate rules, that don't require you to write down all the digits, but are nevertheless exact.

0.333... is approximation of 1/3 No, 0.333... it is exactly 1/3, not an approximation. You can use other rational numbers like 0.3, 0.33, 0.333 or any other in the series as approximations, but those are not the same number as 1/3, which is equal to "0.333..." (represented with infinite digits, and therefore using a different set of rules than "0.3" or "0.333", which have a finite number of digits).

We do not need these numbers. You may not need those for basic calculations, but they're incredibly useful for making mathematical proofs. The set of mathematical rules for limits are useful to demonstrate some properties of the series. For example, in the case of the series $\sum _{k=1}^{n}{\frac {9}{10^{k}}}$, the limit ensures that, no matter how many times you keep adding more '9's, the sum will never by higher than 1. You couldn't tell that from basic calculations through approximation alone.

(m^1/3)^1/n can never be one. Still it is equal to 0.999... You have to be extremely careful to use precise definitions; mathematics are only as useful as the care you take to define your terms without ambiguity. In this case, you can't make n->∞ without taking the limit. A finite n is always smaller than ∞; if you make n->∞ (not equal, but approaching infinite), then you have lim(n->∞) (m^1/3)^1/n. Diego (talk) 16:59, 13 October 2014 (UTC)

How can 0.999 be 1?

(I hope I got the decimal notation for infinitely small but not zero correct... a guess after not being able to find it. 0.000...1 (infinitely zeroes before the 1)

a\ true or false? 0.999... = 1

b\ true or false? 0.000...1 = 0

c\ true or false? 1 - 0.000...1 = 0.999...

d\ true or false? 0.999... - 0.000...1 = 0.99...8

e\ true or false? 0.999... = 0.99...8 = 0.99...7 = 0.99...6

f\ true or false? the difference between every neighbouring real number would be 0.000...1?

g\ what's the total sum of all differences between every neighbouring real number? infinity or zero? - ZhuLien

— Preceding unsigned comment added by 202.168.6.200 (talk • contribs) 04:49, 17 January 2014‎ (UTC)

There is no decimal notation for infinitely (infinitesimally) small non-zero numbers, at least not in the context of the real numbers. In fact there are no infinitesimal non-zero real numbers; that's a consequence of the Archimedean property. Similarly there are no neighbouring real numbers; between every two real numbers a, b there is another real number (a+b)/2. Huon (talk) 22:03, 17 January 2014 (UTC)

I'll also add that 0.999... is a repeating decimal, so by definition it must be a rational number. It also means that it does not have a last (rightmost) digit. — Loadmaster (talk) 00:05, 18 January 2014 (UTC)

There is no notation for numbers between zero and one, at least not in the context of the Natural numbers. 0.999... = 1 is based on using a number set designed to exclude the issue being discussed, so the answer above is like "proving" that fractions don't exist by showing that natural numbers can't accommodate fractions. Algr (talk) 15:30, 18 January 2014 (UTC)

That might be a reasonable analogy if ZhuLien had not explicitly spoken of real numbers. And while there are number systems where there are "neighbouring numbers" (such as the natural numbers), I'm not aware of any such system where "0.999..." is meaningful and different from 1. @Algr: Are you?

The number system with infinitesimals which does have a decimal notation is the hyperreal numbers, but there are, as Algr should be well aware, numerous caveats far beyond the scope of ZhuLien's question. While there are hyperreals which have a decimal representation of 0.999...;...999000... with both infinitely many nines and a last nine, there are infinitely many of those, and there's no particular reason to call any single one of them "0.999..." in preference to the others, just as "0.999..." doesn't usually denote one of the real numbers with finitely many nines and a last nine. Also, those hyperreal numbers do not equal 1, they are not neighbouring 1, and they don't have a "last digit"; the hyperreal number whose decimal representation has nines all the way still equals 1. Somehow I doubt this explanation is helpful for ZhuLien. Huon (talk) 16:50, 18 January 2014 (UTC)

Well I think it is precisely what ZhuLien needed to know. Yes s/he did say "real numbers", but also described characteristics of the hyperreals, so why refuse to acknowledge that? The reason for the entire rejection of .999...=1 is that articles like this treat hyperreals as some kind of forbidden knowledge that readers are unworthy to know. Algr (talk) 18:58, 18 January 2014 (UTC)

OK, I'm going to repeat a point here on the hyperreals that often gets forgotten. Yes, it's true that hyperreals can be considered to have a decimal representation, and the positions in that representation can go "past infinity" (that is, you have an initial segment of positions that looks like the ordinary reals, and then after the end of those, you have something more complicated). If we put a semicolon to indicate the point before which everything looks like the representation of the ordinary reals, then you could have hyperreals that look like, say, 3.14278...;...32813... or something like that.

But here's what you can't have. You can't have 0.999...;...000..., where everything before the semicolon is a 9 and everything after is a 0. That's forbidden, because that would allow the nonstandard model to figure out which were the true natural numbers as opposed to the model's fake infinite natural numbers, and the model can't figure that out.

So that leaves a couple of things. You could have 0.999...;...99990000..., where the 9s go up to some particular nonstandard position, and then switch to 0s afterward. Such numbers are indeed infinitely close to 1, but not equal to 1. What they are not is reasonably described as 0.999..... The reason that you can't describe them that way is that 0.999... should be a unique thing, whereas for the 0.999...;...99990000..., you get a different nonstandard real depending on where you cut off the 9s.

Or, you could have 0.999...;...999..., where you have 9s everywhere. But that number is exactly equal to 1.

So really, with good piece to Tkuvho, the hyperreals are just a red herring in this discussion. --Trovatore (talk) 21:53, 18 January 2014 (UTC)

Thanks first for answering my questions. Although there may not be a notation for what I've described, surely the the numbers must exist? I guess that the issue with an infinite number is I cannot write it down, but it if weren't infinite, lets say there were a few trillion digits before the ending one, although it doesn't directly answer my questions, it does still imply if i had enough time to write down trillion digit numbers I could. The oddity I've asked is the infinite digits in the middle preceeding the last digit can very well be imagined, but I also cannot see any reason why it's impossible to have an infinite number of digits placed anywhere (notation aside). I haven't heard of hyperreals so I will read up, but probably also not a small topic. I wasn't aware that an infinite number made it no longer a real number, perhaps I should have said floating point number instead? I did carefully come up with the original set of questions due to what I had been reading about 0.999... = 1 and there is no way I can imagine that. I can only imagine it is infinitely close to 1, but not quite there. If it were in fact 1, then an asymptote wouldn't exist, or would it? Conceptually we can get as infinitely close to something but never get there (no matter which number it is). Quoting Huon's comments above "I believe you're mistaken about how 0.999... is defined. Correct me if I'm wrong, but you seem to believe that 0.999... is obtained by adding 0.9 + 0.09 + 0.009 + ..., the result of an never-ending summation of ever smaller numbers." which I agree on somewhat. How do you write down in the two seemingly different numbers 0.999... (which represents 1) and 0.999... (which represents the never ending summation of the smaller numbers which obviously isn't 1) so their notations aren't the same? ZhuLien -- [[User:116.240.194.132 (talk) 00:39, 20 January 2014 (UTC)|116.240.194.132 (talk) 00:39, 20 January 2014 (UTC)]] ([[User talk:116.240.194.132 (talk) 00:39, 20 January 2014 (UTC)|talk]]) 11:39, 20 January 2014 (UTC)

"Neighbouring numbers" such as "the largest number less than 1" don't exist in the commonly used number systems that contain at least the rational numbers. You technically could construct a custom-made number system with such a property, but I don't think it would be very useful - for example, you will run into problems with division. There are several different number systems that have something resembling decimal representations with infinitely long strings of digits followed by something else, where numbers might arguably be called 0.000...1 with infinitely many zeros before the "1", among them the hyperreal numbers pointed out above. However, the standard real numbers, the numbers most commonly used, don't have such infinitesimal numbers. Among the reals 0.999... indeed equals 1. An asymptote of a curve, in this context, is a line that gets "arbitrarily close" to the curve, that is, for every positive real number ε there exist points c on the curve and l on the line so that their distance d(c, l) is less than ε. Take for exampe, for positive x, the curve y=1-1/x. The line y=1 is asymptotic to that curve because for every positive ε there exists an x_ε>1/ε, and the points c=(x_ε, 1-1/x_ε) and l=(x_ε, 1) satisfy the property above. (Actually I'd also have to require that curve and line don't just get close anywhere, but "near infinity"; I could modify the definition accordingly, but that would complicate the calculations without helping to clarify the main issue.)

Regarding "the never ending summation of the smaller numbers", how exactly would that be defined? That's a crucial question here. In the context of the real numbers, the answer is a limit: If 0.9+0.09+0.009+... is to be a real number, it's defined as the limit of the sequence (0.9, 0.9+0.09, 0.9+0.09+0.009, ...), and that limit can be shown rather easily to be equal to 1. In the context of the hyperreals, it's possible to define such infinite sums so that 0.9+0.09+0.009+... no longer equals 1 - but at least the way I can think of would lead it to also differ from 0.99+0.0099+0.000099+..., and from 0.9+0.099+0.000999+0.0000009999+..., and so on - even from 0+0.9+0.09+0.009+0.0009+...! Which of those numbers is supposed to be 0.999...?

Some final food for thought: Why should 0.9+0.09+0.009+... be less than 1? If the answer is, "because all the finite sums 0.9, 0.9+0.09, 0.9+0.09+0.009 and so on are less than 1", then I'd like to point out that, however it's defined, 0.999... also is larger than all those finite sums. Huon (talk) 01:59, 21 January 2014 (UTC)

ZhuLien's theory of infinitesimal usefulness.

I am a number. My value is x. I might be positive or negative.

I have many neighbours depending on where they are positioned in my number space.

x+0.0...1 and x-0.0...1 are my nearest valued neighbours yet I cannot see them as they are infinitely far away but I do have an infinite number of neighbours closer than they are.

What I can do is using my value and my nearest neighbours or perhaps nearer neighbours, is work out how crowded we are.

I can compare my crowded space with other number's crowded spaces to see statistics in different ways.

I can zoom in and out indefinitely to my nearest valued neighbours to allow my crowd and other crowds to be visualised at different resolutions.

There may be mathematical operations on both individuals, neighbours and crowds which can be useful in rounding us up, or grouping us together in sets.

These sets can be useful in highlighting patterns or boundaries or just separating us by an arbitrary amount of space. ZhuLien 116.240.194.132 (talk) 05:42, 6 February 2014 (UTC)

• You can't say that without defining the terms you use. If you operate in standard real numbers, then as a matter of fact there is no such number as "next number greater than 1" or "next number lesser than 1". The real numbers have the property that between any two different real numbers there exists a number more than the "left" one and less than the "right" one (and, as a consequence, an infinite amount of such numbers). - Mike Rosoft (talk) 19:38, 7 February 2014 (UTC)
  • But that is exactly what you do say when you describe an exclusive set. Algr (talk)

I have never considered using different number systems until reading more on the threads related to this topic. To me a number has always just been a number. Whether the number was an integer or a floating point number didn't make a number system in my mind. Stating things such as all integers to me was just a set. I guess the term 'real numbers' is more than a set, but also a universally defined set of rules - so I will avoid using such a term as real numbers. Infinitesimal though does have applications and as an example, recently a project I had been working on for a client required numbers to be spacially sorted. For performance reasons I found the easiest way to spacially sort in one direction towards smallness was to take the number which I needed to be next to and add another 5 to the end, ie: 0.555, add a 5 to become 0.5555. Interestingly the programming language I was using (C#) and in fact every other programming language I have used or come across doesn't have a concept of the other direction. In this case I had to create a crude proof of the spacial sort using an artificial maximum number of digits and a smaller sample to reverse the order before slotting in a number - ie: reversing numbers such as 1,2,3,4,5 to be 5,4,3,2,1, and subtracking them from an artificial limit such as 1000 giving 995, 996, 997, 997, 999. Then the adding of a .5/5 would work afterwards removing the artifical limit. -X+1000... I am surprised that no-one else has required to spacially sort numbers before and if they had, the topic doesn't seem common enough to find good algorithms to spacially sort numbers towards infinitesimal. Back to the original argument of 0.999... ~= 1. Although I cannot prove it, I am 100% positive that if I could infinitely add a 9 onto my number for the rest of eternity - I would never reach 1. If someone can prove against that, then I will accept that 0.999 = 1. ZhuLien 27.32.141.11 (talk) 20:36, 7 March 2014 (UTC)

"I could infinitely add a 9 onto my number for the rest of eternity - I would never reach 1". "0.999recurring = 1." These statements are consistent with each other and both true. In forever adding 9s onto 0.999, there would never come a point at which you reach 0.999recurring. MartinPoulter (talk) 20:57, 7 March 2014 (UTC)

Hyperreal math

Trovatore says that "0.999...;...000..." is forbidden. I find that bizarre, as it comes across as a bigger failing then my Base Infinity idea. Also, how many times have people been told that .000...1 is nonsense when all that is needed is a semicolon? So anyway, without "0.999...;...000...", then what is the answer to this?

0.999...;...001... - 0.000...;...001... = ? Algr (talk) 02:58, 9 February 2014 (UTC)

0.999...;...001... is just as ill-defined a hyperreal as 0.999...;...000..., so the question makes no sense. Furthermore, while there are hyperreal numbers with decimal representations of the form 0.000...;...001, there is an entire class of these numbers, and technically they are all of the form 0.000...;...0001000... with varying (but inifinite) numbers of zeroes before the "1" - they do not have a "last digit" or a "1 at the end", which is usually meant by "0.000...1". Furthermore there is no reason whatsoever to single out one of them and to claim that this particular hyperreal, and not any of the other similar ones, equals (1-0.999...). You technically could argue that 0.999... should be a hyperreal of the form 0.999...;...999000... - but then you could just as well argue that 0.999... should be a real number of the form 0.999...999000...; the hyperreals do not provide you with any benefit here. The one hyperreal with nines all the way, 0.999...;...999..., still equals 1.

I find it rather unsurprising that yet another number system fails to do everything you want it to do. There is a reason that the real numbers are the default number system, and that reason is that everything else is even less intuitive. Huon (talk) 20:34, 9 February 2014 (UTC)

What DOES the hyperreal set do then? How is it any better than Base Infinity? Also, the Real set is far from the most commonly used number system. Since computer programmers and accountants vastly outnumber mathematicians, Floating Point and Fixed Point numbers are far more common than Real set. Algr (talk) 07:54, 17 February 2014 (UTC)

I don't think we ever quite finished clarifying what exactly the Base Infinity numbers were supposed to be, but I'd point out three main advantages of the hyperreals:

1. They are a field; in particular, every hyperreal (except 0) has a multiplicative inverse, division (except by 0) is well-defined. That alone is worth a lot, and if I remember correctly, the most likely candidate for the Base Infinity numbers didn't have this feature. This is still but a special case of the second feature:
2. They satisfy the transfer principle, which basically states that whatever "works" for the real will analogously also "work" for the hyperreals. Real numbers have decimal representations? Well, by the transfer principle, so have hyperreals! I can calculate the square root of a positive real number? By the transfer principle, positive hyperreals have square roots too! This is a very neat property that basically assures that the hyperreals will, for the most part, behave the way we expect numbers to behave, with the obvious caveat that when the reals behave counterintuitively, so will the hyperreals.
3. Unlike hyperreals, Base Infinity numbers in general no longer have decimal representations. For example, in your Base Infinity system of numbers, I don't think 1/3 has one any more. Thus, decimals no longer represent all your numbers, but only a tiny and exotic subset.

Of course there's also the obvious problem that the Base Infinity numbers don't have "neighbouring numbers" any more than the reals or the hyperreals do, and thus won't be any better at resolving the issues that led to ZhuLien's question in the first place. Huon (talk) 21:43, 17 February 2014 (UTC)

It's great to see this debate is still raging after all these years. But something needs to be done. 0.999... is not 1, and here is a simple disproof.

About a decade ago, the two sides of the 0.999... debate clashed regularly, and the main article was edited back and forth. Mathematicians and hobbyists on both sides. But one side was particularly more aggressive (the side that thought the two numbers are equal). And now after a decade, it's turned into full-blown suppression. You can't even post a dissenting view in the talk section? That is utterly ridiculous. God forbid you edit the main article. Even after all these years, I simply am shocked that people still buy into these false proofs. The first proof, alone, is simple to see right through. Multiply both sides by 10. It's a little magic trick. It's like something you'll see in a chain email that tricks the uneducated. But this tricks the educated, too. The more accurate version of the "Magic Trick Proof" is thus:

    x = 0.9999...  
    10x = 9.9999...-9/∞ 
    10x - x = 9.9999...-9/∞ - 0.9999... 
    9x = 9-9/∞ 
    x = (9-9/∞)/9
    x = 1-1/∞
    The point of this equation is that you can't multiply 0.999... by 10, it's not a valid mathematical equation.
    I'm not suggesting that we start multiplying and subtracting using infinity, I'm saying we *shouldn't.*

Follow the rules of Wikipedia and allow people to talk about this topic in the talk section. --JohnLattier (talk) 16:51, 18 April 2014 (UTC)

Isn't 1/∞=0? The rules of Wikipedia requires editors to limit conversation to article improvements, which in this case requires that we discuss proofs appearing in reliable sources. Napkin calculations like the above are thus outside the scope of the talk page unless a RS discusses them. Diego (talk) 17:34, 18 April 2014 (UTC)

No, 1/∞ does not equal zero. Hmm, napkin calculations eh? That's how you describe the proofs on the main article? --JohnLattier (talk) 18:25, 18 April 2014 (UTC)

Indeed, ∞ is not a number, and so 1/∞ is not even defined. The second line of this is nonsensical. 10x=9.999..., no extra term needed. If you would like to understand the algorithms for addition and multiplication of repeating decimals see this award winning article Thenub314 (talk) 22:36, 18 April 2014 (UTC)

1/∞ can be defined as a shorthand for a limit: lim 1/x (x->∞). However, this limit is equal to zero, so this won't help your case. You are trying to define 10*0.9999... as 9.9999...90; however, there's no such real number. Each decimal digit corresponds to a natural number, and there's no last natural number (i.e., there's a 1st digit after the decimal point, 2nd, 3rd, 99th, googol-th etc., but NOT an infinity-th digit, as infinity is not a natural number; there's an infinite amount of natural numbers, but every natural number is finite). - Mike Rosoft (talk) 19:43, 10 June 2014 (UTC)

And as I've said a dozen times, if infinity isn't a number, then .999... is undefined because you can't even state how many digits it has. Algr (talk) 16:32, 12 June 2014 (UTC)

Just because infinity isn't a number doesn't automatically cause everything it comes in contact with to become undefined; it's a theoretical concept which creates theoretical results. It can't just be waived, else 1/3 would also be undefined. 0.999... has an infinite number of digits and just because that's not strictly giving you a number doesn't make it undefined. JaeDyWolf ~ Baka-San (talk) 17:46, 12 June 2014 (UTC)

Infinity is NOT undefined; statements: "There is an infinite amount of natural numbers" and "Infinity itself is not a natural number; every natural number is finite" are by no means contradictory. (In order to be able to discuss infinite sets, we'd have to go to set theory; this is a topic I am going to avoid for now.)

You need to understand that a repeating decimal can be expressed without expressly invoking infinity. 0.999... is actually a shorthand for 0.9 + 0.09 + 0.009 + ...; which is itself a shorthand for a limit of a sequence:

• Let A(n) be 9*10^-n, for n being any [finite!] natural number.
• Let S(n) be a sum of the first n elements of A(n). It can be easily shown that S(n) = 1-10^-n.
• Now suppose there exists such a real number L: if you give me any positive ε, I can find a finite N such that for every n>N: |L-S(n)|<ε; in other words, S(n) differs from L by less than ε. (ε can be arbitrarily small, but it can't be zero or "infinitely close" to zero - it must differ from zero by a finite amount.) Then we can call L a limit of the sequence S(n) = {0.9, 0.99, 0.999, ...}, a sum of the series ∑A(n) = 0.9 + 0.09 + 0.009 + ..., and therefore a value of 0.999...
• Turns out, there is a unique such number L, and it is exactly 1. (Not kinda sorta 1, not infinitely close to 1, but equal to 1.) For ε of 0.1 the N is 1; for ε = 0.01 N is 2; and for ε = 10^-x N is x.

See? At no point did I need an infinity-th element, or anything similar; in fact, I have (just for clarity) specifically excluded infinities and infinitesimals from the definition. - Mike Rosoft (talk) 23:39, 12 June 2014 (UTC)

You just hid the infinity, you didn't eliminate it. If every number n is finite, then 9*10^-n must always be a positive value representing the difference between S(n) and 1. Using your words: S(n) can be arbitrarily small, but it can't be one or "infinitely close" to zero - it must differ from one by a finite amount.

— Preceding unsigned comment added by Algr (talk • contribs)

Mike Rosoft is right; his definition doesn't use "infinite" nor any infinite object. It has an infinite process, though: he said "arbitrarily close", which means that his definition of the limit L will work no matter how many (finite) digits you write down or how many (partial) sum of the series you arrive at by doing partial sum. *All* properly defined "infinites" in mathematics are created that way, by defining a "finite process" that can be repeated "arbitrary times" without encountering an exceptional case that breaks the process.

If "n" is finite, then s(n) ≠ 0.999... All repeating decimals have an infinite number of trailing digits - even 1.000... has it. (This doesn't mean that there's an "nth-element at the infinite", but rather that you can always continue adding more digits to the -partial- representation; though you will always need to add the "..." that represents "...and all the other decimals that I can't write down in full").

All rational numbers (i.e. those that can be represented as fractions) have an exact representation as a repeating decimal; if you choose to use decimal representations at all, you have to accept the equivalence between both, otherwise that representation would be useless. They differ from irrational numbers, which don't have such representation because it's not possible to write down all their decimals (as they are infinite and non-repeating), although they can often be represented as the limit of other sequences. Diego (talk) 17:16, 29 July 2014 (UTC)

To some extent communication requires a shared agreement on what is meant by certain things. We usually agree that if x is a "number" then also 10x is a number and that from a+x=b+x we can conclude a=b. What about x=0.999... If you want to say that x is not 1 then you need to breaking the properties of numbers in such a way that x is fairly useless. You suggest that we can multiply x by 10 but we shouldn't. I'd say that EITHER we can multiply and then 10x=9+x OR ELSE x is a thing which can not be multiplied by 10 at all. If you take the second position (or even your stated position) then you have to abandon saying 1/3=0.3333... which means rejecting the possibility of having any decimal equal to 1/3. This leads to a situation where you have 5/8=0.675 but most fractions have no decimal expansion. IF infinite decimals are anything then it would have to be something where we can't add them together or multiply by 10. In your proposal that 10x = 9.9999...-9/∞ what is the purpose of the thing at the end subtracted? All I can see is that we come up with something new mainly because it gives us room to say that 10x is something other than 9+x. But can we then divide both sides by 10 to come back to x=0.999...- 0.9/∞? and then we have x+0=x+0.9/∞. Can we subtract x from both sides? Doesn't look so good. So in sum, saying that x=1 makes things work very smoothly and does not break any of what we do with numbers. Saying x is not 1 means that it is a thing where multiplying by 10 creates trouble and is of no use. Gentlemath (talk) 02:30, 16 June 2014 (UTC)

I don't think anyone would be bothered by the notion that .333... does not exactly equal 1/3, but is just an approximation. In real life, when will this ever make a difference? Algr (talk) 16:53, 29 July 2014 (UTC)

It makes a difference to a mathematician... They couldn't use the "=" sign to represent equal values in their equations, and thus they couldn't reason about complex systems without accumulating errors. Diego (talk) 17:16, 29 July 2014 (UTC)

I think anyone using decimal maths in any context should be shocked by the notion that .333... does not exactly equal 1/3. For one thing, .333... is the opposite of an approximation: every digit is exactly specified. That 1/3=.333... can be confirmed by simple division (and mathematical induction) and checked by a number of simple mathematical proofs (if x=.333... , what is 4x-x?). If you're denying the exact mathematical equality of the two expressions, then you're denying the whole system of maths. MartinPoulter (talk) 12:43, 31 July 2014 (UTC)

From what I have seen, mathematicians avoid decimals like the plague unless they originate from real-world measurements. ".333..." is much like "3.14159..." a pragmatic approximation, but not mathematically exact. Almost all real numbers cannot be written in ANY form because infinite information is required. A few lucky Reals like π and e can be described, but indescribability is a fact of life for real numbers so limits on decimals shouldn't be seen as so tragic. Decimals can only be exact if all the primes in the denominator also appear in the base of your number system. Algr (talk) 13:37, 11 August 2014 (UTC)

Sorry, but this is nonsense. Infinite decimals are exact, and every real number can be represented by one or two infinite decimals. For the claim that "infinite information" is required to describe almost all real numbers I'd like to see a reliable source. While that seems possible, I doubt such a result has been proven. Huon (talk) 16:00, 11 August 2014 (UTC)

Well, the "infinite information" claim is one of those slippery things that are fun to argue about because it seems to be almost well-specified, but not quite, and arguably can never be made completely well-specified. But it is true, with appropriate weaseling, that for any fixed way of specifying a real number, for almost all of them, you'll need infinitely much information to specify the number according to that "way".

I would go further, and assert that every real number actually encodes infinitely much information. It's just that, for a few of them, the information is sort of repetitive. --Trovatore (talk) 16:24, 11 August 2014 (UTC)

There you go Huon, it seems I know something you don't. And I've found the name of it: Uncomputable number. π and e are actually computable, because you aren't limited to using digits to describe them. "The ratio of a circle's circumference to its diameter." is a finite description. Nevertheless, that article confirms that "almost all real numbers are not computable" Algr (talk) 07:11, 12 August 2014 (UTC)

Thank you for providing the proof I asked for; I indeed didn't know that. Computability, however, is rather irrelevant to the issue of uniqueness of decimal representations. Whatever else they may be, the real numbers represented by 3.141..., by 0.333..., by 0.999... and by 1.000... definitely all are computable. In fact, so is every real number with a decimal representation that gets repetitive, in particular every real number with a decimal representation ending in repeating zeroes or (and) repeating nines. Thus the whole issue of 0.999...=1 is an issue within the realm of the computable numbers, making uncomputability something of a red herring here. Huon (talk) 13:16, 12 August 2014 (UTC)

Consider taking the number 1 and dividing it into 10 equal parts, or 10/10. Then consider taking one of those parts (1/10) and dividing that part itself into 10 equal parts, which will give you 10/100. Then consider Taking one of those parts and dividing it up into 10...This process leaves you with 9/10 + 9/100 + 9/1000 and so on...A simple expression of this infinite recursive process can be written out. X = 9/10 + 1/10(X). The algebra trivially resolves to 1 but solving it recursively... 9/10 + 1/10(9/10 + 1/10(x)) will get you 9/10 + 9/100 + 1/100(x) = 9/10 + 9/100 + 9/1000 + 1/1000(x) which we see will carry on indefinitely as 9/10 + 9/100 + 9/1000 + 9/10000...so .999... = 1. 50.136.175.142 (talk) 13:53, 25 October 2015 (UTC)

It can be either true or a type error (but not forced to be either)

Think of "1" as representing a basket of all the different functions that converge at 1. If "0.999..." is interpreted as a series expansion, then it's one of those functions. But under that interpretation, the statement "0.999... = 1" is like saying "apple = basket of all possible apples". Thus if someone writes "0.999... = 1", you can reasonably assume they're comparing baskets (limits) and not apples-to-baskets. It doesn't make any sense otherwise. But that said, you can part ways with that interpretation temporarily. E.g., to justify shifting the decimal point as a shortcut for multiplying by 10, you first reinterpret .999... as (.9 + .09 + ...), distribute the 10, use the associative rule to extract the 9, and convert back to decimal notation. This breaks the convention of immediately "taking the limit" of 0.999..., or its decimal expansion would be (1 + 0 + 0 + ...). So that's the real answer: if someone compares a convergent series to a number, it's convention to assume good faith that they're secretly referring to the limit rather than incorrectly comparing apples to baskets. But you can still temporarily delay "taking the limit" when it's useful (like in, oh, all of differential calculus). You can't "prove" this convention, and all of these supposed proofs are invalid by circular reasoning, temporarily breaking the convention that is supposedly being proven, etc. Maghnus (talk) 11:40, 27 July 2014 (UTC)

How can 0.999... be regarded as either a series or a series expansion? It is a single number. "Taking the limit of 0.999..." makes no mathematical sense. So for that reason, 0.999... is of the same type as 1.000... . With no possibility of a type error, the statement is true. MartinPoulter (talk) 15:32, 29 July 2014 (UTC)

You can interpret the real number 1 as the equivalence class of Cauchy sequences of rational numbers represented by the sequence (1, 1.0, 1.00, 1.000, ...), which happens to consist of all sequences of rational numbers converging to 1. But then you have to interpret 0.999... as the equivalence class of Cauchy sequences of rational numbers represented by the sequence (0, 0.9, 0.99, 0.999, ...), which happens to be the same equivalence class. Huon (talk) 16:28, 29 July 2014 (UTC)

Absolutely agree. My concern is just that some people see 0.999... and read it as meaning a series of numbers, rather than as a single number, and that this is where some of the fallacious arguments seem to come from. MartinPoulter (talk) 12:45, 31 July 2014 (UTC)

Real numbers include two different zeros.

The system of real numbers includes two kinds of zero that I will demonstrate are different. Imagine a machine that generates random rational numbers between 0 and 1. It is 'fair' in the sense that any rational number has the same probability of appearing as any other rational number. What is the chance of this machine generating a 2? Well, it is zero, because we defined it as only going from 0 to 1.

So, what is the chance of this machine generating a .5?

Well normally you would figure (specified outcome) / (All possible outcomes). But in this case the denominator is infinite. One might say that 1/∞ is not a number, but taking the limit of larger and larger denominators gives you an apparent result of zero. Prob(.5) and Prob(2) both seem to equal zero, but they do not equal each other, because one is possible and the other is not! Thus, two different zeros.

This closely relates to the .999... problem because you have an infinite number of elements, (digits for .999..., numbers for Prob(.5)) evaluated with a finite result. The simple solution would be that 1/∞ = the infinitesimal unit, ®1. If you insist that 1/∞ is not defined, then remember that 1/2 is not defined in the set of Natural numbers for the same reason. 2 can't be generated by my machine, but it still exists.

BTW: 1/3 in base infinity is (1/3)® or .1_{base 3}®. Algr (talk) 21:56, 14 August 2014 (UTC)

Please provide the definition of real numbers of your choice and explain how these "two different zeroes" fit into that definition, particularly in light of the Archimedean property which the real numbers satisfy. Also, you may want to think about what 1/((1®)-(®1)) is in your number system. Huon (talk) 01:21, 15 August 2014 (UTC)

What I wrote is a challenge for YOU much as your base infinity question is a challenge to me. Why would you then expect me to answer both challenges? MY answer to the contradictions of the Reals is base infinity: In base infinity, 1/((1®)-(®1)) = 1/(1®-1) = 1®1. Algr (talk) 02:16, 15 August 2014 (UTC)

In that case, the short answer is that you're wrong about the reals, which have only one zero. Events with probability zero may still occur; that does not mean the probability is infinitesimally larger than zero. Equivalently, sets of measure zero need not be empty.

My favorite definition of real numbers is as equivalence classes of Cauchy sequences of rationals, where two sequences are equivalent if their difference converges to zero (within the rational numbers). An immediate consequence of this definition is that if two real numbers differ, there is some rational number smaller than their difference. However, there's no rational number smaller than the difference between the supposed "two zeroes". Thus, the "two zeroes" must be equal.

Also, I would have assumed that

(1®1) * (1®-1) = (1® + ®1) * (1® - ®1)

= 1® * 1® - 1® * ®1 + ®1 * 1® - ®1 + ®1

= 1® * 1® - ®1 * ®1

= 1 - ®1 * ®1.

Am I wrong? If so, what step is wrong? If not, is ®1 * ®1 = 0? Huon (talk) 03:19, 15 August 2014 (UTC)

Your argument fails at the very beginning. The machine you imagine cannot exist. There is no such thing as a "fair" probability distribution over rationals between 0 and 1. Every method for generating those rationals will favor some numbers over others, and therefore is unfair. (there is no uniform distribution over any countably infinite set) -- Coffee2theorems (talk) 21:53, 15 August 2014 (UTC)

Plus, the "two zeroes" aren't really that. One doesn't ask "what's the probability of my machine generating a two", because you've built the machine not to do that; likewise, one doesn't ask "what's the probability of my machine generating an irrational value", because you've built the machine only to generate rational values. The probability of an impossible event is NaN. It's Mu. It's not even wrong. It's a domain error. --jpgordon^{::==( o )} 22:39, 15 August 2014 (UTC)

I have to disagree with Coffee2theorems and Jpgordon here. Firstly, we're doing pure math here, so we can talk about continuous probability distributions. Secondly, given a random variable X with a constant probability density of 1 between 0 and 1 and a constant probability density of zero elsewhere, we can ask about ${\textrm {P}}(X\in \{2\})$, which will be zero, not "mu". (I'm not enough of a statistician to be sure whether such a continuous probability distribution makes sense if we define it only on the set of rationals, but the issue Algr points out would be the same if defined on the reals instead of the rationals.) Huon (talk) 23:22, 15 August 2014 (UTC)

It's not possible to do it on the rationals. That was Coffee2theorems' point, and he/she was completely correct.

Suppose you had a "fair" way of choosing rational numbers randomly. Then, for any two rationals, the probability of picking them is equal, right? That's what "fair" means.

So call that probability p. Then p is either zero, or else it's not.

If p is zero, then by countable additivity (which is one of the requirements on a probability measure), the probability of picking a rational number at all is 0. But that's a contradiction — we assumed that that probability is 1.

On the other hand, if p is not zero, then the probability of picking a rational number at all is infinite, which is likewise a contradiction. --Trovatore (talk) 23:32, 15 August 2014 (UTC)

So it's impossible to define meaningful integration on the set of rationals? Interesting, I've never thought about that. Still, Algr's point remains the same if we do it on the set of reals instead of the set of rationals. Huon (talk) 23:57, 15 August 2014 (UTC)

Right (as to the last sentence), and your analysis of that case is fine. --Trovatore (talk) 01:19, 16 August 2014 (UTC)

I took Algr's wording to suggest that the machine he imagined could be built and used in a sufficiently large universe, and just wanted to point out that it is not so. If you use real numbers, then the machine almost surely produces infinite output and thus cannot be operated in reality, making it a mere fantasy in Cantor's paradise of infinities. Fantasies tend to make less convincing arguments than what is at least in principle possible to build.

That aside, I'm not even sure how his argument would result in two different zeroes. Sure, you can define a probability measure which assigns the number zero to different kinds of events, some possible and others impossible. So what? I can define a function f so that f(-1) = 0 and f(1) = 0, does that mean that there are two different kinds of zeros because one arises from a negative input and another from a positive input? That makes the same amount of sense (i.e. none) to me as claiming that there are two different kinds of zeroes because P(X) = 0 and P(Y) = 0, and X and Y are different kinds of events. -- Coffee2theorems (talk) 02:16, 16 August 2014 (UTC)

So if someone says "There is zero chance of that happening." you can't tell whether or not they think the event is impossible? That seems strangely imprecise for mathematics. Algr (talk) 05:34, 16 August 2014 (UTC)

"On the other hand, if p is not zero, then the probability of picking a rational number at all is infinite, which is likewise a contradiction." How interesting that you would say this, since that is one of the objections to .999...=1. If you add .9+.09+.009... you get an infinite number of positive values added together, thus .999...=∞. And Huon: ®1 * ®1 = ®0;1 Algr (talk) 01:57, 16 August 2014 (UTC)

No, you've missed the point. It's perfectly possible to add infinitely many positive numbers together, and get a finite value. What you can't do is add infinitely many copies of the same positive number together, and get a finite value. The assumption is that the rationals all have the same probability of being picked, and that can't happen, unless that common probability is zero. --Trovatore (talk) 02:04, 16 August 2014 (UTC)

Really? What if we took the average value of the series .9; .09; .009... and added that to itself infinitely? Shouldn't that still equal 1? Algr (talk)

The average value of those numbers (to the extent that it makes sense to speak of an average here) is zero. --Trovatore (talk) 02:31, 16 August 2014 (UTC)

So a set can have an average that is smaller than any element within the set? Algr (talk)

I should know better than to comment on this page. Oh well.

Anyway, there does not properly exist an "average" of these values, in the ordinary sense of the word. For the same reason that there is no uniform probability measure on the rationals, there is also no uniform probability measure on the natural numbers. So you can't get the average in the usual sense, which is just the integral of all values with respect to the probability measure.

However, if you insist on assigning a meaning to the phrase "average of 0.9, 0.09, ...", there are various things you might try, and they all happen to give the value 0. One of them, for example, would be: Let A_n be the average of the first n terms of the infinite sequence, and take the limit as n goes to infinity.

In that sense, yes, the average can be less than any of the terms. --Trovatore (talk) 02:43, 16 August 2014 (UTC)

I agree with Trovatore. But the order of limits cannot necessarily be reversed. In short, we have $\lim _{n\to \infty }A_{n}=0$ and thus $\sum _{k=1}^{\infty }(\lim _{n\to \infty }A_{n})=0,$ but $\lim _{n\to \infty }\left(\sum _{k=1}^{n}A_{n}\right)=\lim _{n\to \infty }0.999...9=1$ and $\lim _{n\to \infty }\left(\sum _{k=1}^{\infty }A_{n}\right)=\infty .$

Also, 1/(1®-1) * (1®-1) = 1 - ®0;1 < 1? If not, why not? Huon (talk) 02:54, 16 August 2014 (UTC)

1/(1®-1) * (1®-1) = 1 and 1 - ®0;1 < 1. I don't see how you got 1/(1®-1) * (1®-1) = 1 - ®0;1 Algr (talk)

As I showed above, here with more details and explanations:

1 = 1/(1®-1) * (1®-1) according your latest comment

= (1®1) * (1®-1) according to your comment here

= (1® + ®1) * (1® - ®1)

= 1® * (1® - ®1) + ®1 * (1® - ®1) according to the distributive law

= 1® * 1® - 1® * ®1 + ®1 * 1® - ®1 + ®1 according to the distributive law

= 1® * 1® - ®1 * ®1 because the middle summands cancel each other out

= 1 - ®1 * ®1

= 1 - ®0;1 according to this comment

< 1 according your latest comment.

If that's wrong, at what step? Or is 1 smaller than itself? Is equality not transitive? Huon (talk) 13:13, 16 August 2014 (UTC) Sorry, I'm to sick to do math this week. Seen two doctors already. Algr (talk)

On some statements concerning equallity of 1 and 0.999...

Here I will try to show that the following well known proof is not correct (or actually is correct concerning only the real numbers, although still with some flaws). I chose it because in it the mistakes are harder to be noticed.

x = 0.999...
10*x = 9.999...
10*x-x = 9.999... + 0.999...
9*x = 9
x= 1 = 0.999...

Let's take 1/3 and try to represent it as a decimal fraction.

1/3 = 0.3 + 1/30 = 0.33 + 1/300 = 0.333 + 1/3000 = ...
or
1/3 = 0.3 + 1/(3*10) = 0.33 + 1/(3*10^2) + 0.333 + 1/(3*10^3) = ... =
= 0.333...3<n times> + 1/(3*10^n) = ... = 0.333... + 1/(3*10^∞)

Now let's substitute the last expression in above mentioned proof.

x = 3*(0.333... + 1/(3*10^∞))
10*x = 30*(0.333... + 1/(3*10^∞))
10*x-x = 27*(0.333... + 1/(3*10^∞))
x = 3*(0.333... + 1/(3*10^∞))

It is the same. Even if we not use the infinitesimal part, which is not a real number (this does not mean it does not exist, but only that it is not a real number), still nothing changes.

What will happen if we get rid of the brackets? Let's try.

10*x = 30*(0.333... + 1/(3*10^∞))
10*x = 9.999... + 30/(3*10^∞)
10*x-x = 9.999... + 30/(3*10^∞) - 0.999... - 3/(3*10^∞)
9*x = 9 - 27/(3*10^∞)
x = 1 - 3/(3*10^∞)
x = 1 - 1/10^∞

1/3 = 0.333...3<n times> + 1/(3*10^n)

So it is all about using or not this infinitesimal part and keep in mind not to break the correlation between the number of the repeating digits and power n in this infinitesimal part. Because if we break this correlation we will have some other fraction and not 1/3 as in the example. The well known proof is a speculation but it is actually not needed, because there is no room for infinite and infinitesimal numbers in the real numbers line (which will make the difference between 0.999... and 1). This is the way mathematicians defined it, so it does not matter I disagree.


Now about some statements. I read some arguments that 9*0.999...is 8.999...91 but they are also wrong. It can be easy understood considering for exemple: 2*0.666... = (2*2)/3 = 4/3 = 1.333... not 1.333...32. Even if we include the infinitesimal part we'll have:

2/3 = 0.666... + 2/(3*10^∞)
4/3 = 1 + 1/3 = 1.333... + 1/(3*10^∞)
which is more than both 1.333... and 1.333...32

S — Preceding unsigned comment added by 95.42.185.235 (talk) 12:40, 11 October 2014 (UTC)

Unless you specify a number system in which you work that's different from the real numbers presupposed by this proof, 3*10^∞ is just a meaningless string of characters, not a number. Furthermore, I expect we can agree that in whatever number system you do use, 0.999...=3*0.333... should hold, without an additional infinitesimal summand. Then the proof from the article will look as follows:

x = 0.999... = 3*0.333...

10x = 30*0.333... = 3*3.333...^[1]

10x-x = 3*3.333... - 3*0.333... = 3*(3.333... - 0.333...)

9x = 3*3 = 9

x = 1

The line marked by ^[1] is the only place where there might be some wiggle room: Here it's assumed that multiplying by 10 will exactly shift every digit of an infinite decimal one place to the left. This proof, wich is not meant to provide utmost rigor but rather easy access for non-specialists, simply takes that for granted, a straight-forward generalization of a behaviour shown by finite decimals. For a more detailed proof we'd need a formal definition of infinite decimals via limits.

Furthermore, I have to note that when you enrich your number system to ensure that the decimal representation 0.999... no longer represents 1, you in the process also ensure that 1/3 is no longer represented by a decimal at all. For the standard real numbers, every number has a decimal representation, and some numbers have two. For your system, no number has more than one decimal representation, and quite a few have none. That makes decimal representations pretty much useless. Huon (talk) 14:48, 11 October 2014 (UTC)

It is true that I didn't specified a number system nor I'm going to invent one. I don't think I can express in mathematical rules a philosophical conceptions explaining infinity and infinitesimals. But infinitesimals exist and real number's definition ignores them. It's ok for practical purposes. I agree there is no difference expressed in real numbers between 1 and 0.999... But this does not mean there is no difference at all. We sacrifice one very important principle if we talk about its non-existance.

Let's consider x = 1/3 = 0.33333 + 1/(3*10^5)

10*x = 3.3333 + 10/(3*10^5) = 3.3333 + 1/(3*10^4) = 3.33333 + 1/(3*10^5)

It is really important to keep the relation between the number of threes and the power in

1/3 = 0.333...3<n times> + 1/(3*10^n),

because if we take

x = 1/3 = 0.333... + 1/(3*10^∞)

10*x = 3.333... + 10/(3*10^∞)

we will make a mistake. Correct is

10*x = 3.333... + 1/(3*10^∞)

9*x = 3.333... + 1/(3*10^∞)- 0.333... - 1/(3*10^∞) = 3

x = 1/3

Main problem with the proof we discuss is that it breaks the relation I mentioned above. There is a difference between 1 and 0.999... and there are strict rules which define it. Actually, it is not needed to define a number system which uses infinitesimals. This is because of the so many times mentioned correlation. It is exactly the same infinity set which is used to define the infinite number of threes and can not be another. Since mathematicians use it when talking about 0.999... and 0.333... I'm sure they have already defined it.
— Preceding unsigned comment added by 95.42.185.235 (talk) 17:08, 11 October 2014 (UTC)

All we care about in this article is the reals; and the string "10^∞" is not meaningful within the reals. Feel free to create your own number system where that string is meaningful, but it has no part in a discussion about a phenomenon within the reals. --jpgordon^{::==( o )} 22:37, 11 October 2014 (UTC)

Already said I'm not going to create another number system. I wrote all above for those who have the inner feeling that something with 1=0.999... is wrong. If you're not one of these people - it's ok. If you don't care about the infinitesimals because they are not real numbers, maybe other people care. — Preceding unsigned comment added by 95.42.185.235 (talk) 08:11, 12 October 2014 (UTC)

The problem is that introducing infinitesimals comes at a steep price, the most obvious part of which is that the way you do it, decimal representations cease to represent useful numbers. Who wants to represent a number a little less than 1/3 when 1/3 no longer has a decimal representation? Also, this is a mathematical topic, not one where "inner feelings" are useful in the absence of rigor. It's possible to introduce infinitesimals rigorously, but then it's not even clear why the number 1/10^∞ (if we can define it) should be called "0.999..." and not, say, 1/100^∞. Huon (talk) 14:03, 12 October 2014 (UTC)

What you refuse to see is that I do not introduce ifinitesimals. I'm only showing a relation which is a fact. The existence of the infinitesimals is just a necessary result of the infinite numbers of nines. There is no need to define the result if the cause for the result (the repeating nines) is well defined. Why the difference between 1 and 0.999... is 1/10^∞ and not 1/100^∞ shold be clear. It was mathematically and logically explained.

Talking about infinitelly repeating nines we have not to forget that every nine includes nine ones. Saying 0.999...=9*0.111... only ignores the problem. The infinitieth nine still has to be composed of nine ones and here arise the necessity of the infinitesimals once again. And they can not be zero because 9*0≠9. I suppose someone might say: there is no infinitieth nine. But then we could not talk about infinitelly repeating nines at all. Which is maybe correct because even the number of the atoms in the physical universe is finite.--212.5.158.204 (talk) 09:07, 15 October 2014 (UTC)

You need to learn to work with formal systems before trying to make novel mathematical proofs (a very good start point is Gödel, Escher, Bach, a must read for anyone logically-inclined). I cannot make sense of sentences like "The infinitieth nine still has to be composed of nine ones" if those words are not defined in unambiguous logical terms, which is what mathematicians implicitly do when they talk about the existence and properties of numbers.

And of course you can talk of infinitelly repeating nines without having an infinitieth nine (in fact I find it hard to do the opposite). "Infinite nines" defines a process, not a number; it's the process where, if you have a finite number of nines in a decimal notation, you can *always, with no exceptions* add just one more nine to the right (if there were exceptions, the process would not be infinite, by definition).

In a process defined this way, what does it mean to have an "infiniteth nine"? It would have to be a "9" where you can't put another "9" to the right by multiplying it by 1/10 and adding it to the first; but that's absurd and a contradiction with the way we have defined the process, which we know is infinite because you can always add more "9"s. Infinity doesn't work that way, and teachers trying to explain it by mentioning objects "in the infinite" are doing their students a disservice IMHO. Diego (talk) 09:35, 15 October 2014 (UTC)

So the number of nines approches infinity but never reaches it. Thus we have a finite number which is only potencially infinite but not realy infinite at any given moment (unless we assume the process has no beginning). I fully agree with such a deffinition. But why then we in this topic compare a process with a number? And is it correct to look at the process as it is finished since it is unfinishable. I hope no one doubts we can always add another nine. Else if we consider one endless process finished, we create a contradiction and will have a wrong base. But if the process is not finished and can not be finished, there will always be difference between the number of nines we have and one. Yes the number of nines will also not be infinite and we will not need infinitesimals. --212.5.158.204 (talk) 11:16, 15 October 2014 (UTC)

That's where confusion usually comes from. When you write "0.999..." you don't represent the process (a series of additions), you represent a finite number, which is the result of the (infinite) process of partial additions. So you never compare a process to a number; the process, if it could be finished, would have a result that is a number. It's like comparing 4^2 and 16; you don't compare 16 with "taking the square" of 4 (an operation), you compare 16 with "the result of" taking the square of 4 (a number).

The act of adding decimals to an intermediate result is an infinite operation; you can always add more decimals, and you can't reach a "final decimal". If you stop the process at any time (which you necessarily need to do if you live in this universe), the result is not the number "0.999...", it's a different number strictly smaller than "0.999..." (you will see many numbers: "0.9", "0.99", "0.999", "0.9999", etc as a result of the process; none of them can be called "the sum of the infinite sequence").

Thus the finite number "0.999..." is defined as a series, i.e. "the sum of an infinite sequence". The only way to reach the end of an infinite process is to cheat; you define a different, finite process and say that the result of that finite process is equal to the result of the infinite process, by definition. In this case, this alternate finite process is "taking the limit" of an infinite operation. The operation (an infinite sum) is endless, but what you call its result ("the limit") is a finite number.

Taking the limit is a simple process; you have to prove that all the partial results in the series get closer and closer to the number (in this case, the number "1"), which we already know is what happens with the series $\sum _{k=1}^{n}{\frac {9}{10^{k}}}$. If you use the definition of the "limit" function, it's simple to see (with few steps of mathematical proof) that the only real number which fits that definition is the number one. If you want to find a different number that fits that definition, you have to redefine the "limit" function to work on a different set of numbers outside the reals. You can do that, but then you're talking of a different mathematical object, which is not the same one that we called "0.999..." when we started. Diego (talk) 12:10, 15 October 2014 (UTC)

Here I can ask if there is no other real number between two numbers does this make them equal? We have the number one and a sum of series with limit one. From the mathemathics point of view the problem maybe is well defined. But as I mentioned somewhere above (and it is actually the main reason I'm writing here) we sacrifice a principle (and I am not even sure why). 5.xxx... will never equal 4.xxx... And no definitions could change it. There is an order in the universe which can not be altered. Even trying to introduce infinite numbers which we can only approximate here they are always hand by hand with the infinitesimals. We can ignore the last but it is our problem. "Universe" will still use them to prevent 5.xxx = 4.xxx.

We may talk about sq root of 2 or three, or 0.999..., but we'll never use them in their ideal form here. Maybe we will in some ideal worlds, but not here. Here even knowing exact algorithms to calculate pi or sq root 2 gives us only approximation for physical world is limited in every aspect - even in time duration.

This probably will be the last thing I write on this topic since it is related with math and if I continue to write have to use non mathematical arguments.

--95.42.185.235 (talk) 14:50, 15 October 2014 (UTC)

Here I can ask if there is no other real number between two numbers does this make them equal? Pretty much yes. I'm not a mathematician, but if I recall correctly that's the definition of the equality of reals in standard number theory; it's the way we have to test if two real numbers are the same.

Here even knowing exact algorithms to calculate pi or sq root 2 gives us only approximation for physical world - that's why don't calculate them, but use algebra... (If you don't calculate numbers but represent them with symbols, you don't loose precision. Mathematical software follows this approach).

Mathematical objects don't "exist in the universe", they're a result of the axioms you choose. If you change the axioms, you're describing a different set of objects. Mathematics is no more (nor less) than a language to make claims with extreme precision. There's no such thing as "number 5.XXXX...", there's the theory of Reals where that string of symbols is given a precise meaning. In this theory, the strings "5.000..." and "4.999..." have the same meaning - i.e. they can be connected to the same object under that theory using the rules of inference of logic.Diego (talk) 15:22, 15 October 2014 (UTC)

If we pretend that the number of nines in 0.999... is infinite, we have to accept that there is a nine in this string which can be defined as 9*1/10^∞. But if accordind to the deffinition for real numbers 1/∞ is not real and zero, we will have zero for 9*1/10^∞ and not 9 at the related position and all positions after it. Thus we will not have an infinite number of nines. If the number of nines only approaches infinity we still don't have an infinite number of nines.

While you are discussing whether to delete this page or not, I read the archives and also realized the problem is far beyond this current topic. But it is actually not my problem to solve. I did what I can to show it is impossible to introduce numbers like 0.999... and 0.333... whithout having the infinitesimals as a consequence and that in this case the infinitesimals can not be zero. We have either to accept the existence of both - the infinitesimals and 0.999... or deny them both.212.5.158.185 (talk) 10:29, 17 October 2014 (UTC)

If we pretend that the number of nines in 0.999... is infinite, we have to accept that there is a nine in this string which can be defined as 9*1/10^∞. Not at all. For me, "the number of nines in 0.999... is infinite" is synonym with "if you try to write down the repeating decimal, you can always add one more 9 to the right". It doesn't imply the existence of a "decimal at the infinite"; that would require the explicit decision of creating (defining) a new mathematical entity, and I don't agree that this entity exists (or rather, that it's needed here: "I don't want to use it in the context of real numbers", because if I use it, I'm changing the context).

Remember, the process of writing a string down (the decimal representation) is different to the number you try to represent with that string (the real number 1). I can define rules for writing down repeating decimals, and those rules work without having separate rules for talking about positions at the infinite. I can do mathematics by accepting the axioms of the first concept and rejecting the axioms for the latter, which is what allows me to say that 0.999...=1 with the axioms I've accepted. Mathematical truth is not absolute, it's always relative to the particular set of rules you accept as valid at any given time; if you change the rules, the truths behind them change as well. This was not well understood before the XIXth century, but that's how modern mathematics are used.

accordind to the deffinition for real numbers 1/∞ is not real and zero According to the definition of real numbers, 1/∞ is not a thing. Although lim x->∞ (1/x) is a number, the number 0.

To relate it to your assertion, we'd have to agree that both 0.999... and infinitesimal exists, or neither of them, if we agree to use rules for infinite positions of repeating decimals. But there's nothing that forces us to use rules for infinite positions; the choice of axioms in a mathematical system is entirely arbitrary, you can choose those that fit your taste and needs. Diego (talk) 11:34, 17 October 2014 (UTC)

It doesn't matter how you formulate it - there are two opportunities: 1) we have an infinite number of nines or 2) we have finite number of nines. The way the topic is formulated we have to accept that the number is infinite. And it means equal, not approching infinity. If we can formulate the n-th nine as 9*1/10^n and if n=∞ (as topic formulation states) it follows that there has to be a nine in the form 9*1/10^∞. It does not matter whether we'll name it last or not and it does not matter that infinity is not defined and not a real number. It is condition stated in the formulation of the problem and not my job to define. About the statement that 1/∞ is not a thing in real numbers - accepting this means that 0.999... does not exist as a real number because as mentioned above there has to exist a nine in the string of reoccurring nines formulated as 9*1/10^∞, which includes 1/∞.--95.43.77.26 (talk) 14:01, 17 October 2014 (UTC)

There's no "equal to infinity" in the reals; if you accept that kind of equality you've already changed talking to something else. And no, it doesn't follow that "if n=∞ it follows that there has to be a nine in the form 9*1/10^∞" - in logic, a new statement only follow from previous ones if there's an inference rule that produces it, and you're not using any inference rule - only saying that things are like that because you say so. That way of thinking is older than the greek themselves, and not how you do proper maths. To convince me, all the steps in your argument need to follow the rules of logic.

And here you're making the fallacy of the excluded middle. There's a third possibility that you didn't take into account - that the process to define the number at the end of the sequence is infinite", but the real number itself described by that process is finite. Doing that, I don't "have to accept that the number is infinite"; number 1 can be described by an infinite process without being itself an infinite amount (in fact, in set theory all reals are defined as sets of infinite elements). Diego (talk) 14:19, 17 October 2014 (UTC)

Whether or not there is an "equal to infinity" in reals is not my problem. It is the formulation of the problem stating that the nines of 0.999... are infinite in number. But we are already going in circle. Also my logic is correct and there are no statements out of nothing. If n corresponds to the number of nines in 0.999... and if this number is infinite, and if the n-th nine in 0.999... is equal to 9*1/10^n, it does follow that there is a nine equal to 9*1/10^∞.--77.85.46.66 (talk) 15:34, 17 October 2014 (UTC)

No, it really doesn't. If there is a nine equal to what you think is 9*1/10^∞, then what is the digit that follows that one? I guess it would have to be equal to 9*1/10^(∞ + 1). But ∞ + 1 = ∞, and we can keep doing that forever, so there are going to be an infinite number of nines after whatever you think is indicated by 9*1/10^∞. Every digit in a decimal expansion has a unique label -- in other words, decimal expansions are countably infinite. There is not a digit in 0.999... that can have that 9*1/10^∞ label; it's really the old "pick a number as big as you can think of. Now add one." game. --jpgordon^{::==( o )} 15:55, 17 October 2014 (UTC)

For a start, you haven't defined what 9*1/10^∞ means. If a term is not properly defined with axioms and theorems, it can not be logically derived. Unless you can give an exact definition of every symbol you use, the process of demonstration is not formal, and anyone can reasonably disagree with your conclusions - by choosing a definition in which your conclusion doesn't follow. Again, those are the principles on which modern mathematics rely upon. Diego (talk) 16:05, 17 October 2014 (UTC)

I think there's some confusion here between cardinals and ordinals. In 0.999..., there is one nine for each natural number: A first nine, a second nine, a third nine, ..., a three thousand seven hundred and fifty-first nine, and so on. How many natural numbers are there? Infinitely many. But the number of natural numbers is not itself a natural number. So if n corresponds to the number of nines in 0.999... (that number is commonly called ℵ₀), there is no n-th nine in 0.999... Huon (talk) 19:36, 17 October 2014 (UTC)

9*1/10^∞ is 9*1/10^n if n=∞. It was explained. It was also explained why n=∞. I do not have to define infinity because it is condition given in the formulation of the problem. Next I will cite something written by Euler just to show the way he defines the things.

"If there is a nine equal to what you think is 9*1/10^∞, then what is the digit that follows that one?"

It does not matter. But answer is: if there are other positions after the mentioned one, they will all be zero, if 1/∞ is zero. Note that ∞ may not be a real number but stands for a number. We just don't know its magnitude. We only know that it is bigger than every other number. The same is with the count of the nines - it is not a real number and we don't know its magnitude, we only know that it is bigger than every other number, so we have full correspondence here (and this correspondence was also explained above).--95.42.178.1 (talk) 19:41, 17 October 2014 (UTC)

Of course it matters. If it didn't matter you'd be correct, but the fact that it matters is the primary thing that makes you incorrect. There is neither a largest integer nor a smallest real number; that's how infinite works. I think you need to study some real mathematics before you attempt any more discussion here; you lack the understanding of series, sequences, natural numbers, and real numbers necessary to make a coherent argument. --jpgordon^{::==( o )} 20:02, 17 October 2014 (UTC)

No it does not matter. Once the string of nines is broken it cannot be infinite anymore. It is not my definition for infinity that an infinite number is such that is bigger than any real number. It's in the main article Infinity.

@Huon, maybe you are right and ℵ₀ is the correct notation instead of n - I'm not familiar with that matter. For me important is the correspondence mentioned and I think my explanations were not hard to grasp.--95.42.178.1 (talk) 21:15, 17 October 2014 (UTC)

Once the string of nines is broken it cannot be infinite anymore. And then it's not 0.999... anymore, either, so it's completely irrelevant to this discussion. There is no "infinity-th" position that "breaks" the string; there's always another nine. Always. --jpgordon^{::==( o )} 21:49, 17 October 2014 (UTC)

I'll add that no number "∞" is given in the problem. It certainly does not appear in "0.999...". As I said above, 0.999... has one nine for every natural number - no less, but no more, either, with the position of each nine given by the corresponding natural number. There's no ∞-th nine in 0.999..., just as there's no 5/3-th nine and no √2-th nine. Even if we used a number system which includes ∞, there's no a priori reason why there should be an ∞-th digit in 0.999...

For more infinite fun, think about this: If ∞ is the number of nines in 0.999..., then what's the number of nines in 9+0.999...=9.999...? Surely it's ∞+1, because there's one additional nine before the decimal separator. However, since you said that there's no number larger than ∞, we must have ∞+1=∞. But then 0.999... with ∞ nines after the decimal separator equals 0.9999... with ∞+1 nines after the decimal separator. Multiplying the latter by 10, however, will shift all of the ∞+1 nines one place to the left, and we'll end up with 9.999... with ∞ nines after the decimal separator - no less, not even some infinitesimal fraction! So we end up with:

10*0.999... = 10*0.9999... = 9.999... = 9+0.999...

Subtracting 0.999... on both sides will give 9*0.999...=9, and the algebraic proof works even if we assume there was some number ∞ that's equal to the number of nines in 0.999... and for which no larger number exists. Huon (talk) 23:20, 17 October 2014 (UTC)

It is a simple logic: if the number of nines is infinite, there has to be an infinieth nine. If there is not an infinieth nine, then the number of nines is not infinite. Also I see you're using arguments which you opposed before. Infinity is infinity. You can not add another nine after you have reached infinite number of nines (the last being also imposible and was stated above). You can add nines before you reach the infinity. But the formulation of the problem states that the number of nines is already infinite, so according to it, you are not allowed to add another nine. Don't try to ask me questions which are consequence of the bad definition of the topic we are discussing. It is not me who has formulated the topic.--95.42.187.202 (talk) 07:22, 18 October 2014 (UTC)

That simple logic is wrong. There are infinitely many natural numbers; infinity is still not a natural number. Similarly, there are infinitely many nines in 0.999..., but no infinitieth nine.

I was trying to show that even if there was an infinitieth nine, 0.999... would still equal 1. I indeed didn't add a nine after the last one for that purpose, but only before. In short, how many nines has (9+0.999...)/10? That still has infinitely many nines, just as many as 0.999..., and it must equal 0.999... because, as you just said, we can't be adding another nine after the nines of 0.999... And as long as x=(9+x)/10, we can conclude x=1. Huon (talk) 15:08, 18 October 2014 (UTC)

If the number of the nines in 0.999... is infinite and if there is no infinieth nine -> there exist at least one position with two nines in it. But this will also break the string of nines at that position. In simple worlds you cannot place n nines in n-1 positions and still have one nine at any position. You need n nines and n positions.

One of the first things I showed in this thread was that the "arithmetic proof" stands in the case we have a difference between 1 and 0.999..., so it actually proves nothing.--95.43.65.53 (talk) 16:06, 18 October 2014 (UTC)

You've moved into IDIDNTHEARTHAT mode. We can probably end this conversation. --jpgordon^{::==( o )} 17:25, 18 October 2014 (UTC)