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Revision as of 13:40, 18 November 2005

Because of their length, the previous discussions on this page have been archived. If further archiving is needed, see Misplaced Pages:How to archive a talk page.

Previous discussions:

• Archive 1 (to 2005-11-16):

0.999... < 1 ?

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 (UTC)

You, also, are wrong. Of course it is correct that it will never exceed 1, and also that it will never exceed 3, but that is not the meaning of the assertion that the infinite expansion equals 1. If it were, then it would also be equal to 3. What is essential is that 1 is the smallest number that the finite truncations will never exceed.

And you are wrong that you learned how to find the limit of an infinite sum. What is taught is how to find the limit of the sequence of finite sums, not the limit of an infinite sum.

And you must have meant, NOT that he was "referring to sequences with positive sums", but to sequences with positive terms.

You wrote that "the fact that you can take the sum as close as you want to 1 does not mean it is equal to 1". You're very confused: you can take the sum of the finite truncations as close as you want to 1, but no one said those are "equal to 1". It is the infinite sum, not the infinitely many finite sums, that were asserted to be equal to 1.

Your points are very childish. If you need help in math, you could ask me or some other professional for such help. Michael Hardy 02:09, 18 October 2005 (UTC)

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 (UTC)

Would you please sign your postings, even if only with an IP number, so that we can know whether two anonymous postings are by the same person or different persons?

I have taught mathematics at five different universities, including MIT for three years, so I am not ignorant of mathematics, and if you want to understand these matters, you would benefit from listening to what I tell you.

You wrote: "I am talking about the 'limit of an infinite sum', not 'limit of the sequence of finite sums'". I don't know what "limit of an infinite sum" means in this context, unless it just means "limit of the sequence of finite partial sums", in which case it's a confused and confusing way of saying that.

Look: The value of an infinite sum IS the limit of the sequence of finite partial sums. They're the same thing.

In particular, the value of an infinite repeating decimal expansion such as 0.33333... IS just the limit of the sequence of finite truncations of it. The limit of the sequence of finite truncations of 0.3333333... is 1/3, so the value of this decimal expansion is 1/3.

You appear not to understand what "Cauchy sequence" means. To say that the terms of a series are getting closer to 0, or even that they are approaching 0, does not imply that anything is a Cauchy sequence. "Getting closer to 0" does not imply approaching 0 as a limit, since the terms of the sequence (1 + (1/n)) get closer to 0 without approaching 0. Moreover, that the terms of a series approach 0 does not imply that its sequence of partial sums or any other sequence associated with it is a Cauchy sequence.

As I said, if you need help in these matters, you should ask me or some other professional. Michael Hardy 18:17, 18 October 2005 (UTC)

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 (UTC)

There is nothing sacred about writing numbers in "bases" like base 10, as opposed to fractions like 1/3. Writing "1/3" or "√2" does represent these numbers finitely. That doesn't say wether they're rational or not. "1/3" is rational; √2 is irrational, but both are represented "finitely" here. Michael Hardy 01:38, 19 October 2005 (UTC)

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

Firstly, to answer one of your questions above: no I did not expect you to be intimidated; I expected you perhaps to be grateful.

Secondly, base 10 is no more sacred than base 3. The number 0.333..., whose expansion in base 10 is infinite, has a finite expansion in base 3, and in that base, the number 1/10 does not have a finite expansion, but an infinite repeating one.

Thirdly, the number 0.9999... with "9" repeating forever, does have a finite base 10 expansion, since it is 1.0, and can be written as a/b, where a and b are integers, since it is 1/1.

Your notions about what is a "rational number" and what is not are merely an example of what many people like to call "mere semantics". Michael Hardy 19:16, 19 October 2005 (UTC)

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

That something is required for rationality does not mean that everything that satisfies it is rational. That confuses a necessary condition with a sufficient condition. A necessary condition for rationality is not a sufficient condition to guarantee rationality.

But I congratulate you on the large number of your words. Michael Hardy 20:26, 19 October 2005 (UTC)

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

You must be a retired lawyer. Michael Hardy 00:01, 20 October 2005 (UTC)

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

Gee, if I'd realized you were Paris Hilton, I'd have recognized your mathematical brilliance. Michael Hardy 01:59, 21 October 2005 (UTC)

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

This is quite trivial... I think the problem stems from the fact that ${\displaystyle \infty }$ CANNOT be regarded as a number with a definitive value. It is just said to be the largest number possible (which, of course, does not exist). I mean, the number 0.999... can be regarded as 1 minus the smallest real number possible (which is ${\displaystyle \infty ^{-1}}$ or ${\displaystyle {\frac {1}{\infty }}}$). But since infinity does not have a defined value, this becomes 0. Therefore ${\displaystyle 0.999...=1-{\frac {1}{\infty }}=1-0=1}$.

The same goes for arguments such as 'does the expression ${\displaystyle \sum _{r=0}^{\infty }{\frac {1}{r}}}$ converge to a limit?' The answer: no. Certainly, the numbers get smaller and smaller and smaller until they become almost equal to zero, but no, THE SUM WILL NOT DIVERGE TO A LIMIT (I will not prove it here).

This debate has gone on for ages; the problem stems from the fact that non-mathematicians will not understand geometric series easily, nor will they understand easily that infinity has no defined value.

I know I've dragged in quite a bit of other problems here, but my main point is that non-mathematicians will never understand this article fully.

But I am a firm believer that tapping '9' on a calculator forever is not within physical limits.  :) x42bn6 Talk 07:21, 8 November 2005 (UTC)

I think you're assuming that all infinite series are divergent. However, there do exist convergent series (like the one used in the proof that 0.999... = 1). Take a look through that article and hopefully it will explain convergence well enough. --BradBeattie 14:46, 8 November 2005 (UTC)

No, I did quite a bit of work on ratio test so I certainly know that convergence exists. x42bn6 Talk 03:25, 9 November 2005 (UTC)

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 (UTC)

Yes, I do understand what the difference is. And I understand that non-mathematicians will never understand this fully until they understand the concept of infinity. And please sign your comments in talk pages using ~~~~. x42bn6 Talk 06:32, 13 November 2005 (UTC)

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

Actually X42bn6s argument can be nicely formalized as a proof using the Archimedean property of the reals:

Assume 0.9999... != 1. We will show that this implies that 1-0.999... is an infinitesimal. For any reasonable interpretation of 0.999... it must be larger than any finite-length 0.999...9. I.e. ${\displaystyle \forall n:\sum _{i=1}^{n}{\frac {9}{10^{i}}}<0.999...}$. Let us call the quantity 1-0.999... for x. To show that x is an infinitesimal we have to show that for all n: ${\displaystyle \sum _{i=1}^{n}|x|<1}$. So let n be given. Let m be the smallest integer larger than ${\displaystyle \log _{10}(n)}$. Obviously ${\displaystyle 0.999...\leq 1}$, so ${\displaystyle |x|=x=1-0.999...>1-\sum _{i=1}^{m}{\frac {9}{10^{i}}}={\frac {1}{10^{m}}}>{\frac {1}{10^{\log _{10}(n)}}}={\frac {1}{n}}}$. But then ${\displaystyle \sum _{i=1}^{n}|x|<\sum _{i=1}^{n}{\frac {1}{n}}=1}$. Thus we arrive at a contradiction: if 0.999... != 1, 1-0.999... is an infinitesimal (or 0.999... < 0.99...9 for a finite-length number, or 0.999... > 1). Since the real numbers possess the Archimedean property and thus possess no infinitesemals, 0.999... = 1.

Rasmus (talk) 19:53, 15 November 2005 (UTC)

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

x is an infinitesimal if and only if ${\displaystyle \forall n:\sum _{i=1}^{n}|x|<1}$. That is the definition, not an assumption. Feel free to use another definition, but unless it is equivalent to this one, you are speaking about something else. The Archimedean property of the reals is, that it does not contain any numbers (except for zero) that can be summed arbitrarily many times and still be finite. You can conceive of fields that does not have this property, but it is not the reals (see Hyperreal numbers).

Calling 1-0.999... for x does nothing for the proof except improve the readability. Feel free to substitute (1-0.999...) everywhere, it makes no difference for the correctness: Let n be given and choose ${\displaystyle m>\log _{10}(n)}$. Then ${\displaystyle \sum _{i=1}^{n}(1-0.999...)<\sum _{i=1}^{n}(1-\sum _{i=1}^{m}{\frac {9}{10^{i}}})=\sum _{i=1}^{n}({\frac {1}{10^{m}}})<\sum _{i=1}^{n}{\frac {1}{n}}=1}$, and we have proven that either 1-0.999... is an infinitesimal, 0.999...<0.99...9 for a finitelength-number, 0.999...>1 or 0.999...=1.

Rasmus (talk) 21:25, 15 November 2005 (UTC)

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

You make a lot of hand-waving here. The definition of infinitesimal I have given is the one used everywhere. You may have a different conception about what an infinitesimal is and be unable to properly define it, but the one I talk about is well-defined and well-understood. Likewise the reals. Likewise most mathematicians have an agreement about what the real numbers are, and which properties they possess. You might have a conception of a number-field that does not include pi, e and sqrt(2), but it is not the same as the one the rest of us call the real numbers.

Anyway: the proof above is interesting because it avoids limits and infinite sums, only drawing on some intuitive correct assumptions about the properties of 0.999... and the Archimedean property of the real numbers (which again is a consequence of the least upper bound property). I just mentioned it since you attacked X42bn6s intuitive understanding. There is really no purpose of us discussing the properties of the real numbers here, since the content of the article is governed by WP:NOR. Unless you can produce reputable sources for 0.999... ≠ 1, the article will continue to assert that 0.999... = 1.

Rasmus (talk) 13:31, 16 November 2005 (UTC)

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 Misplaced Pages's views?! 192.67.48.22

You wave your hands, when you claim that the definition of infinitesimals I used is untrue without supplying references. You claim: "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." From that you must either believe that the reals contain infinitesimals or that pi, e etc. does not belong to the reals. Take your pick. (Btw. it is easy to define a number field that does not include pi, e and sqrt(2). The rational numbers are one example).

You claim that the LUB property is a consequence of the Archimedean property. You have also stated that there is no definition of infinitesimals. What is your definition of the Archimedean property? We obviously have different definitions, since it is easy to see your claim is false using my definitions (The rational numbers obviously have the Archimedean property since Q ⊂ R, but they do not have the LUB property (${\displaystyle \sup _{x\in Q}(x|x^{2}<2)={\sqrt {2}}\not \in Q}$)).

Read WP:NOR for a discussion about reputable sources. In this case it would be a mathematical text-book or a peer-reviewed article. Proving something in the context of Misplaced Pages consists of referring to a reputable source. Disproving is the same. In case of disagreement as to what a reputable source is, we go by consensus. Using this meaning, you have neither disproved that 0.999...=1 or that 0.999...≠1. In the context of mathematics, a proper proof consists of enumerating your definitions and using these to deduce your conclusion. This is done in my proof above and in the (advanced) proofs in the article. Refuting a proof consists of showing that the deductions were incorrect. All you have done is arguing about the definitions. Hand-waving consists of claiming that something is false, but not giving references, counterexamples or proofs. You have done plenty of that.

Rasmus (talk) 14:29, 16 November 2005 (UTC)

I do not need to supply any references to show that your definition of infinitesimal is false. All I have to do is set n = infinity and already I have problems. As far as infinitesimals belonging or not belonging to the reals, it is you who have to take your pick! Firstly, you use the plural form - this is a contradiction in itself. Are there infinitesimals that are smaller than other infinitesimals? Secondly, the completeness principle states that every non-empty set which is bounded from above has a LUB. If you are using this, which you are, then you have to make up your mind whether you are using infinitesimals or not. I am not ignorant of mathematics so your discussion about fields above reveals nothing that I did not already know. Had you read my response in context (which you did not), you would have understood that I am claiming pi, e and sqrt(2) are part of the reals and these numbers have something in common with 0.999..., 0.333... in that they can only be appromixated. You archived the previous posts in which I provided *valid* mathematical proof (nothing hand-waving about this). Go back and read the posts. I pointed out errors in your 'proof' and when you realized that you had in fact written rubbish, you resorted to Wiki's NOR policy. An easy way out for you? 192.67.48.22

Sigh. Obviously n has to be a natural number. But even if we assigned some meaning to n = infinity, the proof would still work. There are no real x≠0, so that ${\displaystyle \sum _{i=0}^{n}|x|<1}$. As there are no infinitesimals in the reals, we cannot really discuss their properties (in a field that contained infinitesimals, however, it would be trivial to prove that there would be at least countable many). You still haven't given me your definition of infinitesimals or of the Archimedean property. You haven't shown how the Archimedean property implies the LUB property. It looks like you didn't really mean that "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", though. There are no formal mathematical proofs in the archive. Feel free to make one here, however. WP:NOR just means that even if you should be able to produce a valid proof for 0.999... ≠ 1 and convince everybody here that it is correct, you do not get to move the article to Proof that 0.999... does not equal 1. You would have to produce a reputable source. It doesn't mean that we can't produce lots of talk-page material, that we can move to another archive, once we are done. Rasmus (talk) 19:43, 16 November 2005 (UTC)

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 Misplaced Pages? 192.67.48.22

Hmmm. If you prefer to google, rather than reading a proper textbook on the subject, let us do that: The two top hits are our own Archimedean property, which uses the exact same definition as me, and Planet Math, which uses a slightly different, but equivalent formulation (Exercise: show how they are equivalent. Hint: use corollary 1, select y=0.5 and find the contrapositive). So now we have established that there are no non-zero real numbers x, for which ${\displaystyle \forall n\in N:\sum _{i=0}^{n}|x|<1}$, we should be able to agree on the validity of the proof!

As for your 'proof', let me see if we can clear up what you mean. First you type:

Sum (i to n as n approaches infinity) = (ar - ar^n)/(1-r) for |r| < 1

I assume you mean (you don't specify the left hand side).

${\displaystyle \lim _{n->\infty }(\sum _{i=1}^{n}(ar^{i}))=\lim _{n->\infty }({\frac {ar-ar^{n}}{1-r}}),|r|<1}$

We can certainly agree on that.

Then you assert that the limit of 9/10 + 9/100 + 9/1000 + ... is 1. I assume you mean

${\displaystyle \lim _{n->\infty }(\sum _{i=1}^{n}{\frac {9}{10^{i}}})=1}$.

That takes some more work to prove, even using the above, but if you accept that ${\displaystyle \lim _{n->\infty }{r^{n}}=0,|r|<1}$, I won't disagree (the proof is not hard, but it takes some work to get all the epsilons and deltas right).

Then you say:

"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.".

When someone says *clearly* it is a clear sign that they are handwaving. Please prove this assertion! Hint: It might be a nice start to define exactly what you mean by 0.999... . Most people would mean ${\displaystyle \sum _{i=1}^{\infty }{\frac {9}{10^{i}}}=\lim _{n->\infty }(\sum _{i=1}^{n}{\frac {9}{10^{i}}})}$

Ouch. You now claim that (R,<) isn't trichotomous. Most people define the real numbers so that (R,≤) is a total ordering. Care to prove your claim? Or just define your ordering. In any case, if you hold to this claim, you are talking about a different set of numbers than what the rest of us call the real numbers (and in that case anything might be true. In Z/2, 1=3).

By the way, you still haven't shown how the Archimedean property implies the LUB property.

Rasmus (talk) 14:32, 17 November 2005 (UTC)

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

You don't need to draw the nice LaTeX formulas to make yourself clear. You do however need to setup the formulas correctly (not being lazy and skipping part of the equations) and be rigorous in how you setup your definitions and how you apply them. You might also want to skip the ad hominem arguments, you can't prove anything in maths using those. That being the case, let us get back to your proof:

You say: ${\displaystyle \sum _{i=1}^{n}{\frac {9}{10^{i}}}={\frac {9/10}{1-10^{-1}}}-{\frac {9/10\times 10^{-n}}{1-10^{-1}}}=1-10^{-n}}$. Obviously ${\displaystyle 1-10^{-n}<1}$ for all ${\displaystyle n\in N}$. But for any reasonable definition of 0.999..., there is no natural number n, so that ${\displaystyle 0.999...=\sum _{i=1}^{n}{\frac {9}{10^{i}}}}$, so you can't prove anything from that. Now it comes down to how you define 0.999... (which you haven't done yet). Most of us define 0.999... as ${\displaystyle \sum _{i=1}^{\infty }{\frac {9}{10^{i}}}=\lim _{n->\infty }(\sum _{i=1}^{n}{\frac {9}{10^{i}}})}$; but I guess you are thinking of some other definition? Anyway, just claiming that it is "*clear*" that because something holds for any finite n, it somehow also applies in the limit, doesn't make it true. You will have to prove it. And you can't really prove anything about 0.999... if you don't first make it clear which number you are talking about. While you are at it, please show me where to find your alternate definition of the Archimedean property, how to show that the Archimedean property implies the LUB property and give me your definition of the real numbers that doesn't include a total ordering.

Rasmus (talk) 20:22, 17 November 2005 (UTC)

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. 192.67.48.22