Talk:0.999...

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• Archive 1 (to 2005-11-16):

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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 $\infty$ 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 $\infty ^{-1}$ or ${\frac {1}{\infty }}$). But since infinity does not have a defined value, this becomes 0. Therefore $0.999...=1-{\frac {1}{\infty }}=1-0=1$.

The same goes for arguments such as 'does the expression $\sum _{r=0}^{\infty }{\frac {1}{r}}$ 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. $\forall n:\sum _{i=1}^{n}{\frac {9}{10^{i}}}<0.999...$. 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: $\sum _{i=1}^{n}|x|<1$. So let n be given. Let m be the smallest integer larger than $\log _{10}(n)$. Obviously $0.999...\leq 1$, so $|x|=x=1-0.999...>1-\sum _{i=1}^{m}{\frac {9}{10^{i}}}={\frac {1}{10^{m}}}>{\frac {1}{10^{\log _{10}(n)}}}={\frac {1}{n}}$. But then $\sum _{i=1}^{n}|x|<\sum _{i=1}^{n}{\frac {1}{n}}=1$. 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 $\forall n:\sum _{i=1}^{n}|x|<1$. 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 $m>\log _{10}(n)$. Then $\sum _{i=1}^{n}(1-0.999...)<\sum _{i=1}^{n}(1-\sum _{i=1}^{m}{\frac {9}{10^{i}}})=\sum _{i=1}^{n}({\frac {1}{10^{m}}})<\sum _{i=1}^{n}{\frac {1}{n}}=1$, 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 Wikipedia'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 ($\sup _{x\in Q}(x|x^{2}<2)={\sqrt {2}}\not \in Q$)).

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 Wikipedia 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 $\sum _{i=0}^{n}|x|<1$. 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 Wikipedia? 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 $\forall n\in N:\sum _{i=0}^{n}|x|<1$, 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).

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

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

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

That takes some more work to prove, even using the above, but if you accept that $\lim _{n->\infty }{r^{n}}=0,|r|<1$, 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 $\sum _{i=1}^{\infty }{\frac {9}{10^{i}}}=\lim _{n->\infty }(\sum _{i=1}^{n}{\frac {9}{10^{i}}})$

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: $\sum _{i=1}^{n}{\frac {9}{10^{i}}}={\frac {9/10}{1-10^{-1}}}-{\frac {9/10\times 10^{-n}}{1-10^{-1}}}=1-10^{-n}$. Obviously $1-10^{-n}<1$ for all $n\in N$. But for any reasonable definition of 0.999..., there is no natural number n, so that $0.999...=\sum _{i=1}^{n}{\frac {9}{10^{i}}}$, 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 $\sum _{i=1}^{\infty }{\frac {9}{10^{i}}}=\lim _{n->\infty }(\sum _{i=1}^{n}{\frac {9}{10^{i}}})$; 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. 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

Calling people ignorant and saying that if people do not agree with you is ad hominem arguments, and generally regarded as poor style. As for the formula, I won't claim there is no mistakes, but I can't really see what you are referring to. If it should be + rather than x, you would get for n=2: 0.9/(1-0.1) - (0.9+10^-2)/(1-0.1) = 1-0.91/0.9 != 0.99.

You wan't to prove something about what happens to the difference between 1 and $\sum _{i=1}^{n}{\frac {9}{10^{i}}}$ as n grows to infinity. Great! How will you do that? Weiserstrass and I would do it by proving that since for any ε>0 we can find a n so that the difference is less than 0, the limit is 0 and thus $\sum _{i=1}^{\infty }{\frac {9}{10^{i}}}=1$. I would guess you are trying to argue that since the difference is >0 for all n, then also $\sum _{i=1}^{\infty }{\frac {9}{10^{i}}}<1$. That is not at all "*clear*". Actually, I claim that it is false, so if you want to convince me, you ought to provide a proof.

As for infinitesimals. I state that if $\sum _{i=1}^{n}|x|<1$ for all x, then x is infinitesimal. That ought to be a clear definition. I also claim that the real numbers contain no such numbers. A different formulation would be that for all real numbers x!=0, there exists a natural number n, so that $\sum _{i=1}^{n}|x|>=1$. That would work just as well for the proof, since we have shown that $\sum _{i=1}^{n}(1-0.999...)<1$ for all natural numbers n, thus either 1-0.999...<0 or 1-0.999...=0.

What are those "holes" you claim I postulate? "as small as you like" and "as close as you like" can be translated into clear logical statements: for a sequence {x_i} to "become as small as you like", it has to hold that $\forall \epsilon >0:\exists n\in N:m>n=>|x_{m}|<\epsilon$. But you might take note of the fact, that the "Archimedean property"-proof did not use any limits or epsilons or deltas, so even if you reject Weierstrass methods, you still have to accept that proof. If you agree with PlanetMath, you should have no problem with the proof (unless you still claim that the real numbers aren't totally ordered).

You still need 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) 14:32, 18 November 2005 (UTC)

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

So you claim this:

$(\forall n\in N:\sum _{i=1}^{n}{\frac {9}{10^{i}}}<1)\Rightarrow 0.999...<1$

I do not challenge the truth of the left-hand side of this deduction. I challenge the "=>" part. You haven't given any arguments (not to mention proofs) as to why you should be able to conclude the right-hand side from the left-hand side. And no: "clearly" isn't an argument, not even if you surround it with asteriskes. You challenge me to find a 'n' for which the above it false. Please read Logical conditional for a discussion about how you prove or disprove a logical implication.

But let us play around with it a bit: If the above holds, then *clearly* also:

$(\forall n\in N:\sum _{i=1}^{n}{\frac {9}{10^{i}}}<0.999...)\Rightarrow 0.999...<0.999...$

We have just substituted 0.999... for 1, and the left-hand side follows easily from the equations we studied earlier. So 0.999...<0.999...?

Ignoring my proofs won't make them go away. You haven't been able to show any incorrectness in any of the versions (beyond trying to redefine the real numbers). You haven't even given a proper definition of 0.999..., not to mention proving how the Archimedean property implies the LUB property and giving me your definition of the real numbers that doesn't include a total ordering. Rasmus (talk) 21:00, 18 November 2005 (UTC)

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 seem to lack some basic understanding about the concept of implications. Consider this. Let me claim:

$(\forall n\in N:\sum _{i=1}^{n}{\frac {9}{10^{i}}}<1)\Rightarrow 0.999...=1$ (note the = on the rhs).

We clearly agree that the lhs is true. Is it then true that you can't challenge the rhs? Or can you provide a counterexample?

It looks like you believe that something like this holds: $(\forall n:a_{n}<b\leq c)\land (lim_{i->\infty }a_{i}=c)\Rightarrow b<c$ Obviously the "0.999... < 0.999..." example I gave is a counterexample to this (and actually, it can be shown that the lhs implies that b = c).

You haven't been able to properly articulate the difference between an infinite sum and the limit of a sequence of finite sums. You keep claiming that they are different, but you haven't explained what you believe an infinite sum is. You keep trying to make a proof using some property of the finite sums, but you haven't provided any link between the finite sums and the infinite.

You should try rereading Least upper bound. Apparently you have forgotten what it means. Obviously any single number has an upper bound (it is limited by itself - you do not need to invoke the Archimedean property to prove that). But to prove that an ordered set S has the LUB-property, you have to prove that any upper bounded subset of S has a least upper bound in S. So your proof-sketch isn't even close. (And since we have already seen that the rational numbers has the Archimedean property, but not the LUB-property, I am still really interested in seeing how you are going to actually prove your assertion).

In trying to find an error with the proof I gave earlier, you claimed (contrary to the common definition) that in the real numbers "The law of trichotomy applies *only* to finitely represented numbers". It is you that are trying to change the subject by not answering this. Do you only answer the easy questions? Rasmus (talk) 15:30, 21 November 2005 (UTC)

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: $(\forall n:a_{n}<b<c)\land (lim_{i->\infty }a_{i}=c)\Rightarrow a_{n}<c$ 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

The lhs doesn't prove that .999... < 1. It proves that $\sum _{i=1}^{n}{\frac {9}{10^{i}}}<1$ for all finite n. That is very much not the same thing. You simply don't understand the concept of limits. The lhs impies the rhs in Rasmus's most recent version due to the concept of limits (and the fact that you can become arbitrarily close to 1 - that is, with any finite number of nines, you can come as close to 1 as you want. .999... and 1 are both the limit of that sum as n approaches infinity.) And your patronizing of Rasmus doesn't make your arguments look any more viable. --HeroicJay 17:36, 21 November 2005 (UTC)

And the law of trichotomy doesn't say anything whatsoever about only applying to numbers with a finite representation. It simply says that, for any two real numbers x and y, exactly one of these three things is true: x > y, x = y, or x < y. That's the whole law. --HeroicJay 17:46, 21 November 2005 (UTC)

For the new little bit that you added, might I point out that it makes no sense. If $lim_{i->\infty }a_{i}=c$, then there exists no b for which $\forall n:a_{n}<b<c$. It's a contradiction in terms. Plus, your rhs pulls an i out of thin air (it's no longer part of the limit.) (EDIT: Scratched out because he replaced it with an n, even though that still doesn't make the assertion he thinks it does. The part about nonsense concerning limits still appplies.) --HeroicJay 17:50, 21 November 2005 (UTC)

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

I know *you* said it. *You* were either mistaken or lying. And, given that, I don't think *you* should be the one talking about reading comprehension. --HeroicJay 18:00, 21 November 2005 (UTC)

*I* did not add it - *Rasmus* added it. He was trying to prove his point. So in saying I am wrong, you are saying he is also wrong. However, from the way *you* (I will continue to do this because it annoys little boys like *you*) write, I can tell you are some immature young punk who has nothing better to do with his time. Have you not heard the saying?

******Speak when you are spoken to boy!******** 192.67.48.22 — Preceding unsigned comment added by 158.35.225.229 (talk • contribs)

You must play by the rules here, which state that the talk page is only to be used to try to improve the article. In particular, you are to remain civil at all times, and no personal attacks are allowed. If you do not obey these rules, you will be blocked from editing Wikipedia. -- Jitse Niesen (talk) 21:38, 21 November 2005 (UTC)

So, you're going to deny what you just admitted that you admitted. Even though I just searched the whole conversation and the only person stating that the law of trichotomy only applied to finitely represented numbers was you, while Rasmus himself insisted that that was not the case after you said it was the first time. Okay, whatever. And I thought this was a public website that anyone could contribute to, so thank you for proving me wrong about that while not responding to any of the facts that I presented just because I wasn't the person you were initially speaking to. I'm sure Rasmus was extremely offended by my addition to this conversation, as I must thank you for pointing out, which is why he said absolutely nothing to or about me when responding to comments that you made after I made my first comments in this conversation. I'm sure his lack of response indicated how angry he was. And thank you for being so much higher and mightier than me, which you apparently deem yourself to be. Now, would you like to discuss facts, or would you like to lie about who said what and insult me some more? --HeroicJay 20:27, 21 November 2005 (UTC)

Anyway, more on limits. The definition of the limit of a sequence of real numbers is as follows: $lim_{i->\infty }a_{i}=c\Leftrightarrow \forall e>0,\exists n\in N{\mbox{ such that }}\forall x>n,|c-a_{x}|<e.$. You're saying that your e is c - b? Well, no matter what, there's an n such that, for all x > n, c - a(x) < c - b, which in turn states that a(x) > b (remember, you flip an inequality sign when multiplying by a negative number. Also, since all a(n) with finite n in this case are smaller than 1 anyway, the absolute value sign has been negated.) Don't even waste your time telling me that this isn't the case; that is the definition of limits. --HeroicJay 18:22, 21 November 2005 (UTC)

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: $(\forall n:a_{n}<b<c)\land (lim_{i->\infty }a_{i}=c)\Rightarrow a_{n}<c$. Well, first of all: what you need is a rule that concludes that b < c (here $a_{n}=\sum _{i=0}^{n}{\frac {9}{10^{i}}}$, $b=\sum _{i=0}^{\infty }{\frac {9}{10^{i}}}$ and c=1. Note that there is no n, so that $b=a_{n}$.) Secondly, even disregarding that, this form of implication would be what is called begging the question: your premises include the conclusion (namely that $a_{n}<c$). Note that the rule I wrote, did not have this problem, since I only assumed that $a_{n}<b\leq c$ not $a_{n}<b<c$ (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 need to provide more context to your challenges. Which formula is it you want me to find an n for? Obviously I am not denying that $(\forall n:a_{n}<c)\Rightarrow a_{n}<c$. But you have not given the faintest argument for how you would use it to argue your claim that $(\forall n:\sum _{i=0}^{n}{\frac {9}{10^{i}}}<1)\Rightarrow \sum _{i=0}^{\infty }{\frac {9}{10^{i}}}<1$. Are you claiming that $(\forall n:\sum _{i=0}^{n}a_{i}<x)\Rightarrow \sum _{i=0}^{\infty }a_{i}<x$?

Please remember that the discussion started out by you trying to refute the proof I gave. I have shown how your claims about the proof being incorrect were founded in an incorrect understanding of the real numbers, but you keep evading my challenges to either confirm that you work with another definition of the real numbers than the rest of us, or to admit that your claims were incorrect. For each step in "my" proof, I can show how it is founded in either some property of the real numbers (Archimedean property, total ordering etc.) or some self-evident property of 0.999... (0.99...9 < 0.999... <=1). You, however, when challenged to explain which property of the real numbers lead you to believe that "sum (i=1 to n) 9/10^i < 1 implies .999... < 1", evade the issue and challenge me to disprove it. Well, since you refuse to generalize it to another rule (such as the one I gave above), the only way to disprove it is by showing that the lhs is true, while the rhs is false. Well, the lhs can be shown to be true by using the sum for the geometric series, and the rhs is false, since by my earlier proof, 0.999...=1 (and since (R,<) are totally ordered, 0.999...=1 => (not) (0.999...<1)). I am sorry if you do not like this argument, but the only way to disprove the statement (A => B) is to prove A and disprove B. Rasmus (talk) 14:25, 22 November 2005 (UTC)

You have not provided any proof and the discussion did not start out with me disproving your 'proof'. You stated that $(\forall n:\sum _{i=0}^{n}{\frac {9}{10^{i}}}<1)\Rightarrow \sum _{i=0}^{\infty }{\frac {9}{10^{i}}}<1$ and asked me if this is what I was claiming to which I responded in the affirmative. And yes, I am claiming that $(\forall n:\sum _{i=0}^{n}a_{i}<x)\Rightarrow \sum _{i=0}^{\infty }a_{i}<x$ 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

I already did. Let $a_{i}={\frac {9}{10^{i}}}$ and $x=\sum _{i=0}^{\infty }{\frac {9}{10^{i}}}$. Then the lhs becomes $\forall n:\sum _{i=0}^{n}{\frac {9}{10^{i}}}<\sum _{i=0}^{\infty }{\frac {9}{10^{i}}}$, which is obviously true, and the rhs becomes $\sum _{i=0}^{\infty }{\frac {9}{10^{i}}}<\sum _{i=0}^{\infty }{\frac {9}{10^{i}}}$, which is obviously false. Rasmus (talk) 14:50, 22 November 2005 (UTC)

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

I did not assume "1 = sum (i to infinity) 9/10^i". I chose $a_{i}$ and x as mentioned above and substituted them in the implication. Surely you are able to substitute? Let us do it slowly: the lhs was:

$(\forall n:\sum _{i=0}^{n}a_{i}<x)$.

We insert $a_{i}={\frac {9}{10^{i}}}$ and $x=\sum _{i=0}^{\infty }{\frac {9}{10^{i}}}$:

$(\forall n:\sum _{i=0}^{n}{\frac {9}{10^{i}}}<\sum _{i=0}^{\infty }{\frac {9}{10^{i}}})$

That is the lhs. Now for the rhs:

$\sum _{i=0}^{\infty }a_{i}<x$

again we insert $a_{i}$ and x:

$\sum _{i=0}^{\infty }{\frac {9}{10^{i}}}<\sum _{i=0}^{\infty }{\frac {9}{10^{i}}}$

Putting both together we arrive at:

$(\forall n:\sum _{i=0}^{n}{\frac {9}{10^{i}}}<\sum _{i=0}^{\infty }{\frac {9}{10^{i}}})\Rightarrow \sum _{i=0}^{\infty }{\frac {9}{10^{i}}}<\sum _{i=0}^{\infty }{\frac {9}{10^{i}}}$

A true statement implying a false. We have made no assumptions beyond the original implication, yet we have an obvious contradiction.

As for your induction "proof", induction works in this way: If you have propositions P(k), and prove that P(1) is true and that P(k) => P(k+1), you have shown that P(k) is true for all natural numbers k. Since $\infty$is not a natural number, you can't use this to show anything about infinite sums. Rasmus (talk) 15:26, 22 November 2005 (UTC)

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

Of course, I can substitute whatever I want. If you claim that $(\forall n:\sum _{i=0}^{n}a_{i}<x)\Rightarrow \sum _{i=0}^{\infty }a_{i}<x$, you must be able to choose $a_{i}$ and x, however you want. I chose them to provide an obvious counterexample to your claim.

As for induction, you need to read up on Mathematical induction. You keep ignoring my reduction ad absurdum examples, but assume that what you claimed was true: Then let P(m)="m is a natural number". Obviously P(1) is true, and since the natural numbers are a group, if m is a natural number, then m+1 is a natural number, too. So P(m)=>P(m+1). But infinity is not a natural number. Rasmus (talk) 16:12, 22 November 2005 (UTC)

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

I told you before to discuss the facts by themselves. It is never necessary to talk about the people you are having a discussion with. Stick to the maths. For your convenience, I repeat the relevant policies I mentioned before: Wikipedia:Civility. and Wikipedia:No personal attacks. Thank you. -- Jitse Niesen (talk) 18:57, 22 November 2005 (UTC)

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

I maintain that your comments were not civil. I would have said the same to Rasmus if I had seen him say something similar.

I don't think your comments are improving the contents. All mathematical textbooks say that 0.999... equals 1. It is the purpose of an encyclopaedia, like Wikipedia, to report what the experts say. So, Wikipedia will say that 0.999... equals 1. If you want to change that, then you have to publish a paper or a book proving that 0.999... does not equal 1, you have to convince some mathematicians so that they will also write that in their papers and books, and then it should be picked up by Wikipedia. This is one of the other policies here: our purpose is not to commit research ourselves, see no original research. -- Jitse Niesen (talk) 15:42, 23 November 2005 (UTC)

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)

It is not only what some academics think, it is what all academics that I've come across, and certainly all mathematicians, think. If you want to try to change the policy, you can discuss this at Wikipedia talk:No original research or Wikipedia:Village pump (policy). I think the policy has served us very well. -- Jitse Niesen (talk) 17:40, 23 November 2005 (UTC)

You are wrong. You speak only your opinion when you say all mathematicians. I would certainly not call you a mathematician or any of Wiki's puppets. It is what you choose to believe. You are arrogant and foolish. I will not waste my time with you anymore. 67.10.167.95 00:06, 24 November 2005 (UTC)

Full circle. I again refer to my 4 < 4 reductio ad absurdum argument below, or, for an infinite summation argument, $\infty <\infty$. I think Mr. 192's biggest problem here is that he's treating infinity as though it is a natural number - an arbitrarily big one, specifically. It's not a number at all. It's a concept, which is really only useful in some concepts in higher math, such as limits and integrals and so forth. --HeroicJay 16:54, 22 November 2005 (UTC)

You claimed that the implication was true. I provided a sequence $a_{i}$ and real number x for which it wasn't true as a counterexample. Ergo, the implication can't be true.

Yes, using induction you can prove the truth of an infinite set of propositions. However, that does not mean you can ignore for which infinite set you prove your propositions. I probably repeat myself now, but using your proposition P(k)="$\sum _{i=1}^{k}a_{i}<1$", induction only proves the truth for $k\in N$. There is no $k\in N$ so that P(k)="$\sum _{i=1}^{\infty }a_{i}<1$".

That (N,+) is a group has a clear meaning, and is sufficient to conclude that $k\in N\Rightarrow k+1\in N$. It is also relevant, since it is a counterexample to your assumption that $(P(1)\land (\forall k\in N:P(k)\Rightarrow P(k+1)))\Rightarrow (\forall k\in (N\cup \infty ):P(k))$

Rasmus (talk) 18:58, 22 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

N is the set of natural numbers. It is infinite in size, but it does not include infinity. Hence, $(N\cup \infty )$ means "The set of all natural numbers, and infinity." Now, it still seems you do not understand the point of reductio ad absurdum. The point is to take a logical construct and find a case leading to a contradiction if it's true, meaning that the original construct was false. As Rasmus was attempting to explain, your main argument was in the form $(\forall n:\sum _{i=0}^{n}a_{i}<x)\Rightarrow \sum _{i=0}^{\infty }a_{i}<x$. If this is true, it must be true for all a_i and x so long as the left side of the implication holds. He was substituting an a_i and x for which the left side held, but the right didn't (because, as you so keenly noticed, it makes a number less than itself.) Thus, $(\forall n:\sum _{i=0}^{n}a_{i}<x)\Rightarrow \sum _{i=0}^{\infty }a_{i}<x$ cannot hold in all cases, and you have yet to provide a special reason why your a_i and x hold but others don't. And re: your comment earlier about assuming what happens at infinity, it's not really an assumption if you can prove it. And finally, can I assume that your "everyone" doesn't include me? --HeroicJay 21:13, 22 November 2005 (UTC)

Do you even know what a counterexample is? If I make some claim, say $x^{2}>y\Rightarrow x>y$, a counterexample is a pair of numbers x,y so that $x^{2}>y$ but not $x>y$. In this case x=-1, y=0 would be a counterexample. I am not sure what you mean by restricting the choice to "true" choices. If you mean restricting to choices for which the implication holds, you wouldn't be able to give a counterexample to something so obviously false as the implication above.

I meant $N\cup \{\infty \}$, not $N\cup \infty$. Sorry. The point was that you try to conclude the truth of the statement $P(\infty )$ from the induction proof. N is infinite, but does not include the element $\infty$. You are not applying the Principle of Mathematical Induction correctly. Rasmus (talk) 21:25, 22 November 2005 (UTC) (I wrote this at the same time as HeroicJay, which is why we make the same points)

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 $b\leq c$. 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 $(\forall n\in N:\sum _{i=1}^{n}1$ $(=1*n)<\infty )\Rightarrow \sum _{i=1}^{\infty }1$ $(=1*\infty )<\infty$, 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)

Really? Do you care to prove what you wrote about 0.999... < 0.999... ? Do you know what QED means? You have demonstrated nothing but your ignorance.192.67.48.22

I was being ironic. Of course I don't believe that 1 + 1 + 1 + 1 < 4; that's just ridiculous. But it's a consequence of the line of logic you're providing. And the fact that 0.999... < 0.999... is a logical consequence of the same, that is, that because .9 < 1, and .99 < 1, and .999 < 1, and so on for all finite numbers of digits, that .999... < 1. But .9 < .999..., and .99 < .999..., and .999 < .999..., and so on for all finite numbers of digits, so you didn't *prove* anything *clearly* at *all*. --HeroicJay 17:36, 21 November 2005 (UTC)

I have another way to 'sort-of' prove that 0.999... = 1. It doesn't exactly prove that 0.999... = 1 but the same principles work. Assume a graph $f(x)=x^{2}$. Then, by differentiating by first principles, ${\frac {\delta y}{\delta x}}={\frac {f(x+\delta x)-f(x)}{\delta x}}$. Then assume we want to find $f'(2)$. We can set $\delta x$ to become closer and closer to 0 in order to find the gradient at that point. Notice that ${\frac {\delta y}{\delta x}}_{\delta x=0.1}={\frac {f(2+0.1)-f(2)}{0.1}}=4.1$, ${\frac {\delta y}{\delta x}}_{\delta x=0.01}={\frac {f(2+0.01)-f(2)}{0.01}}=4.01$,

${\frac {\delta y}{\delta x}}_{\delta x=0.001}={\frac {f(2+0.001)-f(2)}{0.001}}=4.001$... Then we can see that therefore ${\frac {\delta y}{\delta x}}_{\delta x=10^{-\infty }}={\frac {f(2+10^{-\infty })-f(2)}{10^{-\infty }}}=4+10^{-\infty }$. However, if $f(x)=x^{2}$, then we know that as $\delta x\rightarrow 0$ then ${\frac {\delta y}{\delta x}}\rightarrow {\frac {dy}{dx}}$, and then $f'(2)=4$. It must be then that ${\frac {\delta y}{\delta x}}_{\delta x=10^{-\infty }}={\frac {f(2+10^{-\infty })-f(2)}{10^{-\infty }}}=4+10^{-\infty }=4$ as $10^{-\infty }$ 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.

Then assume that as $\delta x\rightarrow 0$ instead of putting $10^{-\infty }$. I just put that there instead of putting .000...1. x42bn6 Talk 01:52, 24 November 2005 (UTC)

intro

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)

This article not only misrepresents 'facts' but is completely biased and is the 'opinion' of the Wiki clowns and authors. If Wiki allows this article to exist, it demonstrates clearly to any astute thinker the dependability of information from Wiki. You are a laughing stock amongst many readers. Any time someone desires a good laugh. all he/she needs to do is search for a Wiki article on a controversial topic. 0.999... has never been equal to 1, is not equal to 1 and will never be equal to 1. This is the fact. What you write here is absolute rubbish. Many of your articles are non-factual and contain serious errors because your editors are idiots. — Preceding unsigned comment added by 158.35.225.229 (talk • contribs) 14:09, 2005 November 21 (UTC)

This article is one of many in mathematics, where the consensus is that proofs are as important as facts. Most proofs have been refactored into their own pages. It is necessary to discuss context for the statement, its meaning, and its proofs, as is routine in mathematics articles. (This is by no means the kind of "discussion" found on a talk page.) Each of the four proofs presented is completely different. This well-established route to mathematical understanding is found in a number of Wikipedia "Proof" articles. Here, it makes the material useful and interesting for a wide audience.

It is my impression that many outside of mathematics think that if a theorem has been proved once, that we can thereafter forget the proof and just use the fact. Among mathematicians, the view is dramatically different. For example, the great mathematician Carl Friedrich Gauß proved the law of quadratic reciprocity in eight ways, and the Wikipedia page of proofs links to a web site listing over 200 variations. If the only interest were in establishing the validity of the theorem, this would be serious overkill!

Wikipedia has a vigorous community of mathematical contributors. Come talk at WikiProject Mathematics if you would like to learn more about the role of proofs, and their importance for Wikipedia. --KSmrq^T 14:42, 20 November 2005 (UTC)

I understand what proofs are and their importance, but asserting that "The purpose of [an] article is to discuss and prove" something is contrary to the intended function of Wikipedia. The purpose of a talk page may be a discussion, and the purpose of the proof(s) is certainly to prove the statemenr, but the purpose of a Wikipedia article is to present facts about the subject based on accepted data and consensus from said talk pages. -- WikidSmaht (talk) 07:25, 21 November 2005 (UTC)

Most likely we broadly agree on the purpose of a Wikipedia article; I can't be sure. If you wish to assert that Wikipedia has no place for proofs, I again suggest that you raise the issue at WikiProject Mathematics. The implications of a decision to ban proofs would go far beyond this one article, so this talk page is not an adequate venue for the discussion. I might also point out that at the Village Pump one regularly sees requests for more discussion in mathematical articles, never less. And again, this is discussion in the sense of presenting facts with their prerequisites, context, and implications, not in the sense of the discussion we're having on this talk page. I think we can agree that a Wikipedia article is not a chat room. :-) --KSmrq^T 13:51, 21 November 2005 (UTC)

I definitely wasn't saying that proofs shouldn't have articles. My point is that I objected to the original intro wording because the purpose of the proof is to prove the concept, whereas the purpose of the article about the proof is only to inform, NOT to "discuss" or "prove" anything. -- WikidSmaht (talk) 19:31, 24 November 2005 (UTC)

About the above discussion

I've been reading the above discussion for a while, and it seems to me that our anonymous friend thinks that: $(\forall n\in N:\sum _{i=1}^{n}{\frac {9}{10^{i}}}<1)\Rightarrow \sum _{i=1}^{\infty }{\frac {9}{10^{i}}}<1$

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 $[0,{\frac {1}{n}}]$ where $n\in N$. 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: $m>n\Rightarrow [0,{\frac {1}{m}}]\subset [0,{\frac {1}{n}}]$

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: $\forall n\in N:\sup(\bigcap _{i=1}^{n}[0,{\frac {1}{i}}])>0$

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: $\sup(\bigcap _{i=1}{\infty }[0,{\frac {1}{i}}])>0$

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:

• $\sup(\bigcap _{i=1}{\infty }[0,{\frac {1}{i}}])\subset [0,{\frac {1}{m}}]$
• $\sup[0,{\frac {1}{m}}]={\frac {1}{m}}$

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: $\sup(\bigcap _{i=1}{\infty }[0,{\frac {1}{i}}])=0$

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)

JIP: I think you have made an error in assuming that the supremum is 'x'.

You stated that:

"Call this supremum x. Clearly x is greater than 0. Now let m be a natural number whose inverse is smaller than x."

Way I see it: if x were the supremum, you would not be able to find an 'm' such that your conclusion is true. I see no problems with her logic. Her method of induction is valid and in my opinion proves conclusively that 0.999... < 1. Don't know why I didn't ever think of a proof by induction. — Preceding unsigned comment added by 71.248.130.218 (talk • contribs) 15:52, 2005 November 25 (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? 71.248.136.206 01:01, 26 November 2005 (UTC)

Mathematicians who think 0.999... and 1 are different numbers

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)

Anon, how about signing your posts? Use four tildas for that, like this: ~~~~

By the way, I am tired chasing after you as you change IP address each time. What if you make an account? Oleg Alexandrov (talk) 22:06, 25 November 2005 (UTC)