![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 | Archive 3 | Archive 4 | Archive 5 |
DO NOT EDIT OR POST REPLIES TO THIS PAGE. THIS PAGE IS AN ARCHIVE.
This archive page covers approximately the dates between 2005-11-16 and 2005-12-07.
Post replies to the main talk page, copying or summarizing the section you are replying to if necessary.
Please add new archivals to Talk:Proof that 0.999... equals 1/Archive03. (See Wikipedia:How to archive a talk page.)
How can he be wrong? He is referring to sequences with positive sums. What are you talking about? He is correct in stating it will never exceed 1. In fact it will never reach 1 either. As for computing infinite sums, I also, know of no formula. All I learned in high school and university is how to compute the limit of an infinite sum. The two are quite different. Finally, the fact that you can take the sum as close as you want to 1 does not mean it is equal to 1. It means that you can take it as close as you like but you will never reach 1. Just sit down and start adding up the terms and I guarantee you that you will sum until your last breath and still you will not have reached 1. Someone can continue to sum after you and he too will die summing the terms because the sum will always be less than 1. Philosophical grounds - hmmm? No, I think he is just using simple high school math. —Preceding unsigned comment added by 68.238.99.105 ( talk • contribs) 00:12, 18 October 2005
Your logic is 'impeccable': would it really equal 3? You appear to be very confused. It cannot equal whatever you like. It will never equal, nor exceed 1 - that is the assertion. I am talking about the "limit of an infinite sum", not "limit of the sequence of finite sums". The formula he quoted is used in determining whether an infinite sum has an upper bound. There is no assertion that it is equal to this upper bound. Your assertion is plain wrong: there is a very easy way to check yourself - start adding up the terms and I can gaurantee you, you will always have a sum that is less than 1. Please don't tell me you are dealing with a finite sum because then your assertion that the infinite sum is 1 is absolute nonsense! You may be confusing yourself with the fact that the terms are getting closer and closer to zero (Cauchy sequence). This does not mean that any term will ever be zero. —Preceding unsigned comment added by 68.238.97.2 ( talk • contribs) 10:49, 18 October 2005
I think you know exactly what I mean when I talk about Cauchy sequence: the distance between the terms is getting closer to zero. That's the definition of Cauchy sequence, i.e. Lt d(p,q) = 0 as min(p,q) approaches infinity. You write: "The value of an infinite sum IS the limit of the sequence of finite partial sums." This is not true. An infinite sum IS *indeterminate*. Get that? The formula used in this article to prove that 0.999... equals 1 proves exactly the *opposite*, i.e. that 0.999... does not equal 1. All the formula shows is that the limit of any of the partial sums of this sequence is 1. To say this is the infinite sum, shows extreme ignorance. The limit of any partial sum is *not* the infinite sum. You taught at Mit? So what, do you think I ought to bow down and be intimidated? You are wrong. You have been taught wrong too. This article is non-sense. The fact that you can write what you do, displays a fundamental lack of understanding. This non-truth of 0.999... = 1 has taken hold because most people don't understand that 0.999... is *not* a rational number. Neither is 0.333... a rational number. Partial sums from these sequences are used to approximate 1 and 1/3 respectively. I can understand using 0.333 to approximate 1/3 in base 10 (only because it can't be represented finitely in base 10) but cannot understand why 0.999 should be used to approximate 1 which has an *exact* representation. Don't you think it's about time you started thinking for yourself? I know how you will respond: You will say a rational number is any number that can be expressed as a/b where a and b are integers (b not 0). This definition of rational number is part of the problem. If a number cannot be represented *finitely* in a base that is well defined, then the number is not rational. Pi, e, sqrt(2), etc are irrational because there is no well defined base in which these can be represented finitely. You can't suggest that Pi, e, etc be respresented in their own base since these numbers cannot be completely determined. i.e. pi is not equal to 1.0 in base pi because the extent of pi is unknown. Similarly for e or any other irrational number. 15H51 18 October 2005 —Preceding unsigned comment added by 68.238.97.2 ( talk • contribs) 21:07, 18 October 2005
Not entirely true. You start with 0.333... and then you try to show that it can be expressed as a/b. Or you start with 0.999..., 3.14... or some other representation and then try to show it can be expressed as a/b. A number is rational if it can be expressed in the form a/b (b <>0) with a,b integers. I maintain this is insufficient, you also need to add that the representation must be *finite* in some radix form. If indeed 0.999... is rational (it's not), then so is pi since pi can be expressed in the form a/b (i.e. 3 + 1/10 + 4/100 + ...) But of course pi is not rational because there is no number system besides pi in which pi can be expressed finitely in radix form. In base pi, pi is *rational*, i.e. pi = 10 (i.e pi + 0 units). 0.999... cannot be expressed as a/b in any number system. Please don't tell me it's 1 or 1/1 - this assumes that it is equal to 1. You need to think really hard about this. 68.238.97.2
You have told me nothing I did not know in your first and second points. In fact, if you read my posts, you would see that I said this. Your third point is false and I pointed this out in my previous response. You cannot prove that 0.999... = 1 because you do not know the difference betwen the limit of an infinite sum and an infinite sum itself. In fact, 0.999... 'is not' equal to 1. You do not understand the formula used to show that the sum of an infinite sequence is bounded from above. Maybe you should sit down and think about it again? You have been unable to refute anything I have said and you have not even tried to understand it. If I am incorrect in stating that finite representation is 'required' for the definition of a rational number, then pi, e and sqrt(2) are all rational seeing these are the sum of their respective expansions. Frankly it has nothing to with semantics, only simple logic that even a ex-professor from MIT can't see or won't see?. 68.238.97.2
Talk about a lot of 'words'.... Could your rebuttal possibly be a little more abstract. You know you are wrong and just can't admit it. ... sour grapes? 68.238.97.2
Wrong again. Retired supermodel. More profitable and unlike teaching/(child minding), no fake power-trips: the runway is a 'real' power-trip. But don't quit your day job. If your posted photo is recent, I can't tell you won't make it. Sorry, don't mean to be rude, just realistic. 68.238.97.2
Did you mean 'can tell he won't make it' ? :-) He is probably still figuring out how to make 0.999... add up to 1. —Preceding unsigned comment added by 192.67.48.22 ( talk • contribs) 13:56, 20 October 2005 (UTC)
Yes, that should have read: "I can tell you won't make it." I see he has not responded to your rebuttal. Instead he chooses to be sarcastic and rude to a lady. Frankly, he skirts the rebuttals and tries to be cunning and humorous. 68.238.97.2
I am sorry but I have no idea who Paris Hilton is? Is he a mathematician? Wait, I can goolge it. Hopefully there aren't too many with that name. Now I know you love chatting but I really wish you would give this subject more thought. I can answer any questions you might have. You ought to be grateful for this. 68.238.97.2
You must be a non-mathematician because evidently you do not understand this at all. You do not understand geometric series or even what the difference is between an infinite sum and the limit of an infinite sum. You are not alone - most mathematicians don't understand this either. Hardy is a fine example. —Preceding unsigned comment added by 192.67.48.22 ( talk • contribs) 2005 November 9
You are obviously a fake. No one understands infinity (not as a concept or otherwise) and you don't have a clue of what you are talking about. You are a good example of the mindset erroneous thinkers have who believe that 0.999.. = 1. You have erred and contradicted yourself several times already: First you state that infinity cannot be regarded as a number, then you proceed to write that 0.999... can be regarded as 1 minus the smallest real number possible which *you* say is 1/infinity. How can you define the smallest real number in terms of a number that is not defined?! Contradiction. Next, you fail miserably with your child-logic: "But since infinity does no have a defined value, this becomes 0." How did you reach this conclusion?! You are way out of your league. Think carefully before you post again! 192.67.48.22
You are assuming that an infinitesimal posseses the property that |x|+|x|+.... < 1 no matter how many |x|s we sum. This is untrue and constitutes your first error. An infinitesimal cannot be quantified. Your second error is you decide to *call* your quantity 1-0.999... some 'x' - you cannot reach a contradiction on a false premise and then assume that your conclusion is true. 192.67.48.22
Your definition of infinitesimal is untrue and it is based on an incorrect assumption. You state that whatever the value of n is, the sum will never reach, nor exceed 1. Both are false. If what you say is true, then why do you not concede that the sum of 9/10^i (from 1 to n) < 1? You state that the sum of |x| (from i to n) < 1 but in the same breath you are trying to show that the sum of 9/10^i (from 1 to n) = 1 ?! You are very *confused* my friend. infinitesimal has never been properly defined. How can you quantify the number that is greater than zero yet less than every positive real number? You can give it a name, which we have: 'infinitesimal'. However, in every other respect, it is exactly like pi, e and sqrt(2), i.e. its full dimensions are unknown. To say that the reals posses no infinitesimals and then claim that pi, e and sqrt(2) (just some examples) is in itself a contradiction. Mathematicians shoot themselves in the head when they make statements such as: 'As small as you like' or 'As close you like'. How small? How close? I know mainstream thought is that the reals contain no infinitesimal. If the infinitesimal does not belong to the reals, then pi, e or any other number with similar properties does not belong to the reals also. This includes 0.999..., 0.333..., etc. 192.67.48.22
Talk about hand waving! Your proofs are all fine examples of hand-waving. Contrary to what you think, the definition you provide of infinitesimal is not used everywhere. How can I have a conception of a number field that does not include pi, e or sqrt(2)? If I did, it would be incomplete and thus erroneous for pi, e and sqrt(2) are all very *real* and finite. You state that the Archimedean property is a consequence of the LUB theorem. Actually, it's the other way round. I have provided sufficient proof that 0.999... is not equal to 1 on the pages you archived. Since you are making the statement that 0.999.. is equal to 1, the onus is on you to provide proof which so far you have been unable to do. As for your *proof* being interesting in that it does not use limits and infinite sums, I would more accurately say that it is not a proof at all but mere hand-waving. And what are *reputable sources* if they don't agree with Wikipedia's views?! 192.67.48.22
Really? So how is it that you understand that sum |x| (i to n) < 1 where x is infinitesimal (and you don't know your ear from your nose either, never mind what an infinitesimal is or is not; you are also unable to define it in any rational way) and you use this in your faulty proof to show that 0.999... = 1 ? Look, the fact that you have a master's or a PHd in Mathematics does not mean anything. You are in many ways more ignorant than someone without any qualification at all. I have not given you a definition for infinitesimal because it is a concept that makes no sense to me. If you cannot define any concept rationally, it is in fact *illogical* (any surprise?) and consequently rubbish that cannot be used to prove anything. The Archimedean property is well known. Your webpage says you have a master's and this is something that is taught in real analysis. Were you not required to take this course? Just google it for crying out loud and you will know what it is.
As for formal mathematical proofs: There is a proof in the archive and I'll state it again:
Sum (i to n as n approaches infinity) = (ar - ar^n)/(1-r) for |r| < 1
We cannot compute an infinite sum but we can investigate whether it has a limit or not. In the case of 9/10 + 9/100 + 9/1000 + ... it can be easily verified that this limit is 1. This states that even if we could sum this to infinity, its value would never reach, nor exceed 1. Thus it is *clearly* evident that 0.999... < 1.
All proofs that try to show it is equal are faulty but the one used most convincingly is a consequence of the Archimedean property:
The law of trichotomy applies *only* to finitely represented numbers, so you can't use an algebraic process that leads to 0.999... not less than 1 and 0.999... not greater than 1 implies 0.999... = 1. 0.999... is not a finitely represented number. Any arithmetic on such a number can only be an approximation (like pi, e, sqrt(2) etc). And yes, there should be a page called "Proof that 0.999... < 1" because contrary to what you think, it is not generally agreed that 0.999... = 1. Except perhaps in the case of the fools who run Wikipedia? 192.67.48.22
My word but you do love yourself, don't you? And you sure know how to use this system. If I knew it half as well as you did, I would draw some nice sigmas, infinity symbols and why, of course beautiful epsilons and deltas to make every Phd green with envy. Now, there is no handwaving in anything I wrote. It is very clear that the sum on the lhs will never exceed 1:
If we split up the quotient as follows: a/(1-r) - ar^n/(1-r) the first term is independent of n and its value is 1. The second term becomes very small (and using Weierstrass's faulty logic - 'as small as you like' but always greater than *zero*). Thus we have 1 - s where s is some value greater than zero. This being the case, when we consider the difference, we always have a value that is *less than 1*. This is very *clear*. Got it? Hey, if you don't get it now, you must be thicker than I thought. Please don't tell me this is not mathematical or robust enough or else you are a disgrace to all the institutions of learning you have ever attended. Look, when I use words like *clear* and phrases like *by definition*, I do not use these in the same ignorant way as most Phds do. So relax. Don't build a brick wall around everyone else when you feel it necessary to do this for yourself. 192.67.48.22
Firstly, I am not attacking you or anyone else and your psychoanalysis is deeply in error just as is your mathematics. Your above formula is incorrect: It's not (9/10 x 10^-n)/(1-.1) but rather (9/10 + 10^-n)/(1-.1). While you are enjoying Latex so much, you may as well do the job right. Okay, so you made a typo. I'll forgive you for this. Now let's move on. You say there is no natural number s.t 0.999... = sum (i to n) 9/10^i Well aside from stating the obvious, what are you trying to say? My proof considers what happens to the difference as n becomes infinitely large. There is nothing strange about this - it's used in limits and calculus and many other branches of mathematics. Regarding my proof: it is very *accurate* and *valid*. The problem is not with my proof but with your *understanding*. You are very confused. You have not answered my question:
You state that sum |x| (i to n) < 1 where x is infinitesimal (yet you are unable to define infinitesimal in any rational way) and you use this in your faulty proof to show that 0.999... = 1 ?
While you are trying to answer this, let me pose some more questions to you: If the real number system has 'holes' (as you claim it does), then how can you use epsilon-delta proofs at all? What does 'as small as you like' and 'as close as you like' mean? How small is small and how close is close?
This is true handwaving mathematics that has been taught the last 100 years. Real analysis is mostly a load of rubbish. Unfortunately you are the product of Weierstrass' ideas and logic that have some serious flaws. In answer to your question: I know the Archimedean principle the same way as it is published on planet math. Finally, please stop regurgitating what they taught you in your real analysis and honor's class and start thinking for yourself. This will be the best thing you can ever do. 192.67.48.22
Your formulas keep changing so I will just ignore these.
Unlike you and Weierstrass, I will not use epsilon-delta arguments because these imply the existence of infinitesimals. You are right about one thing: I am arguing that since the difference >0 for all n, then it is also true that the sum < 1. And yes, it is very *clear*. You claim it is false and I challenge you to find a counter example! So far you have only been able to regurtitate rubbish. I passed my real analysis course a long time ago. It was garbage then and it's still garbage now. My professors were ignorant then and today I find them to be even more ignorant. And saying that you are ignorant is not *ad hominem*. Once again my friend, you are trying to psychoanalyze me. I suggest you stick to proving the math and forget about everything else.
So here's your first exercise: find me an 'n' for which my claim is false and then I will believe you. No more BS (and MS and PHd) please. 192.67.48.22
Nonsense. You made a rigid statement that what I *proved* is false. You then contradicted yourself by stating that you have no objection to the lhs and continue to talk screeds about *proofs*. You do not challenge the lhs of this equation? If you do not challenge the lhs, you cannot challenge the rhs and yes, you cannot challenge the implication either unless you able to provide a counterexample. So far, you have displayed exceptional ability with Latex but your arguments lack any sound proof or direction. You are *clearly* lost and are only deceiving yourself. You have *not* said anything *relevant* or *logical*, much less *proved* anything. Your diversion tactics show that you are trying to save face. Once again, the *implication* is *clearly* that 0.999... < 1. The nonsense you wrote about 0.999... < 0.999... does not deserve a response. I have defined 0.999... *clearly*: Once again: 0.999... = 9/10 + 9/100 + 9/1000 + .... It is an infinite sum. Now you do not know the difference between an *infinite sum* and the *limit of an infinite sum*. I am going to tell you these things are *not* the same: not by *definition* or *otherwise* - I leave this as an exercise for you. The Archimedean property states that if y is any real number, then a number p exists such that y < p. This implies that p is an *upper bound*. It does not take too much intelligence to figure out where this leads. The rest is an exercise for you to show how it leads to the definition of some least upper bound. If you have an MS in mathematics, this is something you ought to be able to prove on your own. Finally, forget about the Archimedes principle and stop trying to coerce me into giving you a definition of a real number system with a different ordering. Please stay with the subject. You will not earn any points for a dissertation on irrelevant topics. 192.67.48.22
You are evidence that an Ms in mathematics is absolutely worthless. You cannot claim that the lhs implies 0.999... = 1 if the lhs specifically states 0.999... < 1. How do you arrive at this logic?! I don't need to challenge your rhs - it is irrelevant and gibberish that only you seem to understand. How can I challenge something that is incorrect? If you provided a correct statement and then drew an implication from this, I would be able to respond sensibly. But I can't respond to nonsense. You need to learn what 'implication' means. I am stating that you cannot find the result of an infinite sum - you can only investigate what its limiting value is (if it has one). This is a clear distinction between a infinite sum and the limit of an infinite sum. Your example that 0.999.. < 0.999... is *not a counterexample* but rather an example of the nonsense you are writing.
LUB: This is the smallest of all the upper bounds a number can have. In mathematics, if S is a set of upper bounds then there exists an element m of S such that m is less than or equal to all the other elements of S.
The law of trichotomy (LOT) applies only to finitely represented numbers: What this means is that if you have a representation of any number in any radix form that is infinite, LOT no longer applies. If this is untrue, *all* your arguments fail without any further support because 0.999... and 1 are two different numbers and yet you claim they are equal. Please don't tell me they are different representations of the same number. They are *not*. Radix systems are designed for *unique* representation. You were not paying attention in primary school when they taught you about 'tens and units'. Perhaps you should go back to primary school and relearn what you seem to have forgotten.
Oh, and if you are trying to put words in my mouth, then at least you ought to try and do it right. I would say: and not what you stated. To say that it implies b < c is nonsense. It is already given that b < c. Boy, you are a sorry MS! 192.67.48.22
HeroicJay: You should add lack of reading comprehension to your list of setbacks: *I* said that the law of trichotomy applies to numbers with a *finite representation*. Why don't you allow Rasmus to defend himself? I am sure he does not need your help yet. 192.67.48.22
192.67.48.22: Obviously your manners are as bad as your math skills. This is not a private forum and it it quite ok for HeroicJay to join the fun. If this is the way you normally behave when confronted with people who argue against you, it might explain your conviction that your arguments are true: Everybody who have attempted to engage in dialogue with you have become disgusted by your lack of manners and ended the conversation, and you have mistakenly taken that to be an admissal of defeat.
HeroicJay: Welcome to the fun! I apologize in advance for repeating some of your arguments, I will probably just go through 192.67.48.22's posting and reply to his points one by one. (And everybody else: let us know if we should take this someplace else. I realize we are way off-topic here).
Back to 192.67.48.22: You claim that "the lhs specifically states 0.999... < 1". It does not. Your implication could be written in english like this :"(all finite representation numbers of the form 0.99...9 are strictly less than 1)=>(the infinite representation number 0.999... is strictly less than 1)". By any reasonable definition, 0.999... is not equal to any finite representation number.
You claim to use the rule: . Well, first of all: what you need is a rule that concludes that b < c (here , and c=1. Note that there is no n, so that .) Secondly, even disregarding that, this form of implication would be what is called begging the question: your premises include the conclusion (namely that ). Note that the rule I wrote, did not have this problem, since I only assumed that not (you might have misread that, judging from your comments).
As for your LUB-paragraph: you simply seem to invoke the greatest lower bound-rule on the set of upper bounds. Since the existence of LUB is equivalent to the existence of GLBs in a field, you are not getting closer to proving that the Archimedean-property implies the LUB-property. You are also completely ignoring the fact the the rational numbers have the Archimedean property but not the LUB property.
If you maintain that it is not true that for any real numbers x,y, either x<y, x=y or x>y; you are not talking about the same numbers as the rest of us. If that is the case, any discourse with you is pointless. (Though I wonder if you realize, that if you do not allow for the law of trichotomy for 0.999..., you cannot conclude that 0.999...<1 => 0.999 ≠ 1 ?)
Your handwaving about the purpose of radix-systems is pointless. Obviously radix-representations aren't unique since 0.1=0.10 . If we can accept these two representations of the same number, we should have no problem with accepting that 0.999...=1, if it is needed to make the system consistent with the underlying math. And as for your comment about primary school: perhaps you are trying to do primary school-math, the rest of us aren't. Once infinity is invoked, things gets a bit more complex and we can't rely on our primary school instincts any more. Rasmus (talk) 21:27, 21 November 2005 (UTC)
HeroicJay: Am sorry I hurt your feelings. However, you have not stated anything that I don't know and you ought to realize that simply commenting without relevance to the topic at hand does not help anyone. And, you were *riling* me with your provocative comments on my use of *asterisks*. I use these for emphasis because many people miss the main point of what is being written. If it annoys you so much, perhaps you should have stayed out of the fray, no? Rasmus has misrepresented so many facts, got so many things wrong and still adamantly insists he is correct. I have enough on my hands dealing with Rasmus without having to respond to you.
Back to Rasmus: Your entire previous paragraph is nonsense and all the accusations you make are baseless. I am taking you to task: find me one n for which the formula is untrue and I will believe you. BTW: Don't talk about LUB and Archimedes - you don't have the remotest idea what these are about and your writing shows that you lack understanding. So, let's deal with one thing at a time. Provide proof that the sum (i=1 to n) 9/10^i < 1 does not imply .999... < 1. The inequality you introduced is baseless and irrelevant. It demonstrates clearly how confused and deceptive you are. 192.67.48.22
You have not provided any proof and the discussion did not start out with me disproving your 'proof'. You stated that and asked me if this is what I was claiming to which I responded in the affirmative. And yes, I am claiming that Show me *one* example where this is not true. A very simple proof is one by induction. Do you need me to help you with this? How did you obtain an MS in mathematics?
Simple Proof by induction:
We have that k is true: sum (i to k) a_i < 1
Is k+1 true? Yes since a_k+1 + sum (i to k) a_i < 1 because no carry is possible.
Thus it follows that we can choose any k and always find that k+1 is true. Q.E.D.
192.67.48.22
You are in error. The rhs does not become what you state. Your reasoning is completely in error. How do you arrive at this? You cannot assume 1 = sum (i to infinity) 9/10^i and then use it in your argument! This proves nothing!! 192.67.48.22
No man, you got it wrong again. You *are* assuming that "1 = sum (i to infinity) 9/10^i" whichever way you look at it. You can't just *substitute* what you like either. As for my proof by induction - it is perfectly valid. Infinity does not have to be a *natural number*, nor does it have any relevance. Mathematical induction works regardless of what value n might assume. You are full of BS. Just eat humble pie and admit you wrong. Anyone with any sense reading this will be laughing at you. It's easy to tell you are a fake by some catch-phrases you use like "Invoke infinity" and by virtue of the fact that you waddle a lot about a lot of irrelevant side topics. Definition of Mathematical Induction: The truth of an *infinite* sequence of propositions P_i for i=1,...,infinity is established if (1) P_1 is true, and (2) P_k => P_k+1 for all k. This is the principle of mathematical induction. 192.67.48.22
No. You cannot claim whatever you want and then use it to prove your point. Whatever you start off claiming must be *true*. So when you make a choice, it must be logical (and true - surprise?). You keep harping on the fact that infinity is not a natural number. So what? As I explained, the theory of mathematical induction deals with the truth of an *infinite* set of propositions. I am not concerned with *infinity* itself but rather about my propositions. I do not even try to deal with infinity. Please quit making loose statements such as "the natural numbers are group". Stay with the subject. HeroicJay makes some statements about the concept of infinity in higher math: even in *higher* math, we do not evaluate what happens at infinity but make assumptions regarding limiting values (if these exist and we know they do not always exist). Look Rasmus, you are wrong and don't have the balls to admit it. Male chauvinist pride? 192.67.48.22
Look cute boy, you may be the hottest mathematician but you don't appear to be the brightest. Please get off my case. I am not attacking anyone. Rasmus and others have been very disdainful to me, yet you have not reprimanded any of them. And I am the only one *sticking* to the math. Finally if anyone is going to take things too personally, he/she should refrain from the discussion. At the end of the day, nothing on this page is going to cause me to lose any sleep. You should be thankful that I am taking some of my valuable time to improve the content of your site. I believe that knowledge should not only be *free* but also *correct*. 192.67.48.22
Why on earth do you have a talk page then? To see who can cite the most references even if idiots are the originators? How does publishing or writing a paper make knowledge more reliable? Thousands of papers have been written that contain junk worth less than the paper it's written on. There is no orginal research here, only logical rebuttal of past ideas that are incorrect and need to be changed. Given that there are many different points of view on this subject, you ought to represent both points of view equally well. To publish only what some academics think, is biased, morally wrong and does nothing to improve the quality of Wikipedia. It only propagates false knowledge and ideas. If this is your policy, then you seriously need to change it! 158.35.225.229 16:47, 23 November 2005 (UTC)
I repeat myself: you have provided no counter-example at any time. You cannot start out by choosing a_i and x as you *please*. I explained this to you. Your choice has to be *true*. You are assuming that sum (i to n) 9/10^i < sum (i to infinity) 9/10^i. This is correct. However, you cannot proceed to argue that sum (i to infinity) 9/10^i < sum (i to infinity) 9/10^i because these are exactly the same thing! My, but you are slow. That (N,+) is a group is *irrelevant*. The last part of your statement is yet again nonsense: You write: For all k is an element of (N union infinity) .... - N is *infinite*. What nonsense are you writing? And what exactly does it mean to have (N union infinity) ? Rasmus, just admit you are wrong. Deal with it and move on. You are trying to save face but everyone who reads this knows you are wrong. You are only making a fool of yourself each time you respond the way you have. 192.67.48.22
Actually it is you who are not applying the principle of induction correctly. Firstly, you are in grave error to write P(infinity) - there is no such thing and we never consider any such proposition. You need to learn what induction really means. I have stated it correctly and you have failed to understand it. I have *proved* conclusively using induction that 0.999... < 1 for any value of n. I do not need to consider P(infinity) - in fact this is impossible to consider in any scenario. As for counterexample - I know very well what it means - even before you were born. You have failed to provide any counterexample and it is evident you don't know what it is. If Newton said that 0.999... = 1, I would call him a fool. How much more I would call less intelligent men like you a fool? You are regurgitating (incoherently) what was brainwashed into your head at whatever institution you attended. So far all you have done is ramble on about irrelevant topics and have not provided a shred of logic to show that 0.999... = 1. And you have been unable to refute my proof in any way. Finally, if my induction is incorrect here, then all mathematical induction is *incorrect* and it cannot be used to prove the validity of any proposition. 192.67.48.22
(This has been detatched from what it replied to, so I'm removing the indentation.) Even if Mr. 192 isn't swayed by that, I now know what he meant by "Rasmus added that" when I went out of my way to show that his a < b < c assertion was meaningless with the limit (yes, it's true when you type . He apparently misunderstood the inequality sign.) I had thought he was still going on about the trichotomy thing. -- HeroicJay 21:47, 21 November 2005 (UTC)
Heh, I was following the conversation and was about to make a similar point to the above by Rasmus until I saw it. In plainer English, while it is true that .9999...9 (with a finite number of nines) is less than one, no matter how large the finite number of nines, it also happens to be less than .999... (with infinite nines), so it doesn't really prove that .999... is unequal to 1. That argument is (loosely) similar to this: 1 < 4, 1 + 1 < 4, 1 + 1 + 1 < 4; thus, 1 + 1 + 1 + 1 < 4. Q.E.D. It's much more obvious that my "argument" is silly, but it follows the same principle: just because a bunch of numbers smaller than .999... are also smaller than 1 doesn't make .999... smaller than 1. (Technically, a better parallel would be , which is absurd to anyone who understands the terminology, but that uses infinite sums and infinity and Mr. 192 whines about those.) -- HeroicJay 01:26, 20 November 2005 (UTC)
... Then we can see that therefore . However, if , then we know that as then , and then . It must be then that as is defined to be zero. But the basis of my proof is that 4.000...1 (4, dot, an infinite number of 0s, 1) = 4. And then 5 - 4.000...1 = 0.999... = 5 - 4 = 1. Therefore 0.999... = 1. If you do not agree with this proof (except for very stupid errors I have made, such as not fixing my LaTeX properly), then, unfortunately, you must disprove all of calculus as it is all based upon it. x42bn6 Talk 07:10, 23 November 2005 (UTC)
Not so. You have not proved anything, nor said anything new. You make statements such as 10^(-infinity) is *defined to be zero*. Nothing could be farther from the truth. Zero is defined as *zero*, nothing else. I can see you are also a good product of the establishment. You need to start thinking for yourself - this is all I can say to you. See page on Mathematical analysis for an example of how you can find derivatives without even considering what an infinitesimal is. 192.
Passed by and changed the intro. No article on Wikipedia is a place of discussion, nor should any article strive to "prove" anything. Articles are here to present the facts in as unbiased a manner as possible. This article, if it continues to exist, should give information on known proofs of the concept. Nothing more, nothing less. Discussion is for talk pages. -- WikidSmaht ( talk) 10:55, 20 November 2005 (UTC)
I've been reading the above discussion for a while, and it seems to me that our anonymous friend thinks that:
Now I've been thinking about this. Suppose we calculate the intersection of closed intervals from 0 to the inverses of natural numbers. (A closed interval means one that includes its endpoints.) Specifically, we investigate the intersection of all where . Now it's clear that if n is a natural number, its inverse is going to be greater than 0. Thus, for all n \in N, the interval is going to include numbers greater than 0. Any smaller such interval is contained in any bigger such interval, so:
Now let's investigate the supremum (least upper bound) of an intersection of such closed intervals. It shouldn't be too difficult to figure out that:
This is because no matter how big n gets, its inverse is always going to be included in that intersection, and is such its supremum. Now according to our anonymous friend, this means that:
Call this supremum x. Clearly x is greater than 0. Now let m be a natural number whose inverse is smaller than x. (Because there's infinitely many natural numbers, we can always find such a natural number.) This would then mean that:
Now we have a case where the supremum of a subset is greater than the supremum of its superset. According to my understanding of suprema, this can't be possible.
So obviously the assumption is invalid, and the correct way should be:
Here we can't "let m be a natural number whose inverse is smaller than" 0, because then its inverse would have to be negative, and that would mean that the natural number itself would have to be negative.
But wait! Didn't we say that no matter how big n is, in the intersection of all intervals from 1 to n, its supremum is always going to be greater than 0? Now we say that in the intersection of all intervals from 1 to infinity, its supremum is going to be 0! Surely this would mean that an infinitely large collection of finitely large things is different from an infinitely large thing!
Well, that's exactly how it is. We all know that, and our anonymous friend should learn that too. — JIP | Talk 16:50, 23 November 2005 (UTC)
Just thinking, can the set you described above have a supremum? If the upper bounds keep getting closer to zero, then zero will be the intersection and the only member of the set. In this case it would be a supremum. I fail to see the analogy with her induction proof and how this proves 0.999... = 1? In her proof you are looking at a sum of things, not a collection of finitely large things. Very confusing. 71.248.136.206 01:01, 26 November 2005 (UTC)
I think I can see what you have done but I still cannot see the analogy and I don't see how reductio ad absurdum in this problem proves anything. If as you say P is true for infinitely many finite sets, it's true for an infinite set, I think you may have contradicted yourself for you stated earlier: Surely this would mean that an infinitely large collection of finitely large things is different from an infinitely large thing! If an infinitely large collection of finitely large things does not imply an infinitely large thing, how do you reach thi conclusion? Sorry, I am wondering if it's me who needs to learn something and not you? 71.248.130.225 19:43, 26 November 2005 (UTC)
Your whole point whatever it is, indicates your argument is confusing and at best blurry? She (192.67.48.22) uses mathematical induction in much the same way as I have seen anywhere else. Reductio de absurdum is Latin for reducing an argument to the absurd and then concluding it is false. You have not done this for your argument is complete with contradictions. I must be frank - I do not have a clue of what you are trying to accomplish and I do not see any connection between your example and her proof. 70.110.85.150 13:18, 27 November 2005 (UTC)
Well, I guess I still don't get your point. You showed that 0 is the intersection of an infinite number of finite sets. She (I think it's a she because one of the comments is about male chauvinism?) is discussing an infinite sum that is not the same as a set. How do you draw conclusions relating your example to her's? It seems (imo) that these are two different things. In mathematical induction we do not try to prove P(infinity) ever, do we? If P(k) implies P(k+1), it seems logical and natural that P(n) is true regardless of the value of n. Can we at any time prove any induction result for P(infinity)? I don't think so because we cannot make any assumptions about P(infinity). If what you are stating is true (you are saying that P(k+1) does not imply P(infinity), aren't you?), then no mathematical induction can be correct because it clearly states that the results hold true for any value of n. It seems spurious to think at all about P(infinity) since there is nothing we can deduce from it. If 0.9, 0.99, 0.999, 0.9999 and so on are less than 1, then what makes you think that there will ever be a term that will make it equal to 1? The terms get closer and closer to zero, so the overall sum remains less than 1. If this is not true, then the Cauchy theorems do not make sense: are you saying that each x_i term does not get smaller as the i_s get larger? Does not m>n imply x_m < x_n ? Does infinity wrap round at some point? 70.110.85.150 15:59, 27 November 2005 (UTC)
I don't think you read the original poster's comments properly. You may want to reread these? She/he/it(whatever), did not state there will never be a term equal to 1. Why don't you reread her post? As far as is concerned, there is no room for this in Mathematical induction. Have you perhaps missed this point? As for examples, you have not been able to convince me of anything. Are you a student of Rasmus's? He appeared to think that his counterexamples were valid too. You first need to present a counterexample that is related to what is being discussed. What you started talking about is material that is commonly found in beginner's courses (0 the intersection of the sets...). This has no connection to the induction proof. It is completely unrelated and anything you deduce from it whether it be by reductio ad absurdum or otherwise is not true. Once you find a counterexample that is connected, then you need to show how it nullifies her proof. Really, you have not been able to prove anything imo. 70.110.85.150 20:36, 27 November 2005 (UTC)
I was reading Jitse Niesen's comment that he knew of no mathematician who might think otherwise. Well, here is one mathematician who questions whether 0.999.. and 1 are actually equal. His name is Fred Richman and the URL is:
http://www.math.fau.edu/Richman/HTML/999.htm
Title of this page is: Is 0.999... = 1?
Another mathematician/computer scientist is Frode Fjeld:
http://www.cs.uit.no/~frodef/frodef.html
This is from the first google page returned for "proof 0.999 less than 1" —Preceding unsigned comment added by 71.248.136.206 ( talk • contribs)
I don't think so. The reals also have serious problems. This is but one of those problems. I think whoever came up with this idea (that 0.999... = 1) did so because it would close the door on those who argue that the reals are but an approximation to actual values where these cannot be expressed as rational numbers. 70.110.85.150 23:17, 27 November 2005 (UTC)
I am willing to bet that you will be surprised how many mathematicians think this is nonsense because it cannot be proved in any way. Out of curiosity, does anyone reading this page know who first came up with the idea that these two numbers are equal? Was it Frechet? Weierstrass? Cantor? Someone else? Would be interesting to know who. 70.110.85.150 20:40, 27 November 2005 (UTC)
I don't have to prove it. I think 192.67.48.22 proved it conclusively by mathematical induction in the above posts. I don't think anyone can refute this proof. It is 100% solid imo. Go ahead and reread it. Yeah, I would think it's safe to say 0.999... is not greater or equal, therefore it must be smaller than 1. By how much is not really important. Is it a real number? Yes, I think it is, but I also think that it is not a rational number. Can it be expressed as a/b or a finite polynomial sum that reduces to a/b s.t b <> 0 and both a and b are integers? I don't think so. 70.110.85.150 23:17, 27 November 2005 (UTC)
Yes, how about it. This is exactly what I wrote in my previous paragraph. 0.999... is not a rational number. Can you find a number between 3.141... and the actual value of pi? 68.238.98.71 11:50, 28 November 2005 (UTC)
What we're going to do is discuss one point at a time. 70.110.85.150, or 68.238.98.71, or wherever you are: I am addressing you. Is there a rational a/b such that 0.999... < a/b < 1? There are two acceptable answers: yes or no. Melchoir 12:05, 28 November 2005 (UTC)
Thank you for your conclusive reply. Now: is it your belief that two Dedekind cuts can partition the rational numbers in the same way, yet still be different? Melchoir 12:16, 28 November 2005 (UTC)
Fair enough. Melchoir 12:20, 28 November 2005 (UTC)
68.238.103.187, are you a different person from 70.110.85.150 and/or 70.247.52.236? Please respond, and additionally, please create an account to avoid future confusion. Seriously, you're killing us. I'm not letting you off the hook: is it your belief that two Dedekind cuts can partition the rational numbers in the same way, yet still be different? The absence of the word "real" in this question was, and still is, intentional. Also note that I have never used the word "definition" on this page. Melchoir 22:26, 28 November 2005 (UTC)
Why would you avoid creating an account? I'm honestly curious. You may have responded to my question, but you haven't answered it. The definition of a Dedekind cut as a partition of the rationals is unambiguous. I will rephrase my question very slightly: can two Dedekind cuts consist of the same partition of the rationals, yet still be different cuts? There is now a second question that you also have not answered: are you a different person from 70.110.85.150 and/or 70.247.52.236? Melchoir 22:45, 28 November 2005 (UTC)
That sounds like a no. I must also assume that you are 70.110.85.150 and 70.247.52.236, since you refuse to say otherwise. You do ask an interesting question: how to represent pi using Dedekind cuts. I am willing to show you, but you must point me to a definition of pi that you agree not to argue with. If I do this, will you agree that all real numbers can be represented by Dedekind cuts? Melchoir 23:29, 28 November 2005 (UTC)
Good, let's use circles. Let A be the set of positive rational numbers x such that a circle of circumference x has diameter strictly less than 1. Let B be the union of A with the set of rationals less than or equal to zero. Then B is easily seen to be closed downward, and it should also be clear that B has no greatest element. Let C be the complement of B in the rationals; it is closed upwards. Let p = (B, C); then p is a Dedekind cut.
Take care to note that the definition of p does not require, assume, mention, or beg for the word pi, the Greek letter pi; the number pi; or any hint of foreknowledge of the existence, in a mathematical, physical, metaphysical, sociological, or anthropological sense, of the concept known as pi.
Now, consider the question: what does it mean for a Dedekind cut to represent a real number? We can answer this question without deciding on a particular definition of "real number"; we need only agree that the set of real numbers is an ordered set which extends the rationals. One says that a cut (X, Y) represents the real number z if X consists exactly of those rationals which are less than z. This paragraph is non-negotiable.
Back to our problem. We all agree that pi is real. Every element of the set B is less than pi, and every rational number less than pi is in B. Therefore the cut p represents pi. Melchoir 06:55, 29 November 2005 (UTC)
Which part do you disagree with? Melchoir 18:49, 29 November 2005 (UTC)
Please do not refer to yourself as "the poster below ". I think I finally understand what you are trying to say about Dedekind cuts, and we can talk about that later. For now, however, I ask you to focus on the question at hand: pi. My argument above proceeds in steps:
Obviously you disagree with at least one of the above. Which is the first point with which you disagree? Melchoir 23:02, 29 November 2005 (UTC)
I have a great deal to say about the "gaps" to which you allude, and I will do so later. However, I would venture that Michael Hardy understands more than both of us put together, which is saying something considering our... differences.
So, you agree with (1) and (2) but not with (3). This is encouraging. Tell me, what does it mean to you to say that a Dedekind cut (X, Y) of rational numbers represents a number z? Melchoir 23:29, 29 November 2005 (UTC)
Please refrain from personal attacks. You claim that we are wasting our time, but I am learning more from you than you realize; you could make an attempt to learn from me. If I read you correctly, you are threatening to leave this discussion. Make no mistake: most folks here at Wikipedia would be happy to see you go, but I would not.
By " Dedekind cut" I mean the definition in that article, and I have already told you what I mean by "represent". I quote:
Since B consists exactly of those rationals which are less than pi, p represents pi. Now, please notice that as far as this argument takes us, p might also represent numbers that aren't pi. Conceivably p could represent lots of numbers very close to pi. Infinitely many, even. But p still represents pi. Do you disagree with this simple statement? Melchoir 00:33, 30 November 2005 (UTC)
That's fine; it's not going to matter how many numbers p, or any other Dedekind cut, represents. Just to be clear, you agree that p represents pi? Melchoir 01:21, 30 November 2005 (UTC)
You once claimed about finding a cut to represent pi: "If you can do this, I will concede that all real numbers can be represented by Dedekind cuts." I cannot proceed until you do exactly that. Melchoir 00:18, 1 December 2005 (UTC)
I have been extremely careful not to claim that p represents only pi. You asked me, "Show me how to represent pi using Dedekind cuts." You never asked for a Dedekind cut that represents only pi, and I never claimed to give you one. I now ask you to acknowledge that, among the numbers that p happens to represent, one of them is pi. Melchoir 01:43, 1 December 2005 (UTC)
You asked me to find a cut representing pi; presumably you thought that I couldn't, and you could be a little more humble and upfront in admitting that I did. And, please do not refer to "common misconception"s among PhDs to try and strengthen your point. I have all the authority in the world behind me, but I have not tried to cram it down your throat. If you want a sociological holy war, start one on your user page. This is about math.
Now, let me make a definition. Let a Melchoir number be a set of "numbers whose limits are the same", whatever that phrase means to you. We can add, subtract, and multiply Melchoir numbers; if we're careful, we can even divide them; and we can compare two Melchoir numbers by comparing their elements. The set of Melchoir numbers is an ordered field. Do you object? Melchoir 18:31, 1 December 2005 (UTC)
I speak of a "sociological war" because you speak of a "common misconception amongst Phds". I do not hold a PhD, and I have never studied number theory. If you're looking for a fight against the Establishment, I can't help you. I never called you inferior. Perhaps you think "my logic" is patronizing because I am bending over backwards to agree with your terminology, along with that of logamath1 at the bottom of the page. Would you really prefer that I cry out every time you defy a universally accepted definition?
I agree that there is such a mathematical object as a formal infinite series; in a topological group, there is also such a mathematical concept as the sum of a formal series. Sure, these are different things. There is a formal series of rationals that perhaps best represents the idea of the string of symbols "0.999..."; let's call it M. There is another formal series of rationals that perhaps best represents the idea of the string of symbols "1.000..."; let's call it N. So M is different from N, but who cares? Both of these formal series are Cauchy, and as luck would have it, they are co-Cauchy; their partials sums can be made arbitrarily close to each other, can they not? Melchoir 23:31, 1 December 2005 (UTC)
Please define the phrases "infinite sum" and "limit of an infinite sum", so that I will be able to decide whether or not they are the same. Meanwhile, by speaking of formal series I have bypassed the whole issue. Do you agree with M and N are co-Cauchy? Melchoir 18:36, 2 December 2005 (UTC)
Okay Hardy. You are correct, it is "for all intents and purposes". However, I have heard very educated people use it this way and since they were not corrected or challenged, I assumed it was acceptable to use. I speak/read/write 6 languages. And I am proud of the way I learned English seeing it it not my mother tongue. You are nothing but an arrogant fool and a miserable man who thinks he has a high IQ. "For all intents and purposes", most people who are not arrogant farts like you, won't even notice I used this cliched phrase incorrectly. So unless you can prove me wrong by focusing on the math, I suggest you save your grammar talents for a more appropriate occasion. Do you know more than one language Hardy? 70.110.81.253 01:05, 30 November 2005 (UTC)
You can't be serious? That's exactly what a Dedekind cut is saying or has it not dawned on you yet? 70.110.81.253 23:16, 29 November 2005 (UTC)
If there is no rational number between 0.999... and 1, and rationals are dense in the reals, seems to me like 0.999... is equal to 1. — JIP | Talk 12:21, 28 November 2005 (UTC)
If you believe that Dedekind cuts can be used to represent any real number, then please show me how to represent pi and e. 68.238.103.187 22:15, 28 November 2005 (UTC)
You may not need to call on any theorems because they won't help you.
I know where this is leading: If 68.238.98.71 says no, then you will try to say that {(-inf,1);[1,inf)} = 1 and {(-inf,1);[1,inf)} = 0.999... therefore they must be the same. However, by saying this you would concede that 0.999... is less than 1 since (-inf,1) does not contain 1 and I don't think you are quite ready to say that 0.999... is a member of [1,inf), are you? 70.247.52.236 18:13, 28 November 2005 (UTC)
I was afraid you would respond this way. You cannot use Dedekind cuts to support your argument because then you are assuming that 0.999... = 1 without any proof. 1 is the upper bound of the infinite sum 0.999... therefore it cannot be a member of [1,inf). Using your argument, one can equally well assume that 0.999... is a member of (-inf,1) for there is no number between (-inf,1) and [1,inf). Dedekind cuts cannot be used to represent all real numbers. How would you represent pi? Please don't tell me: {(-inf,pi),[pi,inf)} because you do not know what pi is. You will have to come up with a far more convincing proof if you are to defeat 68.238.98.71. 70.247.52.236 22:13, 28 November 2005 (UTC)
Please create an account. Melchoir 19:30, 28 November 2005 (UTC)
I'm not really comfortable with this section, how do you guys here feel about changing it?, maybe something like:
Another kind of proof adapts to any repeating decimal. Let the number 0.999… be called c, then 10×c = 10×0.999… = 9.999…; If we substract c, we have 10c - c = 9c and 9.999… - 0.999… = 9, thus 9c = 9. Dividing both sides by 9 completes the proof: c = 0.999… = 1
Please edit this entry, so we can have a 'definite' version of this section. Or am i the only one who has trouble with the current wording? Thank you. Jesushaces 05:05, 30 November 2005 (UTC)
If you knew me better, you wouldn't have provoked me like that...
Ksmrq, I have reviewed the edit history of this article. It has racked up a hundred edits in six months of existence, which is a lot for such a low-visibility page but not completely out of line. It was indeed a "horrible hash" until you rewrote it on 26 October 2005, and in some sense it is a testament to your skill that its content has not changed since then. Although the article's stasis reflects well upon you, it is not necessarily a good thing, and it certainly should not be defended as an end in itself.
Since your 26 October 2005 rewrite was so extreme, we must treat the article as if it were completely new. Indeed, its content was written by one person and has since been revised only cosmetically. Most relevantly: we cannot predict the effect of modifying the new article by extrapolating from the old article. You speak of a "certainty" of "attack". I don't know how you can be so certain.
On to the math. You claim "The algebra proof pretends nothing". In fact, the algebra proof relies on the properties of the real numbers while pretending that it doesn't.
Take a look at this webpage, which was presented by an anon as an example of a mathematician who supposedly thought that 0.999... and 1 are different. The anon was wrong, and you correctly pointed out: "Good grief. Richman is merely having a little constructivist fun looking at an alternative to reals." Richman is sort of playing the iconoclast by referring to "traditional real numbers", an extremely unfortunate phrase given that it is far too late to redefine the reals, even if it were a good idea, and it isn't. However, his mathematics looks sound to me, and he does two important things. First, he names the decimal numbers and takes them seriously. This is important because many people seem to think that the reals numbers are the decimal numbers, and the Elementary section of our article doesn't even suggest otherwise. Unfortunately, Richman does not provide a symbolic representation for decimal numbers (he reuses the notation everyone else reserves for real numbers, another poor choice), so I will invent one: let's agree to write decimal numbers as we would real numbers, except that we replace the decimal point by the letter d. For example, 0d999... and 1d000... are different decimal numbers, essentially by fiat.
Second, he defines addition and multiplication operations on the decimal numbers. This allows him to observe, for example, that 3x 0d333... is not 1d000... but 0d999. Indeed, it is impossible to divide 1d000... by three.
Why am I writing so much about the decimal numbers? Because they provide a counterexample to the reasoning in the "Elementary proofs" section. For the decimal numbers, "manipulations at the digit level are well-defined and meaningful, even in the presence of infinite repetition". Therefore the reader should expect the two elementary proofs given to work for the decimal numbers. But they don't. In fact, we can pinpoint where they break down. In the first proof:
Long division doesn't work on decimal numbers, and 0d3333... is not one third of 1d0000... This proof relies on the numbers discussed being real. In the second proof:
There is no unique way to subtract two decimal numbers, but your argument can be rearranged to avoid subtraction. If you made this rearrangement, you would still need to say at some point
If c is a decimal number, then the second equation does not follow from the first. As Richman might say, 0d999... is not cancellable. This proof also relies on the numbers discussed being real.
The arguments listed under "Elementary proofs" only work if they are applied to real numbers. However, the article currently does not contain the word "real" until the "Advanced" section. This is bad logic and possibly even dishonest. It's not just that the elementary proofs lack "sophistication and rigor"; as stated, they are simply wrong. I see three possible fixes to this problem:
The first two options have the same intent: in a more or less subtle way, tell the readers that something essential is being hidden from them. Sadly, we live in a world filled with disclaimers, and many readers will fail to heed the warning. They will interpret the elementary arguments as purported self-contained proofs and think that we are wrong. This is, in fact, what is currently going on and what you should fear the most. I think my cynical prediction of the future is much more likely than your cynical prediction of the future.
The third option seems like the only honest one to me, and it hides nothing. Just admit that they aren't proofs at all but sleight-of-hand manipulations that are too pursuasive for their own good. In that spirit, if I may quote myself, "The current wording suffers from a pretention to precision that isn't really justified." I stand by it. Melchoir 11:51, 30 November 2005 (UTC)
I am the originator of the 'Induction Proof' that 0.999... < 1. I have read what you have written in the above paragraphs and I think you are only fooling yourselves. The 10x proof does not rely on the unstated properties of real numbers or any established facts and it is entirely false. It seems to me that Dedekind cuts can also be used to prove that 0.999... < 1 as suggested by one reader. The induction proof is solid. You do not need to rewrite the nonsense you have in this article - you need to delete it because it misleads those who do not have enough sense to see it is false. I said I would not post but I hope you will do what is right, not what you think is right, but what I am telling you is right. I know what I am talking about. Am I smarter than most of you? The arrogant answer is yes. Most of you are sincerely deceived. You love ignorance and it shows. 158.35.225.231 12:57, 30 November 2005 (UTC)
Who said anything about disputing the properties of real numbers? It's your misinterpretation of the number 0.999... that makes it appear there is conflict in the real numbers unless it equals 1. 158.35.225.231 14:50, 30 November 2005 (UTC)
I don't want to contribute to your article because it's false. Your main concern is how to remove the contradiction of there being no real numbers between 0.999... and 1. Here's how you can do it: There are infinitely many numbers between 9/10 + 9/100 + 9/1000 + ... and 1 as the following demonstrates.
Let X = 9
X/10 + X/10^2 + X/10^3 + .... [Radix 10]
A= .9 + .09 + .009 + .... = .999...
Let X = 99
X/10^2 + X/10^4 + X/10^6 + .... [Radix 100]
B= .99 + .0099 + .000099 + ... = .999999...
Let X = 999
X/10^3 + X/10^6 + X/10^9 + .... [Radix 1000]
C= .999 + .000999 + .000000999 + ... = .999999999...
A < B < C < 1
The terms of C decrease much faster than B and those of B much faster than A so that no matter how large any of these numbers become, the difference between C and 1 will always be the smallest. Let's call this difference Dc. Likewise let's call Db and Da the differences of b and a with 1 respectively. It is easy to see then that Dc < Db < Da. Therefore no matter how small these differences become, they will always be greater than zero and the ordering will always be preserved, i.e. Dc will always be less than Db which will always be less than Da. Would you be able to represent the number C in Radix 100? Would you be able to represent the number B in Radix 10? Answer is no. Part of this controversy has to do with the way numbers are represented in a radix system. You cannot say (A) 9/10 + 9/100 + 9/1000 +... = 1 for then you must have (B) 99/100 + 99/10000 + 99/1000000 + ... = 1 and we know that A is less than B. Remember we are not comparing limits for this tells us nothing about the size of the number. We can only compare the sum of a finite number of terms, term by term. 158.35.225.231 17:56, 30 November 2005 (UTC)
No. 158.35.225.231 19:33, 30 November 2005 (UTC)
Here you go. Hate to break it to you but I hardly ever use your website. And now you have one more account... Logamath1 20:02, 30 November 2005 (UTC)
Actually that's the problem - they do not all correspond to the same real number 1. The limit of those numbers is 1. Their sum is not 1. Their sum is undefined. And no, you cannot say that at infinity their sum will be 1. If you consider the differences Dc, Db, Da - at infinity these will still not be zero even if these are close to zero. In fact, if you believe in infinitesimals, Dc, Db and Da would be the closest thing to what these might resemble. Of course there is no such thing as infinitesimal (not in non-standard analysis, surreal numbers or any other kind of number system which is a load of rubbish) This is the point and yes JIP, you have missed it. The definition you have for infinitesimal in Wikipedia is really amusing. But then so many of your articles are amusing. logamath1
I think I understand the terminology fairly well. 0.999... does not have to be 1 for the real number system to be logically consistent. There is an obsession that things might fall apart if these two numbers are not equal. As I have shown above, you can have numbers between 0.999... and 1 - this does not violate the Archimedean corollary. In fact the Archimedean property holds for all numbers whether they be real (or not) provided these are treated as approximations. Whether you like it or not, this is the way we have always dealt with real numbers - as approximations. The problem occurs with the interpretation of expressions such as 'finite sum' and 'limit of infinite sum'. These are not the same and should not be treated equivalently. You talk about infinitesimals without having a well-defined meaning. Infinitesimals are a figment of the imagination. Wiki's article on this is a joke (as are several others). In fact having 0.999... = 1 causes the real number system to fall apart. Why? Well, the real number system is ordered. There is an ordering amongst decimals too. This ordering exists so that we can make meaningful comparisons. If you compare a number that differs in the 15 zillionth digit of pi, how would you compare it? You would start from the most significant digit which is to the left of the radix point and proceed until you found the digit that differs. In the same way, our decimal system fails if we break this rule by setting 0.999... = 1 because we compare these in exactly the same way as we would compare pi. I put it to you that whoever came up with the idea that 0.999.. = 1 was a fool. logamath1
You have stated a corollary of the Archimedean property. h is not an infinitesimal but a very small number. An infinitesimal does not exist. logamath1
HeroicJay: It's quite incredible how you agree that infinitesimals are a figment of the imagination and then you call the difference 1 - 0.999... an infinitesimal - isn't it? You are very confused I would say. Logamath1
1-0.999... is undefined HeroicJay. It's as simple as this. We can only approximate its value just like all the other repeating decimal numbers in the decimal system. What is so hard about this to you? Best we can do in this case is to approximate it by 0. However, this does not mean it is exactly equal to 0 but is good enough. logamath1
Not so Melchoir: 1-0.999... would be a very small real number, not an infinitesimal because an infinitesimal does not exist! Therefore 0.999... is less than 1. Do you follow? logamath1
Wrong. There are very small real numbers for if there were not:
logamath1
They do require approximations because you cannot do arithmetic on any numbers that are not finitely represented. This has nothing to do with science but everything with pure math. logamath1
I am only going to deal with Melchoir. It's too difficult because of edit conflicts. Sorry HeroicJay. Try Decimal construction. Otherwise you choose. logamath1
It's not a lame excuse. Don't take anything personally. Besides, it's one against the whole of Wikipedia and most of the Western World's Phds. Cut me some slack please. logamath1
Have fun. I will check in tomorrow or later tonight if I get a chance. Sorry I gotta leave you. logamath1
You could have fooled anyone else but you don't fool me. How small is very small? Please Melchoir, this is about the worst argument you have presented thus far. It's total nonsense! By the way, where is that boy Jitse Niessen? Do you see how HeroicJay is provoking me? He needs to be reprimanded!! Who the hell is he to ask me why I argue?! Why, I would take this up with God if I had to until the very last breath!! I am very tempted now to tell this creep off. Let's see if Niessen will reprimand him or not... logamath1
It seems "very small real number" is a confusing term as well - let me rewrite the whole thing fully to avoid and argument about whether it's reasonable to use that term:
JPD 10:58, 2 December 2005 (UTC)
It's still wrong because:
1) You are referring to a positive real number less than all positive real numbers. There is no such number. How do you know such a number exists? 2) Having assumed that there is such a number, you then state that there is no such number.
Have you thought at all about this? Are you just another brainless PHd? Melchoir's argument even though it is flawed, is far superior to yours. Come on, who do you think you are fooling? logamath1
If you think Melchoir is saying anything different, you obviously don't understand his point, which would explain why you think it is flawed. JPD ( talk) 13:14, 2 December 2005 (UTC)
I am saying that 0.999... is a real number less than 1 such that at least one real number 1-0.999... exists. Although I can't find this number because I believe 0.999... is irrational, does not mean it does not exist. You talk about a reductio ad absurdum proof but there is no such proof. Let 1-0.999... = x and let me show you how this reduces to the absurd if x is equal to 0. Here is a reductio ad absurdum proof:
Let 1 = 0.999... Now add .1 to both sides:
1.1 = 1.0999....
which is clearly false, therefore 0.999... cannot equal 1 and thus it must be less than 1. Don't try to tell me that .1 = 0.0999... - it is not. When we compare numbers in the decimal system, we begin the comparison with the most significant digit.
logamath1
Yes, it is clearly false. Such an x is possible always. Give me any 10^-i and I will simply give you 10^-(i+1). Of course it is possible for any natural number i. And comparing numbers in the decimal system works the same for all numbers including numbers that end in 0.999.... logamath1
I do not have to provide a reason for why decimal numbers are compared this way - this is how the decimal system was designed. You are making a mistake by assuming that x=1-0.999... should be a single positive real number less than 10-i for any i. This is faulty logic. I am making no assumptions about the value of x except that it exists. I do not call it a very small real number and I do not call it an infinitesimal. I only call it a real number that is greater than 0. Finally I did not say at any time that such a positive real number does not exist. Look JDP, the fact that you cannot find the exact value of pi in any radix system does not mean that pi is not a finite value. Yet do you ask: Is there an x such that x-pi = 0? Of course not. Please don't tell me x=pi (in fact it is pi but not by your reasoning. We reach this conclusion because we know pi exists but do not know its exact value just as we don't know the exact value of 0.999... or the value of 1-0.999...) because we don't know the exact value of pi. What you have with 0.999... is very similar: x-0.999... = 0 means x=0.999... and not that x=1. logamath1
Melchoir is completely right. Your comments about pi are a sidetrack, so I will only answer your comments about what you call my faulty logic. The only assumptions about x that I am making are parts of the definition of the real numbers. If x=1-0.999... it must be a single real number, since the reals are a field. If x>10-i for any natural number i, then 0.999... = 1-x < 1-10-i = 0.99..99 with i 9s, which is a contradiction. This is not an assumption, it is simple algebra. Therefore x<10-i for all natural numbers i. I did all this without saying anything about the value of x that didn't follow from it's definition. I didn't call it infinitesimal, or very small. I did, however, show that it is smaller than any positive rational number. Because there are no positive real numbers that are smaller than every positive rational number, x must be less than or equal to 0. If you are dealing with a different number system, like the decimal numbers mentioned above, then you get different results, but it must be 0 in the reals. JPD ( talk) 18:45, 2 December 2005 (UTC)
You are either sincerely mistaken or you are lying. You challenged logamath1 to suggest a construct for the reals. He suggested the Decimal construct. You then tried to show this is true and were not able to do so. 70.110.92.137 19:57, 2 December 2005 (UTC)
Forget about Melchoir and let's see what you are saying: Your logic is strange. You first say that x must be a real number. I agree with you on this. Then you say, "If x > 10^-i for any natural i, then 0.999... = 1 - x < 1-10^-i ..." This is rubbish so I'll stop here for no one will start off an argument by saying that x > 10^-i. You must be a newbie. 70.110.92.137 20:05, 2 December 2005 (UTC)
Please show me how you can do any accurate arithmetic with pi, e or sqrt(2). It is all approximate whether real or decimal and we are talking about decimals here. 70.110.92.137 20:05, 2 December 2005 (UTC)
HeroicJay: As long as you are adding fractions you can obtain accurate answers because these are rational (provided you leave your answers in fraction form). The minute you begin dealing with decimals, all bets are off. As for the part about PhD, I leave this to logamath1 - he can probably answer this better than I. 70.110.92.137 20:05, 2 December 2005 (UTC)
There will be no more edits. I have wasted my time. 71.248.138.198 02:56, 3 December 2005 (UTC)
Heroicjay: I have never met a more obstinate fool than you. Indeed you are duller than most and I see no point in carrying on any exchanges with you.
Wikipedia: In addition to the nonsense you write in this article, there is more nonsense in other math articles. Actually your math articles are referred to as the WikiMath joke articles. Your infinitesimal article states: "...an infinitesimal, or infinitely small number, is a number that is greater in absolute value than zero yet smaller than any positive real number.." - This is such incoherent and illogical garbage for the first part implies it is a positive real number and the second part that it is zero. How can it be both zero and a small positive real number? I leave you to break your stupid heads over this. 71.248.138.198 02:56, 3 December 2005 (UTC)
"There will be no more edits. I have wasted my time. 71.248.138.198 02:56, 3 December 2005 (UTC)" You said you'd stop. Please do, you make no attempt to reach consensus, merely to push your viewpoint time and again on people who aren't going to accept it. You've now moved on to personal attacks, which really have no place on wikipedia. 81.86.207.192 13:04, 3 December 2005 (UTC)
Melchoir writes: "...any careful definition of the real numbers based on decimal expansions SETS 0.999... equal to 1." - Melchoir
Not only is this an outright lie but it is clearly demonstrated to be false:
1) The Indians did not use a decimal point. This was an invention of Scotland.
The notation we use today first appeared in a book called "Descriptio" by the Edinburgh mathematician, John Napier, Laird of Merchiston, in the 1616. He used a decimal point to separate the whole number part from the decimal number part. Known as 'Marvellous Merchiston", he published many other treatises including "Mirifici logarithmorum" (1614) and Rabdologia (1615) on systems of arithmetic using calculation aids known as Napiers Bones.
2) Fractions were invented long before the decimal point.
In the 10th cent. A.D. Arab mathematicians extended the decimal numeral system to include fractions.
Melchoir will try to justify himself by using terminology such as any careful definition of the real numbers - this is entirely subjective and is open to debate.
I asked who came up with the idea that 0.999... = 1. No one on Wiki's board of idiots was able to answer this. Why? The answer is simple: they are all products of today's institutions that are run by a bunch of baboons masquerading as Phds. My apologies to the few Phds who really earned their diplomas but the majority are to be regarded as ignoramuses who are spineless and incapable of thinking on their own.
The definition of the real numbers is not open to debate. Mathematics is useless as a language unless we agree on the meaning of its jargon, and "real number" is too entrenched a concept to change, even if it were a good idea, and it really isn't.
81.86.207.192 158.35.225.228, I wonder if you think the rigor of modern mathematics is somehow a challenge to your creativity. It's a fine intellectual exercise to dream up number systems in which the analogues of true statements for the real numbers turn out to be false. One might even make a living doing this. It doesn't make me a liar.
As the self-appointed dictator of this talk page, I declare that it is closed to personal attacks. Comments insulting other users or the Establishment will be erased. The purpose of the talk page is to discuss improvements to the article. If you have a suggestion, let's hear it. If you're confused but willing to learn, ask a question. If you're here to whine about PhDs, go away. Melchoir 19:12, 5 December 2005 (UTC)
You have not provided one proof or argument that shows why the real number system must have 0.999... = 1. It is easy to strike out what you don't agree with but much harder to admit you are wrong.
...The limit of the series is not the sum of the series and a definition does not make it true. There are real numbers between 0.999... and 1 as has been demonstrated above... There has been a solid proof by induction that has also been rejected... --anon
"The limit of a a sum of an infinite number of terms is an infinite summation, so .999... is the limit is of the sequence .9, .99, .999... and the sum of the series .9 + .09 + .009 + ... (the result is the same either way.)" ...I will let the readers decide on this. The limit of an infinite number of terms IS NOT AN infinite summation. An infinite sum is impossible. And no, you cannot define it this way because it does not make sense... You saying it one million times or claiming that it is true does not make it true. A Definition is useless unless it makes sense AND it is TRUE. —Preceding unsigned comment added by 70.110.81.224 ( talk • contribs) 2005 December 6
Oh please!!! If you are going to restore points, make sure you restore everything... YOu have only restored what makes you look good. The fact that you have final editing say displays clearly that this is not a peoples encyclopedia... Wiki is a view of the world according to YOU... —Preceding unsigned comment added by 70.110.81.224 ( talk • contribs) 2005 December 6
...You cannot show me one reputable source that states 0.999... = 1... Show me an encyclopedia Brittanica or something that has been around for a long time that states this. I put it to you that ... 0.999... = 1 has not been around that long. It was probably conceived in the early or late twentieth century... ...This article should be deleted. It is not encyclopedic... —Preceding unsigned comment added by 70.110.81.224 ( talk • contribs) 2005 December 6
This is exactly the problem: most of the world do not think of 0.999... as the limit of 9/10 + 9/100 + 9/1000 + ... They think of it as an infinite sum or a number. If you talk of the limit of the partial sums of (9/10;9/100;9/1000;...), then its limit is 1. The same principle applies to 1/3, 1/6 and any other fraction that cannot be represented exactly in base 10. 0.333... is not equal to 1/3. 0.333... is an approximation for 1/3 in base 10. Although it cannot be represented exactly in base 10, it can be represented exactly in many other bases, e.g. 1/3 = 0.1 (base 3) 1/3 = 0.2 (base 6) and so on. Now a 1/4 in base 3 is approximated by 0.020202... It is not equal to 0.020202... even though the limit of 0.020202... is 1/4. I do not have a bit of a point, I have a major point. Terminology is extremely important and infinite sum and limit of an infinite sum are not the same thing! Lastly, 0.999... was never defined as a limit, neither were the real numbers defined as a limit until Weierstrass, Cauchy and others decided to 'rigourize' mathematics. In many respects they inhibited the progress in understanding the real numbers. Real analysis is filled with inaccuracies and contradictions as a result. —Preceding unsigned comment added by 158.35.225.228 ( talk • contribs) 18:09, 2005 December 6
0.999... is less than 1 for all the reasons you say it is equal to 1.
It makes perfect sense. You are the one who is having difficulty understanding any of this.
I have just found the most wonderful paper:
Conflicts in the Learning of Real Numbers and Limits by D. O. Tall & R. L. E. Schwarzenberger, University of Warwick Published in Mathematics Teaching, 82, 44–49 (1978). pdf webpage
It's 13 pages long. Whoever finds this sentence in the first place should go skim it; it's got some gems.
I propose to rewrite the article with Tall as our motivation and primary reference. In fact, I would have proposed this before, but without Tall it really would have been Original Research. Additional references would be great, including a more modern one, but it's a good start.
The article currently does not address many common misconceptions; I think we should mention them all and explain why they are wrong, although preferably after the proofs. I want to make it clear that I'm not backing down under fire and "teaching the controversy". There is no controversy, and the article must make that plain. There is, however, a whole lot of confusion that can be productively cleared up in the article itself. In some sense, this effort would be similar to a page on an urban myth.
My questions are:
Melchoir 21:16, 6 December 2005 (UTC)
Also, this earlier paper by the same author is mostly generalized nonsense, but it gets interesting and relevant on page 10. Melchoir 22:12, 6 December 2005 (UTC)
Yes, the papers I link to contain speculation, but they are published, and they contain quotes by students displaying some misconceptions. If we address those misconceptions, such a source is necessary to avoid speculation on our part!
That Hackenstrings bit is very interesting. I've seen nimbers before, but not those. As for the citations currently in the article, the first two don't address any misconceptions, and the third is so confused that if it's preserved it should only be as a bad example! I think we can do a better job.
As for rewriting, I must apologize for what was honestly just a poor choice of words. As a mathematician, I think your proofs are the strongest point of the article, and my proposal does not require that we alter them. For now, I simply want to tack on a section at the end exploring the misconceptions. Nor, of course, do I propose to reproduce Tall, who avoids the structure of the real numbers altogether. His article could be useful mainly in that
If he doesn't address them well enough, we can do better than him, too.
Of course we can't guess what the reader is thinking. The reader, in fact, might agree with us; surely teachers will read our article, not just students! But the encyclopedic voice isn't really appropriate to connect with readers on a personal level anyway. I don't propose that we attempt to open a dialogue with the reader, but that we describe what students are known to think. With references such as Tall, we don't have to guess on that.
As for there being better articles to work on, you're right. Melchoir 04:14, 7 December 2005 (UTC)
The article is neither interesting, nor relative. It has nothing about proofs and talks of infinitesimals as if these were widely understood. There is no such thing as an infinitesimal.... If an infinitesimal is greater than zero but less than every positive number, there can only be one of it.... It would be a good idea to delete this article and if you really want to talk about comparing 0.999... with 1, then you should make it clear that 0.999... is less than 1. Tall refers to some questions posed to students and discusses their response. He tries to infer that students answered incorrectly because they were unsure of the denotation of recurring. ...this is taught in elementary school and those same students claimed that they knew the definition of a limit.... —Preceding unsigned comment added by 68.238.99.174 ( talk • contribs) 02:26, 2005 December 7
In mathematics there is no such thing is correct if such a thing is not properly defined. And you can't dream up fancy words that mean nothing and develop theories out of these. --anon
Care to explain?... --anon
![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 | Archive 3 | Archive 4 | Archive 5 |
DO NOT EDIT OR POST REPLIES TO THIS PAGE. THIS PAGE IS AN ARCHIVE.
This archive page covers approximately the dates between 2005-11-16 and 2005-12-07.
Post replies to the main talk page, copying or summarizing the section you are replying to if necessary.
Please add new archivals to Talk:Proof that 0.999... equals 1/Archive03. (See Wikipedia:How to archive a talk page.)
How can he be wrong? He is referring to sequences with positive sums. What are you talking about? He is correct in stating it will never exceed 1. In fact it will never reach 1 either. As for computing infinite sums, I also, know of no formula. All I learned in high school and university is how to compute the limit of an infinite sum. The two are quite different. Finally, the fact that you can take the sum as close as you want to 1 does not mean it is equal to 1. It means that you can take it as close as you like but you will never reach 1. Just sit down and start adding up the terms and I guarantee you that you will sum until your last breath and still you will not have reached 1. Someone can continue to sum after you and he too will die summing the terms because the sum will always be less than 1. Philosophical grounds - hmmm? No, I think he is just using simple high school math. —Preceding unsigned comment added by 68.238.99.105 ( talk • contribs) 00:12, 18 October 2005
Your logic is 'impeccable': would it really equal 3? You appear to be very confused. It cannot equal whatever you like. It will never equal, nor exceed 1 - that is the assertion. I am talking about the "limit of an infinite sum", not "limit of the sequence of finite sums". The formula he quoted is used in determining whether an infinite sum has an upper bound. There is no assertion that it is equal to this upper bound. Your assertion is plain wrong: there is a very easy way to check yourself - start adding up the terms and I can gaurantee you, you will always have a sum that is less than 1. Please don't tell me you are dealing with a finite sum because then your assertion that the infinite sum is 1 is absolute nonsense! You may be confusing yourself with the fact that the terms are getting closer and closer to zero (Cauchy sequence). This does not mean that any term will ever be zero. —Preceding unsigned comment added by 68.238.97.2 ( talk • contribs) 10:49, 18 October 2005
I think you know exactly what I mean when I talk about Cauchy sequence: the distance between the terms is getting closer to zero. That's the definition of Cauchy sequence, i.e. Lt d(p,q) = 0 as min(p,q) approaches infinity. You write: "The value of an infinite sum IS the limit of the sequence of finite partial sums." This is not true. An infinite sum IS *indeterminate*. Get that? The formula used in this article to prove that 0.999... equals 1 proves exactly the *opposite*, i.e. that 0.999... does not equal 1. All the formula shows is that the limit of any of the partial sums of this sequence is 1. To say this is the infinite sum, shows extreme ignorance. The limit of any partial sum is *not* the infinite sum. You taught at Mit? So what, do you think I ought to bow down and be intimidated? You are wrong. You have been taught wrong too. This article is non-sense. The fact that you can write what you do, displays a fundamental lack of understanding. This non-truth of 0.999... = 1 has taken hold because most people don't understand that 0.999... is *not* a rational number. Neither is 0.333... a rational number. Partial sums from these sequences are used to approximate 1 and 1/3 respectively. I can understand using 0.333 to approximate 1/3 in base 10 (only because it can't be represented finitely in base 10) but cannot understand why 0.999 should be used to approximate 1 which has an *exact* representation. Don't you think it's about time you started thinking for yourself? I know how you will respond: You will say a rational number is any number that can be expressed as a/b where a and b are integers (b not 0). This definition of rational number is part of the problem. If a number cannot be represented *finitely* in a base that is well defined, then the number is not rational. Pi, e, sqrt(2), etc are irrational because there is no well defined base in which these can be represented finitely. You can't suggest that Pi, e, etc be respresented in their own base since these numbers cannot be completely determined. i.e. pi is not equal to 1.0 in base pi because the extent of pi is unknown. Similarly for e or any other irrational number. 15H51 18 October 2005 —Preceding unsigned comment added by 68.238.97.2 ( talk • contribs) 21:07, 18 October 2005
Not entirely true. You start with 0.333... and then you try to show that it can be expressed as a/b. Or you start with 0.999..., 3.14... or some other representation and then try to show it can be expressed as a/b. A number is rational if it can be expressed in the form a/b (b <>0) with a,b integers. I maintain this is insufficient, you also need to add that the representation must be *finite* in some radix form. If indeed 0.999... is rational (it's not), then so is pi since pi can be expressed in the form a/b (i.e. 3 + 1/10 + 4/100 + ...) But of course pi is not rational because there is no number system besides pi in which pi can be expressed finitely in radix form. In base pi, pi is *rational*, i.e. pi = 10 (i.e pi + 0 units). 0.999... cannot be expressed as a/b in any number system. Please don't tell me it's 1 or 1/1 - this assumes that it is equal to 1. You need to think really hard about this. 68.238.97.2
You have told me nothing I did not know in your first and second points. In fact, if you read my posts, you would see that I said this. Your third point is false and I pointed this out in my previous response. You cannot prove that 0.999... = 1 because you do not know the difference betwen the limit of an infinite sum and an infinite sum itself. In fact, 0.999... 'is not' equal to 1. You do not understand the formula used to show that the sum of an infinite sequence is bounded from above. Maybe you should sit down and think about it again? You have been unable to refute anything I have said and you have not even tried to understand it. If I am incorrect in stating that finite representation is 'required' for the definition of a rational number, then pi, e and sqrt(2) are all rational seeing these are the sum of their respective expansions. Frankly it has nothing to with semantics, only simple logic that even a ex-professor from MIT can't see or won't see?. 68.238.97.2
Talk about a lot of 'words'.... Could your rebuttal possibly be a little more abstract. You know you are wrong and just can't admit it. ... sour grapes? 68.238.97.2
Wrong again. Retired supermodel. More profitable and unlike teaching/(child minding), no fake power-trips: the runway is a 'real' power-trip. But don't quit your day job. If your posted photo is recent, I can't tell you won't make it. Sorry, don't mean to be rude, just realistic. 68.238.97.2
Did you mean 'can tell he won't make it' ? :-) He is probably still figuring out how to make 0.999... add up to 1. —Preceding unsigned comment added by 192.67.48.22 ( talk • contribs) 13:56, 20 October 2005 (UTC)
Yes, that should have read: "I can tell you won't make it." I see he has not responded to your rebuttal. Instead he chooses to be sarcastic and rude to a lady. Frankly, he skirts the rebuttals and tries to be cunning and humorous. 68.238.97.2
I am sorry but I have no idea who Paris Hilton is? Is he a mathematician? Wait, I can goolge it. Hopefully there aren't too many with that name. Now I know you love chatting but I really wish you would give this subject more thought. I can answer any questions you might have. You ought to be grateful for this. 68.238.97.2
You must be a non-mathematician because evidently you do not understand this at all. You do not understand geometric series or even what the difference is between an infinite sum and the limit of an infinite sum. You are not alone - most mathematicians don't understand this either. Hardy is a fine example. —Preceding unsigned comment added by 192.67.48.22 ( talk • contribs) 2005 November 9
You are obviously a fake. No one understands infinity (not as a concept or otherwise) and you don't have a clue of what you are talking about. You are a good example of the mindset erroneous thinkers have who believe that 0.999.. = 1. You have erred and contradicted yourself several times already: First you state that infinity cannot be regarded as a number, then you proceed to write that 0.999... can be regarded as 1 minus the smallest real number possible which *you* say is 1/infinity. How can you define the smallest real number in terms of a number that is not defined?! Contradiction. Next, you fail miserably with your child-logic: "But since infinity does no have a defined value, this becomes 0." How did you reach this conclusion?! You are way out of your league. Think carefully before you post again! 192.67.48.22
You are assuming that an infinitesimal posseses the property that |x|+|x|+.... < 1 no matter how many |x|s we sum. This is untrue and constitutes your first error. An infinitesimal cannot be quantified. Your second error is you decide to *call* your quantity 1-0.999... some 'x' - you cannot reach a contradiction on a false premise and then assume that your conclusion is true. 192.67.48.22
Your definition of infinitesimal is untrue and it is based on an incorrect assumption. You state that whatever the value of n is, the sum will never reach, nor exceed 1. Both are false. If what you say is true, then why do you not concede that the sum of 9/10^i (from 1 to n) < 1? You state that the sum of |x| (from i to n) < 1 but in the same breath you are trying to show that the sum of 9/10^i (from 1 to n) = 1 ?! You are very *confused* my friend. infinitesimal has never been properly defined. How can you quantify the number that is greater than zero yet less than every positive real number? You can give it a name, which we have: 'infinitesimal'. However, in every other respect, it is exactly like pi, e and sqrt(2), i.e. its full dimensions are unknown. To say that the reals posses no infinitesimals and then claim that pi, e and sqrt(2) (just some examples) is in itself a contradiction. Mathematicians shoot themselves in the head when they make statements such as: 'As small as you like' or 'As close you like'. How small? How close? I know mainstream thought is that the reals contain no infinitesimal. If the infinitesimal does not belong to the reals, then pi, e or any other number with similar properties does not belong to the reals also. This includes 0.999..., 0.333..., etc. 192.67.48.22
Talk about hand waving! Your proofs are all fine examples of hand-waving. Contrary to what you think, the definition you provide of infinitesimal is not used everywhere. How can I have a conception of a number field that does not include pi, e or sqrt(2)? If I did, it would be incomplete and thus erroneous for pi, e and sqrt(2) are all very *real* and finite. You state that the Archimedean property is a consequence of the LUB theorem. Actually, it's the other way round. I have provided sufficient proof that 0.999... is not equal to 1 on the pages you archived. Since you are making the statement that 0.999.. is equal to 1, the onus is on you to provide proof which so far you have been unable to do. As for your *proof* being interesting in that it does not use limits and infinite sums, I would more accurately say that it is not a proof at all but mere hand-waving. And what are *reputable sources* if they don't agree with Wikipedia's views?! 192.67.48.22
Really? So how is it that you understand that sum |x| (i to n) < 1 where x is infinitesimal (and you don't know your ear from your nose either, never mind what an infinitesimal is or is not; you are also unable to define it in any rational way) and you use this in your faulty proof to show that 0.999... = 1 ? Look, the fact that you have a master's or a PHd in Mathematics does not mean anything. You are in many ways more ignorant than someone without any qualification at all. I have not given you a definition for infinitesimal because it is a concept that makes no sense to me. If you cannot define any concept rationally, it is in fact *illogical* (any surprise?) and consequently rubbish that cannot be used to prove anything. The Archimedean property is well known. Your webpage says you have a master's and this is something that is taught in real analysis. Were you not required to take this course? Just google it for crying out loud and you will know what it is.
As for formal mathematical proofs: There is a proof in the archive and I'll state it again:
Sum (i to n as n approaches infinity) = (ar - ar^n)/(1-r) for |r| < 1
We cannot compute an infinite sum but we can investigate whether it has a limit or not. In the case of 9/10 + 9/100 + 9/1000 + ... it can be easily verified that this limit is 1. This states that even if we could sum this to infinity, its value would never reach, nor exceed 1. Thus it is *clearly* evident that 0.999... < 1.
All proofs that try to show it is equal are faulty but the one used most convincingly is a consequence of the Archimedean property:
The law of trichotomy applies *only* to finitely represented numbers, so you can't use an algebraic process that leads to 0.999... not less than 1 and 0.999... not greater than 1 implies 0.999... = 1. 0.999... is not a finitely represented number. Any arithmetic on such a number can only be an approximation (like pi, e, sqrt(2) etc). And yes, there should be a page called "Proof that 0.999... < 1" because contrary to what you think, it is not generally agreed that 0.999... = 1. Except perhaps in the case of the fools who run Wikipedia? 192.67.48.22
My word but you do love yourself, don't you? And you sure know how to use this system. If I knew it half as well as you did, I would draw some nice sigmas, infinity symbols and why, of course beautiful epsilons and deltas to make every Phd green with envy. Now, there is no handwaving in anything I wrote. It is very clear that the sum on the lhs will never exceed 1:
If we split up the quotient as follows: a/(1-r) - ar^n/(1-r) the first term is independent of n and its value is 1. The second term becomes very small (and using Weierstrass's faulty logic - 'as small as you like' but always greater than *zero*). Thus we have 1 - s where s is some value greater than zero. This being the case, when we consider the difference, we always have a value that is *less than 1*. This is very *clear*. Got it? Hey, if you don't get it now, you must be thicker than I thought. Please don't tell me this is not mathematical or robust enough or else you are a disgrace to all the institutions of learning you have ever attended. Look, when I use words like *clear* and phrases like *by definition*, I do not use these in the same ignorant way as most Phds do. So relax. Don't build a brick wall around everyone else when you feel it necessary to do this for yourself. 192.67.48.22
Firstly, I am not attacking you or anyone else and your psychoanalysis is deeply in error just as is your mathematics. Your above formula is incorrect: It's not (9/10 x 10^-n)/(1-.1) but rather (9/10 + 10^-n)/(1-.1). While you are enjoying Latex so much, you may as well do the job right. Okay, so you made a typo. I'll forgive you for this. Now let's move on. You say there is no natural number s.t 0.999... = sum (i to n) 9/10^i Well aside from stating the obvious, what are you trying to say? My proof considers what happens to the difference as n becomes infinitely large. There is nothing strange about this - it's used in limits and calculus and many other branches of mathematics. Regarding my proof: it is very *accurate* and *valid*. The problem is not with my proof but with your *understanding*. You are very confused. You have not answered my question:
You state that sum |x| (i to n) < 1 where x is infinitesimal (yet you are unable to define infinitesimal in any rational way) and you use this in your faulty proof to show that 0.999... = 1 ?
While you are trying to answer this, let me pose some more questions to you: If the real number system has 'holes' (as you claim it does), then how can you use epsilon-delta proofs at all? What does 'as small as you like' and 'as close as you like' mean? How small is small and how close is close?
This is true handwaving mathematics that has been taught the last 100 years. Real analysis is mostly a load of rubbish. Unfortunately you are the product of Weierstrass' ideas and logic that have some serious flaws. In answer to your question: I know the Archimedean principle the same way as it is published on planet math. Finally, please stop regurgitating what they taught you in your real analysis and honor's class and start thinking for yourself. This will be the best thing you can ever do. 192.67.48.22
Your formulas keep changing so I will just ignore these.
Unlike you and Weierstrass, I will not use epsilon-delta arguments because these imply the existence of infinitesimals. You are right about one thing: I am arguing that since the difference >0 for all n, then it is also true that the sum < 1. And yes, it is very *clear*. You claim it is false and I challenge you to find a counter example! So far you have only been able to regurtitate rubbish. I passed my real analysis course a long time ago. It was garbage then and it's still garbage now. My professors were ignorant then and today I find them to be even more ignorant. And saying that you are ignorant is not *ad hominem*. Once again my friend, you are trying to psychoanalyze me. I suggest you stick to proving the math and forget about everything else.
So here's your first exercise: find me an 'n' for which my claim is false and then I will believe you. No more BS (and MS and PHd) please. 192.67.48.22
Nonsense. You made a rigid statement that what I *proved* is false. You then contradicted yourself by stating that you have no objection to the lhs and continue to talk screeds about *proofs*. You do not challenge the lhs of this equation? If you do not challenge the lhs, you cannot challenge the rhs and yes, you cannot challenge the implication either unless you able to provide a counterexample. So far, you have displayed exceptional ability with Latex but your arguments lack any sound proof or direction. You are *clearly* lost and are only deceiving yourself. You have *not* said anything *relevant* or *logical*, much less *proved* anything. Your diversion tactics show that you are trying to save face. Once again, the *implication* is *clearly* that 0.999... < 1. The nonsense you wrote about 0.999... < 0.999... does not deserve a response. I have defined 0.999... *clearly*: Once again: 0.999... = 9/10 + 9/100 + 9/1000 + .... It is an infinite sum. Now you do not know the difference between an *infinite sum* and the *limit of an infinite sum*. I am going to tell you these things are *not* the same: not by *definition* or *otherwise* - I leave this as an exercise for you. The Archimedean property states that if y is any real number, then a number p exists such that y < p. This implies that p is an *upper bound*. It does not take too much intelligence to figure out where this leads. The rest is an exercise for you to show how it leads to the definition of some least upper bound. If you have an MS in mathematics, this is something you ought to be able to prove on your own. Finally, forget about the Archimedes principle and stop trying to coerce me into giving you a definition of a real number system with a different ordering. Please stay with the subject. You will not earn any points for a dissertation on irrelevant topics. 192.67.48.22
You are evidence that an Ms in mathematics is absolutely worthless. You cannot claim that the lhs implies 0.999... = 1 if the lhs specifically states 0.999... < 1. How do you arrive at this logic?! I don't need to challenge your rhs - it is irrelevant and gibberish that only you seem to understand. How can I challenge something that is incorrect? If you provided a correct statement and then drew an implication from this, I would be able to respond sensibly. But I can't respond to nonsense. You need to learn what 'implication' means. I am stating that you cannot find the result of an infinite sum - you can only investigate what its limiting value is (if it has one). This is a clear distinction between a infinite sum and the limit of an infinite sum. Your example that 0.999.. < 0.999... is *not a counterexample* but rather an example of the nonsense you are writing.
LUB: This is the smallest of all the upper bounds a number can have. In mathematics, if S is a set of upper bounds then there exists an element m of S such that m is less than or equal to all the other elements of S.
The law of trichotomy (LOT) applies only to finitely represented numbers: What this means is that if you have a representation of any number in any radix form that is infinite, LOT no longer applies. If this is untrue, *all* your arguments fail without any further support because 0.999... and 1 are two different numbers and yet you claim they are equal. Please don't tell me they are different representations of the same number. They are *not*. Radix systems are designed for *unique* representation. You were not paying attention in primary school when they taught you about 'tens and units'. Perhaps you should go back to primary school and relearn what you seem to have forgotten.
Oh, and if you are trying to put words in my mouth, then at least you ought to try and do it right. I would say: and not what you stated. To say that it implies b < c is nonsense. It is already given that b < c. Boy, you are a sorry MS! 192.67.48.22
HeroicJay: You should add lack of reading comprehension to your list of setbacks: *I* said that the law of trichotomy applies to numbers with a *finite representation*. Why don't you allow Rasmus to defend himself? I am sure he does not need your help yet. 192.67.48.22
192.67.48.22: Obviously your manners are as bad as your math skills. This is not a private forum and it it quite ok for HeroicJay to join the fun. If this is the way you normally behave when confronted with people who argue against you, it might explain your conviction that your arguments are true: Everybody who have attempted to engage in dialogue with you have become disgusted by your lack of manners and ended the conversation, and you have mistakenly taken that to be an admissal of defeat.
HeroicJay: Welcome to the fun! I apologize in advance for repeating some of your arguments, I will probably just go through 192.67.48.22's posting and reply to his points one by one. (And everybody else: let us know if we should take this someplace else. I realize we are way off-topic here).
Back to 192.67.48.22: You claim that "the lhs specifically states 0.999... < 1". It does not. Your implication could be written in english like this :"(all finite representation numbers of the form 0.99...9 are strictly less than 1)=>(the infinite representation number 0.999... is strictly less than 1)". By any reasonable definition, 0.999... is not equal to any finite representation number.
You claim to use the rule: . Well, first of all: what you need is a rule that concludes that b < c (here , and c=1. Note that there is no n, so that .) Secondly, even disregarding that, this form of implication would be what is called begging the question: your premises include the conclusion (namely that ). Note that the rule I wrote, did not have this problem, since I only assumed that not (you might have misread that, judging from your comments).
As for your LUB-paragraph: you simply seem to invoke the greatest lower bound-rule on the set of upper bounds. Since the existence of LUB is equivalent to the existence of GLBs in a field, you are not getting closer to proving that the Archimedean-property implies the LUB-property. You are also completely ignoring the fact the the rational numbers have the Archimedean property but not the LUB property.
If you maintain that it is not true that for any real numbers x,y, either x<y, x=y or x>y; you are not talking about the same numbers as the rest of us. If that is the case, any discourse with you is pointless. (Though I wonder if you realize, that if you do not allow for the law of trichotomy for 0.999..., you cannot conclude that 0.999...<1 => 0.999 ≠ 1 ?)
Your handwaving about the purpose of radix-systems is pointless. Obviously radix-representations aren't unique since 0.1=0.10 . If we can accept these two representations of the same number, we should have no problem with accepting that 0.999...=1, if it is needed to make the system consistent with the underlying math. And as for your comment about primary school: perhaps you are trying to do primary school-math, the rest of us aren't. Once infinity is invoked, things gets a bit more complex and we can't rely on our primary school instincts any more. Rasmus (talk) 21:27, 21 November 2005 (UTC)
HeroicJay: Am sorry I hurt your feelings. However, you have not stated anything that I don't know and you ought to realize that simply commenting without relevance to the topic at hand does not help anyone. And, you were *riling* me with your provocative comments on my use of *asterisks*. I use these for emphasis because many people miss the main point of what is being written. If it annoys you so much, perhaps you should have stayed out of the fray, no? Rasmus has misrepresented so many facts, got so many things wrong and still adamantly insists he is correct. I have enough on my hands dealing with Rasmus without having to respond to you.
Back to Rasmus: Your entire previous paragraph is nonsense and all the accusations you make are baseless. I am taking you to task: find me one n for which the formula is untrue and I will believe you. BTW: Don't talk about LUB and Archimedes - you don't have the remotest idea what these are about and your writing shows that you lack understanding. So, let's deal with one thing at a time. Provide proof that the sum (i=1 to n) 9/10^i < 1 does not imply .999... < 1. The inequality you introduced is baseless and irrelevant. It demonstrates clearly how confused and deceptive you are. 192.67.48.22
You have not provided any proof and the discussion did not start out with me disproving your 'proof'. You stated that and asked me if this is what I was claiming to which I responded in the affirmative. And yes, I am claiming that Show me *one* example where this is not true. A very simple proof is one by induction. Do you need me to help you with this? How did you obtain an MS in mathematics?
Simple Proof by induction:
We have that k is true: sum (i to k) a_i < 1
Is k+1 true? Yes since a_k+1 + sum (i to k) a_i < 1 because no carry is possible.
Thus it follows that we can choose any k and always find that k+1 is true. Q.E.D.
192.67.48.22
You are in error. The rhs does not become what you state. Your reasoning is completely in error. How do you arrive at this? You cannot assume 1 = sum (i to infinity) 9/10^i and then use it in your argument! This proves nothing!! 192.67.48.22
No man, you got it wrong again. You *are* assuming that "1 = sum (i to infinity) 9/10^i" whichever way you look at it. You can't just *substitute* what you like either. As for my proof by induction - it is perfectly valid. Infinity does not have to be a *natural number*, nor does it have any relevance. Mathematical induction works regardless of what value n might assume. You are full of BS. Just eat humble pie and admit you wrong. Anyone with any sense reading this will be laughing at you. It's easy to tell you are a fake by some catch-phrases you use like "Invoke infinity" and by virtue of the fact that you waddle a lot about a lot of irrelevant side topics. Definition of Mathematical Induction: The truth of an *infinite* sequence of propositions P_i for i=1,...,infinity is established if (1) P_1 is true, and (2) P_k => P_k+1 for all k. This is the principle of mathematical induction. 192.67.48.22
No. You cannot claim whatever you want and then use it to prove your point. Whatever you start off claiming must be *true*. So when you make a choice, it must be logical (and true - surprise?). You keep harping on the fact that infinity is not a natural number. So what? As I explained, the theory of mathematical induction deals with the truth of an *infinite* set of propositions. I am not concerned with *infinity* itself but rather about my propositions. I do not even try to deal with infinity. Please quit making loose statements such as "the natural numbers are group". Stay with the subject. HeroicJay makes some statements about the concept of infinity in higher math: even in *higher* math, we do not evaluate what happens at infinity but make assumptions regarding limiting values (if these exist and we know they do not always exist). Look Rasmus, you are wrong and don't have the balls to admit it. Male chauvinist pride? 192.67.48.22
Look cute boy, you may be the hottest mathematician but you don't appear to be the brightest. Please get off my case. I am not attacking anyone. Rasmus and others have been very disdainful to me, yet you have not reprimanded any of them. And I am the only one *sticking* to the math. Finally if anyone is going to take things too personally, he/she should refrain from the discussion. At the end of the day, nothing on this page is going to cause me to lose any sleep. You should be thankful that I am taking some of my valuable time to improve the content of your site. I believe that knowledge should not only be *free* but also *correct*. 192.67.48.22
Why on earth do you have a talk page then? To see who can cite the most references even if idiots are the originators? How does publishing or writing a paper make knowledge more reliable? Thousands of papers have been written that contain junk worth less than the paper it's written on. There is no orginal research here, only logical rebuttal of past ideas that are incorrect and need to be changed. Given that there are many different points of view on this subject, you ought to represent both points of view equally well. To publish only what some academics think, is biased, morally wrong and does nothing to improve the quality of Wikipedia. It only propagates false knowledge and ideas. If this is your policy, then you seriously need to change it! 158.35.225.229 16:47, 23 November 2005 (UTC)
I repeat myself: you have provided no counter-example at any time. You cannot start out by choosing a_i and x as you *please*. I explained this to you. Your choice has to be *true*. You are assuming that sum (i to n) 9/10^i < sum (i to infinity) 9/10^i. This is correct. However, you cannot proceed to argue that sum (i to infinity) 9/10^i < sum (i to infinity) 9/10^i because these are exactly the same thing! My, but you are slow. That (N,+) is a group is *irrelevant*. The last part of your statement is yet again nonsense: You write: For all k is an element of (N union infinity) .... - N is *infinite*. What nonsense are you writing? And what exactly does it mean to have (N union infinity) ? Rasmus, just admit you are wrong. Deal with it and move on. You are trying to save face but everyone who reads this knows you are wrong. You are only making a fool of yourself each time you respond the way you have. 192.67.48.22
Actually it is you who are not applying the principle of induction correctly. Firstly, you are in grave error to write P(infinity) - there is no such thing and we never consider any such proposition. You need to learn what induction really means. I have stated it correctly and you have failed to understand it. I have *proved* conclusively using induction that 0.999... < 1 for any value of n. I do not need to consider P(infinity) - in fact this is impossible to consider in any scenario. As for counterexample - I know very well what it means - even before you were born. You have failed to provide any counterexample and it is evident you don't know what it is. If Newton said that 0.999... = 1, I would call him a fool. How much more I would call less intelligent men like you a fool? You are regurgitating (incoherently) what was brainwashed into your head at whatever institution you attended. So far all you have done is ramble on about irrelevant topics and have not provided a shred of logic to show that 0.999... = 1. And you have been unable to refute my proof in any way. Finally, if my induction is incorrect here, then all mathematical induction is *incorrect* and it cannot be used to prove the validity of any proposition. 192.67.48.22
(This has been detatched from what it replied to, so I'm removing the indentation.) Even if Mr. 192 isn't swayed by that, I now know what he meant by "Rasmus added that" when I went out of my way to show that his a < b < c assertion was meaningless with the limit (yes, it's true when you type . He apparently misunderstood the inequality sign.) I had thought he was still going on about the trichotomy thing. -- HeroicJay 21:47, 21 November 2005 (UTC)
Heh, I was following the conversation and was about to make a similar point to the above by Rasmus until I saw it. In plainer English, while it is true that .9999...9 (with a finite number of nines) is less than one, no matter how large the finite number of nines, it also happens to be less than .999... (with infinite nines), so it doesn't really prove that .999... is unequal to 1. That argument is (loosely) similar to this: 1 < 4, 1 + 1 < 4, 1 + 1 + 1 < 4; thus, 1 + 1 + 1 + 1 < 4. Q.E.D. It's much more obvious that my "argument" is silly, but it follows the same principle: just because a bunch of numbers smaller than .999... are also smaller than 1 doesn't make .999... smaller than 1. (Technically, a better parallel would be , which is absurd to anyone who understands the terminology, but that uses infinite sums and infinity and Mr. 192 whines about those.) -- HeroicJay 01:26, 20 November 2005 (UTC)
... Then we can see that therefore . However, if , then we know that as then , and then . It must be then that as is defined to be zero. But the basis of my proof is that 4.000...1 (4, dot, an infinite number of 0s, 1) = 4. And then 5 - 4.000...1 = 0.999... = 5 - 4 = 1. Therefore 0.999... = 1. If you do not agree with this proof (except for very stupid errors I have made, such as not fixing my LaTeX properly), then, unfortunately, you must disprove all of calculus as it is all based upon it. x42bn6 Talk 07:10, 23 November 2005 (UTC)
Not so. You have not proved anything, nor said anything new. You make statements such as 10^(-infinity) is *defined to be zero*. Nothing could be farther from the truth. Zero is defined as *zero*, nothing else. I can see you are also a good product of the establishment. You need to start thinking for yourself - this is all I can say to you. See page on Mathematical analysis for an example of how you can find derivatives without even considering what an infinitesimal is. 192.
Passed by and changed the intro. No article on Wikipedia is a place of discussion, nor should any article strive to "prove" anything. Articles are here to present the facts in as unbiased a manner as possible. This article, if it continues to exist, should give information on known proofs of the concept. Nothing more, nothing less. Discussion is for talk pages. -- WikidSmaht ( talk) 10:55, 20 November 2005 (UTC)
I've been reading the above discussion for a while, and it seems to me that our anonymous friend thinks that:
Now I've been thinking about this. Suppose we calculate the intersection of closed intervals from 0 to the inverses of natural numbers. (A closed interval means one that includes its endpoints.) Specifically, we investigate the intersection of all where . Now it's clear that if n is a natural number, its inverse is going to be greater than 0. Thus, for all n \in N, the interval is going to include numbers greater than 0. Any smaller such interval is contained in any bigger such interval, so:
Now let's investigate the supremum (least upper bound) of an intersection of such closed intervals. It shouldn't be too difficult to figure out that:
This is because no matter how big n gets, its inverse is always going to be included in that intersection, and is such its supremum. Now according to our anonymous friend, this means that:
Call this supremum x. Clearly x is greater than 0. Now let m be a natural number whose inverse is smaller than x. (Because there's infinitely many natural numbers, we can always find such a natural number.) This would then mean that:
Now we have a case where the supremum of a subset is greater than the supremum of its superset. According to my understanding of suprema, this can't be possible.
So obviously the assumption is invalid, and the correct way should be:
Here we can't "let m be a natural number whose inverse is smaller than" 0, because then its inverse would have to be negative, and that would mean that the natural number itself would have to be negative.
But wait! Didn't we say that no matter how big n is, in the intersection of all intervals from 1 to n, its supremum is always going to be greater than 0? Now we say that in the intersection of all intervals from 1 to infinity, its supremum is going to be 0! Surely this would mean that an infinitely large collection of finitely large things is different from an infinitely large thing!
Well, that's exactly how it is. We all know that, and our anonymous friend should learn that too. — JIP | Talk 16:50, 23 November 2005 (UTC)
Just thinking, can the set you described above have a supremum? If the upper bounds keep getting closer to zero, then zero will be the intersection and the only member of the set. In this case it would be a supremum. I fail to see the analogy with her induction proof and how this proves 0.999... = 1? In her proof you are looking at a sum of things, not a collection of finitely large things. Very confusing. 71.248.136.206 01:01, 26 November 2005 (UTC)
I think I can see what you have done but I still cannot see the analogy and I don't see how reductio ad absurdum in this problem proves anything. If as you say P is true for infinitely many finite sets, it's true for an infinite set, I think you may have contradicted yourself for you stated earlier: Surely this would mean that an infinitely large collection of finitely large things is different from an infinitely large thing! If an infinitely large collection of finitely large things does not imply an infinitely large thing, how do you reach thi conclusion? Sorry, I am wondering if it's me who needs to learn something and not you? 71.248.130.225 19:43, 26 November 2005 (UTC)
Your whole point whatever it is, indicates your argument is confusing and at best blurry? She (192.67.48.22) uses mathematical induction in much the same way as I have seen anywhere else. Reductio de absurdum is Latin for reducing an argument to the absurd and then concluding it is false. You have not done this for your argument is complete with contradictions. I must be frank - I do not have a clue of what you are trying to accomplish and I do not see any connection between your example and her proof. 70.110.85.150 13:18, 27 November 2005 (UTC)
Well, I guess I still don't get your point. You showed that 0 is the intersection of an infinite number of finite sets. She (I think it's a she because one of the comments is about male chauvinism?) is discussing an infinite sum that is not the same as a set. How do you draw conclusions relating your example to her's? It seems (imo) that these are two different things. In mathematical induction we do not try to prove P(infinity) ever, do we? If P(k) implies P(k+1), it seems logical and natural that P(n) is true regardless of the value of n. Can we at any time prove any induction result for P(infinity)? I don't think so because we cannot make any assumptions about P(infinity). If what you are stating is true (you are saying that P(k+1) does not imply P(infinity), aren't you?), then no mathematical induction can be correct because it clearly states that the results hold true for any value of n. It seems spurious to think at all about P(infinity) since there is nothing we can deduce from it. If 0.9, 0.99, 0.999, 0.9999 and so on are less than 1, then what makes you think that there will ever be a term that will make it equal to 1? The terms get closer and closer to zero, so the overall sum remains less than 1. If this is not true, then the Cauchy theorems do not make sense: are you saying that each x_i term does not get smaller as the i_s get larger? Does not m>n imply x_m < x_n ? Does infinity wrap round at some point? 70.110.85.150 15:59, 27 November 2005 (UTC)
I don't think you read the original poster's comments properly. You may want to reread these? She/he/it(whatever), did not state there will never be a term equal to 1. Why don't you reread her post? As far as is concerned, there is no room for this in Mathematical induction. Have you perhaps missed this point? As for examples, you have not been able to convince me of anything. Are you a student of Rasmus's? He appeared to think that his counterexamples were valid too. You first need to present a counterexample that is related to what is being discussed. What you started talking about is material that is commonly found in beginner's courses (0 the intersection of the sets...). This has no connection to the induction proof. It is completely unrelated and anything you deduce from it whether it be by reductio ad absurdum or otherwise is not true. Once you find a counterexample that is connected, then you need to show how it nullifies her proof. Really, you have not been able to prove anything imo. 70.110.85.150 20:36, 27 November 2005 (UTC)
I was reading Jitse Niesen's comment that he knew of no mathematician who might think otherwise. Well, here is one mathematician who questions whether 0.999.. and 1 are actually equal. His name is Fred Richman and the URL is:
http://www.math.fau.edu/Richman/HTML/999.htm
Title of this page is: Is 0.999... = 1?
Another mathematician/computer scientist is Frode Fjeld:
http://www.cs.uit.no/~frodef/frodef.html
This is from the first google page returned for "proof 0.999 less than 1" —Preceding unsigned comment added by 71.248.136.206 ( talk • contribs)
I don't think so. The reals also have serious problems. This is but one of those problems. I think whoever came up with this idea (that 0.999... = 1) did so because it would close the door on those who argue that the reals are but an approximation to actual values where these cannot be expressed as rational numbers. 70.110.85.150 23:17, 27 November 2005 (UTC)
I am willing to bet that you will be surprised how many mathematicians think this is nonsense because it cannot be proved in any way. Out of curiosity, does anyone reading this page know who first came up with the idea that these two numbers are equal? Was it Frechet? Weierstrass? Cantor? Someone else? Would be interesting to know who. 70.110.85.150 20:40, 27 November 2005 (UTC)
I don't have to prove it. I think 192.67.48.22 proved it conclusively by mathematical induction in the above posts. I don't think anyone can refute this proof. It is 100% solid imo. Go ahead and reread it. Yeah, I would think it's safe to say 0.999... is not greater or equal, therefore it must be smaller than 1. By how much is not really important. Is it a real number? Yes, I think it is, but I also think that it is not a rational number. Can it be expressed as a/b or a finite polynomial sum that reduces to a/b s.t b <> 0 and both a and b are integers? I don't think so. 70.110.85.150 23:17, 27 November 2005 (UTC)
Yes, how about it. This is exactly what I wrote in my previous paragraph. 0.999... is not a rational number. Can you find a number between 3.141... and the actual value of pi? 68.238.98.71 11:50, 28 November 2005 (UTC)
What we're going to do is discuss one point at a time. 70.110.85.150, or 68.238.98.71, or wherever you are: I am addressing you. Is there a rational a/b such that 0.999... < a/b < 1? There are two acceptable answers: yes or no. Melchoir 12:05, 28 November 2005 (UTC)
Thank you for your conclusive reply. Now: is it your belief that two Dedekind cuts can partition the rational numbers in the same way, yet still be different? Melchoir 12:16, 28 November 2005 (UTC)
Fair enough. Melchoir 12:20, 28 November 2005 (UTC)
68.238.103.187, are you a different person from 70.110.85.150 and/or 70.247.52.236? Please respond, and additionally, please create an account to avoid future confusion. Seriously, you're killing us. I'm not letting you off the hook: is it your belief that two Dedekind cuts can partition the rational numbers in the same way, yet still be different? The absence of the word "real" in this question was, and still is, intentional. Also note that I have never used the word "definition" on this page. Melchoir 22:26, 28 November 2005 (UTC)
Why would you avoid creating an account? I'm honestly curious. You may have responded to my question, but you haven't answered it. The definition of a Dedekind cut as a partition of the rationals is unambiguous. I will rephrase my question very slightly: can two Dedekind cuts consist of the same partition of the rationals, yet still be different cuts? There is now a second question that you also have not answered: are you a different person from 70.110.85.150 and/or 70.247.52.236? Melchoir 22:45, 28 November 2005 (UTC)
That sounds like a no. I must also assume that you are 70.110.85.150 and 70.247.52.236, since you refuse to say otherwise. You do ask an interesting question: how to represent pi using Dedekind cuts. I am willing to show you, but you must point me to a definition of pi that you agree not to argue with. If I do this, will you agree that all real numbers can be represented by Dedekind cuts? Melchoir 23:29, 28 November 2005 (UTC)
Good, let's use circles. Let A be the set of positive rational numbers x such that a circle of circumference x has diameter strictly less than 1. Let B be the union of A with the set of rationals less than or equal to zero. Then B is easily seen to be closed downward, and it should also be clear that B has no greatest element. Let C be the complement of B in the rationals; it is closed upwards. Let p = (B, C); then p is a Dedekind cut.
Take care to note that the definition of p does not require, assume, mention, or beg for the word pi, the Greek letter pi; the number pi; or any hint of foreknowledge of the existence, in a mathematical, physical, metaphysical, sociological, or anthropological sense, of the concept known as pi.
Now, consider the question: what does it mean for a Dedekind cut to represent a real number? We can answer this question without deciding on a particular definition of "real number"; we need only agree that the set of real numbers is an ordered set which extends the rationals. One says that a cut (X, Y) represents the real number z if X consists exactly of those rationals which are less than z. This paragraph is non-negotiable.
Back to our problem. We all agree that pi is real. Every element of the set B is less than pi, and every rational number less than pi is in B. Therefore the cut p represents pi. Melchoir 06:55, 29 November 2005 (UTC)
Which part do you disagree with? Melchoir 18:49, 29 November 2005 (UTC)
Please do not refer to yourself as "the poster below ". I think I finally understand what you are trying to say about Dedekind cuts, and we can talk about that later. For now, however, I ask you to focus on the question at hand: pi. My argument above proceeds in steps:
Obviously you disagree with at least one of the above. Which is the first point with which you disagree? Melchoir 23:02, 29 November 2005 (UTC)
I have a great deal to say about the "gaps" to which you allude, and I will do so later. However, I would venture that Michael Hardy understands more than both of us put together, which is saying something considering our... differences.
So, you agree with (1) and (2) but not with (3). This is encouraging. Tell me, what does it mean to you to say that a Dedekind cut (X, Y) of rational numbers represents a number z? Melchoir 23:29, 29 November 2005 (UTC)
Please refrain from personal attacks. You claim that we are wasting our time, but I am learning more from you than you realize; you could make an attempt to learn from me. If I read you correctly, you are threatening to leave this discussion. Make no mistake: most folks here at Wikipedia would be happy to see you go, but I would not.
By " Dedekind cut" I mean the definition in that article, and I have already told you what I mean by "represent". I quote:
Since B consists exactly of those rationals which are less than pi, p represents pi. Now, please notice that as far as this argument takes us, p might also represent numbers that aren't pi. Conceivably p could represent lots of numbers very close to pi. Infinitely many, even. But p still represents pi. Do you disagree with this simple statement? Melchoir 00:33, 30 November 2005 (UTC)
That's fine; it's not going to matter how many numbers p, or any other Dedekind cut, represents. Just to be clear, you agree that p represents pi? Melchoir 01:21, 30 November 2005 (UTC)
You once claimed about finding a cut to represent pi: "If you can do this, I will concede that all real numbers can be represented by Dedekind cuts." I cannot proceed until you do exactly that. Melchoir 00:18, 1 December 2005 (UTC)
I have been extremely careful not to claim that p represents only pi. You asked me, "Show me how to represent pi using Dedekind cuts." You never asked for a Dedekind cut that represents only pi, and I never claimed to give you one. I now ask you to acknowledge that, among the numbers that p happens to represent, one of them is pi. Melchoir 01:43, 1 December 2005 (UTC)
You asked me to find a cut representing pi; presumably you thought that I couldn't, and you could be a little more humble and upfront in admitting that I did. And, please do not refer to "common misconception"s among PhDs to try and strengthen your point. I have all the authority in the world behind me, but I have not tried to cram it down your throat. If you want a sociological holy war, start one on your user page. This is about math.
Now, let me make a definition. Let a Melchoir number be a set of "numbers whose limits are the same", whatever that phrase means to you. We can add, subtract, and multiply Melchoir numbers; if we're careful, we can even divide them; and we can compare two Melchoir numbers by comparing their elements. The set of Melchoir numbers is an ordered field. Do you object? Melchoir 18:31, 1 December 2005 (UTC)
I speak of a "sociological war" because you speak of a "common misconception amongst Phds". I do not hold a PhD, and I have never studied number theory. If you're looking for a fight against the Establishment, I can't help you. I never called you inferior. Perhaps you think "my logic" is patronizing because I am bending over backwards to agree with your terminology, along with that of logamath1 at the bottom of the page. Would you really prefer that I cry out every time you defy a universally accepted definition?
I agree that there is such a mathematical object as a formal infinite series; in a topological group, there is also such a mathematical concept as the sum of a formal series. Sure, these are different things. There is a formal series of rationals that perhaps best represents the idea of the string of symbols "0.999..."; let's call it M. There is another formal series of rationals that perhaps best represents the idea of the string of symbols "1.000..."; let's call it N. So M is different from N, but who cares? Both of these formal series are Cauchy, and as luck would have it, they are co-Cauchy; their partials sums can be made arbitrarily close to each other, can they not? Melchoir 23:31, 1 December 2005 (UTC)
Please define the phrases "infinite sum" and "limit of an infinite sum", so that I will be able to decide whether or not they are the same. Meanwhile, by speaking of formal series I have bypassed the whole issue. Do you agree with M and N are co-Cauchy? Melchoir 18:36, 2 December 2005 (UTC)
Okay Hardy. You are correct, it is "for all intents and purposes". However, I have heard very educated people use it this way and since they were not corrected or challenged, I assumed it was acceptable to use. I speak/read/write 6 languages. And I am proud of the way I learned English seeing it it not my mother tongue. You are nothing but an arrogant fool and a miserable man who thinks he has a high IQ. "For all intents and purposes", most people who are not arrogant farts like you, won't even notice I used this cliched phrase incorrectly. So unless you can prove me wrong by focusing on the math, I suggest you save your grammar talents for a more appropriate occasion. Do you know more than one language Hardy? 70.110.81.253 01:05, 30 November 2005 (UTC)
You can't be serious? That's exactly what a Dedekind cut is saying or has it not dawned on you yet? 70.110.81.253 23:16, 29 November 2005 (UTC)
If there is no rational number between 0.999... and 1, and rationals are dense in the reals, seems to me like 0.999... is equal to 1. — JIP | Talk 12:21, 28 November 2005 (UTC)
If you believe that Dedekind cuts can be used to represent any real number, then please show me how to represent pi and e. 68.238.103.187 22:15, 28 November 2005 (UTC)
You may not need to call on any theorems because they won't help you.
I know where this is leading: If 68.238.98.71 says no, then you will try to say that {(-inf,1);[1,inf)} = 1 and {(-inf,1);[1,inf)} = 0.999... therefore they must be the same. However, by saying this you would concede that 0.999... is less than 1 since (-inf,1) does not contain 1 and I don't think you are quite ready to say that 0.999... is a member of [1,inf), are you? 70.247.52.236 18:13, 28 November 2005 (UTC)
I was afraid you would respond this way. You cannot use Dedekind cuts to support your argument because then you are assuming that 0.999... = 1 without any proof. 1 is the upper bound of the infinite sum 0.999... therefore it cannot be a member of [1,inf). Using your argument, one can equally well assume that 0.999... is a member of (-inf,1) for there is no number between (-inf,1) and [1,inf). Dedekind cuts cannot be used to represent all real numbers. How would you represent pi? Please don't tell me: {(-inf,pi),[pi,inf)} because you do not know what pi is. You will have to come up with a far more convincing proof if you are to defeat 68.238.98.71. 70.247.52.236 22:13, 28 November 2005 (UTC)
Please create an account. Melchoir 19:30, 28 November 2005 (UTC)
I'm not really comfortable with this section, how do you guys here feel about changing it?, maybe something like:
Another kind of proof adapts to any repeating decimal. Let the number 0.999… be called c, then 10×c = 10×0.999… = 9.999…; If we substract c, we have 10c - c = 9c and 9.999… - 0.999… = 9, thus 9c = 9. Dividing both sides by 9 completes the proof: c = 0.999… = 1
Please edit this entry, so we can have a 'definite' version of this section. Or am i the only one who has trouble with the current wording? Thank you. Jesushaces 05:05, 30 November 2005 (UTC)
If you knew me better, you wouldn't have provoked me like that...
Ksmrq, I have reviewed the edit history of this article. It has racked up a hundred edits in six months of existence, which is a lot for such a low-visibility page but not completely out of line. It was indeed a "horrible hash" until you rewrote it on 26 October 2005, and in some sense it is a testament to your skill that its content has not changed since then. Although the article's stasis reflects well upon you, it is not necessarily a good thing, and it certainly should not be defended as an end in itself.
Since your 26 October 2005 rewrite was so extreme, we must treat the article as if it were completely new. Indeed, its content was written by one person and has since been revised only cosmetically. Most relevantly: we cannot predict the effect of modifying the new article by extrapolating from the old article. You speak of a "certainty" of "attack". I don't know how you can be so certain.
On to the math. You claim "The algebra proof pretends nothing". In fact, the algebra proof relies on the properties of the real numbers while pretending that it doesn't.
Take a look at this webpage, which was presented by an anon as an example of a mathematician who supposedly thought that 0.999... and 1 are different. The anon was wrong, and you correctly pointed out: "Good grief. Richman is merely having a little constructivist fun looking at an alternative to reals." Richman is sort of playing the iconoclast by referring to "traditional real numbers", an extremely unfortunate phrase given that it is far too late to redefine the reals, even if it were a good idea, and it isn't. However, his mathematics looks sound to me, and he does two important things. First, he names the decimal numbers and takes them seriously. This is important because many people seem to think that the reals numbers are the decimal numbers, and the Elementary section of our article doesn't even suggest otherwise. Unfortunately, Richman does not provide a symbolic representation for decimal numbers (he reuses the notation everyone else reserves for real numbers, another poor choice), so I will invent one: let's agree to write decimal numbers as we would real numbers, except that we replace the decimal point by the letter d. For example, 0d999... and 1d000... are different decimal numbers, essentially by fiat.
Second, he defines addition and multiplication operations on the decimal numbers. This allows him to observe, for example, that 3x 0d333... is not 1d000... but 0d999. Indeed, it is impossible to divide 1d000... by three.
Why am I writing so much about the decimal numbers? Because they provide a counterexample to the reasoning in the "Elementary proofs" section. For the decimal numbers, "manipulations at the digit level are well-defined and meaningful, even in the presence of infinite repetition". Therefore the reader should expect the two elementary proofs given to work for the decimal numbers. But they don't. In fact, we can pinpoint where they break down. In the first proof:
Long division doesn't work on decimal numbers, and 0d3333... is not one third of 1d0000... This proof relies on the numbers discussed being real. In the second proof:
There is no unique way to subtract two decimal numbers, but your argument can be rearranged to avoid subtraction. If you made this rearrangement, you would still need to say at some point
If c is a decimal number, then the second equation does not follow from the first. As Richman might say, 0d999... is not cancellable. This proof also relies on the numbers discussed being real.
The arguments listed under "Elementary proofs" only work if they are applied to real numbers. However, the article currently does not contain the word "real" until the "Advanced" section. This is bad logic and possibly even dishonest. It's not just that the elementary proofs lack "sophistication and rigor"; as stated, they are simply wrong. I see three possible fixes to this problem:
The first two options have the same intent: in a more or less subtle way, tell the readers that something essential is being hidden from them. Sadly, we live in a world filled with disclaimers, and many readers will fail to heed the warning. They will interpret the elementary arguments as purported self-contained proofs and think that we are wrong. This is, in fact, what is currently going on and what you should fear the most. I think my cynical prediction of the future is much more likely than your cynical prediction of the future.
The third option seems like the only honest one to me, and it hides nothing. Just admit that they aren't proofs at all but sleight-of-hand manipulations that are too pursuasive for their own good. In that spirit, if I may quote myself, "The current wording suffers from a pretention to precision that isn't really justified." I stand by it. Melchoir 11:51, 30 November 2005 (UTC)
I am the originator of the 'Induction Proof' that 0.999... < 1. I have read what you have written in the above paragraphs and I think you are only fooling yourselves. The 10x proof does not rely on the unstated properties of real numbers or any established facts and it is entirely false. It seems to me that Dedekind cuts can also be used to prove that 0.999... < 1 as suggested by one reader. The induction proof is solid. You do not need to rewrite the nonsense you have in this article - you need to delete it because it misleads those who do not have enough sense to see it is false. I said I would not post but I hope you will do what is right, not what you think is right, but what I am telling you is right. I know what I am talking about. Am I smarter than most of you? The arrogant answer is yes. Most of you are sincerely deceived. You love ignorance and it shows. 158.35.225.231 12:57, 30 November 2005 (UTC)
Who said anything about disputing the properties of real numbers? It's your misinterpretation of the number 0.999... that makes it appear there is conflict in the real numbers unless it equals 1. 158.35.225.231 14:50, 30 November 2005 (UTC)
I don't want to contribute to your article because it's false. Your main concern is how to remove the contradiction of there being no real numbers between 0.999... and 1. Here's how you can do it: There are infinitely many numbers between 9/10 + 9/100 + 9/1000 + ... and 1 as the following demonstrates.
Let X = 9
X/10 + X/10^2 + X/10^3 + .... [Radix 10]
A= .9 + .09 + .009 + .... = .999...
Let X = 99
X/10^2 + X/10^4 + X/10^6 + .... [Radix 100]
B= .99 + .0099 + .000099 + ... = .999999...
Let X = 999
X/10^3 + X/10^6 + X/10^9 + .... [Radix 1000]
C= .999 + .000999 + .000000999 + ... = .999999999...
A < B < C < 1
The terms of C decrease much faster than B and those of B much faster than A so that no matter how large any of these numbers become, the difference between C and 1 will always be the smallest. Let's call this difference Dc. Likewise let's call Db and Da the differences of b and a with 1 respectively. It is easy to see then that Dc < Db < Da. Therefore no matter how small these differences become, they will always be greater than zero and the ordering will always be preserved, i.e. Dc will always be less than Db which will always be less than Da. Would you be able to represent the number C in Radix 100? Would you be able to represent the number B in Radix 10? Answer is no. Part of this controversy has to do with the way numbers are represented in a radix system. You cannot say (A) 9/10 + 9/100 + 9/1000 +... = 1 for then you must have (B) 99/100 + 99/10000 + 99/1000000 + ... = 1 and we know that A is less than B. Remember we are not comparing limits for this tells us nothing about the size of the number. We can only compare the sum of a finite number of terms, term by term. 158.35.225.231 17:56, 30 November 2005 (UTC)
No. 158.35.225.231 19:33, 30 November 2005 (UTC)
Here you go. Hate to break it to you but I hardly ever use your website. And now you have one more account... Logamath1 20:02, 30 November 2005 (UTC)
Actually that's the problem - they do not all correspond to the same real number 1. The limit of those numbers is 1. Their sum is not 1. Their sum is undefined. And no, you cannot say that at infinity their sum will be 1. If you consider the differences Dc, Db, Da - at infinity these will still not be zero even if these are close to zero. In fact, if you believe in infinitesimals, Dc, Db and Da would be the closest thing to what these might resemble. Of course there is no such thing as infinitesimal (not in non-standard analysis, surreal numbers or any other kind of number system which is a load of rubbish) This is the point and yes JIP, you have missed it. The definition you have for infinitesimal in Wikipedia is really amusing. But then so many of your articles are amusing. logamath1
I think I understand the terminology fairly well. 0.999... does not have to be 1 for the real number system to be logically consistent. There is an obsession that things might fall apart if these two numbers are not equal. As I have shown above, you can have numbers between 0.999... and 1 - this does not violate the Archimedean corollary. In fact the Archimedean property holds for all numbers whether they be real (or not) provided these are treated as approximations. Whether you like it or not, this is the way we have always dealt with real numbers - as approximations. The problem occurs with the interpretation of expressions such as 'finite sum' and 'limit of infinite sum'. These are not the same and should not be treated equivalently. You talk about infinitesimals without having a well-defined meaning. Infinitesimals are a figment of the imagination. Wiki's article on this is a joke (as are several others). In fact having 0.999... = 1 causes the real number system to fall apart. Why? Well, the real number system is ordered. There is an ordering amongst decimals too. This ordering exists so that we can make meaningful comparisons. If you compare a number that differs in the 15 zillionth digit of pi, how would you compare it? You would start from the most significant digit which is to the left of the radix point and proceed until you found the digit that differs. In the same way, our decimal system fails if we break this rule by setting 0.999... = 1 because we compare these in exactly the same way as we would compare pi. I put it to you that whoever came up with the idea that 0.999.. = 1 was a fool. logamath1
You have stated a corollary of the Archimedean property. h is not an infinitesimal but a very small number. An infinitesimal does not exist. logamath1
HeroicJay: It's quite incredible how you agree that infinitesimals are a figment of the imagination and then you call the difference 1 - 0.999... an infinitesimal - isn't it? You are very confused I would say. Logamath1
1-0.999... is undefined HeroicJay. It's as simple as this. We can only approximate its value just like all the other repeating decimal numbers in the decimal system. What is so hard about this to you? Best we can do in this case is to approximate it by 0. However, this does not mean it is exactly equal to 0 but is good enough. logamath1
Not so Melchoir: 1-0.999... would be a very small real number, not an infinitesimal because an infinitesimal does not exist! Therefore 0.999... is less than 1. Do you follow? logamath1
Wrong. There are very small real numbers for if there were not:
logamath1
They do require approximations because you cannot do arithmetic on any numbers that are not finitely represented. This has nothing to do with science but everything with pure math. logamath1
I am only going to deal with Melchoir. It's too difficult because of edit conflicts. Sorry HeroicJay. Try Decimal construction. Otherwise you choose. logamath1
It's not a lame excuse. Don't take anything personally. Besides, it's one against the whole of Wikipedia and most of the Western World's Phds. Cut me some slack please. logamath1
Have fun. I will check in tomorrow or later tonight if I get a chance. Sorry I gotta leave you. logamath1
You could have fooled anyone else but you don't fool me. How small is very small? Please Melchoir, this is about the worst argument you have presented thus far. It's total nonsense! By the way, where is that boy Jitse Niessen? Do you see how HeroicJay is provoking me? He needs to be reprimanded!! Who the hell is he to ask me why I argue?! Why, I would take this up with God if I had to until the very last breath!! I am very tempted now to tell this creep off. Let's see if Niessen will reprimand him or not... logamath1
It seems "very small real number" is a confusing term as well - let me rewrite the whole thing fully to avoid and argument about whether it's reasonable to use that term:
JPD 10:58, 2 December 2005 (UTC)
It's still wrong because:
1) You are referring to a positive real number less than all positive real numbers. There is no such number. How do you know such a number exists? 2) Having assumed that there is such a number, you then state that there is no such number.
Have you thought at all about this? Are you just another brainless PHd? Melchoir's argument even though it is flawed, is far superior to yours. Come on, who do you think you are fooling? logamath1
If you think Melchoir is saying anything different, you obviously don't understand his point, which would explain why you think it is flawed. JPD ( talk) 13:14, 2 December 2005 (UTC)
I am saying that 0.999... is a real number less than 1 such that at least one real number 1-0.999... exists. Although I can't find this number because I believe 0.999... is irrational, does not mean it does not exist. You talk about a reductio ad absurdum proof but there is no such proof. Let 1-0.999... = x and let me show you how this reduces to the absurd if x is equal to 0. Here is a reductio ad absurdum proof:
Let 1 = 0.999... Now add .1 to both sides:
1.1 = 1.0999....
which is clearly false, therefore 0.999... cannot equal 1 and thus it must be less than 1. Don't try to tell me that .1 = 0.0999... - it is not. When we compare numbers in the decimal system, we begin the comparison with the most significant digit.
logamath1
Yes, it is clearly false. Such an x is possible always. Give me any 10^-i and I will simply give you 10^-(i+1). Of course it is possible for any natural number i. And comparing numbers in the decimal system works the same for all numbers including numbers that end in 0.999.... logamath1
I do not have to provide a reason for why decimal numbers are compared this way - this is how the decimal system was designed. You are making a mistake by assuming that x=1-0.999... should be a single positive real number less than 10-i for any i. This is faulty logic. I am making no assumptions about the value of x except that it exists. I do not call it a very small real number and I do not call it an infinitesimal. I only call it a real number that is greater than 0. Finally I did not say at any time that such a positive real number does not exist. Look JDP, the fact that you cannot find the exact value of pi in any radix system does not mean that pi is not a finite value. Yet do you ask: Is there an x such that x-pi = 0? Of course not. Please don't tell me x=pi (in fact it is pi but not by your reasoning. We reach this conclusion because we know pi exists but do not know its exact value just as we don't know the exact value of 0.999... or the value of 1-0.999...) because we don't know the exact value of pi. What you have with 0.999... is very similar: x-0.999... = 0 means x=0.999... and not that x=1. logamath1
Melchoir is completely right. Your comments about pi are a sidetrack, so I will only answer your comments about what you call my faulty logic. The only assumptions about x that I am making are parts of the definition of the real numbers. If x=1-0.999... it must be a single real number, since the reals are a field. If x>10-i for any natural number i, then 0.999... = 1-x < 1-10-i = 0.99..99 with i 9s, which is a contradiction. This is not an assumption, it is simple algebra. Therefore x<10-i for all natural numbers i. I did all this without saying anything about the value of x that didn't follow from it's definition. I didn't call it infinitesimal, or very small. I did, however, show that it is smaller than any positive rational number. Because there are no positive real numbers that are smaller than every positive rational number, x must be less than or equal to 0. If you are dealing with a different number system, like the decimal numbers mentioned above, then you get different results, but it must be 0 in the reals. JPD ( talk) 18:45, 2 December 2005 (UTC)
You are either sincerely mistaken or you are lying. You challenged logamath1 to suggest a construct for the reals. He suggested the Decimal construct. You then tried to show this is true and were not able to do so. 70.110.92.137 19:57, 2 December 2005 (UTC)
Forget about Melchoir and let's see what you are saying: Your logic is strange. You first say that x must be a real number. I agree with you on this. Then you say, "If x > 10^-i for any natural i, then 0.999... = 1 - x < 1-10^-i ..." This is rubbish so I'll stop here for no one will start off an argument by saying that x > 10^-i. You must be a newbie. 70.110.92.137 20:05, 2 December 2005 (UTC)
Please show me how you can do any accurate arithmetic with pi, e or sqrt(2). It is all approximate whether real or decimal and we are talking about decimals here. 70.110.92.137 20:05, 2 December 2005 (UTC)
HeroicJay: As long as you are adding fractions you can obtain accurate answers because these are rational (provided you leave your answers in fraction form). The minute you begin dealing with decimals, all bets are off. As for the part about PhD, I leave this to logamath1 - he can probably answer this better than I. 70.110.92.137 20:05, 2 December 2005 (UTC)
There will be no more edits. I have wasted my time. 71.248.138.198 02:56, 3 December 2005 (UTC)
Heroicjay: I have never met a more obstinate fool than you. Indeed you are duller than most and I see no point in carrying on any exchanges with you.
Wikipedia: In addition to the nonsense you write in this article, there is more nonsense in other math articles. Actually your math articles are referred to as the WikiMath joke articles. Your infinitesimal article states: "...an infinitesimal, or infinitely small number, is a number that is greater in absolute value than zero yet smaller than any positive real number.." - This is such incoherent and illogical garbage for the first part implies it is a positive real number and the second part that it is zero. How can it be both zero and a small positive real number? I leave you to break your stupid heads over this. 71.248.138.198 02:56, 3 December 2005 (UTC)
"There will be no more edits. I have wasted my time. 71.248.138.198 02:56, 3 December 2005 (UTC)" You said you'd stop. Please do, you make no attempt to reach consensus, merely to push your viewpoint time and again on people who aren't going to accept it. You've now moved on to personal attacks, which really have no place on wikipedia. 81.86.207.192 13:04, 3 December 2005 (UTC)
Melchoir writes: "...any careful definition of the real numbers based on decimal expansions SETS 0.999... equal to 1." - Melchoir
Not only is this an outright lie but it is clearly demonstrated to be false:
1) The Indians did not use a decimal point. This was an invention of Scotland.
The notation we use today first appeared in a book called "Descriptio" by the Edinburgh mathematician, John Napier, Laird of Merchiston, in the 1616. He used a decimal point to separate the whole number part from the decimal number part. Known as 'Marvellous Merchiston", he published many other treatises including "Mirifici logarithmorum" (1614) and Rabdologia (1615) on systems of arithmetic using calculation aids known as Napiers Bones.
2) Fractions were invented long before the decimal point.
In the 10th cent. A.D. Arab mathematicians extended the decimal numeral system to include fractions.
Melchoir will try to justify himself by using terminology such as any careful definition of the real numbers - this is entirely subjective and is open to debate.
I asked who came up with the idea that 0.999... = 1. No one on Wiki's board of idiots was able to answer this. Why? The answer is simple: they are all products of today's institutions that are run by a bunch of baboons masquerading as Phds. My apologies to the few Phds who really earned their diplomas but the majority are to be regarded as ignoramuses who are spineless and incapable of thinking on their own.
The definition of the real numbers is not open to debate. Mathematics is useless as a language unless we agree on the meaning of its jargon, and "real number" is too entrenched a concept to change, even if it were a good idea, and it really isn't.
81.86.207.192 158.35.225.228, I wonder if you think the rigor of modern mathematics is somehow a challenge to your creativity. It's a fine intellectual exercise to dream up number systems in which the analogues of true statements for the real numbers turn out to be false. One might even make a living doing this. It doesn't make me a liar.
As the self-appointed dictator of this talk page, I declare that it is closed to personal attacks. Comments insulting other users or the Establishment will be erased. The purpose of the talk page is to discuss improvements to the article. If you have a suggestion, let's hear it. If you're confused but willing to learn, ask a question. If you're here to whine about PhDs, go away. Melchoir 19:12, 5 December 2005 (UTC)
You have not provided one proof or argument that shows why the real number system must have 0.999... = 1. It is easy to strike out what you don't agree with but much harder to admit you are wrong.
...The limit of the series is not the sum of the series and a definition does not make it true. There are real numbers between 0.999... and 1 as has been demonstrated above... There has been a solid proof by induction that has also been rejected... --anon
"The limit of a a sum of an infinite number of terms is an infinite summation, so .999... is the limit is of the sequence .9, .99, .999... and the sum of the series .9 + .09 + .009 + ... (the result is the same either way.)" ...I will let the readers decide on this. The limit of an infinite number of terms IS NOT AN infinite summation. An infinite sum is impossible. And no, you cannot define it this way because it does not make sense... You saying it one million times or claiming that it is true does not make it true. A Definition is useless unless it makes sense AND it is TRUE. —Preceding unsigned comment added by 70.110.81.224 ( talk • contribs) 2005 December 6
Oh please!!! If you are going to restore points, make sure you restore everything... YOu have only restored what makes you look good. The fact that you have final editing say displays clearly that this is not a peoples encyclopedia... Wiki is a view of the world according to YOU... —Preceding unsigned comment added by 70.110.81.224 ( talk • contribs) 2005 December 6
...You cannot show me one reputable source that states 0.999... = 1... Show me an encyclopedia Brittanica or something that has been around for a long time that states this. I put it to you that ... 0.999... = 1 has not been around that long. It was probably conceived in the early or late twentieth century... ...This article should be deleted. It is not encyclopedic... —Preceding unsigned comment added by 70.110.81.224 ( talk • contribs) 2005 December 6
This is exactly the problem: most of the world do not think of 0.999... as the limit of 9/10 + 9/100 + 9/1000 + ... They think of it as an infinite sum or a number. If you talk of the limit of the partial sums of (9/10;9/100;9/1000;...), then its limit is 1. The same principle applies to 1/3, 1/6 and any other fraction that cannot be represented exactly in base 10. 0.333... is not equal to 1/3. 0.333... is an approximation for 1/3 in base 10. Although it cannot be represented exactly in base 10, it can be represented exactly in many other bases, e.g. 1/3 = 0.1 (base 3) 1/3 = 0.2 (base 6) and so on. Now a 1/4 in base 3 is approximated by 0.020202... It is not equal to 0.020202... even though the limit of 0.020202... is 1/4. I do not have a bit of a point, I have a major point. Terminology is extremely important and infinite sum and limit of an infinite sum are not the same thing! Lastly, 0.999... was never defined as a limit, neither were the real numbers defined as a limit until Weierstrass, Cauchy and others decided to 'rigourize' mathematics. In many respects they inhibited the progress in understanding the real numbers. Real analysis is filled with inaccuracies and contradictions as a result. —Preceding unsigned comment added by 158.35.225.228 ( talk • contribs) 18:09, 2005 December 6
0.999... is less than 1 for all the reasons you say it is equal to 1.
It makes perfect sense. You are the one who is having difficulty understanding any of this.
I have just found the most wonderful paper:
Conflicts in the Learning of Real Numbers and Limits by D. O. Tall & R. L. E. Schwarzenberger, University of Warwick Published in Mathematics Teaching, 82, 44–49 (1978). pdf webpage
It's 13 pages long. Whoever finds this sentence in the first place should go skim it; it's got some gems.
I propose to rewrite the article with Tall as our motivation and primary reference. In fact, I would have proposed this before, but without Tall it really would have been Original Research. Additional references would be great, including a more modern one, but it's a good start.
The article currently does not address many common misconceptions; I think we should mention them all and explain why they are wrong, although preferably after the proofs. I want to make it clear that I'm not backing down under fire and "teaching the controversy". There is no controversy, and the article must make that plain. There is, however, a whole lot of confusion that can be productively cleared up in the article itself. In some sense, this effort would be similar to a page on an urban myth.
My questions are:
Melchoir 21:16, 6 December 2005 (UTC)
Also, this earlier paper by the same author is mostly generalized nonsense, but it gets interesting and relevant on page 10. Melchoir 22:12, 6 December 2005 (UTC)
Yes, the papers I link to contain speculation, but they are published, and they contain quotes by students displaying some misconceptions. If we address those misconceptions, such a source is necessary to avoid speculation on our part!
That Hackenstrings bit is very interesting. I've seen nimbers before, but not those. As for the citations currently in the article, the first two don't address any misconceptions, and the third is so confused that if it's preserved it should only be as a bad example! I think we can do a better job.
As for rewriting, I must apologize for what was honestly just a poor choice of words. As a mathematician, I think your proofs are the strongest point of the article, and my proposal does not require that we alter them. For now, I simply want to tack on a section at the end exploring the misconceptions. Nor, of course, do I propose to reproduce Tall, who avoids the structure of the real numbers altogether. His article could be useful mainly in that
If he doesn't address them well enough, we can do better than him, too.
Of course we can't guess what the reader is thinking. The reader, in fact, might agree with us; surely teachers will read our article, not just students! But the encyclopedic voice isn't really appropriate to connect with readers on a personal level anyway. I don't propose that we attempt to open a dialogue with the reader, but that we describe what students are known to think. With references such as Tall, we don't have to guess on that.
As for there being better articles to work on, you're right. Melchoir 04:14, 7 December 2005 (UTC)
The article is neither interesting, nor relative. It has nothing about proofs and talks of infinitesimals as if these were widely understood. There is no such thing as an infinitesimal.... If an infinitesimal is greater than zero but less than every positive number, there can only be one of it.... It would be a good idea to delete this article and if you really want to talk about comparing 0.999... with 1, then you should make it clear that 0.999... is less than 1. Tall refers to some questions posed to students and discusses their response. He tries to infer that students answered incorrectly because they were unsure of the denotation of recurring. ...this is taught in elementary school and those same students claimed that they knew the definition of a limit.... —Preceding unsigned comment added by 68.238.99.174 ( talk • contribs) 02:26, 2005 December 7
In mathematics there is no such thing is correct if such a thing is not properly defined. And you can't dream up fancy words that mean nothing and develop theories out of these. --anon
Care to explain?... --anon