Talk:0.999.../Arguments/Archive 6

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Publish the source!

Please show me where it is clearly published that an infinite sum is defined as the limit of an infinite sum and I will be the first of the unbelievers to support your article. None of your sources state this explicitly. They are all open to reader interpretation. Once again, you insist no original research - so practice what you preach. You are a laughing stock. Honestly, I warn everyone I know about reading anything on Wikipedia. You are sketchy and unreliable. Go ahead, do the right thing and publish a source - print every word as it appears in your source so that there is no doubt this article is not original research. 41.243.47.226 14:10, 1 December 2006 (UTC)

If you read 0.999...#Infinite series and sequences, your attention will be drawn to note 5, which reads "5. ^ For example, J. Stewart p.706, Rudin p.61, Protter and Morrey p.213, Pugh p.180, J.B. Conway p.31". Full bibliographic details are in 0.999...#References. I recommend trying Stewart first. Melchoir 16:38, 1 December 2006 (UTC)

You could also look at, for instance, this. CroydThoth 21:08, 1 December 2006 (UTC)

Just copying from the archive, there's also:

Advanced Calculus by David Widder, page 285, from 1947

Introduction to the Theory of Fourier's Series and Integrals by H.S. Carslaw, page 49, from 1930.

Advanced Calculus by Sokolnikoff, page 6-7, from 1939.

Calbaer 21:57, 1 December 2006 (UTC)

Sure, here's one from my first year Calculus textbook, Calculus: Concepts and Contexts by James Stewart, second edition (published 2001 by Thomson Learning Inc., ISBN 0-534-37718-1), page 574:

Definition: Given a series $\textstyle \sum _{n=1}^{\infty }a_{n}=a_{1}+a_{2}+a_{3}+\cdots$, let s_n denote its nth partial sum:

$s_{n}=\sum _{i=1}^{n}a_{i}=a_{1}+a_{2}+\cdots +a_{n}$

If the sequence {s_n} is convergent and $\textstyle \lim _{n\to \infty }s_{n}=s$ exists as a real number, then the series $\textstyle \sum a_{n}$ is called convergent and we write

$a_{1}+a_{2}+\cdots +a_{n}+\cdots =s$ or $\sum _{n=1}^{\infty }a_{n}=s$

The number s is called the sum of the series. If the sequence {s_n} is divergent, then the series is called divergent.

Hope this helped! Maelin (Talk | Contribs) 08:40, 3 December 2006 (UTC)

There you go with those pesky facts again. Facts will never sway our favorite skeptics. 72.193.74.36 16:14, 3 December 2006 (UTC)

This theory doesn't work because ultimatly it relies on 'infinity' as if it shares the properties as all other numbers even though it does not considering it is not a set number. Also, even if your convoluted formula did work, there would always be the difference of that .1 or.01 or .001. You can't just assume that there is .9 number with infinite 9's because even so it would be missing that '1'. —The preceding unsigned comment was added by 74.99.150.228 (talk • contribs) 00:36, January 26, 2007 (UTC)

More nonsense. — Arthur Rubin | (talk) 01:12, 26 January 2007 (UTC)

It is very unfair of you to call me a troll. I am your friend and hope that Wikipedia will someday be taken seriously. However, your rules state that no original research is allowed and information must have been published before it can be considered encyclopedic. How long must it have been published for? And by whom does it need to be blessed? The source referenced by Maelin (James Stewart, Calculus and Concepts) has not been around before 1960. What's so special about 1960? Well, a lot. Let me begin by stating that many 'mathematicians' born after this year have (shall we say) not been of a very good caliber in general. Most progress in Mathematics (including differential equations) has been made through the age old method of experimentation. Much experimentation and very little thought. But before I go off on a tangent, let me return to this article. Until 1980, I had never heard any professor of mathematics ever refer to a limit of an infinite sum (obviously a sum that converges) as its definition. There was always a clear distinction between partial sums (which approach the limit) and the actual sum of a series that was considered unattainable. Will I ever state in public that a sum is not defined as the limit of its partial sums? Of course not! Why? Not because it makes sense, but because the powers that be have decided to make it doctrine. Mathematics worked just fine for several hundred years without this ridiculous article and without any real analysis. Is real analysis bad? Well, of course not, but it is by no means beyond reproach and debate. Nothing is beyond debate - ever. You will retard the progress of many aspiring learners who cannot understand this rubbish because indeed that's exactly what it is. Your publication of this article supports the views of individuals such as Prof. Hardy, Kmsrq, etc who are all important contributors in your team. By publishing this article, you are solidifying the viewpoint of these and similar minded individuals. You constantly refer to academia who support your view but do you pay any attention to the hundreds of opposing views? It seems that anything becomes encyclopedic only if it is in line with your views. Repeated comments and posts have shown this article to be so full of holes. Yet you continue to maintain the stance that it is encyclopedic. What an irony! Encyclopedia is a Greek word literally meaning - 'encircling children'. The only real audience you have for such an article is the most foul and obnoxious individuals like Ksmrq and Prof. Hardy. Just because you cannot understand why 0.999... < 1 does not mean that it is untrue. Make sense? You cannot just decide upon a definition and then make it official doctrine. Why, even Newton did not dare publish knowledge just in case it was wrong and he might end up misleading many. Newton was not only one of the greatest mathematicians but no doubt the greatest scientist. Mathematicians today are so full of arrogance and obnoxiousness, they would not be able to recognize a 'truth' even if it were forced up their rear-ends! Back to the article: You discuss so many concepts that do not exist either in theory or practice. A good example is your so-called 'infinitesimal' - this is an oxymoron. Then you try to back your claims up by categorizing 0.999.... Only problem is you misplace this real number in your classification. It is in essence very similar to an irrational number.

The definition of rational number is a number of the form a/b (b not 0) and a,b integers. Has it crossed your minds that for the ancient Greeks a recurring decimal may not have been a rational number? Furthermore, when they designed arithmetic, there was no such concept as a limit. Indeed, how can any recurring decimal be a rational number unless it is actually possible to perform an infinite sum? So the Greeks not knowing anything about limits could not have had the same definition of rational as we have today. What I am saying is that just because a number might have a recurring pattern, this does not mean it is rational.

Once debate is dead, so is knowledge. This is why today so many graduates leave college even dumber than they were when they first started their education. Until there is no doubt about what an infinite sum is, this article should be deleted. You cannot just decide to define an infinite sum as a limit, it is already defined as an 'infinite sum'. 198.54.202.246 12:14, 5 December 2006 (UTC)

That's not the idea at all. An encyclopedia is a place to collect other people's research. There's no magic date something has to have been published and there's no one person or body who needs to "bless" it. It just needs to have been published somewhere, or even just reported upon someplace in the media. Things are reported as "fact" only when the research leaves no reasonable doubt. Even if 0.999... did not equal one (which is plainly does) it would still be wrong (by Wikipedia's policies at least) for Wikipedia to say so because that is not encyclopedic. That would be original research. Wikipedia is a website that can be edited by anyone, with or without any understanding of the topic, and therefore the only way it can ever hope to be respected is by consistently citing reputable sources. That way it can become a huge repository of knowledge. But it was never intended to be a source of knowledge, and that's an important disticntion.

And I keep being told that there are many mathematicians who dispute that 0.999...=1 just as I keep been told there are many scientists who reject evolution, and I have never met even one person from either category. I suspect this is because their numbers are hugely exaggerated.

Lastly, it is entirely reasonable for mathematicians to define terms like "sum" however they please. They're just words. They don't alter the underlying mathematics - you can invent another type of sum if you please, and coin your own term for it. But you can't dispute the ones we have because we've defined them quite precisely already.

Your little paragraph about rational numbers was interesting, though. I think it shed light on some of the problems here. If you'll indulge me, can I just spin the camera on this one and show you it from another angle? I think it would be a useful exercise for all of us...

Let's forget decimals for a moment. Pretend they haven't been invented yet. Let's just look at the maths. A rational number is defined as a/b (with caveats). Let's say a=1 and b=3. So, one third is a rational number. π, on the other hand, is not. Five is a rational number. e is not. The definition works perfectly, and every number we know of falls neatly into the rational or irrational category. There's nothing in there so far the Greeks couldn't have done, and I suspect nothing they didn't do. They were busy.

Now let's introduce decimal expansions. Remember, we've started from the maths and then brought in decimals. They're not numbers; they're not even things. They're just a convenient way of writing the numbers we just defined. Well, five is easy. That's "5". π is a bit trickier. We can Do the first bit, but we need infinite digits to get it right. But because it's irrational we couldn't write it in the old way either, so that's no great loss. The same goes for e — and it turns out that this is true of all irrational numbers. We have a handy second definition here, perhaps.

No. Hang on. One third is infinitely long in decimal too. But a closer look reveals that all of its infinite digits are threes. So let's say that a zero and a dot and an infinite string of threes (which we can just write as "3 reccurring") represents a third. Now we have a handy number system where we can write all rational numbers precisely. It's more writing than the old a/b method, but it's more obvious how large a number is just by looking at it. Let's keep this system and a/b, and from now on we'll use whichever system is more convenient at the time.

Later on, some mathematicians who had the added adantage of a few centuries of extra research formalise this system by introducing concepts such as the limit of an infinite sum. All I did there was start with maths and then simply use decimals as a way of writing that maths. Your "rational numbers" speech implied that decimals are the starting point and that if the Greeks couldn't write a number in decimal then they couldn't use it. We don't know what the exact value of π is, decimal or not, but that doesn't matter: we have a definition for it and we can still use it in algebra. We can even use numbers like i, that patently aren't real. Imaginary numbers can be used in mathematics. Why can't the Greeks use a number, just because they didn't know how to write it down?

What's so unlikely about that? If you look at the decimal system now, and think, "gosh, that's a complex system that involves a lot of things the ancient Greeks wouldn't understand. There's no way that could have come about before this century. There must be some mistake," then you may as well say the same thing about evolution. Only looking at the final step in the process evolution looks unlikely, but looking at each step on its own suddenly it's obvious. Looking at mathematics as it's now defined, yes, it certainly couldn't have been concocted by one Greek guy of an evening. But that's because that's not what happened. It was built up over centuries, and refined, and by now I think we've got a pretty good understanding of how the thing works, I'll thank you very much.

Andrew 14:14, 5 December 2006 (UTC)

The complex number system is also flawed so I don't know what you are trying to say. Having a definition without an exact value is worthless. All our calculations are approximations. Definitions that don't make sense do not belong in mathematics. This article is not supported by any reliable sources. It is only the interpretation of your editors: Meni Rosenfeld, kmsqrq, melchoir, prof. hardy and the like. Your interpretation is just that: an interpretation that lacks proof. Frankly it does not matter what 0.999... equals to, just as it does not matter what pi equals to. Why don't you write an article about what pi is equal to 'exactly'? I fail to see the importance of this article. If any of you had a modicum of common sense, you would delete it. 41.241.187.210 06:44, 18 January 2007 (UTC)

Now complex numbers are flawed as well? Listen, "common sense", as you put it, has no place in mathematics. Maybe you would be happier avoiding math. 72.193.72.226 01:30, 19 January 2007 (UTC)

Above there are references from the 1930s and 1940s which 198.54 chooses to ignore, still maintaining that no one thought infinite sums were defined as their limits before 1960. And he/she is simultaneously claims that the article should be deleted but "Once debate is dead, so is knowledge." Moreover, in this post he/she is declaring himself/herself our friend, yet the archives show numerous instances of personal attacks. And he/she wonders why he/she is called a troll.... Calbaer 17:35, 5 December 2006 (UTC)

So what's wrong with the sources listed above? 131.216.2.83 18:06, 5 December 2006 (UTC)

*Sigh* That's a long reply, Andrew. Surely you have better ways to spend your time than feed trolls? Apparently I don't either, for I made the effort to copy from the main atricle: "...proof (actually, that 10 equals 9.999...) appears as early as 1770 in Leonhard Euler's Elements of Algebra." Endomorphic 20:25, 5 December 2006 (UTC)

If everyone who posts here is a troll or feeding a troll, close the sodding page. It's arguing about a known and repeatedly proven fact, anyway -- clearly it's pointless at best. There's no Talk:France/Arguments. Considering how much stuff is deleted from Wikipedia every day for being "non-notable" this page is indefencible. Sometimes I think Wikipedia just deletes things not because it improves the service but because it considers itself "above" them somehow, but that's quite besides the point. Now, do you want a discussion, or do you just want a page for trolls to post on, safe in the knowledge that nobody will ever read it?

At least, this page should be renamed to something other than "arguments", because the fact is there is no argument. There are just some ignorant people.

On both sides. Andrew 00:27, 6 December 2006 (UTC)

Renaming the page has been suggested before :)

Also note that no less than 9 different sources (one quoted in full) have been provided in response to 189.54's initial "show me where it is clearly published" challenge. Still they disagree. Here's an excersise for the reader: Form a list of all claims made in the second 189.54 post. Then classify each claim as either mathematical, historical, cultural, or ad hominem. We can discuss any *mathematical* claims made. Endomorphic 01:48, 6 December 2006 (UTC)

We said that 198.54.202.246 is a troll (which is a fact), not that everyone who posts here is a troll. Both trolls and genuine questioners are bound to post their question somewhere, and it's better in this page than in Talk:0.999.... Genuines and those that are in doubt should be replied to; obvious trolls should not.

An argument is, by definition, an occasion where people are arguing. For better or worse, this is exactly what is happening here, so I don't really see how the name of the page is inappropriate. It doesn't "create" fruitless discussions, it just makes sure they are made here rather than someplace else. -- Meni Rosenfeld (talk) 07:25, 6 December 2006 (UTC)

Let me add a couple more sources:

"A Course in Pure Mathematics", G.H. Hardy, 1908. Ever heard of him? Also, please note that 1908<1960, though, I still don't understand why that matters.

"What is Mathematics", Courant and Robbins, 1941. 1941<1960.

72.193.74.36 16:33, 6 December 2006 (UTC)

Archive proposal

How about we mark this page as an archive, i.e. no longer used, and refer anyone with actual questions on this subject to Wikipedia:Reference desk/Mathematics? -- SCZenz 02:29, 6 December 2006 (UTC)

And then all the trolls will fill either Talk:0.999... or the reference desk with their garbage. I don't think that's a good option. -- Meni Rosenfeld (talk) 07:11, 6 December 2006 (UTC)

Per Wikipedia:Talk page guidelines, discussion not relevant to the article is subject to removal; arguing about the veracity of properly-cited facts is, in fact, irrelevant. The reference desk, I admit, could be a bit more problematic. -- SCZenz 07:16, 6 December 2006 (UTC)

I don't think so...

.999... is equal to .999... not 1. No offense, but it's actually kind of rediculous to call it 1, when obviously, in plain sight, you read the number as .9999999999999999999999999... and see it as .999999999999999999999999999... I really don't understand how in the world a number is another completely different number. Although, I'm only as teenager so I COULD be wrong, but seriously, .999... is an infinite .11111... away from 1. Maybe someone could explain in simple terms.......... VERY simple. like 14 year old simple.

P.S. My account on Wikipedia is ASHTONZANECKI, but I'm on at school and don't want to log on, so if you were going to leave me a message, leave it instead at ASHTONZANECKI.

$1-.9=.1$, $1-.99=.01$, $1-.999=.001$, etc. So of course $1-.999...$ could not be 0.111...; in fact, it's 0, since it's not negative and it's less than any positive number. That's as simple as it gets. Calbaer 18:47, 15 December 2006 (UTC)

how couldn't it be? if you do the eqation, it would be .1111111... it's as slightly above positive as possible, but still not negative. so it's NOT zero because it's simply not zero... it's .1111... why does "it's less than any other positive number" make it 0? it's still .111111... ASHTONZANECKI 01:42, 16 December 2006 (UTC)

No, you're getting your arithmetic wrong. 0.9+0.111... = 1.0111... and 0.99+0.111... = 1.10111... so how can 0.999... + 0.111... equal one? I can see why you'd think it would, but you've forgotten to carry an infinite amount of ones. Andrew 23:05, 16 December 2006 (UTC)

Hi. 0.999... = 1 in much the same way that 0.5 = 1/2. They are visually distinct notations, but they come to the same value. To understand how 0.999... = 1 might be plausible, you could first try using long division to work out the decimal expansion of 1/3, and then multiply by three to find out the decimal expansion of 3/3... -- The Anome 01:56, 16 December 2006 (UTC)

The article has a number of proofs, and it's not clear what problems the original poster had with them. Moreover, saying that .11111... is less than .01 is an odd use of notation to say the least. Anyway, as Anome indicated, there are many ways to express the same number, and 0.999... just happens to be the same as 1 (or 1.000... or 1/1 or whatever you'd like to call it). Anyway, the archives[1][2][3][4][5] have plenty about why a number that's less than any other positive number cannot be more than zero. See the Archimedean property, or, as indicated in the FAQ, try dividing it by two or ten, which should indicate the contradiction of having a "smallest postive number." —The preceding unsigned comment was added by Calbaer (talk • contribs).

'Limits' of the mind

Although the majority of the mathematical proofs seem perfectly logical at first glance, their implications are troubling. I find it nearly impossible to concieve that an infinitely repeating series of nines is equivalent to one; my logic simply denies it. Maybe my image of .99999... is faulty, for my knowledge of numerics and common sense simply won't accept it. Help? —The preceding unsigned comment was added by 74.135.8.210 (talk • contribs).

It's important to get in your head the right idea of what 0.999... is. Decimal representations are just ways of representing numbers. Numbers are points on the real number line. Much of the intuition/logic conflict in this issue arises because people develop a deep reliance on decimal representations as being fundamental, when in fact they're not. It might help your understanding to think of that point on the number line that is represented by "1", and then consider why 0.999... might also be a valid way to represent that point. Maelin (Talk | Contribs) 01:41, 20 December 2006 (UTC)

...and how it is similar to the 0.333... as a representation for 1/3. Calbaer 01:46, 20 December 2006 (UTC)

Or perhaps a proxy, since it would not have to merely be unlimited but all-encompassing of subsequent thirds to equate, which doesn't really seem to work well with numbers. That is to say it always has a third of an infinitely small unit on the end, making everything inbetween redundant (and merely tagging 3 x 0.1^Infinity on the end isn't sufficient as it loops back on itself and leaves the same problem)... but it works for practical purposes at least. Decimals and fractions don't generally mix too well in my mind (and neither can I express things too well in mathematical terms any more). ~ SotiCoto 195.33.121.133 10:44, 30 January 2007 (UTC)

You say "it always has a third of an infinitely small unit on the end". First, "always" suggests that you're referring to some process. This is not a process, it is a discrete number (with an unchanging value). Second, you say "on the end", but we're talking about an infinitely long decimal expansion, which means it does NOT have an end. You are doubly confused my friend. Tparameter 06:52, 31 January 2007 (UTC)

Never said it was a process. "Always", I suppose, is my casual way of saying "under all perceivable circumstances in which it might occur". First... your confusion. As for the matter of the end... I was told once during a Further Pure Mathematics A-level class... quite randomly... that Infinity wasn't actually boundless, but simply an entity beyond human comprehension... more of an "all numbers" thing rather than "unlimited number". Thats not hugely relevant however as I wish to point that the very fact that it doesn't have an end is why it doesn't work. The very fact that it goes on forever (an effective impossibility in terms of reality) means that eventually it must be submitted to proxy as nothing we can make use of is capable of processing unlimited data. That is to say, relative to certain other values, it always needs something more or less that cannot possibly be tagged on an end that doesn't exist. As such... 1/3 cannot be decimalised with 100% accuracy (only 99.999...% accuracy, ironically).

Incidentally though, I doubt you're going to pay much heed to anything I say here since I'm not a mathematician (and I know how some math folks get about non-math folks); I'm a palaeobiologist by particular academic leaning... but still. ~ SotiCoto 195.33.121.133 12:28, 31 January 2007 (UTC)

You keep bringing up reality. Maths has nothing to do with reality. Numbers do not actually exist, they are mental constructs. Mathematics is a human invention that exists only in our minds. Often, mathematics provides us with a (quite astonishingly accurate) model of the world, but we mustn't forget that the numbers do not really exist. We are not analysing the properties of real world things in mathematics, we are designing mental models from the basis of some very, very simple rules and then we are looking at the results that follow. Maelin (Talk | Contribs) 12:57, 31 January 2007 (UTC)

Oh my. I will have to disagree with my friend Maelin here. I would say that it's the other way around. Mathematics exists in a way which transcends such insignificant things such as humans and the physical universe. If anything, it's reality that doesn't really exist, I view it just as some particular, arbitrary manifestation of some mathematical structure.

We do not care how extensive one's mathematical background is, as long as he comes with a willingness to learn. I could discuss palaeobiology with you; I can assure you that everything I would say will be nonsense, as I know nothing about the subject. You may find it intereseting to explain to me why I am wrong; but interest will turn into contempt should I choose to ignore your explanations and insist that I am right. A single human cannot know everything, but he should have the capacity to learn about things he doesn't know from people who do. -- Meni Rosenfeld (talk) 15:25, 31 January 2007 (UTC)

The key is that this article is about the real number definition. You should study real numbers if you want to know why the real number 0.999... = 1. 72.193.74.36 08:29, 21 December 2006 (UTC)

Correction: "You should accept real numbers if ..." etc. Then again, if you're only referring specifically to 0.999... as a "real number" then its not a problem; the problem arises if you're claiming there is no such thing as a 0.999... that isn't specifically classed as a mathematically conventional "real number". ~ SotiCoto 195.33.121.133 12:28, 31 January 2007 (UTC)

There's no need to "accept" real numbers. It's no more meaningful to discuss accepting number systems than it is to discuss accepting verbs. Nobody shouts, "I reject verbs!" It's meaningless; there is no question of accepting or rejecting them unless you specify particular criteria, such as "I accept verbs as a useful word class!" or "I reject the real numbers as a number system that conforms to my intuition!" Maelin (Talk | Contribs) 13:03, 31 January 2007 (UTC)

It's not exactly true that there is no such thing as a 0.999... that isn't specifically classed as a mathematically conventional "real number". I know of precisely one source which uses the notation 0.999... for something else - Richman's "decimal numbers" (a link to that paper is buried somewhere in the archives). And that "0.999..." is indeed unequal to what the author denotes as "1". Unfortunately, Richman's 1 isn't what you would call 1, either. Barring this abuse of notation (using already well-established symbols such as 1 and 0.999... for something new), the real number 0.999... is all there is. Any sources to the contrary? --Huon 14:03, 31 January 2007 (UTC)

But for the (n+1)th time, the article does explicitly state that it discusses real numbers. Even if that was not the case, there really is no meaningful structure where an object called 0.999... exists. Richman's decimals are irrelevant, I could just as well invent some whimsical structure where 0.999... = 3.47. As a rule, people who say that 0.999... ≠ 1 do not do so because they are sophisticated enough to consider alternative structures, but because they do not understand what decimal expansions mean. -- Meni Rosenfeld (talk) 15:25, 31 January 2007 (UTC)

Number theory

Look out, another teenager who probably doesn't know what they're talking about... It seems to me that really, both sides are just arguing over number systems. If you use standard number theory, all of these proofs work, so I'll agree that .999...=1 in this number system. However, if you don't ignore the significance of an infinitesimal, you have non-standard analysis (there's an article on it) and a non-archimedean number system. It's possible to do this in other number systems because standard analysis includes the Archimedean property as an axiom, something that is assumed for the creation of number theory. The Archimedean property basically states that infinitesimals are insignificant, so it's not possible to argue that they are within the standard number system. Now, if we're basically arguing over number systems, it makes sense to choose the one that most suitably fits our reality. Standard analysis denies the existence of infinitesimals... if this fits our reality, wouldn't that imply that there is a base unit of space, so it can't possibly be a continuum? It makes more sense to me to assume that space is a continuum, and that non-standard analysis fits reality more closely than standard analysis, making it more suitable to humanity's purposes. Is it really valid to use proofs worked out in standard analysis against the statement that non-standard analysis is more intuitive? That's circular logic. Now, is there something I'm missing that would make it all make sense to me? If my logic is valid, the article just covers the argument over which is better, standard or non-standard analysis. As far as I can tell, the proofs are only valid in standard analysis, so none of them carry over. Does this make sense to anyone? Can anyone clear me up on it? Thanks in advance Timeeeee 08:00, 23 December 2006 (UTC)

It does indeed make the most sense to use the number system that best fits our reality. Standard analysis allows for formal definitions of calculus, allowing for formal study of classical and quantum mechanics (physics), probability theory, financial analysis, general optimization, information theory, and countless other theories which form the basis for the man-made world you see in the skyscapers above you, the computer in front of you, the cell phone in your hand, and the financial system you use every day. So you'll find in fact that standard analysis is better, in that it accurately models many phenomena and allows for the use of high-powered mathematics to engineer and understand said phenomena.

However, even brushing that aside, standard analysis does use a continuum. That's what the Archimedean property implies: Between every two numbers, there will always be a third. (See, e.g., Continuum (mathematics).) Everything breaks down into smaller pieces. Of course, some will object that, in the physical world, one eventually runs into atoms on one hand or the speed of light on the other. But even these can be taken into account with calculus; it's just the model that has to change. In any event, the fact remains that the Archimedean property implies a continuum, not the lack thereof.

Note that, as far as I can tell, in most non-standard analysis systems, 0.999.. still equals 1. It's just that, in some, it doesn't. Maybe 0.999.. is undefined or maybe it is ambiguous. But in standard analysis (and, from what I can tell, the useful forms of non-standard analysis), 0.999...=1. The fact that it's non-intuitive doesn't mean we need to switch number systems. It just means that the number systems (and representations) that we use have a bit more than first meet the eye. Calbaer 08:50, 23 December 2006 (UTC)

If you use a system that doesn't have the Archimedean property, and assumes that the infinitesimal is not equal to zero, then even though your answers may change, the only differences will be an infinitesimal, which won't affect things in the practical sense. The only important changes are definitions and perceptions of the standard system. Even calculus can be accurate, you just have to be aware that you never reach the end of an infinite sum, or that h will never actually reach the limit of 0.
Point taken about the continuum. I'd like to add to your explanation of a continuum that if you're talking about space, it's not limited by the size of atoms. If it were, the diameter of an atom would have to be the smallest unit of space possible, and not only are there atoms of different sizes, but the individual parts of atoms have different sizes themselves. Saying that you run into the speed of light on one side is denying the existence of infinite speed, not a continuum of speed. The range of rational numbers between 0 and 1 has a limit on either side, but is still a continuum. I do understand that neither of these are your arguments, but fail to see where calculus helps in either, or how the Archimedean property implies a continuum.
By nonstandard analysis, I don't mean a particular form. If you're arguing that infinitesimals aren't equal to zero, you're arguing against an axiom of standard analysis, an if you had to choose a for of analysis to argue for, it would have to some form of nonstandard analysis. And if you're only changing standard analysis by removing the archimedean property, as I've said earlier, it's probably just as practical, just as useful. Besides all this, do you know if there's a good reason to have the proofs in the article. They use standard analysis, assuming the Archimedean property in order to prove it. This is circular logic, is it not? Timeeeee 04:41, 25 December 2006 (UTC)

The article doesn't address the issue of whether we wish to use real numbers or other systems at all. The proofs only show that the real number represented by 0.999... is equal to 1. As such, we can use any property of real numbers that we wish. How is this circular logic? It should be noted that there is no sense in discussing 0.999..., or any other decimal expansion for that matter, in any context other than real numbers (since decimal expansions are only good for representing real numbers).

On a completely unrelated note, number theory is usually used to refer to the study of integers, so it is probably not a very good header for this thread. -- Meni Rosenfeld (talk) 17:37, 25 December 2006 (UTC)

This is Ridiculous!

This is completely and utterly wrong.

First of all, .3333… does not equal 1/3, it is simply the decimal approximation. This can be proven by multiplying each my 3:

$.333*3=.999$; $(1/3)*3=1$

Furthermore, a variable cannot change when you want it to. It must always be the same. In this algebraic proof, the author adds a 9 at will:

If $c=.999$, $10c=9.99$, not $9.999$. Therefore, $10c-c=8.991$, not 9.

You cannot assume an infinite number of 9’s when you change the variable. If c has infinite 9’s, then 10c has infinity-1 nines. --Nickthehobo 18:07, 24 December 2006 (UTC)

Unfortunately, you are mistaken. A decimal expansion is a sequence of digits, and as such it is determined by the digits in every place. Both 9 + 0.999... and 10 * 0.999... have a 9 as their 1st, 2nd, 3rd, 4th, etc digit, so they are the same. -- Meni Rosenfeld (talk) 18:34, 24 December 2006 (UTC)

There is no last 3. If there is a last 3, then that is not a third. x42bn6 Talk 17:40, 25 December 2006 (UTC)

"A decimal expansion is a sequence of digits..." NOT A NUMBER. You can't use something if you don't even know how many digits are in it! Nickthehobo 21:43, 25 January 2007 (UTC)

We know kow many digits there are in 0.999...: Countably infitely many digits, that is, one for each natural number. --Huon 22:46, 25 January 2007 (UTC)

In terms of counting (since we're counting digits), inf - 1 = inf. In fact, inf - (any finite #) = inf. Tparameter 07:46, 26 December 2006 (UTC)

How convenient. And suddenly Infinity becomes the new "god". Folks can't figure out how to handle it when numbers span off into the unlimited, so they make this number called infinity that can do anything they want it to do. So... just for the sake of humouring me, is "Infinity - (Infinity - 1)" equal to 1 or 0... or perhaps some other number? Oh, and don't get me wrong... I believe that a concept of a terminal number on the other side of the loop from 0 should exist, but that it should exactly equal itself with any finite amount subtracted is just silly: It may encompass all numbers, but thats not the same as it being actually boundless ~ SotiCoto 195.33.121.133 10:52, 30 January 2007 (UTC)

For obvious reasons, "Infinity - (Infinity - 1)" is undefined. Please have a look at Hilbert's Hotel, which explains calculations with the countably infinite more explicitly. --Huon 11:05, 30 January 2007 (UTC)

Here is an example of why infinity does not follow the normal rules of addition and subtraction. Consider the set of all positive whole numbers, that is, the natural numbers. There are infinitely many natural numbers. Now consider just the even ones. There are also infinitely many even natural numbers. Similarly, there are infinitely many odd natural numbers. If we put the even numbers together with the odd natural numbers, we get all of the natural numbers. So $\textstyle \infty +\infty =\infty$. And if we take all the natural numbers and throw out all the even ones, we are left with just the odd ones. So $\textstyle \infty -\infty =\infty$.

Now let's consider all the natural numbers, and then let's think about all the ones that are bigger than 10. There are infinitely many natural numbers that are bigger than 10, but this time if we throw them all out we get only the numbers smaller than or equal to 10, and there are exactly 10 of these. So now, $\textstyle \infty -\infty =10$. But previously, we had $\textstyle \infty -\infty =\infty$. And since clearly $\textstyle \infty \neq 10$, we can then say $\textstyle \infty -\infty \neq \infty -\infty$ For reasons like this, we cannot define typical arithmetic with infinite quantities, because no definition would be consistent with the properties of infinite quantities. Maelin (Talk | Contribs) 12:48, 30 January 2007 (UTC)

In that case, it seems downright insane to even try to use it as a number. What you're basically talking about is an unlimited numerical quantity which (pretty much as just noted) can only be identified in terms of what it is not... rather than what it is, and effectively cannot be manipulated in any way. Its safe to say that my idea of what "Infinity" is happens to be a little different (more of a numerical upper limit representing all numbers, but with a sequence of indeterminate magnitude between 0 and itself: a Yang to zero's Yin), and I won't go into the details here, but I'll safely say its 100% compliant to my own accepted reasoning processes, unlike the notion of infinity you and others have suggested and made use of here. ~ SotiCoto 195.33.121.133 13:04, 31 January 2007 (UTC)

Agreed, which is why it is not a member of any of the usual every day sets of numbers that we use, like the integers, rationals, and reals. Some of these sets have, by various mathematicians, been extended to include infinity-like objects (forming new sets such as the Hyperreal numbers), but those sets are not greatly interesting to non-mathematicians, because the extending process requires sacrificing some quite useful and far more important properties of your set.

Now, with that out of the way, it's very important to realise that none of the proofs in the article nor any proofs here actually deal with infinity. Mathematicians often use infinity as a kind of shorthand for more formal, properly developed ideas, because it's very convenient and often fits intuitively with what they actually mean. Often, our intuitive ways of handling infinite quantities give results that are pretty close to what the more formal methods would do. But anywhere in this discussion that you see someone discussing infinity, there is another, more mathematically formal but probably less convenient, way of expressing what they're saying. Maelin (Talk | Contribs) 13:23, 31 January 2007 (UTC)

There are numerous algebraic structures one can consider, and any statement should be viewed in the context in which it appears. A few examples of structures which involve infinite quantities are ordinal numbers, cardinal numbers, hyperreal numbers, surreal numbers, Extended real numbers and real projective line. To give you a few examples of why the distinction is important:

• In ordinal numbers, every infinite α satisfies α - α = 0 and α - 1 = α (but not α = α + 1!)
• In cardinal numbers, for every infinite κ, κ - κ is undefined and κ - 1 = κ.
• In hyperreal and in surreal numbers, for every infinite x, x - x = 0 and x - 1 ≠ x.
• In the real projective line, ∞ - ∞ is undefined (and in fact, so is ∞ + ∞), and ∞ - 1 = ∞.

Of course, there is the completely unrelated notion, which Maelin has described, of using the symbol ∞ as a shorthand for something which doesn't at all involve an explicit object called "infinity". The obvious example is that the notation $\lim _{x\to \infty }{\frac {1}{x}}=0$ actually means "for every real ε > 0 there is some real number M such that if x > M then $\left|{\frac {1}{x}}\right|<\epsilon$. The ∞ which appears in summations is similar - the notation $\sum _{n=1}^{\infty }{\frac {9}{10^{n}}}=1$ (this is the sum represented by 0.999...) actually means "for every real ε > 0 there is some positive integer N such that if m > N then $\left|\sum _{n=1}^{m}{\frac {9}{10^{n}}}-1\right|<\epsilon$. If you disagree with this statement, you should give an example for some positive real ε such that there is no N which satisfies the requirement. You will not be able to give such an example (I know this, because I can prove that there must always be such an N).

A final distinction should be made between the actual mathematics and the way humans can choose to communicate mathematics. While I cannot read Tparameter's mind, I assume that in saying "inf - 1 = inf" he did not intend to state a fact about some specific structure, but only to make an informal observation with hopes that it will help the other party understand the matter. A request to explain what he meant would be appropriate (for example, he may have meant the provable theorem that in the cardinal numbers, $\aleph _{0}-1=\aleph _{0}$ - an interpretation suggested by "in terms of counting"), but not any other accusation. -- Meni Rosenfeld (talk) 14:06, 31 January 2007 (UTC)

Delete this page

Wikipedia talk pages are not the place for arguments or discussions on the article topics themselves, but rather for arguments and discussions on the articles. This page does not belong on Wikipedia. Andre (talk) 08:47, 25 December 2006 (UTC)

This page is not the article's talk page. It is a mix of lightning rod and excess storage area for discussions that would otherwise clutter that talk page. If you don't believe that, check out the talk page's archives, most of whose discussions concern not the quality of the article, but the question whether the topic is correct, mathematically. --Huon 10:54, 25 December 2006 (UTC)

Yeah. The alternatives would be to simply delete every one of those discussions as soon as they start, or politely explain every time to the editors in question why they cannot talk about their disagreements with this aspect of the real number system on the article talk page. Moving such arguments seems the cleanest solution. --jpgordon^∇∆∇∆ 16:44, 25 December 2006 (UTC)

You also have to consider that, in the minds of dissenters, they are looking to correct a problem in the error by providing a "disproof" of the assertion that 0.999... = 1 in the real number system. A talk page is the appropriate place to discuss possible items for correction that might need to be referenced further for inclusion in the article.

The only difference in this particular case is that almost all the arguments on this particular talk page that 0.999...=1 are incorrect. But if someone thinks (even incorrectly) that an article is wrong and has what they perceive as a valid reason, a talk page is the appropriate place to discuss that. Dugwiki 18:27, 26 December 2006 (UTC)

Algebraic Proof

Although I agree that 0.999... is indeed equal to 1, I feel that the "algebraic proof" given on the article page takes the treatment of numbers somewhat naively. In fact, the exact same logic is used under the section "p-adic numbers" to prove that ...999. is equal to -1, which most of us probably agree cannot be true in real numbers (the proof given by the 7th grader). I think the proof should look something more like this. It uses the formal definition of a series, which is to show that the sequence of partial sums converges:
$0.999...={\frac {9}{10}}+{\frac {9}{100}}+{\frac {9}{1000}}...$
So the sequence of partial sums, is:
$S_{1}=0.9={\frac {9}{10}}$
$S_{2}=0.9+0.09={\frac {9}{10}}+{\frac {9}{100}}$
$S_{3}={\frac {9}{10}}+{\frac {9}{100}}+{\frac {9}{1000}}$
and so on...
So the n^th partial sum will be:
$S_{n}={\frac {9}{10}}+{\frac {9}{100}}+{\frac {9}{1000}}+...+{\frac {9}{10^{n-1}}}+{\frac {9}{10^{n}}}$
S_n is a finite sum, so it can be multiplied by 10:
$10S_{n}=9+{\frac {9}{10}}+{\frac {9}{100}}+{\frac {9}{1000}}+...+{\frac {9}{10^{n-1}}}$
Subtracting S_n from 10S_n gives:
$10S_{n}-S_{n}=9-{\frac {9}{10^{n}}}$
$9S_{n}=9-{\frac {9}{10^{n}}}$
So, the n^th partial sum, S_n, can be written as: $S_{n}=1-{\frac {1}{10^{n}}}$
Again, S_n is the sum of a finite amount of numbers. So we must take the limit as n approaches infinity to show that the sequence of partial sums converges.
$\lim _{x\rightarrow \infty }S_{n}=1$ Therefore, 0.999...=1.
Notice that if this method were used on ...999., it would show that ...999. diverges because the limit of the sequence of partial sums is unbounded. Also, this method does not require 10 to be multiplied by a number with infinite decimal places. Wannger27 05:54, 14 January 2007 (UTC)

The series formulation is purposefully delayed until 0.999...#Infinite series and sequences, which already treats it in two ways: the geometric series formula and the limit of the sequence. The formula for the latter doesn't require so much fanfare to derive. Surely it's obvious that, for example, 0.9999 = 1 - 10^4? Melchoir 08:17, 14 January 2007 (UTC)

Why is it delayed? That is actually the definition of 0.999... It should be first. 72.193.74.36 16:26, 14 January 2007 (UTC)

Because it's mathematically more advanced than the digit manipulation "proofs". --Huon 16:34, 14 January 2007 (UTC)

Yes, Melchoir, I agree that I spent too much time with the derivation for the geometric series formula. This is my first time contributing to wikipedia and I guess I got a little carried away with the math symbols. But my point is that the "algebraic proof" currently given should not be a valid proof since the same logic and process can be used to "prove" that ...999. is a negative number. The only way to avoid this is through the use of partial sums and a limit. Wannger27 21:20, 14 January 2007 (UTC)

There is no problem here. The same logic used to show that the 10-adic number ...999 is equal to -1, is used here to show that the real number 0.999... is equal to 1. -- Meni Rosenfeld (talk) 21:51, 14 January 2007 (UTC)

Maybe I'm overlooking something, but what part of the process doesn't allow me to show that ...999.=-1 in real numbers as well.Wannger27 23:17, 14 January 2007 (UTC)

That the series doesn't converge? — Arthur Rubin | (talk) 02:45, 15 January 2007 (UTC)

Yes, ...999. doesn't converge. But the method shown under algebraic proof can be used to show that ...999. equals -1. So that is why limits should be used in the proof to avoid this.Wannger27 04:10, 15 January 2007 (UTC)

But it's not the same method. It's similar, granted; similar enough for the comparison to be interesting. But it's different.

If you want to apply logic to the methods used in mathematics and claim that this method proves that but not the other, then to be precise you have to start talking about theorems and proofs. The theorem that generalizes the result we're talking about sounds like "Given an infinite repreating decimal 0.dddd…, it represents the rational number blah." It doesn't necessarily apply to everything that someone considers to look the same, including …999. Melchoir 07:08, 15 January 2007 (UTC)

I know I probably sound really annoying now, but I really don't see what the difference between the two methods is. Both begin with a series which is multiplied by 10 and then subtracted from another series. It's true that ...999. is not in the form of an "infinite repeating decimal 0.dddd...," but none of the aforementioned steps (multiply by 10, subtraction) are unique to repeating decimals. It seems logical that a proof like this should be consistent no matter what number is used as long as the steps used within the proof can still be applied. So to sum up my question: If 1 is equal to 0.999... (which I know it does), then why doesn't -1 equal ...999. (which I refuse to accept as true)? Wannger27 08:38, 15 January 2007 (UTC)

The difference is convergence. When you consider ...999 as the sum of a series of reals, it's infinite. And when you try to compute ...999-10*...999, that's in effect subtracting infinity from infinity, which is undefined. --Huon 10:04, 15 January 2007 (UTC)

Yes, I know that ...999. diverges while 0.999... converges. But limits and partial sums need to be used to show this, which is something the current proof doesn't do. Wannger27 10:16, 15 January 2007 (UTC)

Wannger27, it's important to be a little more careful about what exactly is diverging. It's not right to say that …999 diverges. You can say that the series 9 + 90 + 900 + … diverges in the real number system. And since the series diverges, it would be silly to define …999 as its sum! Melchoir 10:48, 15 January 2007 (UTC)

Let's put it this way. If there is a real number with decimal expansion ...999, then you could use the method of our algebraic proof to show that it is equal to -1. But there is no such real number, and we assume that the fact that decimal expansions of real numbers must terminate to the left is common knowledge. 0.999... does, however, represent a real number, which is also assumed to be common knowledge, so we are allowed to do the manipulations. The more advanced proofs come in handy should we choose to question this so-called "common knowledge". -- Meni Rosenfeld (talk) 17:49, 15 January 2007 (UTC)

Meni, I agree with what you said. The requirements to use the manipulations are, for the most part, common knowledge, so a more formal proof would be unnecessary. So I will no longer be pursuing this topic. Thanks for everyone's help. Wannger27 22:02, 15 January 2007 (UTC)

"adjacent 1s"

I cannot understand the following sentence:

"…, and there are infinitely many more representations that include adjacent 1s," (which is in the second paragraph of the section 'Generalizations')

As is written in Golden ratio base#Non-uniqueness, there are many reprensentations of 1 such that 1 = 0.11_φ = 0.1011_φ = 0.101011_φ = ... = 0.10101010...._φ . They are "adjacent 1s" ??

Could someone interpret a phrase "that include adjacent 1s" in plain English? Thank you in advance.--Tsukapee 14:51, 18 January 2007 (UTC)

Common Sense has no place in Mathematics

72.193.72.226 who is a supporter of Wiki's article writes:

"..."common sense", as you put it, has no place in mathematics. Maybe you would be happier avoiding math. 72.193.72.226 01:30, 19 January 2007 (UTC)"

I rest my case. 41.241.187.210 07:27, 19 January 2007 (UTC)

So now you're going into law, then? Anyway, I disagree; I think that common sense has a place in mathematics, but it is not to determine truth. Rather, it is to help guide us to truth. So, while we needn't totally abandon common sense, we need to accept that truth can violate common sense, which I think is more in the spirit of this whole discussion. And, like 72.193.72.226 said, if paradoxes and other seeming violations of common sense disturb you, mathematics is not your field. Calbaer 08:15, 19 January 2007 (UTC)

Common sense is lay-intuition. Math is pure logic. The two only intersect by coincidence. The former has absolutely no value in the latter. The latter is a catalyst that makes people who don't understand invoke the former. 72.193.53.9 04:33, 20 January 2007 (UTC)

Intuition is essential in mathematics. You will never get anywhere without it. As for "lay-intuition" -- I don't know that there's any qualitative difference between a layperson's intuition and that of a working mathematician. The mathematician's intuition has been shaped by coming up against a greater number of difficult cases. (It's a little like the Catholic doctrine that your "properly formed conscience" is infallible. How do you get your conscience "properly formed"? By internalizing Catholic doctrine, of course.) --Trovatore 21:56, 20 January 2007 (UTC)

Intuition is essential, yes -- and even more essential is the understanding that intuition is only the first step; proof comes second (assuming there's room in the margins for it.) --jpgordon^∇∆∇∆ 22:35, 20 January 2007 (UTC)

If intuition is essential, then why do so many conclusions in mathematics defy it? If it were "essential", then it would be present, always. Common sense, intuition, this type of thing - none are either sufficient or necessary, and neither will ever satisfy any standard of proof or disproof. This is the "essential" point.72.193.53.9 15:21, 21 January 2007 (UTC)

Intuition is essential because that is one way new mathematical ideas are formulated. It's intuition -- very informed intuition, certainly -- that leads many mathematicians down otherwise untrodden paths. "Hm, this may be crazy, but what if I try it THIS way?" Most of the time, it's probably "oh, that's why not." But once in a while, "Eureka!" --jpgordon^∇∆∇∆ 15:59, 21 January 2007 (UTC)

I know your case is rested. But first, I have one request. Please cite one serious math paper dependant on, and explicitly citing, "common sense". Thanks so much. 72.193.53.9 05:25, 20 January 2007 (UTC)

From Dictionary.com:

Common Sense: sound practical judgment that is independent of specialized knowledge, training, or the like

Intuition 1. direct perception of truth, fact, etc., independent of any reasoning process;

Not applicable to mathematics. In fact, common sense seems to be the entire basis of irrational mathematical skepticism. In contrast, mathematicians use specialized knowledge, reason, etc. Experience and exploration is not intuition, and definitely not "common sense". Tparameter 17:54, 21 January 2007 (UTC)

Common sense tells us that the world is flat.

Equal to one?

Huh? I highly do not believe that 0.999... is exactly equal to 1. It has to be at least a tiny bit lower than 1. I believe that this article is nonsense. Kamope · talk · contributions 20:29, 20 January 2007 (UTC)

• And which of the provided proofs do you dispute? Belief isn't relevant; this is mathematics, where sometimes counterintuitive results are quite correct and provable. --jpgordon^∇∆∇∆ 20:40, 20 January 2007 (UTC)

Excuse me, but didn't you just say above that "intuition is essential". Then you say that "sometimes counterintuitive results are quite correct". Seems contradictory. 72.193.53.9 15:24, 21 January 2007 (UTC)

But it's not. "Essential" doesn't mean "necessary in every case", or at least I didn't mean it that way. Intuition is essential for discovery, but doesn't play a role in the final result. --jpgordon^∇∆∇∆ 16:06, 21 January 2007 (UTC)

From Dictionary.com:

Essential 1. absolutely necessary; indispensable.

Common Sense: sound practical judgment that is independent of specialized knowledge, training, or the like

Probably not applicable to mathematics. Tparameter 17:52, 21 January 2007 (UTC)

Well, yeah. Absolutely necessary. It's what you do with the intuition that separates the visionaries from the kooks. --jpgordon^∇∆∇∆ 19:38, 21 January 2007 (UTC)

sci.math newsgroup

I would recommend that those who doubt the mathematical truth of this article to peruse the sci.math newsgroup, where this subject comes up quite regularly. There you have the benefit of discussing mathematical "beliefs" with real, honest-to-goodness professional mathematicians and teachers. — Loadmaster 23:32, 20 January 2007 (UTC)

If .999...=1

It does not exactly equal 1 becasue it is obviously not 1. 1=1 not .999...Chessmaster3

Which prooof are you having trouble with? --Maelwys 23:14, 2 February 2007 (UTC)

I think he/she just thinks it's wrong for no particular reason having to do with mathematical logic. The "disproof" given could also be applied to "One is one, not two halves (1=1 not 2/2)," which is similarly false. Calbaer 23:48, 2 February 2007 (UTC)

2/2 is obivously 1 though becasue if you do the math you get 1 similar to i^4 being 1. .999... had not math in it so it simply would equal it self and nothing else.

So what about 1 and 1.000... ? Are they Different? 1, One, Uno? Different? Enough Devil's advocate. 0.999... *does* have "math in it". It means 9/10 + 9/100 + 9/1000 + 9/10000 + ... ad infinitum. When you do the math, you get 1. Endomorphic 22:07, 8 February 2007 (UTC)

Not Equal

Two mathematical objects are equal if and only if they are precisely the same in every way. .99999R and 1 are not the same in every way. http://en.wikipedia.org/wiki/Equality_%28mathematics%29 Chessmaster3 23:23, 7 February 2007 (UTC)

Actually, they are: 0.999... = 1 = 1/1 = cos 0 = ln e = i⁴. Just because two numbers can be written differently doesn't mean they are different. Calbaer 01:00, 8 February 2007 (UTC)

That does not prove they are equal though 0.999... is as close as you can get to 1 with out being 1.209.7.118.199 21:48, 8 February 2007 (UTC)

There are a number of proofs on this talk page and in the article that verify 1 = 0.999... in the real number system. If you aren't sure about one of the steps in one of the proofs, feel free to ask about it. Dugwiki 22:08, 8 February 2007 (UTC)

There is no "closest number". If two numbers aren't exctly the same, you can always find a third between them. How? Take the average. Endomorphic 22:13, 8 February 2007 (UTC)

Decimal Representation

I have a problem with the Digit manipulation proofs. These proofs seem like a failure to be able to represent numbers as decimals and not proofs that .999... equal 1. I would argue (based on long division) 1/3 is not .333... but .333...1/3 or .333...+1/3 of the smallest number a decimal will allow. The second one is a result of this discrepency between numbers and decimels; it would seem like .999... times 10 equals 9.99..., but it is actually 9.99...0 because the shift of the number to the left which means you cannot subtract all the digits because the 9.99...0 has one more digit the .999... . I have no problem with expressing .999... as a limit of 1 because the difference between them is miniscule but to express this as the nature of ultimate reality seems incorrect.--Jorfer 00:31, 12 February 2007 (UTC)

$1/3$ can be proven to be exactly $.333...$:

$.333=.3+.03+.003+.0003+...=\sum _{k=1}^{\infty }3*\left({\frac {1}{10}}\right)^{k}=(\sum _{k=0}^{\infty }3*\left({\frac {1}{10}}\right)^{k})-3={\frac {3}{1-{\frac {1}{10}}}}-3={\frac {3}{\frac {9}{10}}}-3={\frac {30}{9}}-1={\frac {10}{3}}-3={\frac {1}{3}}$ Dlong 00:50, 12 February 2007 (UTC)

I cannot disprove that yet as I am just now in my second semester of Calculus AB. I have a feeling the second and third our not equal in reality. Calculus is based on limits so what you probably proved was .333... is 1/3 by definition in Calculus but it is may not be in reality.--Jorfer 01:02, 12 February 2007 (UTC)

The second and third are not equal? Um, are you saying that $0.3+0.03+0.003+0.0003+...\neq {\frac {3}{10}}+{\frac {3}{100}}+{\frac {3}{1000}}+{\frac {3}{10000}}+...$? Dlong 01:05, 12 February 2007 (UTC)

If that is what it means then it is the fourth and fifth. I am not yet familiar with that notation.--Jorfer 01:10, 12 February 2007 (UTC)

I would suggest reading up on http://en.wikipedia.org/wiki/Geometric_series . Dlong 01:13, 12 February 2007 (UTC)

It seems your basing the equality of the fourth and fifth part on the Infinite geometric series theorems which are considering the limit as n approaches infinite, which is a limit, and like I was saying before, just because they are equal by the definition of limits does not mean that this reflects reality so my complaint stands.--Jorfer 01:28, 12 February 2007 (UTC)

http://en.wikipedia.org/wiki/Mathematics#Relationship_between_mathematics_and_physical_reality Dlong 01:55, 12 February 2007 (UTC)

What is "the smallest number a decimal will allow"? What do you mean by "in reality"? The only reality in mathematics is what is proven from the chosen axioms. This article is about the real numbers - whether they are an accurate description of reality or not is not relevant to this article. Mdwh 02:08, 12 February 2007 (UTC)

That is significant enough to warrant being mentioned in such a controversial article. Most people will automatically assume the article is describing .999... being 1 not just for the sake of mathematics but in reality as well.--Jorfer 02:38, 12 February 2007 (UTC)

So while I don't really agree with Mdwh about the nature of mathematical reality, I have to wonder, just what do you mean by 0.999... "in reality"? Surely you can't mean "in physical reality"; we have no tools in physics to make such fine distinctions. In physical applications, saying that a quantity is 1 usually means something like "it's between 0.998 and 1.002" (adjust depending on your accuracy of measurement in the particular context at hand). --Trovatore 04:10, 12 February 2007 (UTC)

Clearly the article is about mathematics, since it deals with numbers. As for "in reality," it is odd to say it's different than mathematics. Let's say 0.999... was, for all practical purposes, less than 1. Then a particle having inertial mass should be able to go 0.999... times the speed of light given the proper amount of propulsion. And you should be able to calculate the amount of energy that takes and it should be finite. However, if you do the calculations, you'll find that the amount of energy needed is infinite, i.e., you can't do it. That will indicate that 0.999... is in fact no less than 1. While that isn't a mathematical proof (after all, I left out the math!), it illustrates that this is a concept that exists "in reality," not just for the sake of some mathematical convenience or quirk. Calbaer 04:21, 12 February 2007 (UTC)

You have a calculator in which you can imput .999...? I have to get me one of those.--Jorfer 23:04, 12 February 2007 (UTC)

One shouldn't need a calculator to do all calculations; perhaps the idea that any "real" calculation should be doable on a calculator is part of the problem here. Calbaer 23:31, 12 February 2007 (UTC)

Here's what I mean by "doing the calculations": The rest energy of a particular with mass is $E=mc^{2}$. The kinetic energy of the particular is $K=mc^{2}(1/{\sqrt {1-v^{2}/c^{2}}}-1)$, which exceeds $K=E(1/{\sqrt {2-2v/c}})$. That means that, to take a particle at rest to v=0.9c requires more than $E{\sqrt {5}}$, to take it to 0.99c requires more than $E{\sqrt {50}}$, etc. If I want to take the particle to 0.999...9c with n 9s, that would require energy of more than $E{\sqrt {(0.5)(10^{n})}}$. Since 0.999... is larger than 0.99..9 for any finite number of 9s, that means that no amount of energy could take a particle to (0.999...)c. Again, this isn't a real "proof" that that 0.999...9c is no less than c, but rather an illustration of how the number applies to a real situation, and it gives an idea of the concepts behind some of the more complex and rigorous proofs. Calbaer 03:36, 13 February 2007 (UTC)

Still, your basis for claiming .999...*c = 1*c is based on limits. It is likely that in reality .999...*c requires an infinite source of energy, while 1*c is still impossible even with an infinite amount of energy which makes them distinctivly different. It is also possible that in reality .999... is in effect = .99..9 . Just throwing it out there.--Jorfer 04:25, 13 February 2007 (UTC)

Huh? If you don't have a definition for 0.999..., how can you say that it might be equal to 0.99...9 (which isn't a number unless you specify the number of ns) or it might be the fraction of the speed of light which requires an infinite amount of energy but isn't possible? You first need a definition, otherwise "0.999..." is as meaningful "J%#HJ^%~" - just a string of random symbols. You can't say that "there is a difference at least in special relativity between 1*c and .999... *c" if you don't know the definition of "0.999...," whether you are talking about real analysis or special relativity. Moreover, there is no such thing as "an infinite amount of energy" with regards to what you can give to a particle. There is "finite energy" and there is "impossible." Math models the world, and when one is sloppy about the math, one cannot make predictions about the real world accurately. Calbaer 06:08, 13 February 2007 (UTC)

The bottom line: 0.999... = 1 is a valid identity. Therefore, they are one and the same thing. Therefore, if 1 occurs in "reality", then so does the other. 1 is in fact quit useful in "physical reality". 63.224.186.83 14:59, 13 February 2007 (UTC)

It would be different if you define .999... as the closest a number can come to one without actually touching it. This would apply to reality if we describe the closest a particle can get to the speed of light without actually reaching it as .999... .--Jorfer 23:05, 13 February 2007 (UTC)

So it comes down to this:

• Why do you believe that, in reality, there is a closest speed you can get to the speed of light, without reaching it? Why shouldn't there always be a faster speed that's still less than that of light?
• But if there is such a fastest speed (say, if speed is quantized), then why shouldn't it be a real-number fraction of the speed of light, and therefore have only finitely many 9s? --Trovatore 23:13, 13 February 2007 (UTC)

It is because, at least currently, the infinitesimal difference between c and the closest a particle can get to c cannot be measured. According to special relativity it is impossible for a particle to accelerate to the speed of light, but math would imply that a particle would reach the speed of light eventually.--Jorfer 23:26, 13 February 2007 (UTC)

You haven't explained why you think there is such a closest speed, which was the first question above. --Trovatore 23:44, 13 February 2007 (UTC)

Allow me to summarize. You are saying that because a THEORY (Relativity, which is incomplete in ATTEMPTING to explain the workings of the universe, especially on the sub-atomic level) claims that a particle cannot reach the speed of light, that this completely unrelated object, which is provably exactly equal to 1, does not exist in reality. Is that what you're saying essentially? Tparameter 00:44, 14 February 2007 (UTC)

I am saying that a well founded principal such as the limitations for particles with mass to travel at the speed of light might contradict the assertion of this article and may deserve a mention because of that. If nothing else, the sentence I put on the main talk page would do. It point out that math is based on postulates which are not as full proof as the theorems themselves and may contradict physical reality because of this. I phrased in a more NPOV way though by just pointing out what was written in another article.--Jorfer 01:20, 14 February 2007 (UTC)

You are correct that theory is supposed to describe reality. You are incorrect that they contradict in this particular case. In this case, physical reality is fully consistent with the mathematical model, as described above. 0.999...c = c. You think otherwise, but can find no source for this which doesn't require original research to reach your conclusion. I could assert that I might be the rightful King of Scotland, but that shouldn't be included in the corresponding article. Anyway, did you want something like the following?: "Note that there are instances in which expansions like 0.9 and 0.99 do not have meaning; e.g., you can't have 0.99 electrons or 0.99 telephones. In such instances 0.999... is not a useful quantity to study. In fact, for most applications, it is far easier to deal with 1 than 0.999...." I don't think it belongs there, but I'm trying to understand what you would like. Unfortunately, I don't see how a statement you desire could be both true and useful. Calbaer 02:59, 14 February 2007 (UTC)

You are fair in asking for a citation, but I am still figuring out how to go about searching for it. I would like to be able to add: It is important to note, however, that mathematics does not always correspond to physical reality. which demonstrates that number systems work differently in theory then reality. It is interesting to note that this piece of the mathematics article indicates number systems are not infalliable, and though my claim on this topic specifically is currently original research, the general idea still applies. The example of the apples is a good one, as the difference between units varies (unlike in number theory where they are equidistant) as some apples are heavier and some are lighter. Putting this statement in the article raises a reasonable doubt on the main claim of the article against seemingly overwhelming evidence. --Jorfer 04:01, 14 February 2007 (UTC)

No, I don't think that's a good addition. You still haven't given any hint of a physical sense in which 0.999... would be different from 1. It's a defensible position that, in a physical sense, the question of their equality is not well posed, for example because you can't really have infinitely many nines. But it's not a defensible position that there's a physical sense in which 0.999... and 1 both denote, but denote different things. Or at least neither you nor anyone else has come close to explaining such a position. --Trovatore 04:26, 14 February 2007 (UTC)

.999...*c , though I cannot find a source that uses it (probablly because no one has tried to use mathematics to describe the fastest velocity of an object with mass), is the only way that I could think of to mathematically represent the fastest an object with mass can go. It can go faster than .999999995*c but it cannot reach 1*c. I think someone beforehand mentioned that .99...9 is not a number so this must be the only number that could describe this. This seems like the only use for an abstract number like .999... anyways.--Jorfer 04:39, 14 February 2007 (UTC)

No, the above does not "come close to explaining such a position". Sorry, it's just incoherent. --Trovatore 04:53, 14 February 2007 (UTC)

There are infinite numbers between .999999995 and 1. There are exactly zero numbers between 0.999... and 1. Besides that, I haven't a clue what the connection is between the speed of light and 1, and the speed of a particle and 0.999..., which is exactly 1 by the way. 71.32.163.202 04:56, 14 February 2007 (UTC)

Jorfer wrote: "I would like to be able to add: 'It is important to note, however, that mathematics does not always correspond to physical reality.'" However, (s)he hasn't demonstrated a physical reality which does not correspond to the mathematics in question. I only brought up the example of the speed of light because I thought it would help illustrate, that, in fact, this mathematics does correspond to physical reality. However, my purpose for this was only to demonstrate it to Jorfer, not to show anything rigorous. Jorfer doesn't believe my argument, so it's probably not useful to keep talking about it. I can't come up with a physical reality Jorfer believes corresponds to the mathematics, and Jorfer can't come up with a physical reality that anyone else here believes contradicts the mathematics. Thus neither of our arguments should be added to the article. Calbaer 06:01, 14 February 2007 (UTC)

Jorfer's complaint above is actually deeper than just whether or not 0.999... = 1 . His issue, rather, is rather the concept of limits actually "applies" to "real life". He is agreeing that the series (0.9, 0.99, 0.999, ...) has a limit of 1, but he is not agreeing that the corresponding decimal 0.999... should be said to be equal to 1, but that they must instead differ by some as yet undefined infinitesimal amount.

The reason that an infinite series in the real number system has a sum equal to the limit of its corresponding partial finite sums is that the real number system is defined in such a way that two numbers which differ by at most an "infinitesimal" quantity must be equal. The article Archimedean property has a section that explains why the reals have no non-zero strictly positive infinitesimal numbers. Since 0.999... and 1 differ by at most an infinitesimal quantity, and the only such quanity in the reals is zero, 1 - 0.999... = 0 and thus 1 = 0.999... Dugwiki 16:57, 12 February 2007 (UTC)

The article clearly states "In mathematics", and that it is talking about the real number, so there is no problem about confusing this with any other possible meaning for "0.999...". I don't know what you mean by "in reality" though? Mdwh 23:05, 12 February 2007 (UTC)

Further, 0.999... is defined *only within* mathematics. What is 0.999... in physical reality? Jorfer's idea of "the closest number to..." doesn't work at all. For *any* two distinct numbers, you can always find a third number between them, by taking the average. Does Jorfer's reality include pairs of numbers which can't be averaged? Endomorphic 20:46, 14 February 2007 (UTC)

Everything in this example boils down to one question: How do you mathematicly express the closest a particle with mass can get to the speed of light without actually reaching it?--Jorfer 03:01, 15 February 2007 (UTC)

You don't. There is no such thing. The particle can approach c asymptotically. You can get as arbitrarily close as you like. There is always, theoretically, the possibility of adding a finite amount of energy to a particle with a finite amount of energy to give it a greater amount of energy, thus bringing it closer to the speed of light. There is no "greatest possible", unless you include the velocity reached if all the energy in the universe was put into the particle, but that is a limit imposed by practicality, not theory. Maelin (Talk | Contribs) 04:40, 15 February 2007 (UTC)

Not being able to use math to describe the phenomenon would seem like a failure of mathematics.--Jorfer 15:11, 19 February 2007 (UTC)

Huh? It is described already with a very simple equation. E=mc^2. Tparameter 16:22, 19 February 2007 (UTC)

The math does not describe the phenomenon because the phenomenon of "being as close to the speed of light as it's possible to be" does not exist. It's asymptotic behaviour. As the kinetic energy of a particle approaches infinity, the speed approaches c. Any particle that has a finite amount of kinetic energy will have a speed less than c. You can always (theoretically) heap more kinetic energy into your particle to get it closer to c. Since there is no largest finite number, there is no (theoretical) upper bound on the amount of energy the particle can have, so there is no limit to how close the speed can get to c. For every speed that is less than c, there is a greater speed that is also less than c which corresponds to a larger amount of energy. Maelin (Talk | Contribs) 00:56, 20 February 2007 (UTC)

You cannot "heap" more energy into a particle to make it go faster when it theoretically hits .999...*c and thus my point.--Jorfer 03:23, 20 February 2007 (UTC)

It never can hit 0.999...*c for the reasons explained above; there is no "as close to the speed of light as it's possible to be." Calbaer 05:40, 20 February 2007 (UTC)

If you define 0.999...*c as "the closest speed to the speed of light that a particle can have" then it is a non-denoting term - it is a name for a thing that does not exist, just like "the present king of France". The relativistic formula for kinetic energy, E_k as a function of speed, v, is given below. This has a vertical asymptote at v = c, where E_k shoots up to infinity. You can get arbitrarily close to c, and the closer you get, the higher your corresponding E_k will be, but there is no "closest point" because there is no limit to how high the kinetic energy can go. E_k can get arbitrarily large, so v can get arbitrarily close to c. Maelin (Talk | Contribs) 08:07, 20 February 2007 (UTC)

$E_{k}={{mc^{2}} \over {\sqrt {1-{{v^{2}} \over {c^{2}}}}}}-{mc^{2}}$

Try this: stop thinking of the particle in terms of its speed and think of it in terms of its energy. You can work out how fast it's going from its kinetic energy. A stationary particle has no kinetic energy, and a particle travelling at lightspeed has infinite. So if a particle is going at "the closest speed to the speed of light that a particle can have" then logically it must possess "the largest non-infinite amount of energy that a particle can have". That is, since the laws of physics don't impose any theoretical limit on how much energy a particle can have, the largest finite number. There is no "largest finite number"; for all finite numbers you can come up with a larger one by adding seven. And every time you stick another seven Joules into the particle it'll edge that tiniest sliver closer to lightspeed (but won't get there). Andrew 11:39, 20 February 2007 (UTC)

If your particle's speed is "0.999...*c", as you put it, then it's speed is c. Tparameter 15:45, 20 February 2007 (UTC)

I feel that there is a difference between .999...*c and 1*c. It being that .999...*c require an infinite amount of energy and 1*c is impossible even with an infinite amount of energy as it is the assymptote, but we cannot test that due to physical limitations, so it will just have to remain theory I suppose. I would complain that this article is written on the basis that you can reach an assymptote at infinity, an assumption that should be mentioned as it is probablly a postulate due to its theoretical nature.--Jorfer 21:55, 22 February 2007 (UTC)

I wonder: The speed of light is 299,792,458 metres per second. How fast is 0.999...*c? Next question: Can you substantiate your "feeling" about the difference between 0.999...*c and c by any kind of formula, reference or whatever? Is there any reason why even an infinite amount of energy should be insufficient to reach the speed of light? What exactly do you mean by "the asyptote"? The article talks about limits, which are well-defined without using infinity except as notation, but there are no asymptotes that I can see. --Huon 22:26, 22 February 2007 (UTC)

If you plot a graph of energy vs. speed you will get an asymptote at c=1. In math when x(in this case speed) reaches infinity the asymptote is reached, if I am not mistaken. I am debating whether it would reach that asymptote at x=infinity in reality since math is based on postulates which in this case our not testable as far as I am aware and thus unproven as to the reality of .999... = 1. If there is a reasonable doubt as to the validity of this argument outside of mathematics then it should be mentioned in the article. It should also be noted that 299,792,458 metres per second is an approximate number and that the actual speed has an infinite amount of decimal places as do most, if not all scientific measurements.--Jorfer 23:22, 22 February 2007 (UTC)

You can't have an object with an infinite amount of energy, so "0.999...*c" is just as unattainable as c.

"0.9999..." only has meaning in mathematics. There is no meaning to this string of characters "in reality" other than the application of the mathematical definition (where it equals 1) - unless you would like to give us a definition? Mdwh 23:40, 22 February 2007 (UTC)

No, the speed of light defines the metre, and as such the speed of light is exactly 299,792,458m/s. And, your misconceptions about relativity notwithstanding, according to the equations that govern these things, a particle travelling at 299,792,458m/s has infinte energy. A particle travelling slower than that (i.e., any particle) has finite energy. Andrew 15:36, 23 February 2007 (UTC)

I think this would be a "more realistic" definition:

.999... : The closest number <1 to 1.

--Jorfer 23:57, 22 February 2007 (UTC)

If you aren't using standard mathematics and the real numbers (under which 0.999... = 1), you will need to describe what system you are using to give anything in your realistic definition meaning.--Trystan 00:15, 23 February 2007 (UTC)

Regarding x "reaching infinity" (it doesn't): limits are mentioned in the article, specifically in the overview, the Introduction, the Real analysis section, the Scepticism in Education Section, and the Cauchy Sequences section. Also, being testable doesn't count for anything in mathematics. The first few billion natural numbers satisfy Goldbach's conjecture acording to computer calculations, but the conjecture is still open. Further, what's the difference between "P is impossible" and "P is impossible without Q, which is impossible."? Lastly, surely 0.999 \leq (1+0.999...)/2 \leq 1. Endomorphic 00:37, 23 February 2007 (UTC)

And that number does not exist in the real numbers. You could construct a number system where such a number is possible (as mentioned in the article), although there is no reason to consider such number systems closer to "in reality". Mdwh 22:36, 23 February 2007 (UTC)

There is just as much if not more evidence of p-acid numbers representing reality (which I would advocate) as the current number system. The proof for p-acids being wrong is based on an equality which is based on current number theory so of course it would be wrong by the current system. ...999 represents infinite so by proving that infinity (...999) = -1 by the current number system you disprove the current number system and not p-acids. Yes, p-acids are mentioned in the article, but it should be made clear in the article (preferably at the top of the page) that .999... = 1 is a result of the currently accepted number system and may not be as realistic as other number systems.--Jorfer 00:06, 24 February 2007 (UTC)

It is made clear that this article applies to the real number. I'm not sure what you mean by "as realistic" - if you mean than mathematics may only be an approximation of reality (e.g., perhaps spacetime is quantised), do you think that needs to be emphasised on every mathematics article? Mdwh 00:54, 24 February 2007 (UTC)

I would be interested in hearing how 0.999... that is not equal to 1 is somehow "realistic". So far, you have only given a loose theory based purely on very low-level intuition without any sources or even any math whatsoever. The real number system is the basis of the math that is used in engineering and physics to describe everything "real" that we can describe. Tparameter 02:35, 24 February 2007 (UTC)

You may have missed where I seemed to have disproved part of current number theory and proved p-acid numbers when I stated my disgreement with the proof against acid adic numbers given in the article:

The proof for p-acids adics being wrong is based on an equality which is based on current number theory so of course it would be wrong by the current system. ...999 represents infinite so by proving that infinity (...999) = -1 by the current number system you disprove the current number system and not p-acids.

--Jorfer 17:42, 24 February 2007 (UTC)

First of all, it's p-adic numbers, not p-acid numbers. Secondly, we don't prove "p-adics being wrong" or any such thing. They're a beautiful tool, but pretty useless for all real-world purposes. I've never seen "p-adics in physics", for example. Thirdly, there's a set of p-adic numbers for each prime p (and also for non-primes, but those are less beautiful). Which do you propose we all start using? The 2-adics? Or better the 10-adics? Fourthly, that the ...999=-1 proof which is correct for 10-adics fails for the reals shouldn't be a surprise since ...999 is no well-defined real number. The best you could get is a non-converging series, and for those the arithmetic of limits is not as helpful as for converging series. Finally, when you state something like "I seemed to have disproved part of current number theory", you should ask yourself: There are thousands of students learning number theory every year. There are hundreds of professors doing little but math all day. None of them but you found this fundamental flaw of number theory. Sure, that's possible. But isn't it more probable that you made a mistake? --Huon 18:15, 24 February 2007 (UTC)

Huon covered much of what I was going to say (but didn't due to an edit conflict). A few more things: Firstly, Huon pointing out your misspelling isn't being petty; we all ignored your first misspelling, but this misspelling being made over and over reflects a fundamental misunderstanding — "p-adic" is like "dyadic," a mathematical term with a specific linguistic origin. Also, even if you found a "flaw" in p-adics, it's a different number system, so it wouldn't necessarily show that the reals are flawed. Taken together with Huon's observations, these seem to indicate that you lack the mathematical sophistication to judge whether 0.999...=1, and this arguments page cannot make up for years of mathematical education. This isn't an insult — I have a friend with a grammar-school level of mathematical education but a genius IQ — but rather an indication that you need to either learn or review mathematics in order to understand the concepts you are trying to discuss. Many mathematical phenomena seem surprising upon initial observation — that 2 is a prime number, that infinity is not a real number, quadratic reciprocity, the independence of the Axiom of Choice, etc., so you need to judge mathematics based on its own terms rather than on intuition-based and/or informal "proofs." Calbaer 18:32, 24 February 2007 (UTC)

Some suggestions

Um... here're some suggestions on what you could do to stop arguements: 1. Mathematically prove 1/3=0.333... using basic algebra. It'll be tough, and maybe long, but at least everyone'll understand the symbols. Preferably, do this on the article itself, too, if you can find a source that has that. Because most of the people who deny that 0.999...=1 don't really understand what that E-like symbol even MEANS (I know I don't. >_>; ) And 2. An extension of 1... stop using higher math! Allow me to prove why:

Let's assume: Most people who claim that 0.999...=1 know higher math, such as calculus. Most people who know higher math claim 0.999...=1. Many people in an argument assume that, when the other arguer's argument is meaningless to them, it is therefore simply meaningless.

Therefore: Most people who claim the opposite, that is 0.999...=/=1, should therefore NOT know higher math. And when given proof in the form of higher maths, they assume they assume the proof is meaningless, and continue to claim the opposite, that is 0.999...=/=1.

Okay? So, go on. Use nothing more then basic algebra to prove that 0.999...=1. Go on! Oh, and make sure to cite each individual mathematical law you're using to do so. Theorems are OK, as long as they're really basic, and you can manage to find proof that uses nothing more then basic algebra for said theorem. Goodluck!!

Well, here's a basic algebra proof that 1/3 = 0.333...:

$\textstyle {\begin{array}{rrcl}{\mbox{Let}}&x&=&0.333...\\{\mbox{Then}}&10x&=&3.333...\\&&=&3+0.333...\\&&=&3+x\\\Rightarrow &9x&=&3\\\therefore &x&=&{\tfrac {3}{9}}={\tfrac {1}{3}}\\\end{array}}$

Note also that simple results may sometimes require complicated maths to prove. Classic example: Fermat's Last Theorem. Maelin (Talk | Contribs) 09:01, 24 February 2007 (UTC)

I already told you why this proof is faulty. It has to do with the location of the last 3. When you multiply .333... by 10 you do not have 3.333... continuing but have 3.33...3. The difference may seem subtle, but going from an infinite number to a finite number is no small task. By your own logic, in math, all number are equidistant from the decimal but the last finite 3 and the 3 at infinity are not and thus there is "no last three" in the real number system. By this same logic, multiplying .333... by 10 is will not give you 3.333... as the last digit cannot be moved to the left 1 space because of the given logic. Also, if you multiply .333... by 10 you have 1 less decimal place so it is impossible to add another decimal place and thus is impossible to subtract .333... times 10 by .333... because the .333... has one more decimal place.--Jorfer 17:36, 24 February 2007 (UTC)

There. Is. No. Last. 3. --jpgordon^∇∆∇∆ 17:40, 24 February 2007 (UTC)

My point. It is impossible to use such algebra with the current number system. What happens when you multiply a three at infinity by ten?--Jorfer 17:46, 24 February 2007 (UTC)

Yes the proof is not rigorous - but he asked for a simple proof without higher level maths. Unfortuntely higher level maths is needed to prove this rigorously. Mdwh 23:01, 24 February 2007 (UTC)

Just as there's no last digit, there is no digit "at infinity", either. There is one digit for every natural number, and each of them can be moved by 1. The 3 in the n-th place corresponds to 3*10^-n. When you multiply by 10, the shifted 3, now in the n-1-th place, corresponds to 3*10^-(n-1). To say that there is some digit to which no other digit gets shifted under multiplication by 10 is just the same as saying that there's a natural number n-1 to which there's no n. --Huon 17:54, 24 February 2007 (UTC)

.333*10=3.33-.333=2.997 so having a special quality for .333... that allows it to create another 3 at the end when multipled by ten seems to be pseudomath to me and "unrealistic".--Jorfer 22:09, 24 February 2007 (UTC)

• That's ok. Until you can understand the meaning of infinity, none of this works at all. --jpgordon^∇∆∇∆ 22:24, 24 February 2007 (UTC)

It doesn't create another 3 at the end because THERE IS NO END. THERE IS NO LAST 3. There is a 3 that is n places after the decimal point, for every natural number n. There is no largest natural number, because for any natural number, there is another natural number that is one more than it. This is one of the axioms of the natural numbers. Every natural number has a natural number that follows it. In the same way, every 3 in 0.333... has a 3 to its right. When you multiply by 10, every 3 moves to the left by one place. The place it just vacated is then filled by the 3 that was after it. This happens all the way down the line, and the line never runs out of 3s because there is no end to the 3s. Maelin (Talk | Contribs) 22:34, 24 February 2007 (UTC)

I do not see how you can call something with a repeating decimal "real" then. It does not act like a real number at all. If you multiply infinity*2 what do you get? You can't, it is impossible. Such a number as .333... would not follow the same rules as ordinary mathematics.--Jorfer 00:36, 25 February 2007 (UTC)

• Real numbers is what we call anything that can be expressed as an infinite decimal (including, for example, 1.250000...., which we normally notate as 1.25, or 0.333..., or .999...). Is the name confusing you? --jpgordon^∇∆∇∆ 01:01, 25 February 2007 (UTC)

I still think that .333... is just a really close approximation of 1/3. We do not deal with .333... on a daily basis , but we do deal with 1/3. The distinction between .333... and 1/3 is so small that we cannot tell the difference, but even with an infinite line of 3's it will not make it 1/3. I understand it being said that they are equal for convenience (so we can express 1/3 in decimal form), but to (again) say this represents "reality" is false in my opinion. I think you need more than an infinite number of digits to express 1/3 so I do not believe you can "express" 1/3 as exactly equal to .333... . The .333... just keeps coming short of 1/3 forever in my opinion (divide the 1 by 3, again, again...); if .333... goes on forever then how can you say that the final remainder ever gets divided? I think there needs to be a better definition of a real number then it can be expressed by a repeating decimal.--Jorfer 01:27, 25 February 2007 (UTC)

"More than an infinite number of digits" [italics added]? What's more than infinity? There is no number greater than infinity. In fact, there's no number equal to infinity.

You're right, though: Any sum of .3, .03, .003, and so on would never equal 1/3 with any finite number of 3's. However, 0.333... doesn't have a finite number of 3's, it has an infinite number of 3's.

$\sum _{n=1}^{\infty }3({\frac {1}{10}})^{n}=0.3+0.03+0.003+0.0003+...=0.333...$

The "..." at the end doesn't represent a really large number of 3's; it represents an infinite number of 3's.

There's also no "final remainder" because there is no remainder. The number 0.333..., with an infinite number of 3's following the decimal, is equal to 1/3.

So, what is the difference between 1/3 and 0.333...? The difference between 1/3 and .3 (or 3/10) is 1/30; the difference between 1/3 and .33 (33/100) is 1/300; the difference between 1/3 and .333 (333/1000) is 1/3000. So, ${\frac {1}{3}}-\sum _{n=1}^{x}3({\frac {1}{10}})^{n}={\frac {1}{3*10^{x}}}$; therefore, when $x=\infty$, ${\frac {1}{3}}-\sum _{n=1}^{\infty }3({\frac {1}{10}})^{n}={\frac {1}{3}}-0.333...={\frac {1}{3*10^{\infty }}}$. Since $3*10^{\infty }=3*\infty =\infty$, ${\frac {1}{3}}-0.333...={\frac {1}{3*10^{\infty }}}={\frac {1}{\infty }}$, which is an infinitesimal. In other words, the difference is 0.

You also might want to look at this, which shows that repeating digits show up in all bases.—M_C_Y_1008 (^talk/_contribs) 03:14, 25 February 2007 (UTC)

First, I used the word "more" to illustrate why I believe it is impossible to express 1/3 as .333... . Second, we are not talking Marxism-Leninism here; we do not solve problems by simply saying they go on to infinity (alluding to the believed "perfect society" at the the end of time). The fact that their always exists a remainder does not change at infinity and makes it impossible to express 1/3 as .333... because you would need a remainder at infinity, which, as you pointed out yourself, is impossible. Third, just because the difference is infinitesimal does not mean that a difference does not exist. Fourth, recurring decimals occuring in all bases just means (at least to me) that some numbers can only be exactly expressed in their exact form (1/3, ln5, pie); repeating decimals are just approximations.--Jorfer 03:42, 25 February 2007 (UTC)

First, be that as it may, the sentence you typed didn't really mean anything much. Second, what the hell are you talking about? Third, yes it does. The real numbers do not contain infinitessimal numbers other than zero. The remainder is infinitessimal, therefore it is zero. That's what real analysis is for. Fourth, I don't really know what you're saying here. All numbers, by definition, "can only be exactly expressed in their exact form". Oh, and pie is a foodstuff. Pi is a greek letter used to represent a mathematical constant.

The fact is that it doesn't matter what you believe, or even what the clever proofs say. 0.3 recurring is exactly equal to 1/3 because it is defined that way. Infinite decimal strings are defined by their limit, not by their eventual sum (however you want to define that). Saying they're "really" something else is like saying black is "really" called white and white is "really" called black and all this time the entire English speaking world has been misusing both terms. It's wrong because the entire English speaking world made up the terms and they mean whatever it wants them to mean. There is no underlying reality for them to conform to, and the same is true of decimals. Decimal notation is a system mathematicians invented to make calculations simpler, so if mathematicians say 0.3 recurring = 1/3 then 0.3 recurring = 1/3. You might argue that that means the number you think "really" equals 0.3 recurring can't be expressed in decimal, but mathematicians are unlikely to lose sleep over that because that number is not a real number anyway. Andrew 13:47, 25 February 2007 (UTC)

You said, "0.3 recurring is exactly equal to 1/3 because it is defined that way". YES! This is really the whole point of 0.999...= 1. It is by definition. All of these proofs are kind of ridiculous IMO. The definition is all that matters, and definitions do not need to be proved. This article could be very short and elegant. But, I dare not propose such a thing. Tparameter 03:03, 26 February 2007 (UTC)

We have to be careful in saying this that we don't give the wrong impression. 0.333... is not defined as 1/3. It is defined as something that happens to be equal to 1/3. The actual definition of 0.333..., by the general definition of decimal representations, is a limit of a sequence, like this: $0.333...=\lim _{n\rightarrow \infty }{\left(\sum _{i=1}^{n}{3 \over 10^{i}}\right)}$. We can prove that this sequence is convergent (i.e. this limit exists) and that the limit is equal to 1/3 using an epsilon-N proof, which I will not write here. The important thing is that the equalities, 0.333... = 1/3 and 0.999... = 1, are direct results of the definitions. Maelin (Talk | Contribs) 03:44, 26 February 2007 (UTC)

Obviously. Let a=0.999... a=b by definition. b=1 by rules of limits. Therefore a=1. It could be so simple and elegant. Tparameter 05:13, 26 February 2007 (UTC)

Who has claimed this represents reality? And we don't need a better definition of real numbers, because they can be represented exactly by decimals. It's the number system which you are proposing which can't be represented as decimals. Mdwh 06:12, 25 February 2007 (UTC)

Jorfer,

1. As stated above, infinity is not a real number.

2. As stated in the FAQ for this section, "Unfortunately, in order to formally prove many qualities of numbers, one often has to resort to higher mathematics: real analysis in the case of real numbers, number theory in the case of integers, and so forth. The article and arguments page both aim to be understandable to all, but, since many skeptics ask for formal proofs, higher mathematics will inevitably come into play."

3. As stated in the irrational number article, "The first proof of the existence of irrational numbers is usually attributed to Hippasus of Metapontum, a Pythagorean who probably discovered them while identifying sides of the pentagram. However Pythagoras believed in the absoluteness of numbers, and could not accept the existence of irrational numbers. He could not disprove their existence through logic, but his beliefs would not accept the existence of irrational numbers and so, as legend had it, he had Hippasus drowned." Irrational numbers are part of the real number system, so you aren't the first person to label the real number system as not representing reality or "not rational."

4. You're not the first person here to propose nonzero infinisimals. To quote the FAQ, "These is no nonzero constant infinitesimal in the real numbers." More on what infinitesimals are can be found in the article on infinitesimals or in the aforementioned FAQ.

Calbaer 07:46, 25 February 2007 (UTC)

First, infinity is not a real number so why do you treat a number with recurring decimals that go to infinity as such? Second, I already stated in the above section as to why I reject the more rigourous proof, which is the one that uses the convergence theorem. I would question the convergence theorem's precision. The theorem gives a very close approximations to be sure, but is it exact. That same theorem would indicate that infinity (...999) = -1 (an interesting and funny development). You could probably spit out even more elaborate proofs, but they are no doubt founded on the same number theory that states that nonzero infinitesimals do not make a number different.--Jorfer 13:38, 25 February 2007 (UTC)

First, there are indeed infinitely many digits in a recurring decimal. A set is by definition infinite if there is an injective map from the set of natural numbers into that set. Thus, no number "infinity" is needed to talk about infinite sets. "Go to infinity" is, usually, some sort of abuse of notation; there is a (more complicated) concept which does not use infinity and which we speak of as "going to infinity" because it conforms to that intuitive notion.

Second, limits are exact and do not use "infinity" (except in notation, but not in their definition). As I said before, ...999 is not a real number nor a converging series, and for non-converging series the arithmetic of limits is no longer as helpful as for converging series. That's why the ...999=-1 "proof" fails.

And concerning non-zero infinitesimals, they don't even exist in the context of real numbers (nor in the context of p-adics, if I'm not grossly mistaken). And they aren't missed in any of "reality" or physics - or when did you last try to buy an infinitesimal amount of something? --Huon 14:45, 25 February 2007 (UTC)

If someone doesn't understand the math (and won't read the FAQ), there is little we can do here to help him or her. As stated, there are two mutually exclusive kinds of proofs for this: Easy to understand and rigorous. Someone who demands rigor but refuses to use real analysis is going to be disappointed, sort of like someone who demands a "simple" (nonrelativistic) explanation for why an objects can't exceed the speed of light. Finally, as I believe I've said before, the p-adic section doesn't really belong. People who don't understand 0.999...=1 will be confused by it. People who do understand 0.999...=1 will be confused by it. It doesn't address 0.999...=1. I've never understood why it's in here, and I really haven't heard anyone give a good reason for why. I guess I should take that up on the talk page.... Calbaer 19:17, 25 February 2007 (UTC)

Agh, you have all failed me. I requested irrefutable proof that 1/3=0.333..., and got an argument and faultly, inequal proof in return. If you can't prove 1/3=0.333... using algebra properly, then state that in algebra, 1/3=/=0.333..., please. Try again, without arguing. Thank you. 67.83.72.38 18:54, 26 February 2007 (UTC)

Well, you got the one response above that proves it. Here's another one. Try just doing the math, using long division to divide 1 by 3. 3 goes into 1 0 times, with a remainder of 1. 3 goes into 10 3 times with a remainder of 1. 3 goes into 10 3 times with a remainder of 1. 3 goes into 10 3 times with a remainder of 1. 3 goes into 10 3 times with a remainder of 1. 3 goes into 10 3 times with a remainder of 1. 3 goes into 10 3 times with a remainder of 1. Repeat ad naseum, no matter how many times you do it, you're still getting the same remainder, doing the same calculation, and writing down the same number. There is no chance that 3 will ever go into 10 any number of times but 3, nor will it ever have a remainder besides 1, so the number continues infintely as 0.333... --Maelwys 19:09, 26 February 2007 (UTC)

Defining and defending the "proper" rules of algebra is going to lead to more advanced mathematics than you seem willing to engage in. Describing the foundation of even the most basic mathematics is a complex subject.--Trystan 19:14, 26 February 2007 (UTC)

I proved 0.333... = 1/3 above, right after you requested it, at the top of this section. The proof may rely on known mathematical results that you are not aware of, but this is not the same as being invalid. If any steps of the proof seem unsupported to you, just say which ones, and I or someone else can prove them for you as well, and so forth backward until we reach either the axioms or statements that you are willing to accept (within reason, of course). Note that vague or intuitive disputes do not count. If you think the proof is invalid, you must be able to state which line of the proof does not follow from the previous line. Maelin (Talk | Contribs) 23:42, 26 February 2007 (UTC)

I apologize then. Specifically what I want proven is how, if x=0.333..., how 10x-x=9 rather then 9.000~001, since after all as far as I'm aware, a typical rule for multiplying things by 10 simply results in shifting every number to the left, which would mean 10x=9.999~990. Maybe I don't understand basic arithmetic correctly? If that's the case, I'd like you to educate me on how that's impossible. Maybe you could say there's an infinite number of 9s in 0.999..., but what's after that infinite, is the question. Another nine? To put it simply, I don't think 10*0.999... is 9.999.... Sorry for bothering you so, but it just seems to me that if you say only higher math can prove something, then it's not really true unless you wish to use said higher math. That is, what I'm saying is... if you can't prove 0.333...=1/3 on algebraic levels, then I don't see why you should eve consider 1/3 to be 0.333... in algebra. Well, aside for the fact that many consider 1/3 to be defined as 0.333....

Additionally, could someone please tell me why 0.999...=1 can't somehow be abused to effectively debunk mathematics and get away with saying 1=0 or 1=42 or etc? But that's not really the subject at hand, so if you don't want to, you don't have to. Either way.. thanks! 67.83.72.38 06:21, 27 February 2007 (UTC)

Let's think about the "multiplication by ten" issue a little differently; it might do away with one easy point of confusion. Instead of thinking about "shifting every number to the left", which is perfectly valid but still leaves one with the image of something like a zero having to fill in the rightmost space (said rightmost space doesn't exist, but one might wrongly think it does), think of multiplying by ten as shifting the decimal point one place to the right. Certainly that works for all cases. So 10 * .999999999... simply shifts the decimal point one over to 9.9999.....; same infinite string of nines we had in the first place; no need to worry about conceptually shifting any digits to the left, since you're just manipulating that pesky decimal point.

As far as considering the equality "in algebra", well, you don't have the tools at the level of high school algebra (if that's what you mean) to work with limits and such, and without them, you can't rigorously prove the equality. But you also can't prove that 1 + 1 = 2 without tools you don't have at the level of algebra (since they're not teaching number theory until college these days.) Sorry if I'm misconstruing your meaning. --jpgordon^∇∆∇∆ 07:03, 27 February 2007 (UTC)

Allow me to give another explanation of that multiplication. A real number in the interval (0,1) can be given as a=0.a₁a₂a₃... There is one digit a_n for every natural number n. This string of symbols represents 0*1+a₁*1/10+a₂*1/100+a₃*1/1000+..., the n-th digit corresponding to a_n*1/10^n. If we multiply this number by 10, the result can be written as b=b₀.b₁b₂b₃..., where, obviously, b_i=a_i+1. Now since every natural number i has a successor, all of b's digits are given by this rule. If b had a zero in some place, a would have had the zero, too - and in our example that can't be, because all of 0.333...'s digits are 3's. There is also no digit "at infinity", since infinity is not a natural number (nor is there anything "after infinity" - there are no digits which don't correspond to natural numbers).

Also note that I don't claim that only numbers in the interval (0,1) can be given in the form specified above - obviously 0 has this form, too, and by 1=0.999..., so has 1, but neither is needed for 1/3=0.333...

Concerning your question as to why the equality of 0.999... and 1 doesn't lead to contradictions, that's harder to answer without referring to advanced math (equivalence relations and the quotient of a set by such a relation, for example). But consider this analogy: You're probably used to several fractions representing the same rational number, such as 1/2 = 2/4 = 3/6 =..., So when you write rational numbers as fractions z/n with an integer z and a (non-zero) natural number n, you have to identify some of those pairs (z,n) that give the same rational number. Similarly, when you want to write real numbers as decimals, you have to identify some decimal expansions that give the same real number. If you try to construct a contradiction, you will probably end up dividing by 1-0.999... (which is zero) or some such - look at the following section, where an attempt at such a contradiction is made but fails because the infinite power and the infinite root are not inverse to each other (or, more to the point, because it is claimed that the infinite root of 0=1-0.999... should be anything but 0). Huon 10:23, 27 February 2007 (UTC)

So what you're saying is: 0.333... is not merely extremely close to 1/3, but is 1/3... and is actually capable of technically increasing more then other numbers, on the same unnoticeable level that is considered by those who don't believe 0.999... to be 1 to be the difference between two said numbers (which would effectively mean the increase doesn't actually exist, according to what your saying). Then, should there be a number x, which when multiplied by any natural number, it effectively shifts every number to the right of the decimal point so that it's to the left of the decimal point (yes, I know, such a number is probably impossible, but maybe higher maths will someday spawn such a number >_> ), then the number will basically be ...999.999..., and have effectively gained +1?

Additionally, what does 0.999... mean for numbers higher then 1? Is 1.999=2, or is only 1.999...998 equal to 2 (or maybe both?)? Which is it, because that seems important too. After all, if it IS in fact the latter, then that means as natural numbers go, their decimal values decrease, until they reach the hypothetical number ...999 (as many nines on it the left side of the decimal point as 0.999... has on the right), which would be equal 999...998 because of it.

Of course, I imagine I could see why it's not true 0.999...*2=1.999...998. It even go against what I wrote above. But does that mean 0.999...+0.999... is 1.999...? All in all, the only problem I have with .999... is that in order for it to function, it requires that all it's digits past the decimal point after any decimal point additions (like 0.4999... (0.999/2)) have to be continuous digits, usually 9 but not always (0.999.../9=0.111...) I guess it all works out if that's true, but before I finish this 'post' thingy, I must ask how IS that true? (Remember please, when explaining, that my mathematical knowledge is limited to Algebra 2. ^_^; ) 67.83.72.38 17:04, 27 February 2007 (UTC) (PS: one more question if no one minds, am I incorrect in assuming all higher math is in some way derived from algebra, just in very difficult manners that no one has time to thoroughly explain without actually trying to teach such mathematics? Thank you for your responses, as well as any continued responses...!)

The easiest way to think about it is that there is no last digit therefore you cannot manipulate it. Kind of like an infinite line of drink cans - it is impossible to reverse its order because you cannot grab the last can. If you could, you contradict yourself because there is a can after that one. x42bn6 Talk 22:05, 27 February 2007 (UTC)

Though you cannot "grab" the last one, I feel it is safe to assume that their exists one down the line. The problem people really have (me included) with this equality is not the math, but the number theory it is based on. The argument all boils down to an understanding of infinite, which is beyond man's mind to truly comprehend.--Jorfer 22:51, 27 February 2007 (UTC)

It can be as well-defined as any other concept, and can then be used in ways which are consistent with that definition. All math has the potential to be confusing and counterintuitive, but that's no reason to throw up one's hands and proclaim it unknowable.--Trystan 23:01, 27 February 2007 (UTC)

Man may not be able to comprehend infinity, but many men can abstract the idea of infinity and obtain some amazing things from it (engineering, theoretical physics, the Internet, etc.). If there's a first something and a last something and you can count down everything in-between, clearly that's not infinite. But you're saying that digits are an infinite series with a first 9 and a last 9. Can we not count all the 9s in-between? Is there some "unreachable 9"? If it's unreachable, how can it have any meaning?

As a matter of fact, you can eventually, in theory, count up to every decimal place. Each decimal place has an integer index associated with it. But there's no last one, so there aren't a finite number of them. As pointed out in a prior thread, you should probably start out with Hilbert's paradox of the Grand Hotel. If you can't understand the concepts therein, you'll have trouble rigorously understanding repeating (and other nonterminating) decimals.

And, of course, it bears repeating that if there were some way of showing that 0.999... = 1 that was both completely rigorous and simple, there would be no need for this article. Calbaer 23:09, 27 February 2007 (UTC)

By the way, to answer your last question, the higher math used here is indeed derived from algebra and arithmetic, in much the same way that algebra is derived from arithmetic (by allowing for unknowns). It's just adding new concepts and definitions, but it's ultimately all consistent. As for your other questions, it would help to focus on what you don't understand about the concepts at hand, rather than on what else you might be curious about. And a lot of these things are answered in the two FAQ, in the article, and in the discussion (though no one will fault you for not reading the last of these, seeing as how it's just about the longest discussion I've seen for a Wikipedia page). Calbaer 23:20, 27 February 2007 (UTC)

Something wrong with this?

1=0.999...
1-0.999...=0
(1/10)^infinity=0
infinity root of ((1/10)^infinity)=infinity root of 0
1/10=0
1/10*10=0*10
1=0

Usually, the maps $x\mapsto x^{\infty }$ and $x\mapsto {\sqrt[{\infty }]{x}}$ are undefined. Give precise definitions for these maps, and you will probably see that they're not inverse to each other. --Huon 10:22, 26 February 2007 (UTC)

You could use the same logic to say that since $\infty =\infty +1;0=1$. Basically, you can't really do math based on equalities of infinity, since any formula with infinity in it is basically equal to any other formula with infinity in it.

Just to clarify something, though, $x^{\infty }$ can be definied in the real number system be defined as $\lim _{z\rightarrow \infty }x^{z}$, just as $\sum _{z=1}^{\infty }=\lim _{n\rightarrow \infty }\sum _{z=1}^{n}$. You're simply looking at the infinite limit of a constant product series.

Similarly, you could attempt to define an "infinite root" of a number x as that number y which satisfies $\lim _{z\rightarrow \infty }y^{z}=x$. However, in the real number system, 0 and 1 are the only real numbers with infinite roots (they are their own infinite roots). Dugwiki 16:46, 26 February 2007 (UTC)

P.S. In fact, under that definition, 0 has an infinite number of infinite roots, since for all y such that -1 < y < 1 you have $\lim _{z\rightarrow \infty }y^{z}=0$. 1 is its only infinite root. Dugwiki 16:50, 26 February 2007 (UTC)

Actually $x^{\infty }$ can only be defined in that way on the half-open interval (-1,1], and there it maps 1 to 1 and everything else to 0 - outside that interval the sequence doesn't converge. And I would define the "infinite root" as $\lim _{n\to \infty }{\sqrt[{n}]{x}}$, which is defined on [0,oo), where it maps 0 to itself and everything else to 1. Then I have well-defined functions on subsets of the real numbers, but they're not inverse to each other. Huon 17:37, 26 February 2007 (UTC)

No, it doesn't really map 1 to 1. 1^∞ is undefined (an indeterminate form). --Trovatore 18:50, 26 February 2007 (UTC)

Thanks for the clarification, Huan. Obviously what I talked about would only be defined when the limit converges. If you extend the reals to include positive and negative infinity, then you could also presumably use the same definition for positive reals greater than 1 and say that the limit is positive infinity. (Negative reals less than or equal to -1 would still have no convergent limit since the signs constantly alternate).

As far as $1^{\infty }$ goes, though, it seems to me that it would be 1, since $\lim _{z\rightarrow \infty }1^{z}=1$. There's nothing indeterminate that I can see about that - it's the limit of a constant value function. Dugwiki 20:50, 26 February 2007 (UTC)

However, $\lim _{z\rightarrow \infty }\left(1+{\frac {1}{z}}\right)^{z}=e$. — Arthur Rubin | (talk) 21:23, 26 February 2007 (UTC)

In the sense of l'Hôpital's rule "$1^{\infty }$" is indeed an indeterminate form, where "$1^{\infty }$" abbreviates $\lim _{x\rightarrow \infty }f(x)^{g(x)}$ where $f(x)\rightarrow 1$ and $g(x)\rightarrow \infty$ as $x\rightarrow \infty$. However in Huon's case "$1^{\infty }$" abbreviates $\lim _{x\rightarrow \infty }1^{g(x)}$ with a constant base 1 replacing the (mostly) arbitrary function f(x). l'Hôpital's rule does not apply; $1^{\infty }=1$ in the specific sense Huon introduces above. To make l'Hôpital applicable, you have to be able to fiddle the comparative rates at which f(x) converges and g(x) diverges, which is not an option Huon's function gives you. Endomorphic 21:45, 26 February 2007 (UTC)

How do you figure? Nothing in what Huon wrote implies that the base is to be held constant. --Trovatore 21:52, 26 February 2007 (UTC)

Striking out the previous comment since it was already addressed. Basically the point is that the definition I posited is a constant raised to a power, not an arbitrary function raised to a power. Dugwiki 21:56, 26 February 2007 (UTC)

But you can't tell that from the expression 1^∞. And anyway you weren't the one making the initial claim that I objected to. --Trovatore 22:04, 26 February 2007 (UTC)

I did say that the infinity-th power could be defined in that way, referring, of course, to Dugwiki's definition. I would have supposed that to be obvious from his post preceding mine, my reference to a non-converging sequence, and my analogous definition of the infinite root. So $x^{\infty }:=\lim _{n\to \infty }x^{n}$, with the special case $1^{\infty }=\lim _{n\to \infty }1^{n}=1$. Huon 22:23, 26 February 2007 (UTC)

You mean the unsigned boxquote above your text? I didn't read it. It looked like part of your text. --Trovatore 22:28, 26 February 2007 (UTC)

Uh, no, the unsigned boxquote is not from either me or Dugwiki. That was by some anon (whom we probably have confused completely by now) arguing that 1=0.999... leads to contradictions. I only mentioned that his infinity-th power and corresponding root won't behave as he expects, and then Dugwiki and I engaged in a discussion about how they could be defined at all. My first answer to Dugwiki is the first place where anybody explicitly mentions the infinity-th power of 1, and I believed that was what you didn't agree with. Huon 22:40, 26 February 2007 (UTC)

<unindent> We're evaluating a function of h(x) at x=1. That function is $h(x)=\lim _{z\rightarrow \infty }x^{z}$. You evaluate h(x) at x=1 and you get $h(1)=\lim _{z\rightarrow \infty }1^{z}=1$. We're not discussing continuity or differentiability or anything, so you have no reason to vary x. The domain of h(x) doesn't involve sequences converging to $t\in (-1,1]$ or continious paths terminating at $t\in (-1,1]$. The domain is just $(-1,1]$, and you can't vary x when you're evaluating h(x) at x=1. Endomorphic 23:44, 26 February 2007 (UTC)

The fact that it's discontinuous at 1 suggests that it's not what we should be discussing, and the infinity^th root is defined only at 0 and 1, at best, and is clearly not the inverse function of the infinity^th power. — Arthur Rubin | (talk) 00:18, 27 February 2007 (UTC)

That's a much simpler discussion. $(1/10)^{\infty }=0=(1/5)^{\infty }$, demonstrating the function is not one-to-one and therefore has no inverse, discontinuous or not. Endomorphic 00:38, 27 February 2007 (UTC)

Obviously we're all getting waaaay off on a tangent away from the main purpose of this talk page. But I think we're actually all in agreement on the mathematical facts, namely that -

1. $\lim _{z\rightarrow \infty }x^{z}$ doesn't exist for constants x less than or equal to -1
2. For the interval (-1,1), $\lim _{z\rightarrow \infty }x^{z}=0$. Thus in a sense every number in (-1,1) is an "infinite root of 0".
3. $\lim _{z\rightarrow \infty }1^{z}=1$, so as above 1 is by that sense its own infinite root.
4. For constants x greater than 1, $\lim _{z\rightarrow \infty }x^{z}$ doesn't exist (or if you prefer the limit is $\infty$). Dugwiki 16:36, 27 February 2007 (UTC)
5. The above describes an "infinite power" function which is not invertible (its only possible values are 0, 1 and $\infty$), and likewise the corresponding concept of "infinite roots" shows that only 0 and 1 have infinite roots and the infinite roots of 0 are the elements of the interval (-1,1). Dugwiki 16:36, 27 February 2007 (UTC)

You guys are going about this entirely the wrong way, let me show you why .999... does not equal to 1

The original statement states:

c = .999...

10c = 9.999...

-c -.999...

This is where the problem is. Although it is stated that c = .999... you can't just choose to not take away c from the other side, you must go through the equation according to the rules of mathematics, no matter what it is, you still have to take away the same exact thing. Hence,

10c = 9.999...

- c - c

9c = 9.999... - c

add to even it out

+c +c

10c = 9.999...

____ _____

10 10

c = .999...

Or simpler:

c = .999...

10c = 9.999...

____ _____

10 10

c = .999...

Or more vividly:

c = .999...

[10c] 10 (.999...) = 9.999...

this means multiply by .999... not 10.999...., so if you try taking away .999... from each side you would yield 10 = 9, which is absurd, hence it must be 9 = 9, but the unknown variable is not used.

[9c] 9 (.999...) = 8.999...

[1c] 1 (.999...) = .999...

c = .999...

In conclusion, no you can't pick and choose to use an unknown variable on one side and use a known variable on the other side, you must adhere to the rules and take away the EXACT same thing, not assumptions of what it is. But that doesn't even matter because the other proofs show that .999... is not 1. Anyone who thinks otherwise has their head too far up their ass to see the clear picture by using simple elementary math. —The preceding unsigned comment was added by Xasthuresque (talk • contribs) 01:46, 28 February 2007 (UTC)

Moved from talk by Melchoir 02:33, 28 February 2007 (UTC)

If you're uncomfortable with algebra, the first proof is not the one you cite but rather an arithmetic one. In the algebraic proof, c is not an unknown. c=0.999... is a given. What's left to prove is 0.999...=1. If we couldn't use givens and axioms to prove things, mathematics would go nowhere. If instead your objection, very similar to others, is that 9.999...-0.999... is not 9 due to some "missing little bit," please read the article, FAQ, and above discussion. Considering the way in which you claim to be the one to overturn common mathematical knowledge, I suspect you might not, in which case you will be ignored as someone not truly interested in logic. If you do read and attempt to comprehend it, contributions regarding concerns you have will be addressed. Calbaer 03:00, 28 February 2007 (UTC)

I think the editor is attempting to argue that from a=b, and c=d, I can't get a-c=d-b. Hence the comment that " Although it is stated that c = .999... you can't just choose to not take away c from the other side, you must go through the equation according to the rules of mathematics, no matter what it is, you still have to take away the same exact thing" What the editor apparently does not understand is that they are the exact same thing, since c is .999... JoshuaZ 03:06, 28 February 2007 (UTC)

All right fine, I might not be that knowledgeable in math, but let me propose this then. I don't believe that .999... is a real number, it it was we could denote .999...8 (with 8 at the end of an infinite chain) and so forth, because we can't, we must work with finite numbers, it must be some sort of special case of limits of whole numbers, such that 0 > .5 > [.999... >1] > 1.5. Basically where the line would "meet" the x axis on a graph of perfect infinite slope. This is how infinity should be defined, as a limit of whole numbers. Xasthuresque 1:38, 28 February 2007

• the end of an infinite chain ... if it's infinite it has no end. That's the whole point of infinite. There's no there there. --jpgordon^∇∆∇∆ 06:50, 28 February 2007 (UTC)