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Slope counterproof

The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

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I ran across a description of this on another forum and decided to draw it up.

Lines A and B are parallel and at distance 1. Red lines are drawn to progressive powers of ten, resulting in a familiar sequence. The limit of this sequence is 1. But if you try to use 1 at the limit, you get a line BZ with slope zero - identical to B, and yet intersecting A by definition. Contradiction! Therefore .999... cannot equal 1. Algr (talk) 09:04, 6 April 2009 (UTC)

By definition, the limit is not a point on the graph. Oli Filth^{(talk|contribs)} 10:10, 6 April 2009 (UTC)

On projective plane lines A and B intersect at infinity. If B is line y=0 and A is line y=z, then they intersect at (the equivalence class of) point (1,0,0). Line A projected to affine plane z=1, is of course y=1. Tlepp (talk) 10:27, 6 April 2009 (UTC)

If 0.999... were less than 1, you'd still end up with lines with an intersection at infinity, wouldn't you? Or would you end up with non-parallel lines which don't intersect? Either way, it's no better than what happens when 0.999...=1, so it's not a proof either way, but an indication that "has an intersection in Euclidean geometry" may not be preserved by limits. Huon (talk) 10:37, 6 April 2009 (UTC)

No. You would not end up with intersecting lines because (all together now...) "Infinity is not a number.". Although this is not a proof, it is a brilliant illustration demonstrating that your concept of limit and real number(Cauchy sequence or Dedekind cut) is not well-defined. 98.201.123.22 (talk) 23:43, 7 April 2009 (UTC)

I have given a definition of real numbers using Cauchy sequences (and definitions of Cauchy sequences and limits not using real numbers) above. Please point out what, precisely, isn't well-defined. Huon (talk) 01:47, 8 April 2009 (UTC)

Whether 0.999... is less than 1 or not, on projective plane the line y=(1-0.999...)x intersects the line y=z at the point (1, 1-0.999..., 1-0.999...). Thus it's not a proof either way. Btw I can draw too :) Tlepp (talk) 13:10, 6 April 2009 (UTC)

The contradiction only exists if you insist that the slope of BZ is zero. If it is not, then BZ is simply not parallel to the other lines, and so there is no problem with intersections.Algr (talk)

If there's no problem, then give the Euclidean coordinates of the limit point Z. Huon (talk) 20:57, 6 April 2009 (UTC)

It is the same as the coordinates where .999... = 1. 1 per digit. Algr (talk)

0.999... = 1 is a true statement regardless of the coordinates. It's like saying "the coordinates where 1+1=2." Please give coordinates in the form (x,y), where x and y are unique real numbers. Heck, you don't even have to tell me what they are, as long as you can prove that they exist, are real, and are unique. --Zarel (talk) 22:10, 6 April 2009 (UTC)

No Zarel, by insisting that .999... = 1, YOU are the one who is insisting that that limit point exists, so YOU are the one who needs to be able to provide it's coordinates. If the limit point does not exist, then infinite processes can never reach their limits, and .999... can never equal one, even if calculated infinitely. This is familiar territory, it's "Infinity only works when we want it to." again. Algr (talk) 22:43, 6 April 2009 (UTC)

By insisting that 0.999...=1, we claim that the limit of the sequence of slopes exist. We do not claim that the limit point exists, and the existence of one limit would imply the existence of the other only if the map from a slope to its point of intersection were continuous - but it isn't even defined at 1. In effect, you're taking a sequence converging to zero and then divide by it - and then act surprised when, while you can divide by all non-zero elements of the sequence, you can't divide by the limit. Are you saying that no sequence can ever converge to 0? As an aside, if 0.999... were not equal to 1, wouldn't you still say there's a limit point for slope 0.999...? If not, why should a limit point exist when 0.999...=1? If so, give the x-coordinate of that point. And no, "1 for every digit" would be infinity, which is not a viable coordinate in Euclidean space. Huon (talk) 23:37, 6 April 2009 (UTC)

This discussion is exactly the sort that caused me to suggest we delete this page last week. We're just re-treading the same old ground again and again. Do we really need yet another thread where we carefully explain that a geometric "process" never reaches its limit? Oli Filth^{(talk|contribs)} 22:54, 6 April 2009 (UTC)

That is precisely the point I am trying to make. Algr (talk

Albeit from the "appearing to refuse to comprehend the implications of limits" side of the argument?! If you're averse to the "familiar territory", why bring up yet another fatally flawed "counterproof" which fails to distinguish between the process and the limit? Oli Filth^{(talk|contribs)} 23:17, 6 April 2009 (UTC)

You hit me with this abuse and then complain when I don't respond, and that makes me a troll?

If it comes across as "abuse", then that's unintentional; it's just my honest observation. You yourself complain that this is "familiar territory", but that's only due to you starting yet another thread which, unfortunately, is based on the same flawed assumptions that were being brought up a year ago. Time and again, people have clearly pointed out that the process and the limit are not the same thing, but you still post again as if it were not the case. Whether it's wilful ignorance, a genuine inability to comprehend basic real analysis, or actual trolling, this strikes me as a clear indication that this page no longer serves any useful purpose. Oli Filth^{(talk|contribs)} 08:14, 7 April 2009 (UTC)

Actually, by insisting 0.999...=1, I'm saying that the limit point should be at a point where 1-slope = 0.999... = 1 - in other words, where the slope = 0 = rise/run. Since rise = 1 and run = limit, so 1/limit = 0, so the limit point does not exist in $\mathbb {R} ^{2}$. You're the one who insists it exists. --Zarel (talk) 06:41, 7 April 2009 (UTC)

I note that you're talking about a projective plane here. Did you know that lines with the same slope still intersect in a projective plane? Therefore, parallel lines do not exist in a projective plane (it may help to think of them as spheres), and the fact that two lines are not parallel says nothing about their slope. Your "counterpoof" does nothing except prove that 0.999... = 1. --Zarel (talk) 22:19, 6 April 2009 (UTC)

I'm NOT talking abut a projective plane, others brought that up. (to complicate things?) And in any case, such projections are never considered to be equal to the original object being projected. A photo of a cube is not a cube. Algr (talk) 22:43, 6 April 2009 (UTC)

See above. Most people assumed you were talking about a projective plane, since the limit point doesn't exist in an $\mathbb {R} ^{2}$ plane.

(You accidentally deleted this message here) --Zarel (talk) 13:13, 7 April 2009 (UTC)

Notice that the image is not drawn to scale. This, from a practical perspective, makes sense, because if it were, the lines would get very flat, very quickly, and they would be obscured by each other. However, by drawing it as above, (and including the limit line as they did), they made what essentially amounts to an optical illusion that the limit line should not be flat.

Once you start drawing it to scale, you will see how quickly it gets flat, which is all the more reason to believe that the limit should ultimately be absolutely flat (ie. slope = 0). --72.177.97.222 (talk) 11:26, 6 April 2009 (UTC)

Oh, I see you (Algr) yourself drew it. Well, you didn't draw it to scale, so you made (perhaps inadvertently) an optical illusion. --

72.177.97.222 (talk) 11:47, 6 April 2009 (UTC)

Surely a perspective drawing with vanishing points would be the illusion. Right angles aren't even 90° ect. Algr (talk)

Good try, Algr. For future reference, there's no such thing as a "counterproof" in mathematics. Gustave the Steel (talk) 14:35, 6 April 2009 (UTC)

Well, you can have a proof that contradicts another proof, but at that point you generally realize that one of those proofs is flawed, and try to determine which one.

In this case, the proof is flawed because it's assuming that R^2 is non-Euclidean. --Zarel (talk) 22:04, 6 April 2009 (UTC)

No, this is worse than non-Euclidean.

It's non-real, too. --72.177.97.222 (talk) 11:56, 7 April 2009 (UTC)

What's wrong with the term counter-proof? We have counter example. It's not standard, and I've never heard of it, but why not use it? It might be taken so that a counter-proof of A = B would be a proof that A ≠ B. A counter-proof of "If X then Y" would be a proof that "If X then not Y. I don't see why people are being so snobbish; saying things like "For future reference, there's no such thing as a "counterproof" in mathematics." How does Gustave the Steel know that for sure? This page is awlays full of the same: people trying to get one up on other people, and show themselves as mathematically superiour. People should act and reply in good faith: you understand what Algr was trying to say, so answer that point and don't play semantics.  Δεκλαν Δαφισ   (talk)  11:43, 27 July 2009 (UTC)

I'm not sure where Gustave the Steel decided that; I'll let him (I assume?) defend himself on that point.

Algr's point was addressed. Several answers were given, including, but not limited to: his proof depended on a warped image, his proof assumes certain things about a limit that simply are not true, his proof refers to an undefined concept of limit (not critical, since it's trivial to define a limit of lines if they all go through the same point), his proof assumed that the limit point Z should exist, etc. etc. Don't tell us that no one made a serious attempt to show Algr the errors in his proof.

But now, allow me to show you how I am mathematically superior to you: A counterproof of "If X then Y" would not be a proof that "If X then not Y". It would only be a proof that there exists a scenario (here I'm using "scenario" as a rather vague term; usually, it will refer to the set of variables that X and Y depend on) in which "If X then Y" fails. There may still be scenarios, however, where "If X then Y" still technically succeeds, even if it doesn't do so every time.

But, of course, that was just a mistake. You're human. We all understand that, and can forgive it. Unfortunately, that's something you seem to have difficulty accepting and admitting to yourself. Historically, you have been quite the snobbish person yourself on this page. I think you are quite capable of contributing some very good points yourself whenever people try to bring up faulty points here, but you always seem to go blind with rage (or possibly just snobbishness) whenever someone with their heart in the right place makes a minor error in failing to state their points precisely enough for you. --COVIZAPIBETEFOKY (talk) 12:47, 27 July 2009 (UTC)

I thank you for giving a weight to my previous comments. You reply was offensive, petulant, and arrogant, not to mention incorrect. As I made clear in my statement: I gave ideas as to what a counter-proof might be. Your show of "mathematical superiority", was in fact an attempt to show logical superiority. Contapositives, converses, and negations are formal logic, not mathematics (although they are used therein). Besides, you speak of counter-examples! A proof shows that something is true always, so a counter-proof might be taken to show that something is always false, just as I said above. Showing that scenarios are incorrect means you are offering counter examples. I have no need to enter into petty arguments, you have given a good example of the arrogance I speak. I made a nice helpful post and you replied with abuse, nice! If you chose to judge me by a handful of post almost a year ago, and please feel free. But I think you should reply to a post in the tone of that post. I was friendly, and tried to be helpful. You were just abusive. Grow up! As for "your mathematical superiority", just google me.  Δεκλαν Δαφισ   (talk)  13:12, 27 July 2009 (UTC)

Awww, you're still every bit as arrogant as you ever were! And, certainly, no matter how much you refuse to admit it to yourself and others, unfriendly. I never questioned your credentials, sir; I was merely trying to point out something about a pot calling a kettle black. Tell me that I don't need to go into silly pointless analogies in order to communicate my point. --COVIZAPIBETEFOKY (talk) 13:38, 27 July 2009 (UTC)

What I would like you to tell me is how your abuse, in any way, shape or form, added to the comment thread. You wasted your own time and effort abusing another human being. You seem to be iterating your own "pot calling a kettle black" argument. Please stop. If you have something good nature to add that will aid the discussion thread then please do so. If you want to abuse and insult me then do it to my face...  Δεκλαν Δαφισ   (talk)  14:00, 27 July 2009 (UTC)

Stop it. Maelin (Talk | Contribs) 14:17, 27 July 2009 (UTC)

Well, I would like to think that I was trying to inform you of how arrogant you were being in your comments.

I would also like to think that you would have listened and realized that what I was saying is true, that you tend to flaunt your credentials in everyone's face and use them as a license to attack other people, and perhaps you could learn to loosen up a little and stop criticizing the people who are trying to show Algr his mistakes.

But since that is never going to happen, I see no point in taking this discussion any further. I apologize for wasting your time. --COVIZAPIBETEFOKY (talk) 14:18, 27 July 2009 (UTC)

You seem to be carrying a grudge, and to have a chip on your shoulder. Please cut and paste my so-called arrogant comments I have made in this "Line Counterproof" thread. You won't because there aren't any. I was being helpful and friendly; this you cannot deny, and that's why you hark back to my comments a year ago. I was not criticizing the people that are trying to show Algr his/her mistakes; I was criticizing the way people were doing it, like Gustave the Steel for example: "For future reference..." That's not going to help anyone. That's an attempt at mathematical authoritarianism, made all the more unpalatable by the clear lack of mathematical authority held by those making such statements. And I haven't flaunted my credentials, but simply used them in defence of your attacks, and used them to highlight the weaknesses of other attacks: my mantra is this: don't bang on like an expert when you're clearly not. It's all in black and white, just scroll up. I tried to be helpful and you just turned up out of the blue and attacked me. I am going to bring this thread to the attention of the admins. You have been abusive, rude, and clearly not willing to discuss the point in hand due to some other hidden agenda.  Δεκλαν Δαφισ   (talk)  14:46, 27 July 2009 (UTC)

All this happened in one morning! Wow. First of all, I couldn't find any definition of "counterproof" in Dictionary.com, Wiktionary, or any paper dictionary in my posession. You must therefore admit that my earlier statement is factually correct, at least according to my own available knowledge.

Furthermore: if a mathematical theorem has been proven, and the proof is sound, then there literally cannot be a "counterproof" or even a counter-example. Thus, my statement holds both linguistically and mathematically.

Finally, you assume I was being rude for some reason. Understand this: I meant "Good try" sincerely, and "For future reference" is not a phrase I associate with rudeness. It's easy to read whatever emotion or tone you want into text, but you should consider that I may not have intended my statement the way you took it. Gustave the Steel (talk) 17:48, 27 July 2009 (UTC)

Like I said: I have never heard of a counter-proof. My point was, why not use it? I was trying to say that a counter-proof to a statment would be a proof that a statment is false. For example a counter-proof of "3 is an even number" would be to use the Chinese Remainer Theorem to show that gcd(2,3) ≠ 2, and so 3 is not divisable by 2, and so is not even. Your attempt to show that your statement holds both linguistically and mathematically assumes that a given statment is correct and as such a counter-proof could never be given, since it's true, so what you say is a tautology.  Δεκλαν Δαφισ   (talk)  18:35, 27 July 2009 (UTC)

It seems we are working on two different assumed definitions of "counterproof". As I interpreted it, the word refers to a proof which shows that another proof is false; you and I agree that no such proof can exist. You interpret "counterproof" as an argument which disproves an assertion. However, by that logic, all proofs would be considered "counterproofs" of statements which disagree with their conclusions! "Proof" and "counterproof" would therefore be perfect synonyms. Your definition is better than mine, since mine describes an impossibility while yours is merely a redundancy. Still, I'm not convinced the word should exist at all. Gustave the Steel (talk) 02:49, 28 July 2009 (UTC)

Algr, there's no contradiction at all! The notation 0.999... defines a limit, given by the usual sequence (a_n) := (0.9, 0.99, 0.999,...). No-one ever claims that this sequence reaches 1. We say that the limit of the sequence is 1. The elements of the sequence get closer and closer to 1 without going past 1, and we can choose an element of our sequence so that it and all succesive elements are within a given distance from 1: There exists N such that for 0 < ε < 1, we have 1 - ε < a_N < a_N+1 < 1. That's just the idea of a limit: our sequence gets closer and closer to 1, although we never claim that it reaches 1, it just gets closer and closer, in fact a close as you like.

It's just the same with your lines. The sequence of lines will never be the line B since they must always intersect A. But as the intersection point on A moves out to infinity the gradient of the line gets more and more shallow. No-one ever claims that the red lines will become the line B, only that their gradients will become as close to 0 as you choose.

In the case of the limit of (a_n) we can always zoom in to see clear space between any a_N and 0. Likewise, for the line example, we could always zoom out to see that no single line in the sequence of lines is actually the line B.  Δεκλαν Δαφισ   (talk)  12:41, 27 July 2009 (UTC)

tango tried to close this thread. This was because people were more personal arguments than mathematical arguments. As I have said, please stick to the mathematics, and stay away from personal attacks... I mention no names. I removed Tango's archiving because he is not an Admin (although he once was) and has no right to block further conversation on what could be a very interesting topic; personal attacks aside. I look forward to hearing about some interesting mathematics.  Δεκλαν Δαφισ   (talk)  15:10, 27 July 2009 (UTC)

There is no need to be an admin to close a thread, it's an informal thing. I'll allow your reversion to stand for now but, if things don't improve on both sides, I will close it again and will go and find an admin to enforce the closure if necessary. --Tango (talk) 16:12, 27 July 2009 (UTC)

Feel free, then the Admin will see that I have been subject to attacks and insults, and that I have been trying to defend myself using reason.  Δεκλαν Δαφισ   (talk)  16:17, 27 July 2009 (UTC)

Declan, I don't know if you are just oblivious to it, but you really are coming across as an insufferable, pompous wanker. I'm not saying you are, but that is really how your posts are being interpreted. Please try to allay the apparent self-righteousness and maybe we will be able to have some more mathematical discussion and less of this tiresome bickering. Maelin (Talk | Contribs) 10:23, 28 July 2009 (UTC)

Maelin, thanks for proving my point. There seem to be a core of rude, offensive, arrogant people on this page, and in your case: vulgar! Every time I try to bring an end to this someone - you in this case - pops up out of the blue and insults me. Why did you have to write that post? Did it add to the mathematical discussion? No! Just give it a rest, please.  Δεκλαν Δαφισ   (talk)  10:33, 28 July 2009 (UTC)

Oh no! You've brought back the Algr! Hi Declan! I just wanted to point out the irony that I've so often been accused of not understanding limits or not having read that article, but when another word comes up, others actually take pride in choosing not to understand the obvious intent. That kind of behavior doesn't inspire trust that you know what you were talking about. Imagine if a judge invalidated a contract because someone wrote "irregardless" instead of "regardless", and thus allowed one side to take unfair advantage of the other. Algr (talk) 15:32, 28 July 2009 (UTC)

2 (Two)

Well the abuse is familiar. Huon, where did I divide by zero? If the limit can't be used as the conclusion to the original sequence of lines, then how do you justify using 1 as the conclusion to a sequence .999... YOU seem to be saying that the value of an infinite sequence is not equal to it's limit - and that is the point of the original illustration. Algr (talk) 07:00, 7 April 2009 (UTC)

I asked for the x coordinate of that limit point you claim should exist for a reason. (I note you once again avoided saying anything about that coordinate if 0.999... were less than 1.) What your example demonstrates is this:

The points on line A can be parametrized by their x coordinate. The coordinate of the point of intersection of a line of slope x is given by 1/x. You consider the sequence of slopes given by 1-0, 1-0.9, 1-0.99, 1-0.999, and so on. This sequence converges to 1-0.999...=0. The sequence gives rise to a sequence of points parametrized by coordinates 1/(1-0), 1/(1-0.9), 1/(1-0.99), and so on. If there were a limit and the construction were continuous, that limit would have to be 1/(1-0.999...)=1/0. So what you've shown is that when a sequence converges to 0, the sequence of reciprocal values does not converge at all. Undisputed and well-known, but irrelevant to the discussion at hand.

If 0.999... were less than 1, things would not get better at all. You'd have to settle for one of a couple of bad options:

1. The sequence (0, 0.9, 0.99, 0.999, ...) does not converge at all. That would mean that there's no limit slope, and it would beg the question what 0.999... is at all and why that non-converging sequence is important to 0.999...
2. The sequence converges, but the line with slope 1-0.999..., while non-parallel to A, does not intersect line A. Euclid would not like this.
3. The sequence converges and the line with slope 1-0.999..., while parallel to A, is not equal to B. That's two lines parallel to A through the origin, something Euclid wouldn't be happy with either.
4. The sequence does converge and there's a point of intersection whose x coordinate you mysteriously can't give, while it can be shown to be larger than any integer.

Finally, define "value of an infinite sequence". That sounds as if you have some serious misconceptions about series and limits, but maybe I just misunderstand you. Huon (talk) 10:14, 7 April 2009 (UTC)

The sequence of slopes is actually .1, .01, .001, ect... I added the "1-slope" detail so that the sequence would produce .999... and not .000...1, which offers too many ways for people to evade the point. So there is no division by zero. If we use base infinity, then the description is simple. Point BZ is at 1_0®0 and the slope of the line is ®1. (I changed the notation for base infinity slightly, the ® acts like a decimal point, separating the reals and infinitesimals place.) Algr (talk)

So you choose option 4 (and to give credit where it's due, you do give the intersection point's coordinate in a way - though I believe you still have to leave Euclidean geometry to do so). But I wonder: Why is that point of intersection 1_0®0 and not, say, 0.5_0®0? I'll argue as follows: Take lines with slope 2*(1-0) intersecting A at 1/2, with slope 2*(1-0.9) intersecting A at 10/2, with slope 2*(1-0.99) intersecting A at 100/2 and so on. The sequence of slopes is twice the sequence of slopes you gave, so the limit of the sequence of slopes should be twice the limit of that other sequence, or 0®2, shouldn't it? And the point of intersection of the limit line should be 1/(0®2)=0.5_0®0, shouldn't it? Do those sequences of points really converge to different infinite points?

Now I'll get even meaner and consider the sequence of lines with slopes 1/5*(1-0), 1/5*(1-0.9), 1/5*(1-0.99) and so on. By analogy this sequence of lines should produce a limit point at 5_0®0. But wait - it's the previous sequence of lines with the first one omitted! Does it change the limit of a sequence when I omit the first element?

So obviously the calculus of limits suffers a severe breakdown in your setting. Actually I believe the best solution to this problem would be to use the topological notion of limit I gave above, by which limits cease to be unique when we have infinitesimals. Thus, in your setting, I'd say that (0, 0.9, 0.99, ...) does converge to 0.999...≠1, but it converges to 1 as well. If you don't use that topological definition, I'd like to see yours. Huon (talk) 11:50, 7 April 2009 (UTC)

Algr, you are just continuing to prove that you do not know the definition of a limit. The sequence of intersection points, as Huon points out, is (1, 10, 100, 1000...). It is easy to look at that sequence and make the mistake of thinking "well, that limit is clearly infinity!" However, in the definition of a limit, the points have to converge onto each other. You may be confused by this common notation:

$\lim _{n\rightarrow \infty }a_{n}=\infty$

However, suffice it to say that, in the strictest sense, that is nothing more than an abuse of notation. At best, it has a special definition treated quite separately from the standard definition of a limit. That statement, in fact, necessitates that the limit doesn't exist. --72.177.97.222 (talk) 11:59, 7 April 2009 (UTC)

I just noticed that things aren't as bad as I thought - they're worse. I wrote that using the topological definition of limits, they'd cease to be unique in the presence of infinitesimals. That's not necessarily true; it of course depends on the topology of our set. I don't see any useful topology for the proposed ring structure; we can choose between an incompatible topology, the trivial topology, and one which has the constant sequences as the only sequences of real numbers converging at all. Huon (talk) 12:58, 7 April 2009 (UTC)

In reply to:

But the POINT of the graph is that a set of logical steps produces a contradiction if you assume that 0.999...=1, but works fine of you don't. There is no logical reason why this process should have no limit, but the limit can't be one for the reasons you, Zarel, state. So the absence of the limit point that you are looking for proves that .999... cannot equal 1. (Lets continue below the pic, it is getting very tangled here.)Algr (talk)

The logical reason why this process should have no limit is because you're dividing by zero.

You say it "works fine if you don't [assume 0.999...=1]". So if it "works fine", then where is the limit point if you don't assume 0.999...=1? You have repeatedly evaded this question.

I mean, the most you do is mention base infinity. Base infinity is not a real number system, and suffers from the same problems the projective plane suffers from: Lines with the same slope will still intersect, so proving that a line intersects a line of slope 0 does not prove that the slope of the first line is nonzero. --Zarel (talk) 13:13, 7 April 2009 (UTC)

I don't think that in Algr's base infinity number system different lines of the same slope intersect. The problem with his number system is that for all I can tell it's not a field and consequently non-parallel lines need not intersect either. For example, there's no point of intersection for a line through the origin and (1_0®1, 1) with a line through (1,0) and (1,1), though geometrically there should be. Maybe I still don't understand his number system correctly; in that case please correct me (and give the y coordinate of that point of intersection I don't see). Huon (talk) 14:55, 7 April 2009 (UTC)

This is a good question. You've created a right triangle with an infinitesimal angle, and another just short of 90°. This seem to point to a fractional place in between the reals and first infinity, but now I have to figure out where the information is that points this line to 0,0 and not 1,0, and how to describe the result. Still working on it. Algr (talk) 18:07, 10 April 2009 (UTC)

Well it seems that Base Infinity has some problems that I don't know how to resolve. But it is beside the point anyway, so I'm going to start another subsection to try to get this back on topic. Algr (talk)

What is the limit of those red lines?

What is the limit of those red lines? The limit can't be (edit) a line of slope 0, because that would produce a line that never intersects line A.

I recall some talk about parallel lines intersecting 'at infinity', but I have a disturbing suspicion that the proof of this involves calculating the slope of line BZ, which means we are spinning around in yet another logic loop. And please no more calling on the limit fairy. Limits are just an organized way of ignoring infinite and infinitesimal values. That is great for practical applications, but I am asking how you can justify ignoring infinite and infinitesimal values on a theoretical basis. Are we really SURE that there is no way that an infinite or infinitesimal quantity could ever have a real world effect? How many black holes fit in your world view? Algr (talk) 08:02, 15 April 2009 (UTC)

The limit can't be 1 because 1 isn't a line in the first place: Type mismatch.

Actually I can't think of any topology whatsoever on the set of lines in R² which would allow us to speak of a "limit of the red lines". The closest I can come is by measuring the distance between lines as the angle between them - but that would mean that parallel lines have distance 0, and I'd consider not all lines, but lines up to parallel translation. If we use that metric and its induced topology, the set of red lines converges to B - and to A simultaneously, because that topology doesn't distinguish between A and B.

And yes, we're pretty sure that infinitesimals don't have any real-world effect whatsoever. After all, we wouldn't be able to measure the infinitesimal, would we?

Finally, I'm still looking forward to meaningful definitions of limits, convergence and continuity on your number set. Without those, we can forget practically all of analysis, making your number set extremely useless. Huon (talk) 10:48, 15 April 2009 (UTC)

If we can't measure the infinitesimal then don't worry about that black hole that I aimed at your house. I'm sure it won't have any real-world effect whatsoever. Algr (talk)

That black hole has non-infinitesimal (and non-infinite) mass and a non-infinitesimal event horizon. What infinitesimal value should I be concerned about? Not its "infinitesimal size", surely, since anything beyond the event horizon doesn't really matter from a real-world point of view. Huon (talk) 10:48, 16 April 2009 (UTC)

It might matter to someone who actually WANTS to know how a black hole works. I doubt I'll be seeing the core of the sun any time soon, but physicists don't dismiss that as irrelevant. Algr (talk)

Then show me a peer-reviewed physics article using infinitesimals to describe black holes. Huon (talk) 20:38, 16 April 2009 (UTC)

Well, I presented everyone with the "1-slope" series that inescapably produces .999.... Huon acknowledged (in the edit summery) that "there is no limit". Since "No limit" conflicts with "Limit = 1", I think you should concede that this is a case where .999... does not equal 1. I constructed this series with the most basic geometry, and approached infinity in exactly the same way that .999... does within real numbers. So I don't see why any strange topology is called for. Parallel lines don't cross. A line segment with a vertical dimension of 1 cannot possibly have a slope of zero no matter what the horizontal coordinates are. I don't need to define what those coordinates are or what .999... is in order to demonstrate what they can't be. So there it is. Algr (talk) 18:28, 16 April 2009 (UTC)

Please explain what you mean by the limit of a sequence of line segments. Then prove that the limit of the given line segments must intersect A. Eric119 (talk) 19:22, 16 April 2009 (UTC)

Please explain what you mean by "Please explain". The red lines are DEFINED as intersecting A. Algr (talk) 20:36, 16 April 2009 (UTC)

You use the term "limit of a sequence of line segments". How is that defined? And while the red lines intersect A, why should that property be preserved by taking the limit? For example, the red lines all have non-infinitesimal positive slope - shouldn't the limit have non-infinitesimal positive slope as well? If the property "has non-infinitesimal positive slope" isn't preserved by taking limits, why should the property "intersects A" be? Obviously there's something to prove. Huon (talk) 11:34, 17 April 2009 (UTC)

Algr, you produced a sequence of lines and a related sequence of numbers. The sequence of numbers converges to 1 within the real numbers. For sequences of lines, as Eric119 points out, we'd need a concept of convergence to determine whether there is a limit, what that limit is and whether the process relating the lines to the numbers is respected by taking limits - that is, whether the limit line is still related to the limit number. The "strange topology" is called for because without it "convergence" isn't even defined, and asking about limits of lines is void.

I notice that you now speak of line segments. For line segments there is another meaningful topology, that generated by the Hausdorff distance. In that topology, the sequence of red line segments doesn't converge at all; rather, they grow ever farther apart. But that tells us nothing about 0.999..., it just tells us that the map from 1-slope to line segment isn't continuous at 1 (it isn't even well-defined).

Anyway, doesn't "no limit" conflict with "limit = 0.999..." as well? So wouldn't it by your logic contradict the entire existence of 0.999...? Huon (talk) 20:38, 16 April 2009 (UTC)

Your point is absolutely correct, that his logic also dictates that the limit is not = to 0.999... - moreover, nay, tangentially (if you will), I might add that mapping his "parallel" lines onto the Riemann sphere will show that they do in fact intersect at point infinity. Tparameter (talk) 21:03, 16 April 2009 (UTC)

Well if you can't even assume that the limit of .9; .99; .999, is .999... then when would .999... ever occur? There is no mechanism that produces .999... in a way that is unambiguously 1. So why define .999... as 1 and not the limit of .9; .99; .999, ? It seems like you have conceded that .999... simply doesn't mean anything. Algr (talk)

We don't need to assume the limit of .9,.99,.999,... is .999... because that's how we define .999... --Tango (talk) 23:06, 1 May 2009 (UTC)

More related to the previous discussion, we seem to agree that .999... is the limit of (0.9, 0.99, 0.999, ...). But what I meant was: You seemed to state that my claim that a certain sequence of lines doesn't have any limit at all was a contradiction to 1 being the limit of (0.9, 0.99, 0.999, ...). By parallel reasoning, if that were so, my claim would also contradict that 0.999... is the limit of (0.9, 0.99, 0.999, ...). So either 0.999... doesn't even exist, or my claim contradicts neither that 0.999... is the limit nor that 1 is the limit of (0.9, 0.99, 0.999, ...). Of course the latter is true. Huon (talk) 01:24, 2 May 2009 (UTC)

Now that I have read everything on this page (whew, that took a while!) I have decided to comment. Algr, you seem to agree that 0.999... is the limit of the sequence: 0.9, 0.99, 0.999, 0.9999, etc. regardless of whether we are talking about line segments or anything else. In addition, after reading other sections, it seems to me that you agree with the fact (correct me if I'm mistaken) that, by using the formula for the sum of a geometric sequence, the limit of said sequence is 1. Your argument is that the result of a process does not always equal its limit. So it appears that you believe that the limit of the sequence has two values (0.999... and 1). However, an infinite sequence can only have one limit (or no limit, if it diverges). So doesn't that mean that the two limits are equal, meaning that 0.999... = 1? Algr, or anyone for that matter, if you have any objections, please say so, and I will attempt to revise my statement. 71.191.190.200 (talk) 19:49, 12 May 2009 (UTC)

What do spheres have to do with anything?

...and what do spheres have to do with anything? You can define parallelism so that lines never intersect even on a sphere: They would simply be circles on plains that don't intersect. Hence lines of latitude on a globe are parallel, but lines of longitude are not. Algr (talk) 22:46, 1 May 2009 (UTC)

Stereographic projection maps the sphere with the north pole removed to the Euclidean plane. Under this map, images of circles on the sphere are lines and circles in Euclidean space, with the Euclidean lines corresponding to circles through the north pole. As you get closer to the sphere's north pole, the corresponding points on the plane will get ever farther from the origin. Thus, the north pole corresponds to "infinity". Then parallel lines intersect "at infinity": Their corresponding circles intersect at the north pole; moreover, they not only intersect, but are tangential to each other.

There are also maps from the Euclidean plane (or, more generally, from an n-dimensional vector space) to the corresponding n-dimensional projective space - say, by mapping point (a₁, ..., a_n) to projective point [1:a₁:...:a_n]. Now you'll find that the pre-images of projective lines in Euclidean space are once again either circles or (Edit: My mistake, no circles!) lines, and when the pre-images of projective lines are parallel Euclidean lines, the projective lines intersect in a projective point with coordinates [0:p₁:...:p_n]. The obvious image is that the projective points without Euclidean counterparts are "at infinity" (with different such points being at infinity in different directions), and parallel Euclidean lines will correspond to projective lines intersecting "at infinity".

Note that these are two different models of "Euclidean space with infinity added". In one case you add just a single point "infinity", in the other one point "at infinity" corresponds to each family of parallel lines. These models are useful because in both cases you have a bijection between the points not "at infinity" and the Euclidean plane, and the images of Euclidean lines in these models become something almost-useful - useful if you add the correct point at infinity. So both of these generalizations' geometries are intimately linked to Euclidean geometry. If you use your given definition of parallelity ("circles on a sphere are parallel if they lie on parallel planes"), how is that linked to Euclidean geometry? Huon (talk) 01:14, 2 May 2009 (UTC)

Okay, I think I roughly understand stereographic projection. But why use it here? It seems to me that if you want to make ∞ and -∞ mean the same thing, then mapping the plane to a torus makes more sense then using a sphere, and is far easier to do. That way the X and Y coordinates have separate circles at infinity and one can be finite while the other is infinite. And why do you want to make ∞ and -∞ mean the same thing anyway? BTW, [Asteroids (video game)] and similar games happen in toroidal space. Algr (talk) 09:05, 2 May 2009 (UTC)

The problem is that toric geometry once again is not related to Euclidean geometry. I don't see a canonical map from the Euclidean plane to the torus minus its "infinite" circles, and a line in the Euclidean plane wouldn't correspond to anything useful. If it's not parallel to either the x or the y axis, what infinite point would correspond to it? The generalization of lines on the torus are two distinct classes: Lines of rational slopes, and lines of irrational slopes, which are dense subsets of the torus!

As to why we'd want to make ∞ and -∞ mean the same thing: We don't really, that's not our aim but a consequence of building "a nice model of the Euclidean plane with points at infinity added". The sphere and stereographic projection are nice because there the absolute minimum - a single point - is added; the projective space is nice because it's easy to use, follows simple rules and is connected to Euclidean space by a simple map (By the way, if you identify the Euclidean plane with the complex numbers, the sphere cooresponds to the complex projective space, so these models are somewhat related.) You could add a circle at infinity to the plane with different infinite directions; that would give you a disc as model for the Euclidean plane and infinity. But the images of Euclidean lines on the disc will become something complicated, and I don't see any immediate advantages for such a model. Huon (talk) 11:44, 2 May 2009 (UTC)

Spheres are simply different from planes, and aren't a good way of learning anything about them. When a plane is projected, the sphere looses information. The torus does not. In the sphere, if one coordinate X,Y is infinite, the other becomes irrelevant. ∞,1 and -7,∞ both describe the identical point. This never happens on a torus, and to me such a loss seems to totally be opposed to the concept of two dimensional coordinates. In four dimensions you can bend a finite plane into a torus without even stretching the paper. That makes the conversion much more topographically valid then a sphere, in which you both stretch space and poke a hole in an arbitrary position. So what purpose does this Euclidean sphere serve except to eliminate unwanted information? What do spheres have to do with .999...? Algr (talk) 06:02, 3 May 2009 (UTC)

In the plane, a coordinate is never infinite. Every point on the plane exists a finite distance away from every other point and exists as a finite sum of basis vectors. So the information that you claim to lose is never there in the first place. Stereographic projection is one way of converting the infinitely large plane into a finitely "sized" space, it has the side effect of adding a point that, in the plane, would be considered infinitely distant from every other point. Depending on your point of view, this can be an advantage or a disadvantage. But regardless, the plane is neither a sphere nor a torus. All three spaces are topologically distinct. Maelin (Talk | Contribs) 07:53, 3 May 2009 (UTC)

I was afraid of this. It looks like Euclidean space is just another way to avoid looking at the issue. It is like using whole number analysis to prove that fractions don't exist. Why would a coordinate never be infinite in a plane? You can't help but find infinity if you deconstruct a limit. I built my graphic proof around that, and ultimately your answer is to tell me that 1 = 0 and parallel lines aren't parallel. Generally when your assumptions produce such a result, it is time to go back and reevaluate what you believe. Algr (talk) 13:15, 3 May 2009 (UTC)

We keep coming back to the same issue - definitions. A plane doesn't include any points at infinity because of how it is defined. You can define it to have one if you like, and you get (topologically) a sphere. Or you can define it to have a different point at infinity for every direction (modulo signs) and you get the projective plane. There are probably various other ways to do it too, but those are the most useful and, thus, commonly used. If you want to talk about a plane with an infinity, do so, but don't call it the Euclidean plane because it isn't. --Tango (talk) 13:29, 3 May 2009 (UTC)

I never even mentioned Euclid, Huon just assumed he was invited. Algr (talk)

You started this section with a geometric example concerning lines in a plane. If that's not Euclidean geometry, you'll have to give lots of definitions before that example makes any sense. Feel free to do so. Huon (talk) 21:16, 3 May 2009 (UTC)

"Plane" without qualification always means "Euclidean plane". If you mean some other sort of plane, you need to say so. --Tango (talk) 13:24, 4 May 2009 (UTC)

I think I'm missing something - how do you map a plane to a torus? (In a useful way.) --Tango (talk) 13:29, 3 May 2009 (UTC)

A torus is a circle revolved along another circular path. The first circle is the Y coordinate, and the second is the X. Imagine a sheet of graph paper rolled into a tube. Then bend the tube into a ring. Algr (talk) 19:57, 3 May 2009 (UTC)

That's not a map. A point on a circle can be described by an angle φ; a point on the torus can thus be described by a pair of angles (φ, ψ). Please give an explicit map from the plane to the torus that sends a point (x, y) on the plane to some point (φ(x,y), ψ(x,y)) on the torus. Then give the image under your map of the lines y=0, y=1, y=x and y=2x, and tell us what infinite points (if any) belong to the toric equivalents of these lines. That should clarify why toric geometry isn't all that useful - at least for these purposes. Huon (talk) 21:16, 3 May 2009 (UTC)

You are absolutely correct. The geometry of a torus is sometimes used in computer graphics, but it's a finite space. There is no analogy to the Reimann Sphere and planes. Tparameter (talk) 21:26, 3 May 2009 (UTC)

I can map a square (or rectangle) to a torus, that's easy (I might even take that as the definition of a torus in some circumstances), but a square is not a plane; it is a finite region of a plane. --Tango (talk) 13:20, 4 May 2009 (UTC)

Any finite line segment can be mapped to an infinite line segment using tanh(x) and tanh^-1(x) Do this for x and y, and you have an infinite toroidal plane. Algr (talk) 17:15, 4 May 2009 (UTC)

Yes, that allows you to map the plane to the torus, but your map isn't surjective. In much the same way as you add a point at infinity to turn a plane into a sphere, you would need to add two circles at infinity to turn a plane into a torus. You can add a point at infinity while keeping most of the vector space structure (which is essential for anything more than very basic geometry), I'm not sure you can add two circles at infinity without breaking that structure. You can add one and get the projective plane [ish], but I've never seen it done with two. If you can describe how it would work, I would be very interested to hear it. --Tango (talk) 17:30, 4 May 2009 (UTC)

Yes, I always intended for there to be two circles at infinity - they mirror the two circles of x=0 and y=0. To create this torus, start with a line segment from -1 to 1. Apply tanh^-1(x) to make the segment from -∞ to ∞. Fold the segment into a circle, making -∞ and ∞ the same point opposite zero. Now move that circle in another similar circle to create the torus. The result is a plane where parallel lines never intersect, even if they reach and pass infinity.Algr (talk) 18:42, 4 May 2009 (UTC)

(Outdent) Some parallel lines will still intersect. Have a look at the images on the torus of the lines y=x and y=x+1. Their images on the torus will both point to the same infinite point, namely (∞,∞). You have made the lines parallel to your coordinate axes special. Just have a look at the original problem where you asked for the limit of the red lines. If you try that on the torus and take the "naive" limit (formally, the limit set in the Hausdorff topology of subsets of the torus), the result will not be the image of any line on the plane, and it will include more than one infinite point. And rotation is a nightmare in your toric geometry. Huon (talk) 19:04, 4 May 2009 (UTC)

Sure, you can do it topologically, but you haven't given a coordinate structure on the torus. Unless you can transfer the additional structure from the plane to the torus, you can't use it for much. The natural coordinate structure induced from considering the torus as the product of the real projective line with itself (which, I think, is what you've done) probably won't be very useful. As Huon points out, lines in that coordinate space don't behave as you want them too. --Tango (talk) 19:24, 4 May 2009 (UTC)

The collision of y=x and y=x+1 is an illusion caused by the lack of a way to describe finite deviations from infinity. With base infinity I was trying to create a number system where numbers and geometry near infinity would behave locally just like they do near zero. Alas it doesn't work yet, but as I've said, calculus wasn't built in a day. Tango, I'm not sure what you are asking for. The obvious system would have y being the small circle that goes through the doughnut hole, and x being the larger circle. (For example the contact points with a plate that the doughnut is resting on.) Algr (talk)

It's not an illusion, it is a simple consequence of your construction. The Riemannian sphere does what you describe - you just define a different coordinate chart centred on infinity (which would then be called "zero" in the new coordinates). As long as you only want it to work locally, then it is easy. It is trying to get it to work globally that is difficult (if not impossible), and I think that is actually what you are after. What I want is for you to explicitly give a coordinate system which does what you say it does. What you describe is a pretty standard coordinate system for the torus, but it doesn't have non-intersecting parallel lines, which seems to be your goal. --Tango (talk) 12:46, 5 May 2009 (UTC)

This is seriously beyond satire. Tparameter (talk) 05:34, 5 May 2009 (UTC)

Pah. Ask Einstein what 1/3 of the speed of light times three is. Algr (talk)

Why would he answer anything other than "the speed of light"? Einstein was a very good mathematician, I'm pretty sure he knew what a third times three was. --Tango (talk) 12:46, 5 May 2009 (UTC)

Those who know everything, learn nothing. Algr (talk) 18:56, 5 May 2009 (UTC)

He who learns nothing can prove nothing. Wise-sounding aphorisms are fun but they don't really make you correct. Maelin (Talk | Contribs) 02:38, 6 May 2009 (UTC)

Are you guys serious? You honestly don't understand the Einstein reference? Or are you just choosing to be obtuse? (I'll explain it if you ask.) Algr (talk) 03:45, 7 May 2009 (UTC)

Sigh. Let's hear it, then. Maelin (Talk | Contribs) 06:22, 7 May 2009 (UTC)

Ok. Two spaceships are both headed to Earth from opposite directions at 2/3 the speed of light (Relative to the Earth.) Their speed relative to each other is not 4/3 C, but something like .8 C. What does Euclid have to say about that? Algr (talk) 21:08, 7 May 2009 (UTC)

Einsteinian relativity uses a four-dimensional manifold as spacetime, with a Lorentz metric. Note that the manifold locally looks like a subset of the R⁴, with no infinitesimals involved. If you ignore general relativity, you may still consider space (without time) as Euclidean. Where's the connection to anything discussed before? (As an aside, unless I'm very much mistaken, when your frame of reference is Earth's, you will observe the distance between the spaceships decrease by 4/3 c.) Huon (talk) 21:58, 7 May 2009 (UTC)

Picard. Is there no end to the things you will grossly misinterpret in your futile crusade to disprove this inconsequential equality? Maelin (Talk | Contribs) 01:54, 8 May 2009 (UTC)

Well, everything under the words "What do spheres have to do with anything?" above is a total waste of time, as I never intended the original graph to be warped into any kind of three dimensional space. I only came up with the torus because I thought it would satisfy you somehow. Algr (talk) 08:00, 8 May 2009 (UTC)

Back to the proof

I should clarify this. My proof is based on the contradiction of a line of slope (1-.999...) intersecting a line of slope 0. You guys brought up spherical geometry to say that parallel lines converging was not a contradiction. My response was that even if you insist on mapping the plane to a three dimensional shape, (why?) this still doesn't make it mandatory that parallel lines converge. On my infinite torus, Y=0 and Y=1 don't converge, so a line of slope (1-.999...) intersecting both still conflicts with the assumption that .999... = 1. On the Euclidean plane, at the point were you show that .999...=1, zero also equals one. So how does that negate my proof? Algr (talk) 16:59, 9 May 2009 (UTC)

Your "proof" is very simply flawed. That point you've labelled "Z" simply doesn't exist. The reason we branched off into a discussion of non-Euclidean geometry is because you wanted to make it exist and using projective space is the only way to do that. You need to chose a space in which parallel lines (at least those ones, if not all) converge in order for "Z" to exist since "Z" is the intersection point of two parallel lines. --Tango (talk) 17:06, 9 May 2009 (UTC)

That makes no sense whatsoever. Why would Z have to violate the definition of parallelism to exist? Normally when an idea violates the rules of math of physics, that proves that it doesn't exist. Algr (talk) 17:43, 9 May 2009 (UTC)

Not really. When something violates your present theory, that means that the thing does not exist, in your theory. Your point Z does not exist in the theory of Euclidean geometry. It exists in plenty of other theories, including even some well-studied ones which we looked at when this topic branched out. You can, of course, invent a theory with whatever properties you desire, but it will not necessarily be interesting or useful. Maelin (Talk | Contribs) 17:59, 9 May 2009 (UTC)

Because it is the intersection point of two lines with slope zero. Two lines with the same slope are parallel. If you can prove that Z exists in the Euclidean plane, then that would be a proof that 0.999... isn't equal to 1. However, we know that isn't the case, so we know Z doesn't exist, that's why we went off on a tangent to explain in what spaces it would exist, but none of that was actually relevant. The simple fact is that you haven't proven that Z exists, so you haven't proven that 0.999... doesn't equal 1. (Whereas we have proven that 0.999...=1, so we have proven that Z doesn't exist.) --Tango (talk) 18:10, 9 May 2009 (UTC)

Why would Z not exist? Z is based on a construction identical to .999... If Z doesn't exist then neither does .999... That would prove that .999... is not one. Algr (talk) 18:18, 9 May 2009 (UTC)

There is a big difference between the constructions. The difference between each successive term in the sequence 0.9, 0.99, 0.999,... gets smaller and smaller, the difference between each successive term in the sequence 10,100,1000,... gets bigger and bigger. The former series converges, the latter does not, so the former limit exists (and we called it 0.999...) and the latter does not. Your point Z is the point whose x-coordinate would be the limit of the latter sequence, so it doesn't exist. --Tango (talk) 18:23, 9 May 2009 (UTC)

If you define the red lines by their angle with line B, they also get smaller. They converge on 0, but can't reach it because they are defined as intersecting A. As a result, you can either accept an infinitesimal angle for B and BZ or make 1=0 so that the A and B intersect. Making .999...=1 comes at the cost of rendering the whole Real set meaningless. Saying "Z doesn't exist" is just denial of the obvious. Algr (talk) 19:01, 9 May 2009 (UTC)

If you are defining them by their angle then you can't also define them to intersect A. A line starting from the origin is uniquely determined by its angle, so that is all the definition you are allowed, if you introduce an additional definition you risk a contradiction (which is what happens in this case). This is one of many examples where a limit does not satisfy a property satisfied by all members of the sequence. All members of the sequence intersect A, the limit (B) does not. It's a difficult concept to get your head around and is, as such, the cause of many mathematical misconceptions, but there is simply no requirement for a limit to satisfy a property satisfied by all the terms in a sequence tending to it. --Tango (talk) 19:10, 9 May 2009 (UTC)

The problem in Algr's reasoning gets clearer if we substitute the line A by a line C through the point (0, 1) with slope 1-0.999... (whether or not that's slope 0 doesn't matter right now). Then all the red lines intersect C, with the points of intersection between the red lines and C infinitesimally close (that includes the "distance 0" case of 0.999...=0) to the points of intersection with A. Wouldn't then the limit line also have to intersect C, despite having the same slope? On the other hand, if the limit line doesn't intersect C, why should it intersect A? Huon (talk) 00:34, 10 May 2009 (UTC)

Only the red lines with finite length would intersect C. Since you have defined C as having the same slope as BZ, I don't see why you would expect them to intersect. The C intersections and A intersection would no longer be infinitesimally close once the lines are infinitely long. An infinite number of infinitesimals (or even zeros) can add up to a real number. Algr (talk) 20:50, 11 May 2009 (UTC)

What do you mean by "once the lines are infinitely long"? There is no point in the Euclidean plane that is infinitely far away from the origin, that's simple not how the definition works. When we talk about a line being infinitely long we just mean it is unbounded, there is no real number that it is shorter than. There isn't some point on it that is actual infinite in any sense. --Tango (talk) 22:40, 11 May 2009 (UTC)

So you get half way through the exercise and start making assumptions that conflict with the premise. Oi. Algr (talk)

We've been saying this since that beginning, it's not a new assumption. It is the definition of the Euclidean plane. There are no points at infinity in the Euclidean plane. Accept it. --Tango (talk) 14:11, 12 May 2009 (UTC)

If only the red line segments of finite length intersect C, why don't just the red line segments of finite length intersect A? That would obviously have to be shown. Huon (talk) 00:38, 12 May 2009 (UTC)

Because the Archimedean Property would not allow that. Any two lines with different slopes must have a line between them with an intermediate slope. (In the case where C and A are the same, you have assumed your premise and the proof is invalid.) If C has a different slope then A, then there must be a line between them that crosses A but not C. Algr (talk) 06:06, 12 May 2009 (UTC)

I don't understand what you mean. You were trying to construct a point of intersection between A, a line of slope 0 and another line by a limit process, thus arguing that the other line does not have slope 0, right? We agree that the limit process does not produce a point of intersection with C, a line of slope 1-0.999..., so how do you know that it does produce one with A without assuming as a prerequisite that 0.999...≠1? Assuming A≠C amounts to assuming 0.999...≠1, so that would be circular reasoning, invalidating the proof. Huon (talk) 12:43, 12 May 2009 (UTC)

I think you've lost me here too. Your line C would be identical to BZ. All the other red lines would intersect it at 0,0. Maybe another drawing would help?Algr (talk) 21:46, 14 May 2009 (UTC)

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

New proof by contradiction?

Consider the function f(x)=-1/(10^x). The function is asymptotic to 0 as x->infinity (that is, for every value y<0, no matter how close y may be to 0, there exists a value x such that y<f(x)<0).

Now, add 1 to the function: f(x)=1+(-1/(10^x)). Observe that f(x) is now asymptotic to 1 as x->infinity. For every value y<1, no matter how close y may be to 1, there exists a value x such that y<f(x)<1.

IF 0.999... < 1
THEN there exists x such that f(x) = 0.999...
(AND there exist infinitely many values a such that a>x and 0.999... < f(a) < 1)
HOWEVER no such x exists; f(x) < 0.999... for all values of x
Therefore 0.999... is not less than 1.

Do you see any flaws? Also, has this already been shown by anyone else? I anticipate your feedback eagerly. Gustave the Steel (talk) 03:31, 8 April 2009 (UTC)

Incidentally, this should work for any f(x) asymptotic to 1. I chose 1+(-1/(10^x)) because, given a positive integer value n, f(n) returns n 9's after the decimal. Thus, f(x) clearly cannot return infinite 9's on a finite value of x. With this particular function, doubters may more easily verify the "however" step without needing more rigorous justification. Gustave the Steel (talk) 03:41, 8 April 2009 (UTC)

I think that this is just a very fancy way of asking what is the average of .999... and 1. But the concept of .999... as "The highest value less then one." applies only from the perspective of any one place within base infinity. So when x = 1_-1®, f(x) = .999... (BTW, Huon, I'm still working on your other question above.) Algr (talk) 05:58, 8 April 2009 (UTC)

I'm not that happy with this proof. It relies on f being asymptotic to 0 (or to 1), which is itself not explained in more detail. Why shouldn't f be asymptotic to a negative infinitesimal (or to 0.999...) instead? If people were ready to accept that such negative infinitesimals don't exist, they'd also have to accept directly that 0.999...-1 equals 0, wouldn't they?

By the way, in Algr's number system, 0.999... was defined to be 1®-1, that is, 1 minus a certain infinitesimal. But we have f(1_-1®)=1-1/(10^1_-1®). If that were equal to 0.999..., we could conclude that 1/(10^1_-1®) = 0®1 = 1/1_0® and, by taking reciprocal values, 1_0® = 10^1_-1® = 1/10*10^1_0®. Multiplying by 10 we get 10*(1_0®)=10^1_0®. That doesn't look good to me; apparently f(1_-1®) is not equal to 0.999... I believe the reason is that powers of 10 aren't defined for the non-real numbers in Algr's system. Huon (talk) 11:30, 8 April 2009 (UTC)

Wikipedia explains the concept of asymptotes as this: "as one moves along the graph of f(x) in some direction, the distance between it and the asymptote eventually becomes smaller than any distance that one may specify." I have shown that my f(x) is asymptotic to both 0.999... and 1 - and, in any case, the concept of asymptotes is more widely understood and less disputed than other means of informal proof (such as showing how repeating decimals become rationals, etc.).

Besides, many people often claim that 0.999... "gets closer to" but "never reaches" 1. One of my goals here is to give an example of something that really does get closer to 1 and never reaches it, and show that it also fails to reach 0.999..., thus ridding them of the notion that 0.999... represents a function instead of a number. (They'll never state this notion out loud, but it is often reflected in their language.)

As for Algr's number system, well, I need some time to think about it. Gustave the Steel (talk) 17:42, 8 April 2009 (UTC)

A different digit manipulation proof

Moved to main talk page... -- The Anome (talk) 10:49, 14 April 2009 (UTC)

MFD

The discussion at Wikipedia:Miscellany for deletion/Talk:0.999.../Arguments (2nd nomination) has resulted in this page being kept.--Aervanath (talk) 18:59, 13 April 2009 (UTC)

Side question about 3.1416...

In these examples:

3.1416...

0.999...

Do both "..." mean the same thing? This has to do with me defining limits above. Algr (talk) 06:52, 19 April 2009 (UTC)

Not quite. Both effectively mean "and further digits (not all zero) in every subsequent decimal place", however in the 0.999... case, there is the additional meaning of "and this pattern continues throughout (nines all the way down)" whereas in the case of π there is no pattern to continue. Maelin (Talk | Contribs) 07:44, 19 April 2009 (UTC)

Well, technically, there is a pattern - namely, the pattern of "digits of pi". It's just that there's no pattern other than the trivial one (i.e. "digits of pi"). --Zarel (talk) 08:57, 19 April 2009 (UTC)

Technically, 3.1416... isn't pi, whose decimal representation is 3.14159... - we wouldn't round when giving just the first few digits. So all we know is that 3.1416... is the decimal representation of some real number, and context doesn't indicate which real number. Huon (talk) 11:24, 19 April 2009 (UTC)

The theme here is don't use ... to define anything. There are probably arbitrarly long sequences of 9's in the decimal expansion of pi; there's six 9's in a row starting at the 762nd place, so 3.1416...4999999... (skipping the middle 757 digits) might represent pi, or it might represent the rational number 3.1416...5 or it might represent 3.1416...49999994882 (also rational) or perhaps some other irrational number which happens to have the same first 800 decimal digits as pi. Endomorphic (talk) 16:42, 1 May 2009 (UTC)

Yes, and on top of that, we were previously using "..." in the middle of digits to signify proposed infinitesimals like this: 0.000...1. So "3.1416...4999999..." might actually make people mad at you! Algr (talk) 21:35, 1 May 2009 (UTC)

It's been said many times before: The meaning of "..." depends heavily on context. In this context it is perfectly clear (because it was explicitly stated) that the dots are representing a finite number of missing digits. --Tango (talk) 23:04, 1 May 2009 (UTC)

New Proof: Flip function

Consider a function Flip(x) that has the result of pivoting all base ten digits around the ones place. This could be seen as a decimal equivalent of taking the reciprocal, or misreading base ten as base one tenth. (Doing this in another base would be a different function.)

In finite numbers, the following properties are self evident:

Flip(1) = 1

Flip(10) = .1

Flip(1.23) = 321.

Flip(Flip(x)) = x. The function is self reversing, like reciprocals.

Flip(x)= has only one answer for any x. (Unlike √x, for example.)

The result of Flip(x)ing a negative number is unexpected. We will discover it below.

So, what is Flip(0.999...)? Well, it is obviously ...9990. But for various interesting reasons within standard mathematics, that number is actually negative ten! A totally different answer from Flip(1). So .999... can't be equal to 1 because its flip value isn't even close to Flip(1).

Flip(-7) = 3.999...

Now you could insist that 3.999...=4, and claim that Flip(4) has two answers, 4 and -7, but flipping a negative number doesn't always result in training nines:

Flip (-8/9)= .888...

What would be the other answer to Flip (.888...)? If it doesn't have one, then how is it different then 4?

So flipping a negative number gets you a number with an infinitesimal component. This only makes sense if they are not the same as the numbers that they are near. So .999... is not the same as 1. Algr (talk) 22:40, 11 May 2009 (UTC)

I marvel at the mental gymnastics required to accept that ...9990=-10 while rejecting 0.999...=1. The proofs are analogous - except that the first doesn't work in the real numbers (where ...9990 isn't defined at all). Anyway, what you define is not a function on the real numbers but on decimal representations. In order to show that Flip is a well-defined function on the real numbers, you'd have to show that either no real number has more than one decimal representation (ie that 0.999...≠1, and avoid circular reasoning!) or that all different decimal representations of a single real number get mapped to the same image (which is obviously not the case).

You can build easier examples of this type: Just define a function Swap to swap all 7's for 9's in a given decimal representation, and vice versa. Swap(7)=9, Swap(1.9)=1.7, and so on. Since once again it's a function on decimal representations and not on real numbers, it maps 0.999... and 1 to different images. But that proves nothing. Huon (talk) 00:57, 12 May 2009 (UTC)

But that is YOUR gymnastics that you are marveling at. ...9990=-10 is based on the digit manipulation proof that appears in the main article. If ...9990 isn't defined, then neither should .999... be. If it IS defined, then I've shown that it conflicts with .999...=1. BTW, your Swap_7;9 function produces no problems at all unless you try to reconcile it with .999...=1. You'll probably say that .999... is different from ...999. because it converges to a real number - but that is again assuming your conclusion, and we've already discussed two non real interpretations of .999... Algr (talk) 02:45, 12 May 2009 (UTC)

If you are defining ...9990 as $\lim _{i=1}^{\infty }9\cdot 10^{i}$, then that expression is meaningless as the series doesn't converge. This is quite obviously different to the case of 0.999....

Once again, I contend that you are nothing but a troll, and frankly I'm surprised that people (myself included) continue to bite. Oli Filth^{(talk|contribs)} 08:31, 12 May 2009 (UTC)

No, ...9990 is not my idea, and I had thought it was accepted mathematics. I describe why it equals -10 below. (Which I notice you didn't comment on.) And names will never hurt me. Algr (talk) 09:01, 12 May 2009 (UTC)

I've commented on it implicitly: the expression "...9990" is meaningless, and therefore isn't a real number, and therefore can't possibly equal -10. Oli Filth^{(talk|contribs)} 13:04, 12 May 2009 (UTC)

How on earth can you possibly accept ...9990=-10 when you don't accept .999...=1? The former appears to be equating an infinite number to a finite one. The latter is just missing the ghostly infinitesimal quantity that so many people yearn to keep there. --COVIZAPIBETEFOKY (talk) 02:56, 12 May 2009 (UTC)

Is this a cross post? Read what I said above. I don't accept ..9990=-10, but it is an inevitable consequence of a "proof" that .999...=1. I use it to show a contradiction that occurs within those assumptions. ======== BTW, when I first encountered ...999.=-1, the way it was presented lead me to believe that the equality was standard mathematics. I'll have to find that again, it seems to have been pulled off this page. Algr (talk) 03:02, 12 May 2009 (UTC)

Exactly how did you come to the conclusion that that was a cross post?

And the link you're looking for is 0.999...#p-adic numbers. As in, it has nothing to do with real numbers.

And you really need to get a life. --COVIZAPIBETEFOKY (talk) 12:24, 12 May 2009 (UTC)

Get a life? I'm having fun and learning new things. What are you doing? Algr (talk)

Please address the other parts of my response. And I would love to hear of the things that you have "learned" here, because what I have seen you do cannot possibly be described as learning. Perhaps the word you were looking for was "arguing"? --COVIZAPIBETEFOKY (talk) 22:11, 13 May 2009 (UTC)

So... please explain exactly how ...9990 = -10 is "an inevitable consequence" of a proof that 0.999... = 1. I'd wager that you are deliberately stretching some particular notion past its defined extent, as usual. Show me where ANY of our proofs implies that ...9990 = -10. Maelin (Talk | Contribs) 05:38, 13 May 2009 (UTC)

I will hereby use an analogous proof to "prove" that the rational number 1/2 is not equal to the rational number 2/4. I first define a function, Fiddle() which swaps all the 2's in a fraction for 3's. Then Fiddle(1/2) = 1/3, but Fiddle(2/4) = 3/4. Since 1/2 and 2/4 have different Fiddle results, they can't be the same!

Why is this argument invalid? Because my function is not a map on rational numbers, it is a map on their fraction representations. All I have proven is that the REPRESENTATION 1/2 is different to the REPRESENTATION 2/4. But that's obvious, we could tell that just by looking at them. If your function relies on the uniqueness of representations to be applicable to your desired set (as Fiddle relied upon the uniqueness of fraction representations to be applicable to the rationals, and Flip relies on the uniqueness of decimal representations to be applicable to the reals), then your function is going to fall apart if it turns out those representations AREN'T unique (like fractions, or decimal expansions). This, I think, is your key misunderstanding, Algr: you are conflating real numbers with their decimal representations.

Also, I am perplexed as to how ...9990 is equal to -10. Maelin (Talk | Contribs) 03:19, 12 May 2009 (UTC)

Maelin, you have made the mistake of applying a function to both sides of a fraction and assuming that the results would yield the same fraction. But that never works. Lets say f(x)= x+1. If we apply that to 1/2, we get 2/3. But if we apply it to 2/4, we get 3/5. Is x+1 somehow an invalid function? It fails the same test you applied to Fiddle and Flip. Algr (talk) 04:21, 12 May 2009 (UTC)

Uh, what? If f(x) = x + 1, then if I apply f to 1/2, I get f(1/2) = 1/2 + 1 = 1/2 + 2/2 = 3/2, and if I apply f to 2/4, I get f(2/4) = 2/4 + 1 = 2/4 + 4/4 = 6/4. And we know that 3/2 = 6/4. The x + 1 function doesn't care about the representation, it just adds 1 to a number. My year seven student can work out how to add 1 to a fraction, man, come on. Surely you at least know how to do that. The flip and fiddle functions are different from this add-one function, because they are representation dependent. Did you perhaps have in mind the function that adds 1 to the numerator and adds 1 to the denominator? Because yes, that IS representation dependent. Therefore it is NOT a function on the rationals. Maelin (Talk | Contribs) 05:58, 12 May 2009 (UTC)

But that ISN'T how you applied Fiddle before. If x+1 means that you look at the whole fraction's value, then Fiddle(1/2) must also recognize that "1/2" as a whole is not 2, and is therefore left alone. BTW, Fiddle and Swap_7;9 are different from Flip because they both contain conditional operations. There is nothing conditional about Flip - every number is processed the same as every other. Algr (talk) 06:17, 12 May 2009 (UTC)

Look, Flip is a function R -> R iff every x in R is mapped to a unique value. In other words, iff whenever Flip(x) = y and Flip(w) = z, then NOT (w = x) (which I will write as w != x). Now, is your Flip a function? We notice that Flip(0.999...) != Flip(1) (indeed, the left hand side is not in R under the usual interpretation of digit strings), so Flip is a function iff 0.999... != 1. Sadly, that is what you're trying to prove. You cannot show that Flip is a function unless you first show 0.999... != 1. You have begged the question. Just as all the other respondents have said. Phiwum (talk) 12:01, 12 May 2009 (UTC)

I hereby define Rapture(x) as adding 1 to both the numerator and the denominator, UNLIKE Algr's f(x) = x+1. Notice that there are no conditional operations; the function adds 1 to both the numerator and the denominator no matter what the values of the numerator and the denominator are.

Now for a proof that 1/2 != 2/4:

Rapture(1/2) = 2/3

Rapture(2/4) = 3/5 != 2/3

QED

Here's an even worse example:

Rapture(3/1) = 4/2

Rapture( (-3)/(-1) ) = (-2)/0

RED ALERT! RED ALERT! --COVIZAPIBETEFOKY (talk) 12:43, 12 May 2009 (UTC)

The original argument was that if you have ...999. and you add 1, you get ...000 because all the nines cary over, and because there is no "first digit" in the sequence, ...000 is simply zero. So since (-1) +1 = 0, ...999. must equal -1. (This is NOT what would happen in base infinity, but it is the same approach to infinitely repeating decimals that makes .333... = 1/3 and .999...x10=9.999...) While I don't agree with this logic, I do understand it, and I extended the idea to calculate ...9990= -10, and ...888= -8/9. Also, computers handle negative numbers like this: The highest bit is the sign bit, so all bits at '1' indicates -1. From there, all the binary functions work the same and there is no need for any special instructions on how to deal with negative numbers. Algr (talk) 04:39, 12 May 2009 (UTC)

I've made a lot of money in my career as a programmer cleaning up the messes left behind by programmers who make that last assumption. 32767 + 1 = -0 is not always the desired result. --jpgordon^∇∆∇∆ 04:57, 12 May 2009 (UTC)

Oops! Perhaps I should have said no need to build special instructions into the microchip. The humans still need to know what they are doing!  :) Algr (talk)

Yes, we all know the reasoning behind such statements. They generally aren't applied to real numbers because they don't make sense under the standard metric (under which ...999. diverges, so doesn't exist as a real number), but we can turn a blind eye to that. The real problem with your Flip function is, as has been explained in detail above, it is only a function on the real numbers if 0.999...!=1, so you are begging the question. --Tango (talk) 14:22, 12 May 2009 (UTC)

"Diverges" simply means rises or falls infinitely, right? Why is infinity acceptable as a process, (how many nines in .999...?) but not as a result? It seems strange that an infinite number of nines can be meaningful on one side of the decimal point, but not the other. In base infinity, ...999. = 1_-1®, and all the other variants (...888.) also hold true. Algr (talk) 20:48, 13 May 2009 (UTC)

It only seems strange if you've paid literally no attention to articles such as Limit (mathematics) or Real number. Oli Filth^{(talk|contribs)} 21:54, 13 May 2009 (UTC)

AGF Algr (talk)

This discussion is a total waste of everyone's time unless everyone involved understands the basics of the real numbers and convergence. You've been playing on this talk page for long enough to have learnt this stuff, so what else should I assume? Oli Filth^{(talk|contribs)} 22:11, 13 May 2009 (UTC)

This has nothing to do with assuming good faith. After arguing with us on this talk page for more than a year, you STILL have not even bothered to learn the most basic and fundamental of notions required to understand the issue. You still have no idea what you are talking about. You obviously avoid any article we link to avoid the possibility of discovering you are wrong. You have no interest in finding or understanding the truth and are, at this stage, only arguing because you can't bear the thought of conceding defeat or admitting that you don't understand. Am I wrong? Prove me wrong by explaining what an epsilon-delta proof is, and how it pertains to 0.999.... You won't, of course. What can you POSSIBLY hope to gain from this idiocy if you won't bother learning anything about it? Maelin (Talk | Contribs) 01:19, 14 May 2009 (UTC)

If you work in the extended real numbers then infinity is acceptable as a result, however if you do that you introduce problems - the extended real numbers aren't a field, which is very inconvenient. However, even if you do that you wouldn't get ...999.=-1. If we define ...999. as the limit of the sequence 9, 99, 999, ... then that clearly isn't -1 since the terms are getting further and further away from -1, not closer and closer. To get it to tend to -1 you need to introduce a different metric (the 2-adic or 5-adic metrics would do it - in fact, I think they are the only ones that would), but when you do that you end up with a completely different number system (the rational numbers are the same but when you take the completion you get something very different from the real numbers). --Tango (talk) 22:13, 13 May 2009 (UTC)

Note: When Tango talks of "the completion" of the number set, that is referring to the same concept of completion that you have previously refused to understand. --69.91.95.139 (talk) 22:16, 13 May 2009 (UTC)

To answer another part of Algr's question: Infinity is not "acceptable as a process". When we say the number of nines is infinite, we mean that there's (at least) one for every natural number. There is no process whatsoever involved, nor a number "infinity". Even when we write stuff that uses the symbol "∞", such as $lim_{n\to \infty }a_{n}$, that's usually shorthand for something defined without any reference to infinity.

On a completely unrelated note, Algr, does your "base infinity" system allow numbers with infinitely many digits such as 0®9_9_9_...? (See above for what I mean by "infnitely many".) Huon (talk) 23:07, 13 May 2009 (UTC)

"Yes, we all know the reasoning behind such statements."

If this is the case, (referring to the explanation for ...999.=-1), then why did everyone act like they had never heard of such a thing before? Why wait for me to post the explanation before admitting that you already knew it? I've noticed this tactic in use several times in this discussion, and find it frankly rather dishonest. When I first started posting here, no one was willing to refer to infinitesimals or hyperreals until I discovered those words for ideas that I had been struggling to propose. When I did something similar by avoiding directly referring to the x coordinate of point Z, you took me to task for it. If you are trying to tell me that something doesn't exist, it doesn't help your credibility if you are also hiding things that do exist. And please don't bring in Jack Nicholson saying "You can't handle the truth!", because he was the BAD guy. Algr (talk) 22:03, 14 May 2009 (UTC)

Who said they hadn't heard of it? Plenty of people said it wasn't a real number but I don't remember anyone saying they didn't understand where you were coming from. --Tango (talk) 22:05, 14 May 2009 (UTC)

Given that the article contains the ...999 example, it can hardly be said to be hidden. We all know the article; we all know that within the 10-adic numbers, one may argue that ...999=-1. The article also mentions infinitesimals and non-standard analysis since at least October 2006 - before you started discussing it. We "hide" nothing. Huon (talk) 00:22, 15 May 2009 (UTC)

I asked him where his ...9990 = -10 thing came from, because I genuinely didn't know where he got that idea from. There was no dishonesty involved. Maelin (Talk | Contribs) 03:12, 15 May 2009 (UTC)

Flip(0.999...) equals: infinity - 10

Flip(0.999...) equals ...9990, which equals to infinity minus 10. 68071 (Talk) —Preceding undated comment added 02:31, 21 May 2010 (UTC).

The Cyclical Paradox of the First Proof

→ Discussion topic moved to here from the talk page. || Loadmaster (talk) 14:20, 22 May 2009 (UTC)

I do not inherently disagree with the concept. What we must look at is the inherent cyclical nature of .33333333... = 1/3 transcending via a three multiplier to .999999999... = 1 (3/3). The cyclical nature of the paradox rules it out. One must first prove that .33333... does in face equal 1/3 for it is based off of the same principle as .999999999999... = 1. I do not believed that it should be removed because of its inherently simple nature and that it can be proved by other methods. But perhaps a note to that measure? —Preceding unsigned comment added by Martin.feetofclay (talk • contribs) 08:10, 9 January 2009 (UTC)

I am not quite sure I understand you. At the level of the discussion we have here, I think pointing out that long division of 3 into 1 gives the repeating decimal 0.333… is enough of the a proof for this fact. What do you feel is missing, or what exactly do you want to note? Thenub314 (talk) 09:38, 9 January 2009 (UTC)

I thought is was understood that 1/3 is merely an estimation of .333...(?) —Preceding unsigned comment added by 67.85.127.127 (talk) 20:28, 19 April 2009 (UTC)

It is only an estimate, just in the same way that 0.999... is only an estimate to 1. But the retards in wikipedia don't seem to comprehend this simple idea and there's no way to make them understand it (they're unwilling to even talk about it - just read this discussion page). —Preceding unsigned comment added by 93.173.26.6 (talk) 16:02, 20 May 2009 (UTC)

No, it is not an estimate. Demonstrably, the unique limit of the sequence of partial sums is 1/3. Michael Hardy (talk) 16:18, 20 May 2009 (UTC)

I will try and explain your misconception if you are only willing to listen. You are entirely right that one third is the unique limit of the sequence 0.3, 0.33, 0.333, etc. But we are talking about a single number here, and not of a limit of a series. Yes, the limit is one third; but that is only an approximation. You may find it useful to read further explanations at Limit_(mathematics). —Preceding unsigned comment added by 93.173.26.6 (talk) 18:06, 20 May 2009 (UTC)

The limit is "a single number". If it's an approximation, then what is it an approximation to? Oli Filth^{(talk|contribs)} 20:31, 20 May 2009 (UTC)

It is an approximation to the series. I will give you an easier example. Take the series 1, 0.5, 0.25, 0.125 etc (this is called the hypergeometric series). It is obvious that the limit is 2 but we would be dumb to say that the series equals 2. —Preceding unsigned comment added by 93.173.15.129 (talk) 21:01, 20 May 2009 (UTC)

That's a standard geometric sequence. The series formed by the sum of these indeed has a limit equal to 2. No one is saying that "the series equals 2", because that is meaningless; the series isn't a number. Equally, saying that "1 is an approximation to the series" is meaningless. Oli Filth^{(talk|contribs)} 21:13, 20 May 2009 (UTC)

That's exactly my point. In the same way that it is nonsensical to say "the hypergeometric series equals 2" it is meaningless to say 0.999..=1 because 0.999.. is just an approximation to 1. Don't you see the analogy? —Preceding unsigned comment added by 93.172.160.227 (talk) 13:05, 21 May 2009 (UTC)

No! 0.999... is defined as the limit of the series 0.9, 0.99, 0.999, .... The limit of this series is 1. Therefore 0.999... = 1. Which part do you disagree with? Oli Filth^{(talk|contribs)} 13:13, 21 May 2009 (UTC)

I disagree with your first statement. I don't see how you define 0.999.. as a limit while it is an approximation. You can't go mix numbers and sequences or approximations and exact values. —Preceding unsigned comment added by 93.172.65.202 (talk) 17:13, 21 May 2009 (UTC)

What I mean is; the meaning of the symbol "0.999..." has been defined to denote $\lim _{n\rightarrow \infty }\sum _{i=1}^{n}9\cdot 10^{-i}$. It's a convention. Oli Filth^{(talk|contribs)} 17:20, 21 May 2009 (UTC)

You've messed up your notation there - that last n should be an i or some other dummy variable. --Tango (talk) 17:22, 21 May 2009 (UTC)

Good point! Corrected now. Oli Filth^{(talk|contribs)} 18:42, 21 May 2009 (UTC)

Yes I understood that's what you mean on how it should be defined. What I'm saying is that this is a nonsensical definition which makes no sense. —Preceding unsigned comment added by 93.173.54.150 (talk) 22:15, 21 May 2009 (UTC)

Why? It's the only sensible definition (can you think of a better one?). Oli Filth^{(talk|contribs)} 22:21, 21 May 2009 (UTC)

Then you wouldn't mind if we decided to change the definition so that every converging series equals its limit? —Preceding unsigned comment added by 93.173.54.150 (talk) 09:22, 22 May 2009 (UTC)

That doesn't make sense. 0.999... is not a series, it's a number. Oli Filth^{(talk|contribs)} 09:47, 22 May 2009 (UTC)

I don't think you've been following our discussion. Try and read up what I've said in my very first messages. That was my initial claim - you must treat it as a single number. You said it is the limit of a series, and to ridicule your claim I gave the example of the hypergeometric series. This is an approximation to the number 1 - I do not deny that. But when you say "it is not a series, it's a number" - what number is it? Don't say it is the number 1, because that's what you're trying to prove! —Preceding unsigned comment added by 93.173.54.150 (talk) 13:22, 22 May 2009 (UTC)

It's the limit of the series, that limit is a number. A simple (for anyone that has taken a first year course in real analysis) calculation shows that that number is 1. --Tango (talk) 13:46, 22 May 2009 (UTC)

If you refuse to accept that the number represented by a decimal expansion is equal to the limit of its finite expansions, then you are forced to conclude that 3.14159... is not pi. After all, the two are only equal in the limit, which you say is only an approximation. --COVIZAPIBETEFOKY (talk) 13:53, 22 May 2009 (UTC)

How does that have to do with our discussion? Pi is an irrational number, and there is no way to repressent it by a finite decimal expansion. —Preceding unsigned comment added by 93.173.54.150 (talk) 16:32, 22 May 2009 (UTC)

It's related because 0.999... and 3.14159... are exactly the same animals, namely, infinite decimal expansions. Why shouldn't they behave the same? --COVIZAPIBETEFOKY (talk) 21:58, 22 May 2009 (UTC)

I am great at math but terrible at biology so I don't know what animals you speak of. However, 0.999.., if you claim it is equal to 1, is a rational number, while Pi is irrational. So, Pi can't be written by any finite or repeating decimal expansion, while 0.999.. (if it equals 1) can.  — [Unsigned comment added by 93.173.165.222 (talk • contribs).]

Any non-zero real number, rational or not, can be written as an infinite decimal expansion. If the number is rational, that expansion happens to be a repeating one, if it's irrational, then the decimal expansion doesn't repeat. When we consider decimal expansions in full generality, there's no reason to make a distinction between the repeating ones and the non-repeating ones; in each case the corresponding real numbers are the limits of the sequeces of finite decimal expansions. As an aside, even if 0.999... were irrational, it could obviously be written as a repeating decimal. Huon (talk) 00:54, 24 May 2009 (UTC)

I am fully aware of everything you wrote here, but I'm entirely clueless as to how this relates to our discussion above. 93.172.72.250 (talk) 22:19, 24 May 2009 (UTC)

(dedent)

Anon here appears to refuse to accept 0.999... = 1 because this is only true when you take the limit of the former, which is, apparently, an "approximation". Similarly, anon would be forced to reject 3.14159... = pi, since the former is a decimal sequence, and the latter is defined as a ratio of the circumference of a circle to the same circle's diameter, and hence, are only equal when you take the limit of the first sequence.

Do you see why this is relevant now? You have not provided any good reason that 0.999... and 3.14159... should be treated differently in this respect; after all, they're both infinite decimal sequences. And if you treat them both the way you are suggesting, then it would appear that it is not possible to represent pi exactly as a decimal expansion. In fact, any infinite decimal expansion appears to fall short of its real counterpart by a tiny, infinitesimal amount. Why are we wasting our time using a decimal representation which appears to be so useless when it comes to describing the numbers we want it to describe?

But suppose we want a decimal expansion to represent pi exactly. In fact, why don't we take on 3.14159... to represent pi, even though it's only an approximation. We know it actually falls short of pi a tiny bit, but we'll just ignore that infinitesimal error, and take the closest, most reasonable real number it could conceivably represent. That way, instead of having a seemingly useless decimal representation in which only numbers of the form n/10^m for integers n and m can be represented exactly, we can, in fact, represent all real numbers. Gee, it would really be nice if there was some kind of mathematical tool for ignoring that pesky infinitesimal error which is making these infinite decimal expansions so clunky and useless... --COVIZAPIBETEFOKY (talk) 02:53, 25 May 2009 (UTC)

0.999... (and 3.14159...) is not a sequence; is no "limit of 0.999...". Rather, 0.999... is itself the limit of the sequence (0.9, 0.99, 0.999, ...). Thus, 0.999... is a real number, and since the limit of the sequence can be shown to be 1, we conclude 0.999...=1. Especially, there are no approximations going on anywhere. If at all, you might say that the finite decimals making up the sequence are approximations of 0.999..., not the other way round. Huon (talk) 10:12, 25 May 2009 (UTC)

I'm trying to speak their language Huon. Sure, I slipped a little into imprecise language in the first couple sentences, but I doubt there was any misunderstanding with that part.

The point is that the anon(s) are pointing out the appearance that 0.999... = 1 is only an approximation, and they have good reason to believe that. They're wrong, of course. All you guys do is try to bury the issue of the infinitesimal error in the ground. I'm trying to point out precisely why we do that. You guys are saying: what approximation? It's exact! I'm saying: yes, we ignore that infinitesimal error. Here's why.

In short, I'm being honest. --COVIZAPIBETEFOKY (talk) 12:19, 25 May 2009 (UTC)

Wow. Everyone suddenly went very quiet.

To those who don't believe this article: Why did you get quiet? Did I actually convince you? Or are you just going to lie low for a few months, and jump in when another discussion gets hot?

To those who believe 0.999... = 1: Perhaps my response to Huon seemed a little too personal. I really don't mind people posting criticisms like Huon did. I kind of expected it.

I hope one of two things can happen here: 1) Those who refused to believe 0.999... = 1 post admitting defeat, or 2) The discussion continues. Since I know the former is against all laws of physics (it would also be less satisfying for those of use who enjoy a good discussion), the latter has to happen. I guess it just needs some time. I really didn't mean to kill the thread. I'm sorry. --COVIZAPIBETEFOKY (talk) 17:38, 8 June 2009 (UTC)

I'd like to get back to the new opening proposal that stalled out. (On the discussion page.) It was strange. At first it looked like we had found a solution that everyone could agree with, but then people started getting all ideological and insisting that nothing change and it all stalled out. Algr (talk) 06:16, 9 June 2009 (UTC)

(ec) What do you define 0.999... to be, then? I define it as $\sum _{i=1}^{\infty }{\frac {9}{10^{i}}}$ and infinite sums are defined as the limit of their partial sums. --Tango (talk) 17:21, 21 May 2009 (UTC)

To Oli_Filth: you said '"0.999..." has been defined to denote $\lim _{n\rightarrow \infty }\sum _{i=1}^{n}9\cdot 10^{-i}$. It's a convention.' Of course this is true, and I completely agree. Let me ask, however, why the article suggests that the equality is some sort of discovery. It has been "long accepted by mathematicians", and so forth, with all the "proofs" and whatnot. 0.999... very simply is defined as the limit. The limit equals 1. End of story. Really, seriously, this article is embarrassing. Tparameter (talk) 01:49, 22 May 2009 (UTC)

The proofs in this article aren't there for the benefit of mathematicians. No mathematician questions the equality because no-one can become a mathematician without taking a basic course in real analysis. Once you know some real analysis the equality is blindingly obvious (I guess it was "discovered" when limits and real analysis in general were first formalised, before then mathematicians did sometimes talk about infinitesimals as if they actually existed in the real numbers so I guess they might have questioned the equality). This article is an attempt to convince people that don't know real analysis and aren't willing to just trust the experts that the two symbols really are equal. --Tango (talk) 02:12, 22 May 2009 (UTC)

The "proofs" in the article are not proofs at all. 0.999... has a definition, not a proof. I realize this has been covered before, but I bring it up only because for the first time in a long time I have seen someone point out the truth. The article could be very simple - detailing the definition of 0.999..., show a few intuitive so-called proofs that match the results of the definition, and be done with it. Really, it is neither here nor there. I'm only fantasizing. But regarding your last point, people without formal higher math training would be more easily convinced if it were truthfully declared right away that the DEFINITION of 0.999... is the limit, and that limit = 1. Instead, it is suggested that this is a theorem of some sort, and many of the "proofs" are obviously missing rigor. Tparameter (talk) 04:08, 22 May 2009 (UTC)

It is a result that must be proven. It is defined to be the limit, you must then prove that that limit is 1. The rigorous way to do that is with a simple calculation involving epsilons, but that isn't understandable to the people this article is aimed at. This isn't a mathematical paper intended to prove it rigorously, it is intended to convince laymen. You have a sacrifice some rigour in order to do that. --Tango (talk) 13:46, 22 May 2009 (UTC)

I disagree. I posit that the situation is this: we have a general definition of a decimal expansion. E.g., something like

$n.a_{1}a_{2}a_{3}...=n+\sum _{i=1}^{\infty }{\frac {a_{i}}{10^{i}}}$

Then for $n=0$ and $a_{i}=9$ for all $i$, we can show that the expression on the right equals 1.

I think describing the general setting first is a better way to approach the issue. Any definition specifically for 0.999... is bound to seem ad hoc to skeptics. Eric119 (talk) 19:42, 23 May 2009 (UTC)

Absolutely, 0.999... isn't defined to equal 1, it is defined as an infinite sum in the same way are any other decimal expression is. That's why it doesn't make sense to decree that is equals anything else because you would be breaking the definition of decimal expressions. --Tango (talk) 19:55, 23 May 2009 (UTC)

I agree completely. The article, as it is currently, is misleading at least. Tparameter (talk) 21:27, 24 May 2009 (UTC)

Why don't people believe?

Every time I visit this page it seems to devolve more and more into insanity. The examples given are becoming so contrived! If people don't accept that $\lim _{k\to \infty }\sum _{n=1}^{k}9/10^{k}=1$, i.e. $\sum _{n=1}^{k}9/10^{n}\to 1$ as $k\to \infty$, then how are they going to be convinced by inventing functions like ${\mbox{Flip}}(x)$? Surrounding an idea by smoke and mirrors is not going to convince anyone!

Here's the bottom line: The notation 0.999... is shorthand for

$\lim _{k\to \infty }\sum _{n=1}^{k}9/10^{k}=0.9+0.09+0.009+0.0009+0.00009+\cdots$

If we accept that the shorthand 0.999... is shorthand for a formally definable limit then we can progress. The idea of a limit is a subtle thing, it says that a quantity tends towards something. The notation 0.999... = 1 means that the sum 0.9 + 0.09 + 0.009 + 0.0009 + ... tends towards 1 as we add more and more terms. And this is clearly true:

$S_{k}:=\sum _{n=1}^{k}{\frac {9}{10^{n}}}=0.9+0.09+0.009+\cdots +{\frac {9}{10^{k}}}$

gets closer and closer to 1 as $k$ gets bigger. In fact $S_{k}<S_{k+1}<1$ for all $k\in \mathbb {N} .$ Given any positive real number ε, taken to be as small as you like, you will always be able to find a k such that $S_{k}-1<\varepsilon$, i.e. you can get as close to 1 as you like. Moreover, you can increase k as much as you like, but $S_{k}$ will never be greater than 1. In fact, for finite k,

$S_{k}={\frac {9(r-r^{k+1})}{1-r}}\ {\mbox{ where }}\ r=0.1$.

So we are left with these facts:

1. $S_{k}$ gets closer and closer to 1 as k gets bigger.
2. We can chose a k large enough so that $S_{k}$ is as close to 1 as we like.
3. There is no k such that $S_{k}>1$.

The conclussion is that $S_{k}\to 1$ as $k\to \infty$. In shorthand this can be written as 0.999... = 1. Although, to be honest, I dislike this notation because it is lazy and it leads to confussion (see above!)  Δεκλαν Δαφισ   (talk)  18:35, 14 July 2009 (UTC)

The flip function was invented by someone trying to disprove the equality. Of course you have to get more and more convoluted if you want to continue finding arguments to support something that isn't true. The problem with your proof (which is, of course, the standard proof) is that is relies on a few technicalities of real analysis so isn't very convincing for people that haven't studied maths at degree level (only first year degree level, really, but still degree level). People without that training are often inclined to think there are infinitesimals. In a number system with infinitesimals your conclusion doesn't really follow from those three facts, so you need to prove the Archimedean property (which is a one-line proof for anyone that understands the construction of the real numbers, but is rather more complicated for the kind of people that dispute the equality). --Tango (talk) 18:56, 14 July 2009 (UTC)

Proof? Who said anything about a proof? Once we accept the notation 0.999... to mean the limit as given above then it is a simple check. Why do you mention all of the problems that other people may have, in a voice that sounds like you don't have them? You seem to have replied for the sake of it, and made an argument for the sake of it. Please don't put words into the mouths of those that are yet to speak.

My point was clear: the notation 0.999... is a limit, and means that the partial sums tend to 1 as we add more and more of them. My three facts cannot be contested; with or without infinitesimals. Once we accept the definition of a limit and the shorthand 0.999..., then the equality holds. No ifs, no buts!

If there is a problem with the mathematics then please tell me. If you have a problem in your understanding then please tell me. Otherwise, let's leave this space clear for people to voice real questions, and let's not fill it with the same old tired arguments. If people want that then they can read the previous posts, or the miles of archives.

To be honest Tango, I think you have said enough on the issue. Your ideas are crystal clear. Why not leave the floor open to others to voice their concerns? If you have anything to add then please add it to my own talk page. Don't clog up this space with the same old tired arguments. (That don't even seem to be your own true beliefs, yet an unquenchable thirst for argument)  Δεκλαν Δαφισ   (talk)  19:20, 14 July 2009 (UTC)

My understanding of real analysis is just fine, thank you. You asked why people don't believe, so I explained. As someone that has spent a lot of time on this page I think I have a pretty good idea of where people's confusions come from. --Tango (talk) 21:19, 15 July 2009 (UTC)

It's a limit

I pointed this out a couple years ago and thought I'd bring it up again. The "..." by definition is referring to a limit. So in other words, it is true that the process of adding smaller and smaller 9s over and over to the end of the decimal will not ever add up to 1. Never, no matter how long you put them on there, it will always be less than 1. However, the number, as written, doesn't end in a 9. It ends in "..." That "..." is a symbol that has a specific mathematical meaning, just like 9 has a meaning and + has a meaning and so forth. By definition, those little dots are saying, "we know adding 9s over and over won't ever equal 1. But, in the process of adding those 9s over and over without end, what is the smallest possible number that we know we'll never reach?" Or, rather, "what's the limit of adding those 9s over and over?" And the answer is 1. Again, this the very definition of what those dots mean. They don't mean that lots of 9s somehow eventually become 1. People say "an infinte number of 9s" and this is part of what confuses the issue. The dots are asking specifically to take a limit, to specify that very next number, the smallest number that we know we won't ever reach by simply adding 9s. They are not saying that if you keep adding smaller 9s you eventually get 1.

This is partly why the bar notation is preferable. Because sometimes the ... also just means "this decimal doesn't end" like in the case of 3.14159..., but because it isn't a repeating decimal, we're not really asking for a limit. Often repeating decimals have a little bar over the top instead of the dots. Think of that bar as the line on the graph that we know the numbers will never reach. The value that they are "trying" to get to, but can't. That line is the actual answer we're talking about, not the numbers under it; that's what the symbol means. The limit. And the limit of adding all those 9s, the smallest number that you won't ever finitely reach by adding 9s, is 1. --The Yar (talk) 20:28, 14 July 2009 (UTC)

Sorry, after more careful review, it appears Declan was making basically the same point above. I completely agree. As I've stated before, part of the problem in this debate comes from those who are correct about it but insist on using proofs and other arguments that don't acknowledge the fundamental misunderstanding. --The Yar (talk) 20:33, 14 July 2009 (UTC)

Yes, I've made this point several times on the talk page as well. The article itself is flawed because it suggests there are proofs and whatnot. The truth is that by definition, this is a limit, and from there it is evident that this particular limit equals 1. End of story, you are correct. Tparameter (talk) 11:20, 15 July 2009 (UTC)

For the most part, I agree. However, 3.14159... still denotes a limit. It just isn't as uniform a limit as 0.999... is. But all the digits exist and can be calculated, leading to the sequence (3, 3.1, 3.14, 3.141, 3.1415, 3.14159...) of which pi is a limit. --COVIZAPIBETEFOKY (talk) 11:45, 15 July 2009 (UTC)

Aww, man, I was just gonna say that.

For reference, all real numbers are equal to the limit of the sequence of their digits. This might be unintuitive, at first, since each digits contains zero information about the next, but it makes sense once you look at the definition of a limit: "for each real ε > 0 there exists a real δ > 0 such that for all x with 0 < |x − c| < δ, we have |f(x) − L| < ε." --Zarel (talk) 19:53, 15 July 2009 (UTC)

Is it really known that each digit of pi conveys no information about the next digit? I thought that was a widely accepted but unproved conjecture, but I sure could be wrong. (All of this is idle curiosity, unrelated to the subject at hand, of course.) Phiwum (talk) 21:10, 15 July 2009 (UTC)

Not to steal your fire twice in a row or anything, but I think the definition of a limit of a sequence would be a bit more appropriate here... --COVIZAPIBETEFOKY (talk) 21:53, 15 July 2009 (UTC)

COVIZAPIBETEFOKY, you make a good point. The notation 0.999... and the notation 3.1415... mean quite different things. And, as you say, the bar notation $0.{\overline {9}}$ is preferable to 0.999... beacause it may cause less confussion. Even if we renamed the article $0.{\overline {9}}$ I think that we may still have some people questioning the validity of the statement $0.{\overline {9}}=1.$ But I feel that my previous comments (in the above post) should resolve that issue. One thing that I might like to ask is that people use more mathematical language. We need to fix this discussion in rigour. If you happen not to be a particularily strong mathematician then please don't worry! Just tell us that you're not and make your point.  Δεκλαν Δαφισ   (talk)  20:08, 16 July 2009 (UTC)

You seem to be crediting me with quite a few things that should actually be attributed to The Yar. I certainly did not mean to make the point that the "notation 0.999... and the notation 3.1415... mean quite different things". In my opinion, both are non-rigorous notations indicating an implied sequence, which, presumably, is easy enough to guess from the first few digits (which, in these two examples, is certainly the case). The first sequence starts with 0, and then every subsequent element is a 9. The latter sequence represents the digits of pi in base 10.

As for your comment above, it's been done. Numerous times. In fact, your comment is nothing new. Other people have come in here before and written up similar well-thought-out arguments for 0.999... = 1. Guess how well that went? Other people have pointed out that it's a simple matter of calculating the limit, no real 'proof' involved. Guess how well that went? These people will always find a problem with any argument that you throw at them, although it sometimes seems like they don't even read the argument, and they just say something similar to "but you're just burying the issue... somewhere... Don't ask me where, you just are!" Remember, we aren't dealing with PHD mathematicians here. These are amateurs who refuse to learn anything about real mathematics, and tend rely only on their own intuition. Need I even say that intuition is a distinctly non-rigorous tool? --COVIZAPIBETEFOKY (talk) 23:59, 16 July 2009 (UTC)

One important skepticism reason is missing...

I understand that 0.999... = 1 in mathematics, but I see that you're missing an important criticism on the "skepticism" section of the article, and the only one that actually makes sense (and with which I agree).
Again, let me say that I DO understand exactly why 0.999... = 1.
BUT, one reason many people don't accept it is simply because in "real life" (as opposed to "mathematics") there's no such thing as 0.9999...
In "real life", meaning reality, in order for a number like 0.999... to exist it would have to be constantly adding more 9s to itself.
In other words, in real life there are no infinitely large numbers (be it in length or size).
So in real life 0.999... doesn't even exist to begin with, let alone to be "equaled" to 1.  — [Unsigned comment added by 143.166.226.60 (talk • contribs) 00:34, July 18, 2009.]

This is true, but 0.999... exists in real life, because it is equal to 1, and 1 exists. ;)

1 does not equal 0.999..., but 0.999... = 1. I'm looking at one LCD Monitor, but there's no such thing as 0.999... LCD monitor. There may be flawed monitors, or monitors missing parts, but it's always one monitor. Gpia7r (talk) 14:44, 6 August 2009 (UTC)

There are no infinitely large numbers in real life, but there are certainly infinitely precise numbers. What is the width of a book, in feet? The decimal never stops. Heck, it never even repeats (the Greeks were horrified when they learned this, which is why the cult of Pythagoreans practiced math with irrational numbers in secret).

By the way, welcome to Wikipedia. If you continue contributing, feel free to register an account. In addition, remember to sign your messages with --~~~~ (you can press the Signature button, between nowiki and horizontal line, to insert this automatically) --Zarel (talk) 19:59, 18 July 2009 (UTC)

I will agree that there are no infinitely large numbers in real life if you will agree that there are no integers in real life either. Nobody has ever shown me a two, just some symbols on paper or sets of objects that they say have this mysterious "twoness".66.30.14.175 (talk) 02:51, 6 August 2009 (UTC)

There are no numbers in real life. There are real life things which we model with numbers. See Mathematical model. 0.999... and 1 only exist in mathematics and they are equal in mathematics, so it is meaningless to say they are ever different things. --Tango (talk) 16:34, 6 August 2009 (UTC)

Have you ever spoken to another human about these ideas? Please track down a current/past collegue from Durham. They will surely shed some light upon your madness. Here's an example of numbers in real life: Given that you continually put forward such half baked ideas as "There are no numbers in real life" it follows that you can't be a very sucessful mathematician. Given that you spend all of your time writing on this maths page, without being a sucessful mathematician, you must have employment problems. Now here's an example: if a hypothetical editor, say Sprite, were to owe the bank £20,000.99 then is that a real number? ~~ Dr Dec (Talk) ~~ 01:25, 9 August 2009 (UTC)

Oh, go away... --Tango (talk) 01:52, 9 August 2009 (UTC)

I'm sorry, but didn't you just get through with crying like a baby to the authorities at the incidents board? And now you barge in here like a sore loser trying to win back his credibility by attacking another person's credibility?

No, it doesn't work that way. A PHD does not automatically mean you get respect from everyone in your life. --COVIZAPIBETEFOKY (talk) 02:00, 9 August 2009 (UTC)

Who mentioned a PhD? Not me! I really don't care about getting respect from half of the editors on this page, who have repeatedly demonstrated their blind stupidity. Who's trying to win credibility? Not me! I'm just making a simple statement about half baked ideas. COVIZAPIBETEFOKY, you're coming across like a wanker, but I'm sure you're not. ~~ Dr Dec (Talk) ~~ 10:19, 9 August 2009 (UTC)

No, you're being a hypocrite for doing exactly what you accused us of doing: writing insulting posts that contribute nothing to the mathematics of the discussion. And you're also being dishonest with yourself and us as to why you made that post. That was most certainly not a "simple statement" about "half baked ideas".

I think maybe you should take Tango's response to heart. --COVIZAPIBETEFOKY (talk) 12:04, 9 August 2009 (UTC)

I found your behaviour, and the behaviour of other editors to be petty, evil and petulant. I tried to reason with you and you wouldn't listen. I tried to ask the admins to help and they wouldn't. So, as the old saying goes: If you can't beat 'em, join 'em! And for the record: I'm not being dishonest to myself. When I read a comment as clearly ridiculous (yes, worthy of ridicule) as "There are no numbers in real life" then I have to say something. If you think the idea that "There are no numbers in real life" is not half baked, then you really need help. It's a joke! ~~ Dr Dec (Talk) ~~ 12:57, 9 August 2009 (UTC)

No, it's not a joke, and as a PHD mathematician, I would think that you would have read some of the literature that addresses this issue. The number "3" is an abstract entity which does not exist on its own in real life. It only exists to model, for example, "how many" letters there are in the word "the". The same is true of the number used to describe how much money one person owes another, or what the ratio of the length of the hypotenuse of an isosceles right triangle is to that of its legs. None of these real-life instances are actual numbers; they're just instances where numbers are useful for understanding and describing the situation. --COVIZAPIBETEFOKY (talk) 14:05, 9 August 2009 (UTC)

Yes, I have read a lot of mathematics, but you need to decide what to take on board and what not to. You speak of philosophy, and not mathematics. Some things have to be taken as given or else you go nowhere. As an undergraduate I shared a house with two philosophy undergraduates. If you don't draw a line in the sand at some point you'll find yourself regressing infinetly. I'll stick to common sense: If I have three bananas then I have three bananas. I'm not going to try and complicate things. Abstraction is quite different to over complication. If you want to progress in your mathematical adventures then I suggest you take note. ~~ Dr Dec (Talk) ~~ 14:13, 9 August 2009 (UTC)

Yes, you do have three bananas. If numbers exist as a non-abstract entity in real life, why did you have to pick a scenario in which you had to specify a unit? After all, units complicate things by bringing with them a whole new set of rules that the real numbers never had to deal with: in order to add two quantities with units, you have to first make sure they match units. When you multiply quantities with units, the units get multiplied as well. And don't get me started on exponentiation.

Just the fact that you had to introduce units in order to specify a quantity in real life using numbers should clue you in that the numbers have no independent existence on their own in real life.

And don't start telling me that there are dimensionless quantities in real life; there are ratios of dimensioned quantities, which is in essence no different than assigning the unit to be the "amount of" denominator (or length, or volume, or whatever may be appropriate in a given scenario). This isn't philosophy; this is fact. Every quantity in real life has a unit. --COVIZAPIBETEFOKY (talk) 14:34, 9 August 2009 (UTC)

I don't disagree with you. My point is: ...and so what? How does this philosophical approach aid you in anyway? How is it helpful or informative? It is, basically, mental masturbation. You say that it's not philosophy and that it's fact. Oh really? I am going to follow your philosophical approach to the n-th degree. As Descartes said: I think therefore I am. That means that you can be sure of your existance because you are able to think. What about my existance? Do I exist? Did Descartes exist? Does Tango exist? Or are we, as Douglas Adams said, figments of a deranged imagination? You might find yourself saying "Don't be silly Declan", but why? You need to draw a line in the sand, take things as given, and build upon that. If you want to spend your time saying that numbers don't exist then maybe you should start worrying about the existance of those who taught you about the non-existance of numbers... maybe even the existance of this very Wikipedia website. It's a waste of time, man. In fact, I don't even know why I'm wasting my time replying to someone who may or may not exist. (p.s. Tango's statement that numbers don't exist was clearly untrue. You have crystalised what s/he was trying to say, so well done on that one) ~~ Dr Dec (Talk) ~~ 15:24, 9 August 2009 (UTC)

There was, you know, an actual point being made: 0.999... is no more abstract than 1 is. Tango made that statement in response to the OP, who was suggesting that 1 exists in real life and 0.999... doesn't. So, no, it wasn't a total waste of time.

Again, this is not a philosophy; it's an observation. You're right; while I didn't exactly scoff or say "don't be silly" like you suggested, I do believe that the world I perceive exists as I perceive it. And it's because I choose to believe that. However, this is a whole separate issue, unrelated to the belief that I am talking to a real person over the internet. I observe that numbers are never used in real life except with the aid of units, therefore, I conclude that the numbers themselves are merely a tool for understanding and modeling the real world. And a very useful tool at that.

That wasn't philosophy. Philosophy begins when we suggest that numbers have their own existence, in, say, some separate "platonic" universe. Have people suggested that? Sure! Do I believe it? I don't know. I don't consider it extremely important to decide whether a platonic universe exists, because it doesn't affect my day-to-day life. Either a platonic universe exists, and mathematicians are digging from it, or it doesn't, and mathematicians are inventing things that may never have been conceived of before. There's not really a difference, practically, either way.

But I do consider it important to recognize that the numbers that I use to describe the real world are not themselves present in the real world. In all the great sciences, it is emphasized that a theory is not the real world: it is only a model of the real world, which may be replaced by a better model in the future. That is the whole point of science. In modern physics, people are researching the discrepancies between the current models and observations of the real world, in the hopes of doing two things: showing how the current model is insufficient to describe the world, and demonstrating what a future model should take into account if it should be expected to replace the old one. The real numbers only serve to model quantities in the real world, and we shouldn't be so arrogant as to suggest that they are, indisputably, the right model for that purpose. --COVIZAPIBETEFOKY (talk) 15:51, 9 August 2009 (UTC)

Like I said: I don't disagree. You make some good points. ~~ Dr Dec (Talk) ~~ 16:13, 9 August 2009 (UTC)

I do appreciate you giving credence to my points, but you then proceeded to call it "basically, mental masturbation" and, yet again, merely a philosophy. I would appreciate it if you took that back, as I have explained why it is an important difference of attitude in science, and not merely a philosophy (although I do agree, to a point, that it is a philosophical viewpoint; philosophical does not automatically mean unimportant, though). --COVIZAPIBETEFOKY (talk) 16:26, 9 August 2009 (UTC)

I'm curious. You disagree with me but don't disagree with COVIZ. but, as far as I can tell, we're saying the same thing. Could you explain what you see to be the difference between what we're each saying? --Tango (talk) 16:59, 9 August 2009 (UTC)

COVIZAPIBETEFOKY, I didn't mention anything about mental masturbation in my last post, so there's nothing to take back. Also, I most certainly never once said that philosophy is unimportant thought. Maybe I didn't explain myself properly. There's got to be a point when you assume some things. If you don't make assumptions then axiomatic mathematics fails to exist. ~~ Dr Dec (Talk) ~~ 17:38, 9 August 2009 (UTC)

Tango, you made the silly statement that "numbers do not exist in real life". What COVIZAPIBETEFOKY did was to qualify the doubt. In fact what COVIZAPIBETEFOKY said also disagrees with the exact statment that "numbers do not exist in real life". It's a subtle difference, but it's key. ~~ Dr Dec (Talk) ~~ 17:38, 9 August 2009 (UTC)

COVIZ., Declan and I seem to be interpreting your comments differently. Could you clarify? --Tango (talk) 17:44, 9 August 2009 (UTC)

Tango, instead or writing a reply six minutes after my comment, why don't you go away and think about what's been said. That way you will answer your own questions. COVIZAPIBETEFOKY will say something along the lines that "numbers in themselves do not physically exists and that they are an abstract concept created to describe the world around us..." and you will say "Oh, look, he says that they don't exists! That's what I said! Ner-ner-ner in you face Declan!" It's the "..." (just like in the 0.999...) that are the key. Read the thread, inwardly digest, and then come back to speak. ~~ Dr Dec (Talk) ~~ 17:54, 9 August 2009 (UTC)

I'll wait for COVIZ. to tell me what he is going to say, thank you. --Tango (talk) 17:58, 9 August 2009 (UTC)

That reply took you four minutes, wow! This is a good example of this 0.999... issue: within time your reply time will tend to a limit of zero seconds. ~~ Dr Dec (Talk) ~~ 18:06, 9 August 2009 (UTC)

For, the existance of numbers is not an issue: they clearly exist. The question is the nature of their existance. This is the point I conceed to COVIZAPIBETEFOKY: numbers do not exist physically, but they exist on a metaphysical level. ~~ Dr Dec (Talk) ~~ 18:11, 9 August 2009 (UTC)

(dedent; not a complete one, because I don't want the post following this one to be confused as a reply to this one)

I would say that a quantity of a something definitely exists on a metaphysical level like Declan says, although I hasten to add that we have no iron-clad reason to assert that these quantities are well-modeled by the real numbers.

And, Declan, I'll accept that as a take-back. You did mention mental masturbation earlier, but it seems now you've changed your mind. --COVIZAPIBETEFOKY (talk) 00:02, 10 August 2009 (UTC)

Uh.. no. There are quantities in real life. I can count one bottle, two bottles... I can even agree to count half a bottle as 0.5 if I cut a bottle in half. But no one can count 0.999... That's what I'm trying to get at. Saying "there are no numbers in real life" is silly because it's redundant and it's obvious that I'm not talking about that. I'm talking about quantities. Again: I can count 1 bottle, 2 bottles (I agree that the numbers are convention, but that does not matter one bit), 3 bottles, I could even count blorgstrek bottles if we had agreed to say blorgstrek instead of "four". The words do not matter. What matters is I can perceive different quantities. And it's not just integer, I can perceive halves, thirds, quarters... But I cannot perceive 0.999... or any other "infinitely ongoing number" like that. THAT is the point, ladies and gentleman. Spare me of the strawman fallacies, I'm not a kid. —Preceding unsigned comment added by 143.166.226.59 (talk) 07:41, 7 August 2009 (UTC)

0.999... is one, so if you can have one bottle you can have 0.999... bottle. Numbers only exist in mathematics, this distinction of yours in nonsense. --Tango (talk) 20:31, 7 August 2009 (UTC)

"There are no infinitely large numbers in real life, but there are certainly infinitely precise numbers. What is the width of a book, in feet? The decimal never stops." - This is just not true, of course the decimal stops, books are made of a finite amount of atoms, so even though there might be thousands of digits after the decimal, it will stop because a book is finite. 143.166.226.59 (talk) 07:45, 7 August 2009 (UTC)

HAHA no. You are assuming that decimals are the ultimate authority on "finite" precision vs "infinite" precision, and then assuming that a finite number of atoms must mean "finite" precision. If you decide that the latter must determine finite precision, then the former cannot determine finite precision. For example:

Have you ever cut a pie or a pizza into 6 slices? That's not exactly what you would call a process which would require infinite precision, is it? However, if you do it perfectly, each slice is exactly 1/6 part of the whole, or 0.1666..., which is an infinite decimal.

Of course, that's an idealized example, based on the perfect mathematical entities of circles and arcs which cannot be imitated perfectly in real life. Let's go back to our book example. For the sake of argument, let's say that it takes 30,000 atoms to make an inch (this is probably a horrible underestimate, not to mention there are many other factors that go into deciding the length of something besides just the number of atoms across, but so be it). And suppose the width of the book is 70,000 atoms. How much is that in inches?

2.333... inches. --COVIZAPIBETEFOKY (talk) 12:44, 7 August 2009 (UTC)

Easier example: How many quarts in a pint? One-half, or 0.5. How many yards in a foot? One-third, or 0.333.... Is the former any more "real" than the latter? Not really. It's just that some people have a problem with infinity. The nature of the problem can have dozens of nuances, but we highlight the primary categories in the article. It's not going to be an exhaustive list, just a list of those most commonly cited. Calbaer (talk) 19:49, 7 August 2009 (UTC)

Okay this is 143.166.226.59; I just registered.

replying to COVIZAPIBETEFOKY:

I don't think we have to laugh at each other. It's still too early into the discussion. :)

You're still thinking in mathematical terms.

Yes I can cut a pizza into 6 slices.

Each slice is 1/6 of the area of the pizza, yes.

Each slice is therefore 0.1666... times the area of the pizza, yes.

That's all math. Coming back to reality, you have:

Each slice is 1/6 of the area of the pizza, still.

Each slice is therefore 0.1666...667 times the area of the pizza.

Each slice is the area of the pizza divided by 6, and that will not yield a number with infinite decimals because it's still a finite piece that can be measured finitely.

See, when you say "each slice is exactly 1/6 part of the whole, or 0.1666..., which is an infinite decimal.", you're already assuming mathematics.

You have to realize that we humans invented that. That's all convention. I know it works perfectly, but it is convention.

I am not discussing math here. I'm discussing quantities in reality.

Using what Algr just said here, 1/6 won't be an infinite decimal if we use base 6. Do you see now?

But if each slice is 0.1666...667 times the area of the pizza, then the number of pizzas you have exceeds one. Gustave the Steel (talk) 06:12, 8 August 2009 (UTC)

I believe I already stated exactly the problem with the logic you are employing here: finite number of atoms does NOT demand finite base 10 decimal representation.

That is all. --COVIZAPIBETEFOKY (talk) 12:31, 8 August 2009 (UTC)

COVIZAPIBETEFOKY is right. It's tempting to think that if you have something physical that you can measure then it must have a nice length, i.e. its length will be a rational number. The ancient Greeks struggled with this idea. Take a right angled triangle whose oposite and adjecent sides both have length 1. Then the hypotenuse has length ${\sqrt {2}}$. ~~ Dr Dec (Talk) ~~ 11:44, 9 August 2009 (UTC)

Hi Anon! Sign up and get a name, it makes things easier. Your idea sounds like the "infinity only works when we want it too." problem that I have spoken about here for a while. In the real set, infinity is not a number. But to state that .999... = 1 requires infinity as an accomplished value, because a finite number of 9 digits would never equal 1. If infinity doesn't exist in the real set, then how can an infinite number of digits be meaningful? As for the example about .333... above, I'd say that one can't precisely define 1/3 in decimal because base 10 = 2x5. You'd need a base with 3 as a factor. Algr (talk) 20:43, 7 August 2009 (UTC)

That the number set doesn't contain infinite values doesn't mean the notation can't have infinite sequences. The two are completely separate issues. --Tango (talk) 21:09, 7 August 2009 (UTC)

But it does mean that you can't talk about the Real set as if it were the only game in town. If you don't acknowledge numbers outside the Real set, it becomes like using whole number analysis to "prove" that fractions don't exist. Algr (talk) 22:48, 7 August 2009 (UTC)

You have yet to provide a viable alternative to the real numbers where 0.999... is both meaningful and not 1. The fact remains that the number set both most useful and most closely linked to decimal representations is the set of real numbers. Huon (talk) 00:20, 8 August 2009 (UTC)

Hackenstrings, remember? Algr (talk)

Two problems with hackenstrings: a) They're too many. There are hackenstrings which don't correspond to any decimal representation. b) There is no 0.999... hackenstring. There's a 0.111... hackenstring in base 2, but that's not the same. That's just the most obvious problems about their relationship to decimal representations. Huon (talk) 14:33, 8 August 2009 (UTC)

What is this? Are you saying that .111..._{base 2} does not equal .999..._{base 10}? Then which of them equals 10? And why is too many hackenstrings a problem? There are decimal representation that don't correspond to any rational expression. If that weren't true, reals wouldn't be a separate set from rationales. Algr (talk)

I may be reading too much into this, but I think what Huon is trying to say is that hackenstrings aren't what you've been looking for all your life, any more than hyperreal numbers or p-adic numbers are. --COVIZAPIBETEFOKY (talk) 19:11, 8 August 2009 (UTC)

Indeed. I am saying that a 0.111..._{base 2} hackenstring exists and that it is indeed not 1; but "0.999..." simply does not denote any hackenstring. And "too many" is a problem because then your numbers are no longer represented by decimal representations. That makes the entire concept of "representation" useless. If the hackenstrings were the natural choice of number system, wouldn't you miss those hackenstrings that can't be represented? Huon (talk) 00:12, 9 August 2009 (UTC)

Huon, I'm giving you the benefit of the doubt, and suggesting you might want to re-read what you just wrote here and perhaps edit it? Algr (talk) 01:11, 11 August 2009 (UTC)

What part of that should require reconsideration? Do you claim that "0.999..." denotes a hackenstring? Give that hackenstring, please, plus an explanation why that hackenstring should legitimately be called "0.999...". The hackenstrings are closely related to base 2 representations of real numbers, and the numbers 1/2, 3/4, 7/8 and so on are finite-length hackenstrings which in a rather obvious way "converge" to a hackenstring of infinite length (though I doubt that convergence is actually defined for hackenstrings, or that that problem could easily be overcome). The numbers 0.9, 0.99, 0.999 and so on all are hackenstrings of infinite length, and I don't see a definition of convergence that would make them converge to anything - especially not to the 0.111..._{base 2} hackenstring. Huon (talk) 02:11, 11 August 2009 (UTC)

As you have been told many times before, the real numbers are very useful for modelling the real world. There is no other set I know of that could usefully replace it. --Tango (talk) 00:53, 8 August 2009 (UTC)

Nothing could replace the Earth, but pretending Mars doesn't exist would still hurt your ability to understand Earth. Algr (talk)

People do study other number systems, but they don't claim they those systems should take precedence over real numbers. --Tango (talk) 01:24, 8 August 2009 (UTC)

Take precedence? Are numbers sets competing in a popularity contest? Algr (talk)

Well, yes, I suppose they are. Maths is full of conventions and they become conventional because they are popular. Unpopular things don't get used and don't become accepted as standard. It is conventional to consider a sequence of digits, perhaps with a dot in it, to be a decimal expansion of a real number, unless stated otherwise, because real numbers (and its subsets) are the most popular number system. --Tango (talk) 04:00, 8 August 2009 (UTC)

But then numbers like 1/5 or 1/2 would no longer have finite representations.

Even if you choose a base with all of those as factors (30), that still leaves 7, 11, 13, 17, 19, 23, 29...

And surely there should be a number filling the (two) gaps between {X: X²<2} and {X: X²>2}, which cannot achieve a finite representation in any integer base. And, even if you reject the gap argument, this particular example can be constructed geometrically as a ratio of two lines, so you cannot deny its existence as a ratio of lengths, at the very least. --COVIZAPIBETEFOKY (talk) 01:50, 8 August 2009 (UTC)

Moved Nothingist's response and Gustave's reply to Nothingist to the post they were actually in response to. --COVIZAPIBETEFOKY (talk) 12:27, 8 August 2009 (UTC)

0.999... does not require infinity for it to be a number nor to be equal to 1. In fact, we can discuss it without ever referring to any concept called 'infinity' at all. The decimal expansion has a 9 in every digit after the decimal point. That is, for each number $\textstyle n\in \mathbb {N}$, the nth digit is a 9. Whilst little sideways eight symbols might appear as a convenient shorthand in our notations for limits there is no entity called or representing "infinity" involved in the actual definition. So the fact that infinity is not a number is never a problem. Maelin (Talk | Contribs) 03:56, 8 August 2009 (UTC)

There are infinitely many natural numbers, though. A strict finitist wouldn't allow non-terminating decimal expansions. Algr seems to be trying to use finitist arguments when he clearly isn't a finitist... --Tango (talk) 17:40, 8 August 2009 (UTC)

That's right. Maelin has simply hidden the infinity within assumptions about $\mathbb {N}$. That is simply ignoring the problem, not solving it. I'm not a finitist, but I don't see how anyone who says "infinity is not a number" can then choose NOT to be a finitist. Algr (talk)

No one said infinity is not a number. It's just not a real number.

And, historically, using infinity explicitly leads to trouble, because infinity is such a poorly behaved number. That's not to say that it can't be done, but that it has to be done extremely carefully. Calculus was not put on a rigorous basis until mathematicians decided to drop the infinitesimals (replacing them with limits) in the 19^th century, and non-standard calculus (ie. calculus with infinitesimals) was not fully developed for a long time after that. --COVIZAPIBETEFOKY (talk) 19:26, 8 August 2009 (UTC)

Because there is more to maths than the members of the real number set? Infinity is not a member of the real number set however there are infinite values in plenty of other sets (and proper classes), for example the cardinal numbers has infinite values, one of which is the size of the real number set and another of which is the size of the natural number set. There is no reason for the size of a set to be a member of that set. --Tango (talk) 19:28, 8 August 2009 (UTC)

You'll have to excuse me, but even though I started this discussion and I'd like to keep on participating, I won't be able to because of a technical problem. I can't imagine how you guys can keep track of a discussion so confusingly formatted and dry on the eyes. I guess I'm too used to forums. Anyway, I can't even begin to understand who said what. Sorry. Nothingist (talk) 03:48, 15 August 2009 (UTC)

That's what signatures are for. To find your post that was in response to me, which I moved, do a find on "I don't think we have to laugh at each other. It's still too early into the discussion." I moved it to the post which it was in response to. The format here is you indent directly after the post you're responding to. If it helps any, I usually find it easier to follow the discussion through the history, since that gives it chronologically. Of course, that also gives you the errors that people make and then correct afterward, so it's not perfect. --COVIZAPIBETEFOKY (talk) 16:08, 16 August 2009 (UTC)

here's an easy way to visualize .999.. = 1 "in real life" without having to rely on abstracted mathematical concepts. The only assumption you have to make is that you can divide a pizza into smaller and smaller units without running in to atoms or whatnot. Take a pizza and cut it into 10 slices. Now take one of those slices and cut that into 10 slices. Now take one of THOSE slices and cut that into 10 slices. Now keep doing this forever. If you look at the divisions youve made in your pizza youll see they are: 9/10, 9/100, 9/1000.... etc. But you still have 1 (one) whole pizza. You're just making infinite divisions of it in ratio of 1/10. Hope this helps in your future understanding of the geometric series. 76.103.47.66 (talk) 16:43, 22 October 2009 (UTC)

without having to rely on abstracted mathematical concepts. The only assumption you have to make is that you can divide a pizza into smaller and smaller units without running in to atoms or whatnot. AHA! But if you never run into atoms then you ARE relying on abstraction... —Preceding unsigned comment added by 143.166.226.61 (talk) 11:36, 21 January 2010 (UTC)

True, a better term to use would be advanced mathematical concepts. But the point still holds, it's not as if it's unintuitive or difficult to imagine a pizza (or any imaginary circle) that you can keep cutting smaller and smaller without worrying about running out of pizza/circle, and if it is, then that's probably where the core confusion is for people who can't grasp why .999... = 1. 76.103.47.66 (talk) 01:51, 14 February 2010 (UTC)

The Remainder

.999 is close to one, but even stretching it to infinity it will always have a remainder, falling short of absolute 1. Add .999 a thousand times, add 1 a thousand times, is the gap big enough for you? Add them a million times, the gap is bigger, still consider it the same? Add to infinity if you must. Of course it takes longer to see the farther you stretch the 9s, but there is always a remainder of 1 at the end of the day, and so it will always fall short. No, 1/9 is not equal to .111 to infinity. It's convenient for solving problems closely enough, but not equal because once again there is a remainder. There's also the issue of significant digits. Anyway, a remainder is relevant, no matter how far you move it. Mukanil (talk) 20:47, 8 December 2009 (UTC)

How often do I have to add 0.999... and 1 until the gap between the multiples becomes 1 or larger? For any positive real number there's a positive rational number (a fraction) smaller than it, can you give a fraction smaller than 1-0.999...?

There is no end to the nines in 0.999..., there's one for every natural number. Also, there is no process involved in 0.999..., we don't just approximate 1 by a sequence of numbers with longer and longer strings of nines. Rather, 0.999... is the shorthand for the limit of such a process of approximation, not any of the numbers involved in that process. And that limit is 1. Huon (talk) 21:07, 8 December 2009 (UTC)

Or to put Huon's point more simply: what is 1 minus .999 repeating?—Chowbok ☠ 00:17, 9 December 2009 (UTC)

Mukanil has a good suggestion. Try it! But a thousand times is a lot. start with two. So let y=1 and x=0.99999... (infinitely many 9's). Let's agree that the difference d=y-x is not negative. I suggest it is 0 and you suspect it is positive. Whatever that difference is it would be the same as the difference between 1+y=2 and 1+x=1.99999... or 2+y=3 and 2+x=2.99999... . So, following the suggestion, let's double the difference by comparing y+y=2 and x+x=1.99999...(still infinitely many 9's). We would both agree that the difference has doubled from d to d+d=2d and yet it has stayed the same. As soon as you accept that d+d=d you can subtract to get that d=0 (unless you decide that the usual arithmetic rules are broken). Adding three or four times: x+x+x=2.999... x+x+x+x=3.9999... . If you add x to itself a thousand times you will get 999.999.... (again, infinitely many 9's). How far is that from 1000?--Gentlemath (talk) 22:20, 13 December 2009 (UTC)

That's cool. Go write a mathematical paper, have it peer reviewed by mathematicians and published in a respected journal, and then we might add it to the article. Regardless, this is not the place to discuss that.

We have a reliable sources policy because if we ourselves had to argue about whether each suggested addition was correct or incorrect, it would never end. Better to leave it to experts in the field. --Zarel (talk) 22:10, 15 December 2009 (UTC)

What's the remainder when subtracting .999... from 1? If they're supposed to be the same thing, then why does my calculator consider them different? --Neptunerover (talk) 09:37, 25 January 2010 (UTC)

Because your calculator, only being able to contain a finite number of digits, can't actually subtract 0.999... from 1 at all. It can subtract 0.99 or 0.999 or 0.9999999999999999999999999999 from 1, but those are all non-repeating-decimal numbers. Fried Gold (talk) 18:19, 27 January 2010 (UTC)