Talk:0.999.../Arguments/Archive 5

message: Like I said I would, I engaged in email correspondance with two professors regarding 0.999... and 1. We were unable to come to any sort of agreement. I could see through the proofs they suggested. I'm happy they responded, and in case they read this: thanks. But I was disappointed not to come to any kind of conclusion. The only new thought I gained from the experience is that perhaps both 0.999... and 1 can be the limit for the same summation because as the number of summations increases, both 0.999... and 1 are approached and neither are reached. That still does not mean they are the same exact number. In the end, I of course am aware the professors I corresponded with are of very high intelligence and far superior to myself in mathematics, but I remain of the mindset that a majority of even the most intelligent individuals can still be wrong (aka Big Bang = beginning of time and matter; dimensions are bendable objects; string theory; many philosophical topics). It's an interesting debate, really shows a lack of understanding of infinity/infinitesimal/limits in our current culture, and a need for more work. This is one of the rare debates where the uninformed commonfolk, using dictionary definitions and human intuition, while not necessarily having a sharper understanding of the situation and being easily swayed, have more accurate immediate impressions as to the identity of the infinitesimal because of less exposure to mathematical trickery. --JohnLattier 18:52, 8 November 2006 (UTC)
replies: text: Of course you could not come to an agreement with math professors. They told you that you were wrong, and you obviously don't agree! Do you really think two independent professors both gave you bad proofs? Could it possibly be that you're simply incorrect? Keep working on it! -- Schapel 19:00, 8 November 2006 (UTC)

text: John: You say both 0.999... and 1 can be the limit for the same summation because as the number of summations increases, both 0.999... and 1 are approached and neither are reached. That still does not mean they are the same exact number. Not true. Any real sequence cannot have two distinct limits (this is quite easy to prove). You have indicated several times that you want to work in some number system different from real numbers, and have implicitly stated some desired (by you) properties of that system. Please give a formal construction of the number system you want to work with. Others have also raised this point, but you have never replied to it. --Kprateek88(Talk | Contribs) 11:25, 9 November 2006 (UTC)

text: You are going on in circles again. What do you mean by "common number system"? For most people it would be the real numbers. For you evidently it isn't. You have not defined your number system. number system I have just described No, that is a very vague description of a supposed "number system". If your system (whatever it is) is "universally understood" as you claim, could you please cite some sources where it is described in detail? Does base 10 have any special role to play in your number system? If we were born with six fingers on each hand would your system be different (ideally it shouldn't)? --Kprateek88(Talk | Contribs) 13:59, 9 November 2006 (UTC)

text: Ask your self: who has more right? the two math professors, or you? →AzaToth 20:01, 8 November 2006 (UTC)

text: A good friend of mine, have suggested that you should contect Gene Ray about your ideas. →AzaToth 20:07, 8 November 2006 (UTC)

text: * Have \lim_{x \uparrow 1} x = 0.999..." this tells me you are way off. --JohnLattier 00:22, 9 November 2006 (UTC)

text: John, having varified that your ideas are *without any doubt* Original Research, does this mean Wikipedia won't be hearing any more from you? Untill you're published, that is? —The preceding unsigned comment was added by 210.54.148.98 (talk • contribs) 20:29, 8 Nov 2006  (UTC)
John, what good is a number system if we can't even guarantee that we can multiply numbers? What good is a number system in which perfectly well-defined calculations are "illogical"? What good is a number system where before we do some calculations, we have to come and ask you, personally, JohnLattier, whether each step we perform is "logical"? You certainly haven't given any formal rules nor explanations about why certain calculations are "illogical", as you claim.
You want to reject huge, deeply-developed branches of mathematics because they don't sit with your intuitions about 0.999.... If mathematicians had always rejected ideas because they didn't square with intuition, we'd never have discovered negative numbers, we'd never have discovered irrational numbers, we'd never have discovered imaginary numbers. If scientists in general did it, we'd never have realised that the Earth was spherical, we'd never have formalised gravity, we'd never have discovered relativity. We'd certainly never have come up with quantum mechanics. Your inability to accept ideas despite all the evidence because they are mildly counterintuitive, and your refusal to formalise your objections to them, have no place in mathematics nor any other science. Maelin (Talk | Contribs) 01:16, 9 November 2006 (UTC)
thread_title:  Spoke with 2 professors 

message: 41.etc. is clearly contributing nothing to the discussion at this point, and is essentially just trying to waste our time.  Therefore posts that are primarily insults should be reverted on sight rather than responded to.  I've blocked the current IP for 3 hours for this behavior, but it appears the user hops IP's and will be back. -- SCZenz 17:16, 9 November 2006 (UTC)
replies: text: Well, it's his abuse and reverting that are the main problem.  Others are contributing little to nothing, but they're not abusive so we aren't blocking them. Calbaer 17:35, 9 November 2006 (UTC)
replies: text: You are correct.  It is the abuse that makes his contributions unacceptable, and which justify removing/blocking on sight if they continue as they have.  -- SCZenz 17:54, 9 November 2006 (UTC)
thread_title:  Rudeness by 41.etc 

message: *The infinitesmal is a construct invented to help in the development and understanding calculus.  However, for calculus to work, a nonzero infinitesmal can't exist, and, since calculus is the basis of most advanced applied mathematics, it doesn't exist in practice.
*You can't say real analysis is bunk without bothering to learn it.  Read a book or take a class.  Make sure you finish it, too.
*You can't propose a new number system on Wikipedia.  If you reject all referenced number systems and lack a reference to the number system you're discussing, you are proposing a new number system.  Read Wikipedia:NOR and find a reference defining the number system you're referring to, or accept that your intuition and the mathematical systems in use contradict.  Move on to quantum mechanics, special relativity, the Monty Hall problem, and other topics which contradict intuition but are still true. Calbaer 17:35, 9 November 2006 (UTC)
replies: text: The infinitesimal does NOT exist.  You cannot define junk and then claim it to be true. 41.243.26.131 10:23, 30 November 2006 (UTC)
thread_title:  For the few who continue to reject the generally accepted definitions of mathemetics 

message: I'm not arguing the fact that 0.999... = 1 (I believe it), but if it does, what is/are the closest number(s) to 1 without being 1? Is there any way to express it/them? (sorry, forgot to sign this earlier...) Qhiiyr
: There is no closest number to 1, as illustrated in the above arguments.  Simply put, for any real number a (other than 1), (1+a)/2 is closer to 1 than a, (3+a)/4 even closer than that, and so forth.  So a cannot be "the closest" to 1 or even closer than a finite number of numbers.  This is thus like asking what the "largest integer" is.  For any integer k, I can find a larger k+1.  So even though we have concepts like "the closest" or "the largest," they need not exist for a particular set of numbers (the reals and integers, respectively). Calbaer 17:13, 12 November 2006 (UTC)
thread_title:  closest possible number to 1 

message: I doubt JohnLattier is talking about real analysis. If I understood him correctly (correct me if I'm wrong!), he concedes that in real analysis 0.9999...=1, but that leading to such an equation should be considered a disadvantage of real analysis and that there is a better "system already in place" where infinitesimals exist and 0.9999...<1. But what is the "system already in place"? I believe John will have difficulties to give some sort of reference, because the system in place usually is real analysis. There's a reason why any approach using infinitesimals is called "non-standard". Nevertheless, John, you might try to go ahead and describe what your number system should satisfy. In particular:
# Are the rational numbers a subset of your numbers? How about the standard real numbers?
# I assume your system will include infinitesimals, but will the existence of numbers such as "the largest number less than 1" be required?
# Will every number have a decimal representation? If you want to include numbers with descriptions such as 0.000...1 with infinitely many zeroes, you should also define (or should I say re-define?) "decimal representation".
# What kind of arithmetics is allowed? That's a tricky question; feel free to leave any arithmetics you are not sure about undefined: We can't divide by zero, so why should we be able to, say, substract an infinitesimal?
# For questions of continuity and convergence, a notion of "open set" would be nice.
The problem is that adding infinitesimals leads to other counter-intuitive results and/or makes arithmetics so limited as to make the number system pretty useless. If you answer as many of the above questions as you can (and provide any additional properties you deem necessary for your numbers), I'll try to conjure up a number system satisfying what you want and show some of the strangeness it leads to - or I'll try to show that your prerequisites contradict each other. Yours, Huon 15:47, 13 November 2006 (UTC)
replies: text: I see you acting in good faith Huon, but must have better things to do than try to explain a new topology to someone who believes that \sum_{i=0}^{\infty} a_{i} \neq \lim_{N \rightarrow \infty} \sum_{i=0}^{N} a_{i} Endomorphic 22:51, 13 November 2006 (UTC)

text: There is no such thing as the "largest number less than 1" in the reals.  Reals are a subset of Complex Numbers, and your statement isn't true there either.  So, name this mythical number system!!! Actually, what you say is impossible.  There is no way for the reals to be contained by an ordered number system.  Tparameter 23:32, 13 November 2006 (UTC)
replies: text:  Such expressions get thrown around, JohnLattier, because mathematicians like to be able to do maths with their numbers. If $x$ and $y$ are real numbers, we should be multiply them without the universe imploding. You would otherwise exile the humble parabola $y=x^2$ - is x an infinitesimal? what if x is infinite? is the quadratic valid for some Reals and not others? Quagmire! Endomorphic 02:37, 14 November 2006 (UTC)
thread_title:  What should a number system satisfy? 

message: Just wanted to reply quickly to a specific comment John made in a recent thread above. He said:
"Thank you for addressing my concerns. I agree with #1, but I don't have the higher mathematical training to produce a rigorous proof. My reasoning goes as follows. I start with a foundation that all numbers are represented by one and only one decimal expansion. Excluding any zeroes you may add. As it is a foundation, I don't see any way to prove it right or wrong."
In fact, although it's counter-intuitive, it follows from the fundamental axiom of the real numbers, the Least upper bound axiom, that in fact all rational numbers with a terminating decimal expansion (one that ends in all zeroes) also have another decimal expansion that ends in all nines. 
The reason is as follows.  The Least Upper Bound axiom says that all upper bounded subsets S of the reals have a real number which is a least upper bound.  That least upper bound does not have to be in S, but it does have to be a real number.  For example, the subset of all numbers strictly less than 1 has a least upper bound of 1, but 1 is not included in that subset.  Using that axiom, you can directly prove the Archimedean property that there are no positive infinitesimal numbers c such that for all n>0 nc>1 (the proof is included in that article).  From that result, it follows that if two real numbers differ by at most an infinitesimal number c, then since c must equal 0 the two numbers likewise must be equal.
So from the Least Upper Bound axiom it follows that since 0.999... and 1 differ by at most an infinitesmal amount, and there are no non-zero infinitesimals in the real numbers, it follows that in fact 0.999... and 1 must be equal.  And, in fact, you can show that all real numbers with a finite representation also have an equal alternate representation where the last digit is reduced by one and all zeroes are replaced by 9's.  For example, 0.25 = 0.24999... .  
Thus, although it's not intuitive, decimal representation in the real number system under the Least Upper Bound axiom is not unique. The only way to have truly unique decimal representation for all numbers is to remove that axiom, in which case you are allowing subsets to exist that have no least upper bound.  It is possible to construct such a number system, but as above that is not the standard axiom system used for the reals. Dugwiki 16:27, 13 November 2006 (UTC)
replies: text: Excellent post.  This is irrefutable.  Tparameter 17:09, 13 November 2006 (UTC)

text: You say an upper bound should be included in the set, but this isn't even true in the rationals!  What's the least upper bound of the set {x|0 <= x < 1, x is rational} ? -- SCZenz 23:27, 13 November 2006 (UTC)
replies: text: The least upper bound of the set of rational numbers in [0,1) is 1.  1 is not a member of that particular set, but it is a member of the set of real numbers, which is what is assumed under the Least Upper Bound axiom for the reals.  In other words, every bounded subset of the reals has a least upper bound that is a real number.  That least upper bound doesn't have to be in the subset in question, but it does have to exist in the reals.  Hope that clears up your question. Dugwiki 16:01, 14 November 2006 (UTC)
replies: text: It was more of a question for JL, to help him understand the phallacy of his line of arguments. -- SCZenz 16:14, 14 November 2006 (UTC)

text: You can't define an open-interval on the 1 side, [0,1), and then claim it's closed!!!  It's an open set, which means the (least) upper bound is not in the set, by definition.  Please learn basic math. Tparameter 23:29, 13 November 2006 (UTC)
replies: text:  Guys, sort out your labels. Subsets of real numbers never have unique upper bounds. If $x$ is an upper bound for a set $S$ then every $y>x$ is also an upper bound of $S$. You guys mean least upper bounds, which (for the Reals) always exist and are always unique. JohnLattier further confuses bounds with maximal elements, which is where a least upper bound for a set $S$ happens to lie within $S$ itself. Endomorphic 23:34, 13 November 2006 (UTC)
replies: text: You're right.  I corrected my remark above. -- SCZenz 23:37, 13 November 2006 (UTC)

text: +1. Tparameter 23:46, 13 November 2006 (UTC)
thread_title:  Real number system does not have unique decimal representation 

message: This 'problem' (okay okay you qualified mathemeticians out there I know it's not a problem for you, just my intuition ;p) has fascinated me since I read about it on a forum I frequent. I have flash-learned math just to be able to fully grasp what the issues are and in doing so very much enrichened my understanding of numbers.
During the thread I came up with (with the help of a pure math graduate) a different way of describing the problem I would have to solve were there to be a proof that 0.999.. does not equal 1. Just wondered what you people 'in the know' on here thought of it and whether you thought it was solvable/provable. (I suspect I already know the answer, but here we go anyway...)
\{x: 1 > 1/x > 0.\overline{9}\ and\ x > 0\} \neq \emptyset \Rightarrow 0.\overline{9} \neq 1
Now I'm about a week new to anything over basic GCSE math, but I've been reading like its gonig out of fashion on the subject and as far as I can see, if you can prove the above it follows that the hypothesis that  0.\overline{9} = 1 must be false. 
I get the feeling I must be wrong somewhere, but I don't know where... any pointers? Archgimp 08:12, 14 November 2006 (UTC)
replies: text: Well, there's two things there. If you want it to be a valid disproof, you need to prove firstly the premise (that is, prove that the set you described is nonempty), and then you also need to prove the logical implication (i.e. prove "if this set is nonempty, then 0.999... is not 1"). It seems like the logical implication is fairly obviously true, although I can't see how to prove it at first glance. But nonetheless, your set is empty. I can't think of a good proof that doesn't rely on the fact that 0.999... = 1 (which is hardly going to be convincing for you), but someone else might be able to whip up a proof of the emptiness of your set. Maelin (Talk | Contribs) 08:30, 14 November 2006 (UTC)
replies: text: I don't mind being proven wrong at all, after all the only thing at the moment which tells me that  0.\overline{9} = 1 is intuition, and intuition can get your legs broken. (I wonder how many people laying in traction in hospital say to themselves "I know I checked both ways before crossing, I always check both ways"). The interesting part for me is that the method described above could lead to a new understanding of the issue, and I came up with it all by my lonesome (although undoubtably, someone else has also come up with it before me, I haven't seen it written anywhere else). I think of it this way, even if the mountain doesn't have a summit, it's satisfying to climb higher up than you managed last attempt. Archgimp 08:55, 14 November 2006 (UTC)
replies: text: It could be useful, yes, if understanding that set was easier than understanding the problem... and, of course, if what you want to prove were true. -- SCZenz 09:10, 14 November 2006 (UTC)

text: So let's see... if x is in the set you define above, then 1/x < 1 and 1/x > 999...9/100...0, where the fraction has as many 9's and zero's, for any number of repeated 9's and 0's.  We can rewrite the fraction, more concretely, as \frac{10^n-1}{10^n} for any n.  so we have (by flipping the inequalities) that 1 < x < \frac{10^n}{10^n-1} for all n.  Define y = x-1, so that we have 0 < y < \frac{10^n}{10^n-1}-1 = \frac{1}{10^n} for all n.  Thus we have an x in your set only if there's a positive number y that's less than \frac{1}{10^n} for all n.  There is no such number, so the set is empty. -- SCZenz 09:18, 14 November 2006 (UTC)

text: Continuing my musings... If what I said above is true (i.e. if your set is non-empty), then from the field axioms we have that a number y-1 exists such that. y*y-1 = 1.  Multiplying the inequality by y-1*10n, we get that y^{-1} > 10^n for all n.  This is effectively equivalent to saying there exists a largest number, or at least a number bigger than every integer; it should be even more clear that this is false (it's a direct consequence of how the natural numbers are defined, in fact).  -- SCZenz 02:13, 15 November 2006 (UTC)

text: You're welcome.  Thanks for asking, and listening; it ended up being a useful illustration of how trying to stuff numbers between 0.999... and 1 requires infinity, which I think most of us can agree shouldn't be a number.  (And indeed isn't, in the usual real analysis.) -- SCZenz 17:08, 15 November 2006 (UTC)
thread_title:  An alternative way of defining the problem? 

message: John L. Bell was sourced by John Lattier as someone who said agreeable things about infintesimals.
Things gleaned directly from Invitation to Smooth Infinitesimal Analysis (PDF file linked at the bottom of SIA):
# SIA is introduced as an extension to the Reals. Which means that if decimal representations are still valid in SIA, and convergent series still converge, then 0.999... = 1 holds for SIA, because it holds in the Reals.
# Function continuity falls directly out of SIA axioms: equation 4, f(\epsilon) = f(0) + \epsilon D, "is taken as axiomatic". So in SIA, every function is linear at every point at which it is defined. Sure. Discontinuities (even points of non-differentiability) either prevent function eligibility, or are surrounded by enough little infintesimals to force linearity, regardless of the implications for convergence of limits. Doens't sound like an advantage.
# In the third paragraph: "in standard mathematical analysis... 0 is the sole infantesimal". The claim 0.999... = 1 is about standard analysis. SIA is very much *non-standard*; Bell has to break a bunch of sensible properties to get SIA.
# Straddling pages 5 & 6, "in SIA no microquantity (apart from zero itself) is provably not equal to 0" - so what microquantities does SIA have? Introduce an axiom to the effect of "There exists a non-zero infintesimal", otherwise you're just playing dress-up with the Reals.
# Top of page 6: "microquantities do not have multiplicitave inverses, and R is a field" - what? Are these microquantities Reals in SIA or not? Recall from page 2: the infintesimals \Delta = \{x:x \in Reals and x^2=0 \} supposedly reside in the Reals. As do the other infintesimals in the sets I and J defined on page 8. They don't all have multiplicitave inverses. But the Reals are still a field. How?
# Top of page 7: "In some models of SIA, R satisfies the Archimedian property" - no kidding! That's when 0 is the only infintesimal and you're doing standard real analysis, but without the tools that SIA breaks.
To be honest, SIA looks like a poor attempt to do calculus in the spirit of o(h) without explicitly using o(h), breaking the law of the excluded middle, the total ordering of the Reals, and limit convergence along the way. The derivative f'(x) can be defined in standard real analysis (if/when it exists) as the unique D for which f(x+h) = f(x) + Dh + o(h^2) holds for suitably small, non-zero h. SIA just invents some new objects to sweep o(h^2) under the rug. Says so in the very first paragraph. There may be some justification for SIA, but John L. Bell doesn't provide it. The "L" in John L. Bell doesn't stand for Lattier, does it? Endomorphic 22:28, 15 November 2006 (UTC)
thread_title:  Smooth infintesimal Analysis 

message: It clearly says at the top that "this page is for mathematical arguments..."  So, how do we characterize the actual dissident arguments below?  The "autistic kid" doesn't believe it argument, or the "Lattier Demanipulation Theorem", or the similar arguments denying very basic rules of calculus or laws of real numbers - they're not "mathematical" by nature.  They are purely intuitive arguments presented by laypeople who have not even mastered arithmetic, much less Calculus.  Is this the right forum, or should this introductory statement be changed to allow for delusional intuitive arguments? 131.216.2.83 20:20, 18 November 2006 (UTC)
replies: text: And this comment is a mathematical argument? Vantucci 22:05, 18 November 2006 (UTC)
replies: text: Perhaps we should name this the personal attack forum? People get very sensitive when their magic tricks are exposed as fraud. For newcomers such as 131.- I suggest you read back into the archives for mathematical arguments. I don't feel like repeating myself, especially when greeted with disrespect. --JohnLattier 03:18, 20 November 2006 (UTC)
replies: text: It's not fraudulent magic tricks, it's mathematics as universally accepted by mathematicians. People actually find it annoying when other people accuse formally defined and universally accepted mathematics of being "trickery". Particularly when those other people consistently fail to justify their claims, and when they ignore the thrust of long, carefully constructed arguments made specifically to help them understand, choosing instead to latch onto some small and irrelevent side-point. Consider Meni Rosenfeld's incredible but futile attempts here. Maelin (Talk | Contribs) 03:25, 20 November 2006 (UTC)
replies: text: Exactly!  I have yet to see one sound mathematical argument by JL.  Wiki isn't the place to publish the ideas of dreamers and skeptics.  It is an encylopedia meant to convey well-sourced facts.  Until JL has some sources for his CLAIMS, there is no reason to even read his drivel. 131.216.2.83 03:33, 20 November 2006 (UTC)
replies: text: This has been discussed already. Being an arguments rather than edit discussion page protects JohnLattier who repeatedly violates 3 of 4 Wikipedia founding principles - WP:OR, WP:VERIFY and WP:WWIN. Personally I blame the naming of this page. "Arguments" makes it sound like we would welcome people who stubbornly insist on refuting established, accepted, well documented, and rigerously proven facts - sorry - theorems (which are even more true than facts). "Justification" or "Interactive Proof" are perhaps preferable. We're to help people who want to know *why* 0.999... = 1. We shouldn't be discussing whether it is or not. Endomorphic 03:18, 23 November 2006 (UTC)
thread_title: The Purpose of this Page