Misplaced Pages

Revision as of 17:57, 30 January 2006

Because of their length, the previous discussions on this page have been archived.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 (2005-05-06 to 2005-11-16)
• Archive 2 (2005-11-16 to 2005-12-07)
• Archive 3 (2005-12-07 to 2005-12-09)
• Archive 4 (2005-12-09 to 2005-12-20)

Rules of engagement

Dear all,

I assume all of us are well-meaning people, wanting a good result for this article. That makes me ask the following favor to anonymous contributors:

• Please make an account. It is not productive for us to deal with a person who always uses a different IP address. It is impossible for us to keep in touch with you this way. Making an account will take you five secods. Just choose an imaginary username which has nothing to do with your real name, and a password. No more. But so much gained.

• Please sign your posts. Use four tildas for that, like this: ~~~~. I would really, really ask you to give it a try. You leave a lot of unsigned comments and nobody can tell which is what you wrote, and which is somebody else, and which is another anonymous user.

I truly appreciate you taking your time to read all this, and do us a the small favors I asked which will take you a very small amount of time (not infinitesimal, but close :) and will make it so much more pleasant for us to have constructive discussions. Sincerely, Oleg Alexandrov (talk) 06:04, 18 December 2005 (UTC)

As an aside, a discussion about this very topic has been hotly debated by an on-line society for people with a high IQ. Keep it going SM! — Preceding unsigned comment added by 68.148.229.166 (talk • contribs) 06:30, 2005 December 18 (UTC)

What?! You mean Hardy is discussing this without me?! How can this be? I too have a high IQ (over 140) -not that it means anything. The IQ concept was developed by a human who I believe had to wipe his arse every time he had a BM. Do you think Hardy wipes or washes? I wash only. Toilet paper is for those who have bad hygienne and stinky butts. Although I don't care to0 much for Islam, this is something we could learn from them. Wait, I think the ancient Greeks invented the bidet if I am not mistaken? Oops, I think Melchoir is going to censor this when he wakes up. Hopefuly a few people will get to read it before he does and have a good laugh if nothing else. Not that it's any of my business but I am infinitesimally (singular ONLY) curious, do you wash or wipe Oleg? 71.248.130.143 14:14, 18 December 2005 (UTC)

I am not going to argue about infinitesimal. As far as I am concerned, this (notice I will not even consider the plural form - it makes no sense whatsoever to me) does not exist and you most certainly cannot prove that x+x+x+... < 1 for an infinite number of terms in x unless x is zero. You can only show in a similar manner to Rasmus that n*10^(-n) < 1 using induction. Now as I stated, one can show that 0.999... < 1 and that 0.999... = 1 depending on how you approach the proof. I believe that this anomaly exists because 0.999... is not a finitely represented number. This is a problem with the decimal system and all other radix systems. What this means is that the Archimedean property applies only to reals that can be finitely represented. Rasmus's proof is no better than the induction proof that 0.999... < 1 since it uses a result of induction to arrive at the conclusion that nx < 1. Rasmus justifies his argument by stating that because the Archimedean principle cannot be applied, x must be zero. However, I maintain that the Archimedean property can only be applied to finitely represented numbers. For x > 1/n, n can take on the value of a suitable natural number but not infinity. To use the fact that 0.999... has an upper bound in a proof such as Rasmus's defeats the purpose. So what should be believed? I think that 0.999... should be considered less than 1 because it has to be considered in the context of the decimal system. If the full extent of 0.999... were known, there would be no problem with the Archimedean property or any of its corollaries. 158.35.225.229 18:49, 21 December 2005 (UTC)

Infinitesimals

I have not yet given up on the "infinitesimal" proof, especially since anon agrees there are no infinitesimals in the field of real numbers. (Compare the 0:16 post of 18 December 2005. By the way, meaningful concepts of infinitesimals in larger number sets can be found in Winnig Ways for your Mathematical Plays, part 1, which is written on a rather basic level.)

0.999... and 1 are real numbers. Let x = 1-0.999... be their difference. Note that I do not care whether I can give a decimal representation for x. Note also that I do not make claims about the existence of numbers between 0.999... and 1. All I claim is that I can subtract, and that the difference of two real numbers is again a real number. And that is due to the fact that the real numbers are a field. By definition, a number y ≠ 0 is an infinitesimal if every sum |y|+...+|y| of finitely many terms is less than 1, no matter how large the finite number of terms. We agreed such a thing does not exist in the reals. Now form any sum |x|+...+|x| of finitely many terms, say n terms. Obviously, |x| = 1-0.999... < 1/n. Thus, |x|+...+|x|<1. So if x were greater than 0, it would satisfy the definition given above. That can't be, since we agreed that no real number is an infinitesimal. Our only way out is x=0. Thus, 0.999...=1.

If there are problems with this proof, please be precise in denoting them.--Huon 00:00, 20 December 2005 (UTC)

The proof would be flawless if your definition of infinitesimal is true. Only problem is it is not true because it does matter how large the finite number of terms become. If you feel comfortable that the sum of these terms will always be less than 1, how is it that you do not feel the same way about the sum of 9/10+9/100+9/1000+... ? I can make the same statement here, i.e. for finitely many terms, this sum will always be less than 1. So what?! 71.248.130.208 02:54, 20 December 2005 (UTC)

Hmmm, I had sort of given up on this discussion, but one last try: Let x = 1-0.999... as above. Consider the set S = {x, 2x, 3x, 4x, ...} = {nx|n in N}. (If you don't agree with me setting x = 1-0.999... , just consider the set S = { (1-0.999...), 2(1-0.999...), 3(1-0.999...), ... } = { n(1-0.999... ) | n in N} instead ). Could you answer these questions?

1. Does S have an upper bound?
2. Does S have a least upper bound?
3. If S has a least upper bound, what is it? If you can't give an exact answer, can you give an interval (ie. 0.5 < sup S < 1)?
4. If 1-0.999... != 0, then 1/(1-0.999...) must be a real number. Can you describe the properties of 1/(1-0.999...)? For instance is there any natural number n, so that 1/(1-0.999...) < n ?

Rasmus (talk) 07:23, 20 December 2005 (UTC)

Answers:

 1. S does not have an upper bound therefore it cannot have a least upper bound.
    So questions 2 and 3 are not relevant.

 4. 1/(1-0.999...) is a real number. Properties: all we can say is that it is a
    very large indeterminate number comparable with infinity. There is no natural
    number n, so that 1/(1-0.999...) < n.

So now you are going to conclude that since 4 is true that 1/(1-0.999...) is not a real number - yes? What about 1/(3.15-pi)? Is there a natural number n so that 1/(3.15-pi) < n ? 158.35.225.231 13:09, 20 December 2005 (UTC)

All members of S are of the form nx. Consider one such member, nx. As we saw a month ago, we can choose a natural number ${\displaystyle m>\log _{10}(n)}$, and use the fact that ${\displaystyle 0.999...>\sum _{i=1}^{m}{\frac {9}{10^{i}}}}$ to show that ${\displaystyle x=1-0.999...<1-\sum _{i=1}^{m}{\frac {9}{10^{i}}}={\frac {1}{10^{m}}}<{\frac {1}{10^{\log _{10}(n)}}}={\frac {1}{n}}}$, and thus ${\displaystyle nx<n{\frac {1}{n}}=1}$. So all members of S are less than 1, yet you claim it has no upper bound?

As for 4, you claimed here that you accepted Planet Maths definition of the Archimedean property (or was that another anon?). You don't feel this is a contradiction? Planet Math claims: "Let xbe any real number. Then there exists a natural number n such that n>x".

And finally, of course there is a natural number n so that 1/(3.15-pi) < n. pi<3.142, so 1/(3.15-pi) < 1/(3.15-3.142) = 125.

Rasmus (talk) 15:28, 20 December 2005 (UTC)

You have just proved that 0.999... < 1: nx < n*1/n = 1 => x < 1. If x is a real number greater than 0, there exists a natural n such that 0 < 1/n < x or nx > 1. So if x = 0 (which is what you would require for having 0.999... = 1) then no n exists such that nx > 1. Hence x must be greater than 0 and if x is greater than 0, then 0.999... must be less than 1. Now do the right thing and delete this garbage article. ] 17:35, 20 December 2005 (UTC)

Could you clarify that argument please? In the above I showed how for all natural numbers n, nx < 1. If you accept that "If x is a real number greater than 0, there exists a natural n such that 0 < 1/n < x or nx > 1", the conclusion must be that x is not a real number greater than 0.

You also didn't comment on the Archimedean property (are you the same person as 192.67.48.22?)

Rasmus (talk) 22:28, 20 December 2005 (UTC)

You showed nx < 1. The Archimedean property says there exists an n s.t. nx > 1. There is no x that satisfies nx < 1 and nx > 1. So how do you reach the conclusion that x = 0? You are looking only at nx < 1 and thus drawing the conclusion that x must be zero? Okay, let me try to understand what you are saying:

The Archimedean property shows the relationship between a natural number n and a number x greater than 0 such that nx > 1. This means that x and n must be greater than zero. Your proof demonstrates that a number x and some natural number n have the property that nx < 1. The only n that satifies this is n=0 for otherwise x must be zero. How do you associate the Archimedean property with your proof? They both state different facts. So what I am trying to say is this: if you are to draw any conclusion that is backed by the Archimedean property, then your proof must result in a form that resembles it, i.e. nx > 1 and not nx < 1. You cannot arrive at the conclusion that nx < 1 and then state by the Archimedean property that x is not a real number greater than 0. By demonstrating that x (1-0.999...) < 1/n for any n, you have proved conclusively that x is greater than 0 because for whatever 1/n you give me, I can always find an x that is smaller. This x is greater than zero and sounds very real to me. 00:26, 21 December 2005 (UTC)

What I showed before were that for x=1-0.999... and all natural numbers n: nx<1. (I actually only wanted to use it for showing that S had an upper bound, since you had earlier rejected the application of the Archimedean property). Since the Archimedean property state that for all real x>0, there exists a natural number n, so that nx>1, we have a contradiction unless x is not a real number greater than 0.

I can't make much sense of your last argument. You claim that "x (1-0.999...) < 1/n for any n" => "x is greater than 0"? I assume the parenthesis is just a clarification and not a multiplication, so that it is actually (for all n in N: x < 1/n) => (x > 0) ? Your argument for this seems to imply that you can change the x as you go?! Anyway x=-1 (or even x=0) is a counterexample, which, frankly, you ought to have been able to see for yourself.

Rasmus (talk) 07:32, 21 December 2005 (UTC)

Fine. I see your argument now. It's always been confusing because for any 1/n, I can always find an x that is smaller but not zero. In an earlier discussion, you maintained that the induction proof was incorrect because it does not show P(infinity). Do you realize that one can say the same to you regarding this argument? You may say that the lowest x one can find is zero but then you are assuming P(infinity) is true. So although your argument is valid, you have not shown P(infinity). It seems to me that one can show equally well by induction that 0.999... < 1 and using your method that 0.999... = 1. How can 0.999... be both less than and equal to 1? This is strange... 158.35.225.229 13:14, 21 December 2005 (UTC)

Well, the difference is that I don't need to go to the limit. To use the Archimedean property, I only need to show that for all finite natural numbers n: nx<1. I do not need to show that "${\displaystyle \infty x<1}$" (whatever meaning one would assign to that statement). Rasmus (talk) 14:07, 21 December 2005 (UTC)

One can say exactly the same for the induction proof, i.e. only need to show that 0.9999xn < 1. Same thing. Let me get one thing straight: you are also saying that if the Archimedean property does not apply, then x cannot be a real number, right? If your answer is 'yes', then the Archimedean property only applies to finitely represented reals in any radix system. 0.999... is not finitely respresented. 158.35.225.229 14:20, 21 December 2005 (UTC)

Rasmus is going to say that

• the Archimedean property is a property for the entire set of numbers, and it applies.
• x is indeed a real number, but not one that is greater than 0.

After all, what we are trying to show is just x=0. If you are now willing to sacrifice the Archimedean property for your brand of "real" numbers, you will probably agree that yours are not what mathematicians usually call the real numbers.

Concerning the definition of infinitesimals I gave above: That was the Misplaced Pages definition; I just copied it. If you don't believe that definition to be correct, look it up in, say, Winnig Ways.

Finally, of course 0.999...9 with a finite number of nines is less than one - by 10^{-n}, if n is the number of 9's. Now if you truly were going to use a limit argument for the case of an infinite number of nines, then the difference between 1 and .999... would have to be ${\displaystyle 10^{-\infty }}$, whatever that is. I personally do not endorse the following reasoning, but you might still find it interesitng: In order to show that x:=1-0.999... is an infinitesimal or zero, I can also show that x+x+x+... < 1 for an infinite number of terms: For every natural n, n*10^{-n} < 1. Thus, by your own methods, ${\displaystyle \infty \cdot 10^{-\infty }<1}$. Thus, even with a stronger (and more strange) definition and with your methods of reasoning, x is an infinitesimal or zero, and infinitesimals don't exist. Thus, x=0 and 0.999...=1. What now? If you still doubt that x is either zero or an infinitesimal, please give a definition of infinitesimal you are willing to accept (keeping in mind that in the reals, there are no infinitesimals). --Huon 17:15, 21 December 2005 (UTC)

I am not going to argue about infinitesimal. As far as I am concerned, this (notice I will not even consider the plural form - it makes no sense whatsoever to me) does not exist and you most certainly cannot prove that x+x+x+... < 1 for an infinite number of terms in x unless x is zero. You can only show in a similar manner to Rasmus that n*10^(-n) < 1 using induction. Now as I stated, one can show that 0.999... < 1 and that 0.999... = 1 depending on how you approach the proof. I believe that this anomaly exists because 0.999... is not a finitely represented number. This is a problem with the decimal system and all other radix systems. What this means is that the Archimedean property applies only to reals that can be finitely represented. Rasmus's proof is no better than the induction proof that 0.999... < 1 since it uses a result of induction to arrive at the conclusion that nx < 1. Rasmus justifies his argument by stating that because the Archimedean principle cannot be applied, x must be zero. However, I maintain that the Archimedean property can only be applied to finitely represented numbers. For x > 1/n, n can take on the value of a suitable natural number but not infinity. To use the fact that 0.999... has an upper bound in a proof such as Rasmus's defeats the purpose. So what should be believed? I think that 0.999... should be considered less than 1 because it has to be considered in the context of the decimal system. If the full extent of 0.999... were known, there would be no problem with the Archimedean property or any of its corollaries. 158.35.225.229 18:51, 21 December 2005 (UTC)

We agree that I can't show x+x+x+... < 1 for infinitely many summands unless x=0. But I definitely can "show" that (1-0.999...)+(1-0.999...)+(1-0.999...)+... < 1 for infinitely many summands. Let's do it step by step:

• 1-0.9 = 0.1 < 1
• (1-0.99)+(1-0.99) = 0.02 < 1
• (1-0.999)+(1-0.999)+(1-0.999) = 0.003 < 1 ...

Similarly, for every n, the sum of n terms of the form (1-0.999...9) (n nines) is less than 1. Thus,

• (1-0.999...)+(1-0.999...)+(1-0.999...)+... < 1 for infinitely many summands (using methots not endorsed by me).

Thus, we have 1-0.999... = 0. To be precise, this "proof" is not mathematically rigorous (that's why I employ all these quotation marks), but it is just as good as the "induction proof" claimed to show that 0.999...<1. If one of these "proofs" is correct, then so is the other. Thus, in a way, I have disproved the induction proof, since using its methods leads to a contradiction. --Huon 19:42, 21 December 2005 (UTC)

I am not sure this method works but this is not relevant to what I said. Anyway, I agreed one could prove this but I stated that it is a result of induction. I said that Rasmus's proof is also a result of induction. 158.35.225.229 19:51, 21 December 2005 (UTC)

Actually, none of the proofs here are using induction. Induction is a special technique to show that a statement is true for all natural numbers. An induction-proof is easily recognized by being split into two parts: The basis (showing that the statement is true for n=0) and the inductive step (showing that if the statement is true for n, it is also true for n+1). Neither mine, nor your or Huons have this form.

As for infinitesimals, they are not defined by any infinite sums. Rather they are defined by the property that any finite sum is less than 1. Ie. if for all natural numbers n, nx<1, we say that x is an infinitesimal. But it is just a name, if you don't like it, we can call them for very-small-numbers. We also found out that if 0.999...<1 then 1/(1-0.999...) is a illimited number very-large-number. The existence of very-small- and very-large-numbers is in contradiction with the Archimedean property and the Least-upper-bound-property. What we call the real numbers is (uniquely) characterized by being a complete ordered field that has the LUB-property. You can of course define another set of numbers without the LUB-property and choose to call them the real numbers (but then you need to convince everybody else to follow your naming-convention...). Let us take a look at your options:

You can follow Fred Richman and use the Decimal Numbers. Then by definition 0.999...<1, but you not only lose LUB-property, you also lose negative numbers (since 0.999... + 0.999... = 1.999... = 1 + 0.999..., 0.999... doesn't have an unique additive inverse) and division (since there is no Decimal Number x, so that x(1-0.999...)=1).

You can extend the real number field with infinitesimals very-small-numbers (using the transfer principle). The only important property you lose is the LUB-property (and thus the Archimedean property). Of course you cannot express these numbers using decimal numbers, and most people don't like the concept of very-small-numbers. Also, even here there is no real reason not to define 0.999...=1. But these numbers are actually rather interesting. They are called the hyperreals and are the subject of non-standard analysis.

Rasmus (talk) 22:17, 21 December 2005 (UTC)

Actually all the proofs here use induction. You have also used induction whether you like it ot not. How did you arrive at x < 1/n ? You started by assuming a finite sum and then continued to show how it is always less than 1. This is true and it proves that 0.999... is always less than 1. However, you took this result and then tried to explain it away with the Archimedean property. You are defining real numbers using the Archimedean property. What you don't seem to understand is that the Archimedean property does not allow for very small numbers or very large numbers. It just so happens 0.999... is a number that is misunderstood because it is very close to 1. However in the decimal system, there is no way to represent many numbers exactly, so you resort to the LUB property to reach conclusions about numbers. In my opinion it is just as easy to have 0.999... < 1. You do not lose any of the properties you mentioned. Look, you could easily define the real number system in terms of the decimal system if you were to use finite representation of numbers. In fact, this is how we use the decimal system. By doing so, all the properties of the real numbers hold, including very small and very large numbers and the Archimedean property.71.248.139.119 23:01, 21 December 2005 (UTC)

Rasmus and I do not use induction, since we do not use that the result holds for n in order to show it is true for n+1. Rasmus arrived at x<1/n by an explicit calculation; he did not use x<1/(n-1) in order to show it. (By the way, x was taken as 1-0.999... How does x<1/n imply 0.999...<1 ?)

Using only finite decimal representations as numbers sacrifices several properties:

• Without division (such as 1/3, which has no finite representation, if I understand you correctly), the reals are no longer a field.
• Completeness is also lost, since there are Cauchy sequences of finitely represented numbers whose limit has no fintite representation.
• That's not one of Rasmus' properties, but surely an orthogonal triangle with two sides of length one should have a third side whose length is again a real number? After all, to the ancient greek mathematicians numbers were objects of geometry, be it lengths, areas or volumes.

Finally, then our whole discussion would be rather empty, since 0.999... is not finitely represented and would not even be a real number (whether it is less than 1 cannot be answered using a set which does not even contain 0.999...). Probably I misunderstood the statement about "defining the real number system using only finite representations of numbers"; please clarify it. --Huon 00:13, 22 December 2005 (UTC)

Very well. I concede that Rasmus's proof is quite solid. I can't argue against it. I still don't think you can rule out the design of the decimal system contributing to these anomalies, i.e. you have Rasmus's proof on the one hand and a simple proof by induction on the other hand that says exactly the opposite. Maybe you should include both proofs in an article about whether 0.999... equals 1 or not. Remember the decimal system is a model of the reals. 71.248.139.119 01:39, 22 December 2005 (UTC)

Explanation for removal of good-faith insertion

An anonymous IP added a proof idea in the "Elementary" section that essentially duplicated a proof in the "Advanced section". While a good-faith edit (thanks), it doesn't seem to offer any improvement. The "squeeze play" idea is a good one, and does sometimes appear in informal explanations. The point here is that we already cover the same ground more rigorously. --KSmrq 18:56, 10 January 2006 (UTC)

The debate resumes - Problems with Rasmus's Proof

The article is still biased and incorrect. It shows only one view of this problem - Rasmus's proof that 0.999... = 1. This is an induction proof contrary to what Rasmus states. Rasmus also states that the difference between his proof and the opposing induction proof is that he does not need to run the limit to infinity. Well, the opposing proof does not require this either. To say that 0.999... > Sum (i=1 to m) 9/10^i is a result of induction. Is this still true if we run m through to infinity? The answer is no. All of Rasmus's remaining proof is based on this first induction result. Seems like Rasmus's proof is not as solid as once thought. I am inclined to have 0.999... < 1. 158.35.225.229 18:55, 11 January 2006 (UTC)

That 0.999... is greater than ${\displaystyle \sum _{i=1}^{m}{\frac {9}{10^{i}}}}$ for all natural numbers m might indeed be proved by induction; one can also prove it directly. I assume Rasmus believed it to be obvious. And indeed this result becomes wrong if we proceed to the limit for m tending to infinity. But Rasmus need not do that. On the other hand, the "opposing proof" states quite analogously that ${\displaystyle \sum _{i=1}^{m}{\frac {9}{10^{i}}}<1}$ for every finite m (which is undisputed), and then "shows", supposedly by induction, that the same must be true for the limit - just the argument which we now see to be false. That "proof" makes an even more general (false) claim; it can be found in the Archive.

Still, this gave me just another idea for a proof that 0.9999...=1. I will stop making any assumptions about the nature of 0.9999... but the following:

• For every natural number m, ${\displaystyle \sum _{i=1}^{m}{\frac {9}{10^{i}}}<0.9999...}$. Here I make no claims about limits.

Now we use two facts of analysis:

• Let a_m be a convergent sequence of real numbers with ${\displaystyle \lim _{m\to \infty }a_{m}=a}$. Let b be any real number with a_m < b for all m. Then ${\displaystyle a\leq b}$. Note that not necessarily a<b; a proof should be found in almost every undergraduate textbook on analysis.
• We have ${\displaystyle \lim _{m\to \infty }\left(\sum _{i=1}^{m}{\frac {9}{10^{i}}}\right)=1}$. This is a statement about limits only; it does not depend on (or even use) the definition or properties of 0.9999... A proof would be a slight generalisation of the proofs used to show convergence of the geometric series.

Now let us combine these facts. Take ${\displaystyle a_{m}=\sum _{i=1}^{m}{\frac {9}{10^{i}}}}$, b=0.9999... Then by using the second fact of analysis, the sequence of the a_m's converges to a=1. Now using the first fact of analysis and the assumed property of 0.9999..., we conclude that ${\displaystyle 1\leq 0.9999...}$. Since 0.9999... is not greater than 1, we conclude 1=0.9999... --Huon 14:10, 20 January 2006 (UTC)

By the second fact on analysis, the sequence of the a_m's converges to b=1 and not a=1 as you have written. So you end up concluding that ${\displaystyle 0.999...\leq 1}$ and not ${\displaystyle 1\leq 0.999...}$. So in essence you were unable to show that it is equal. Finally since we can show by induction that ${\displaystyle \sum _{i=1}^{m}{\frac {9}{10^{i}}}<1}$ for any m (except infinity), it would be correct to say that ${\displaystyle 0.9999...<1}$. 68.238.101.241 22:17, 22 January 2006 (UTC)

I'm sorry, but I can't follow any of your arguments. I defined a to be the limit of the a_m's. Since you agree that the a_m's converge to 1, we have a=1. By comparison, b is a number which is larger than all the a_m; 0.9999... satisfies that condition.

On the other hand, I agree to ${\displaystyle \sum _{i=1}^{m}{\frac {9}{10^{i}}}<1}$ for any m (except infinity). But why does that imply anything about 0.9999..., which has an infinite number of nines? By comparison, I could state: ${\displaystyle \sum _{i=1}^{m}{\frac {9}{10^{i}}}}$ is rational for any m (except infinity). Is 0.9999... still rational? If so, please give numerator and denominator; if not, then why should "being less than 1" be preserved if "being rational" is not?

By the way, please create an account. --Huon 19:51, 23 January 2006 (UTC)

No, since the a_ms converge to 1, you have that b=1, not a=1. You do not know what the last a_m is, do you? Rationality is preserved with 0.999... < 1. There is no problem with any facts of analysis. 68.238.108.20 11:52, 24 January 2006 (UTC)

Excuse me? a is defined to be the limit of the a_m's. Thus, when the a_m's converge to 1, we have a=1. Or, as a formula: ${\displaystyle a:=\lim _{m\to \infty }a_{m}=1.}$ Why should a not be 1? Why does b even enter this part of the discussion?

The "last a_m" comment is rather confusing. There is no "last a_m"; how should I know it? Why should I even want to know it?

Finally, if 0.9999... really is rational, give numerator and denominator. --134.76.82.144 13:13, 24 January 2006 (UTC)

Oops, I was in a hurry and forgot to sign in; that was me. --Huon 15:07, 24 January 2006 (UTC)

It matters a lot that b enters the discussion because you have made several erroneous assumptions based on b being the LUB: you stated that a_m < b for all m. Now since b=1 it must follow that any of the a_ms are less than 1 for a_m < b. You are getting yourself horribly muddled up. I did not say 0.999... is rational. I said there is nothing irrational about 0.999... < 1. Please do not misquote me. 158.35.225.231 17:35, 24 January 2006 (UTC)

I made no assumptions based on anything being an LUB. I don't even talk or care about LUBs here. All I assumed about b was that it is greater than all of the a_m's. I did not claim b=1; but indeed 1 would fit that description, and if I had chosen b=1, then I would have proved 1 being less than or equal to 1, which is correct. But 2, for example, would also fit that description, and if I had chosen b=2, then I would have proved 1 being less than or equal to 2 (which is, of course, also correct). But 0.9999... is also greater than all of the a_m's (or do you doubt that? It was the one property of 0.9999... I assumed). Thus, b=0.9999... is a valid choice. And then I prove that 1, which is the limit of the a_m's, is less than or equal to 0.9999...; that's what I wanted to show.

On the other hand, concerning rationality: You said, more or less: "All of the a_m's are less than 1; thus, 0.9999... is also less than 1." That is the best "proof" we have for 0.9999... being less than one, but it contains a serious gap when you say that a property which is shared by all the a_m's (being less than 1) must also be shared by 0.9999... Now I chose another property all of the a_m's have in common - being rational numbers. By an analogous argument, 0.9999... would also have to be a rational number, would it not? If you claim this kind of reasoning is sound, then you claim 0.9999... to be a rational number, and you should be able to give numerator and denominator. If, on the other hand, you do not believe that this kind of reasoning is sound for the property "being a rational number", then why should it be true for "being less than 1"?

I am sorry for misinterpreting your remark about rationality of statements instead of numbers, but I had assumed it to be an answer to my argument about rational numbers (which, maybe, was a bit too short itself).

Finally, I would ask you to be more precise. I made erroneous assumptions? Maybe, but which ones? You say b=1? Why, if I never stated it to be so? --Huon 23:31, 24 January 2006 (UTC)

Actually you wrote: "Let b be any real number with a_m < b for all m." So you made an assumption that b is a LUB. In your last post you claim that b=0.999... is a valid choice. It is not a valid choice because you do not know what it is. I think you need to leave analysis aside because it is confusing rather than helping you. You also write: "By an analogous argument, 0.9999... would also have to be a rational number, would it not?" To which I respond: No, this reasoning is in error, 0.999... would not have to be a rational number. Just consider the set of all rationals less than srqt(2). Would sqrt(2) have to be rational? Analysis is full of errors and contradictions. Rather than change it, most prefer to keep it along with the confusion and difficulties it poses for all learners. Math Professors (even those who have taught analysis over 30 years) are still unable to demonstrate flawless proofs. How is it that poor students who are complete novices are expected to learn and master this nonsense in just one course? Almost every student I have talked to who has passed an analysis course is still unable to completely grasp what it's all about. Most will gradually forget everything after the course because they 'learn' it parrot-fashion. Just ask a graduate if he/she can tell you exactly what the completeness property is - you will be surprised to find that most will not know. In fact many professors do not quote it correctly either. 158.35.225.231 15:35, 25 January 2006 (UTC)

Thanks for these detailed remarks. First of all, b need not be the LUB. It is an upper bound, but not necessarily the least. And indeed 0.9999... is an upper bound for the sequence of the a_m's. Or do you doubt that? If it were not, there would be a number 0.9999...9 (with a finite number of nines) which would be larger than 0.9999...! For this proof, I don't have to know what 0.9999... is; all I need to know is that it is greater than all of the a_m's.

Now you seem to agree that just because all of the a_m's have a certain property (being rational numbers, in my example), 0.9999... need not share this property. Thus, even though all of the a_m's are less than one, why should 0.9999... be? In all these discussions I cannot remember a single argument for this step.

Finally, concerning your remarks about analysis, I agree that it's an intricate subject and that mathematicians, professors included, may sometimes make flawed "proofs". But that's rather off-topic here (unless you find flaws in my proofs), and I invite you to discuss it on my talk page instead. Huon 17:55, 25 January 2006 (UTC)

I'm going to try and break user:Huon's proof down into a few steps to try and clear up the misunderstandings, because it makes perfect sense to me:

Lemma 1: Let a_m be a convergent series of real numbers with limit a. Let b be a real number such that a_m < b for all m, ie. b is an upper bound for the set {a_m} (but not necessarily the LUB). Then ${\displaystyle a\leq b}$. (Proof omitted, but definitely standard university-level).

Lemma 2: ${\displaystyle \lim _{m\to \infty }\left(\sum _{i=1}^{m}{\frac {9}{10^{i}}}\right)=1}$, ie. the limit of the series (0.9, 0.99, 0.999, ...) is 1. (Again, proof omitted but again university-level at most.)

Theorem: 0.999... = 1. Proof:

Let ${\displaystyle a_{m}=\sum _{i=1}^{m}{\frac {9}{10^{i}}}}$, so ${\displaystyle a=\lim _{m\to \infty }\left(\sum _{i=1}^{m}{\frac {9}{10^{i}}}\right)=1}$ (By Lemma 2). Also let b=0.999...

Then a_m < b for all m. (As user:Huon has pointed out, b is not necessarily the least upper bound.)

By Lemma 1, ${\displaystyle 1=a\leq b}$.

The only tricky bit here is proving that ${\displaystyle 0.999...\leq 1}$, and to be frank I can't quite work it out myself and end up proving myself in circles, and about the only proof that seems to work is the rather pointless "by inspection". I think the whole discussion about stating that it's rational or not is based on the fact that while it's easy to see that a_m < 1, there's nothing to say that 0.999... should possess any of the properties common to the a_m, be it "rational" or "less than 1".

Oh, damn. Okay, Huon, here's a problem in your proof: As far as I can tell, you can only make your statement that a_m < 0.999... by the same reasoning that ${\displaystyle 0.999...\leq 1}$ - by inspection of the digits. Explain to me how else that works and the proof works.

But then again, I'm still trying to work out how 0.999... has any meaning unless it's defined as the infinite series ${\displaystyle \sum _{i=1}^{\infty }{\frac {9}{10^{i}}}}$ which, as far as I can recall, is defined as the limit in any case. How do you get around that? (Bleh, in trying to clarify something I've gotten myself even more confused)Confusing Manifestation 18:44, 25 January 2006 (UTC)

ConMan, of the steps you want clarification on, neither requires inspection of digits. They both rely on the general principles, which are theorems in any order topology:

• If a sequence is eventually >= some number, so is its limit.
• If a sequence is eventually <= some number, so is its limit.

Does that help? Melchoir 19:23, 25 January 2006 (UTC)

The history of this series of proofs is that an anon disagreed with defining ${\displaystyle \sum _{i=1}^{\infty }{\frac {9}{10^{i}}}=lim_{n\rightarrow \infty }\sum _{i=1}^{n}{\frac {9}{10^{i}}}}$. Various proof using constructions of the real numbers (Dedekind cuts etc.) had also failed to convince her. So I tried a proof that relied on as few properties of 0.999... and of the real numbers as I could manage. That proof used the Archimedean property rather than the property of limits that Huon uses; but otherwise the principle used is similar: If we refuse to define an infinite sum as the limit of the partial sums, we don't really have much else to fall back on. We might define 0.9999... = 2, and we wouldn't really have much of a contradiction. However, Huan and I argue, any meaningful definition of 0.999... would need to have ${\displaystyle \forall n\in N:\sum _{i=1}^{n}{\frac {9}{10^{i}}}<0.999...}$ and ${\displaystyle 0.999...\leq 1}$; and from that we conclude that 0.999... = 1. Rasmus (talk) 20:49, 25 January 2006 (UTC)

Indeed I don't prove that 0.9999... is either larger than all of the a_m's or less or equal 1. While I explicitly stated that I assume the first property, I was a bit sketchy on the question of 0.9999...>1. Rasmus sums up my position quite nicely. I believe the anon would disagree with every definition of 0.9999... (possibly claiming we "cannot know what it is"); thus Rasmus and I try to give proofs using not a definition, but only "obvious" properties of 0.9999... Yours, Huon 21:41, 25 January 2006 (UTC)

Rasmus wrote: "If we refuse to define an infinite sum as the limit of the partial sums, we don't really have much else to fall back on." I completely agree. It is this definition that I cannot accept. In my understanding an infinite sum is not computable. All we can do is determine whether or not an infinite sum has a limit. To define an infinite sum as the limit of the partial sums is completely illogical. Whatever 0.999... is does not matter that much as long as we know it is less than 1. If you accept Dedekind cuts, then you should have no problem seeing that 0.999... < 1 by 'definition'. So all we can tell about 0.999... is that it is less than 1. Nothing else. 71.248.144.149 22:02, 25 January 2006 (UTC)

Unfortunately, your understanding of infinite sum seems to be at fault. Please define what, for you, an infinite sum is. If you cannot do that, it might be wisest to consider an infinite sum as a series, and thus the limit of a sequence of finite sums. By the way, I also see no reason at all why 0.9999... should be an infinite sum in the first place. And unless you want to state that there is a number 0.9999...9 with finitely many nines which is greater than 0.9999..., my proof holds no matter what 0.9999... really is.

Concerning Dedekind cuts: If I understand you correctly, you mean that the Dedekind cuts given by the pairs of sets ${\displaystyle (-\infty ,\alpha ],(\alpha ,\infty )}$ and ${\displaystyle (-\infty ,\alpha ),[\alpha ,\infty )}$ respectively correspond to different real numbers. That is not the case. In effect, Dedekind cuts are used as follows: Let X and Y be sets of real numbers such that the union of X and Y contains all real numbers, and that for elements x, y of X and Y, respectively, x<y holds. Then there is exactly one real number z such that for all elements x of X and y of Y we have ${\displaystyle x\leq z\leq y}$. That number z would be the number corresponding to the Dedekind cut (X,Y). It does not matter wether z is an element of X or of Y. You can check that the two Dedekind cuts above both correspond to the same ${\displaystyle \alpha }$.

If you are not convinced, have a look at the Dedekind cut gven by the sets of negative and nonnegative numbers, and the Dedekind cut given by the sets of nonpositive and positive numbers. At least one of them should correspond to 0. Which one? And to what does the other correspond? --Huon 23:16, 25 January 2006 (UTC)

The smaller set you described (X) does not include alpha, the larger one (Y) does. There is no number such that x <= z <= y for there are no numbers between X and Y. 158.35.225.231 18:45, 26 January 2006 (UTC)

So you say that the sets ${\displaystyle (-\infty ,\alpha ],(\alpha ,\infty )}$ do not form a Dedekind cut? Of course there is a z such that x <= z <= y; such a z need not lie somewhere between X and Y; it may be an element of one of them. In the example given above, X={x real with x <= alpha}, Y={y real with y > alpha}, alpha does satisfy just that: We simultaneously have x <= alpha for all x in X and y > alpha for all y in Y, both by definition of X and Y, respectively. Thus, x <= alpha < y, which is even more than required. Since every real number is either greater than alpha (and thus an element of Y) or not greater than alpha (and thus an element of X), the sets X and Y given above also form a partition of the reals. Thus, (X,Y) is a Dedekind cut, and alpha takes the role of z.

Besides, if you do not accept Dedekind cuts where the "smaller set" includes alpha, then I seem to have misunderstood the remark about 0.9999...<1 by 'definition'. Maybe I miss something; please give more details. --Huon 19:10, 26 January 2006 (UTC)

You are saying the z is alpha. So if this is true, then (-oo,1] (1,oo) is the cut that represents 1. If you say this cut represents 0.999... also, then you are mistaken. 70.110.87.205 23:20, 26 January 2006 (UTC)

Why am I mistaken? And what other Dedekind cut represents 0.9999...? I am getting weary of asking for the same things again and again. Please be more precise; give more details. Please create an account. Is either so difficult? Besides, I still wait for any detailed challenge to my latest proof, as detailed by Confusing Manifestation. --Huon 13:58, 27 January 2006 (UTC)

This cut represents 1. So does (-oo,1)

I arrive at the assumption that 0.9999...=1 by the proof I gave at the beginning of this subsection. ConMan gave a more detailed version of this proof a little later, pointing out the one gap it contains: I did not explicitly state that I assume 0.9999... to be not greater than 1 (but on the other hand, nobody ever proposed 0.9999...>1).

Somehow, we seem to be running in circles. I am still waiting for the opponents of 0.9999...=1 to answer the following questions:

• If you claim to have a proof that 0.9999... is not equal to 1, please give it in full detail. Above I pointed out where I saw a gap in previous attempts at a proof.
• Give a detailed account why my proof above is not accepted.
• Give a Dedekind cut corresponding to 0.9999...

Without either of the first two, I see no need to continue these discussions. Yours, Huon 22:14, 27 January 2006 (UTC)

The onus is on you to prove that 0.999... = 1, not the other way around. Neither you nor anyone else has been able to do this satisfactorily. As for a Dedekind cut for 0.999..., I could just as well say: (oo,0.999...] (0.999...,oo) - this corresponds to 0.999... Now if you can give a Dedekind cut for pi or e (Melchoir's Dedekind cut for pi is a joke), then I will give a better Dedekind cut for 0.999... 71.248.128.176 12:09, 30 January 2006 (UTC)

As I said before, I proved 0.9999...=1, and there still has been no detailed account of why that proof should be false.

Concerning Dedekind cuts, first of all I must admit that I was careless in reading up the definition; I missed the provision that the "smaller" set may not contain a largest element. Thus, when I wrote about (-oo, alpha], (alpha, oo) being a Dedekind cut, that is indeed false. Unfortunately, you seem to have fallen in line with my faulty notation (in retrospect I see you tried to warn me on this point). And since it does not really matter wether the "smaller" set does not contain a largest element or the "larger" set does not contain a smallest element, I won't change notation in mid-discussion.

Now on to your Dedekind cut. You will probably claim it is not the same as (-oo,1] (1,oo), but failed to give any arguments. Then we could form the intersection ${\displaystyle (-\infty ,1]\cap (0.9999...,\infty )}$. This intersection should be non-empty, since else we would have 0.9999...>=1. Does it contain elements besides 1? If so, how many? Give at least one example. Anything you could say about the properties of these elements would also be appreciated.

I still do not see why this Dedekind cut should lead to the result that 0.9999...<1 "by definition", as was claimed above. If I assume that 0.9999... and 1 are distinct, then I have found distinct Dedekind cuts, but if they are the same, then so are the cuts.

Finally, concerning Dedekind cuts representing pi or e, Melchoir's cut in the archive is not at all a joke. Melchoir explicitly shows why his partition of the rationals (he did not speak of Dedekind cuts of the reals, as I do here) is a Dedekind cut, and it is obvious that it indeed represents pi. The problem at that time seemed to be that the cut "says nothing about the value of pi". Indeed it does not, but why should it?

Please remember the entire discussion about Dedekind cuts (including my mistake) is beside the point. I gave a proof of 0.9999...=1 above. --Huon 14:45, 30 January 2006 (UTC)

So now that we agree that Dedekind cuts say nothing about the value of a number, we can continue to discuss your proof without these. As I stated, your proof is incorrect from the very first step, i.e. you are first assuming that sum (i=1 to m) 1/9^i < 1 for finite m and then using this to show 0.999... = 1. See, if the first step in your proof(in fact it is Rasmus's proof) were correct, the rest of the proof would be true. Unfortunately, you cannot assume this is true for finite m and then proceed to arrive at the result that 0.99... = 1. It is a contradiction.

As for your second proof above: Again your proof fails badly: In Lemma1 you write that a_m < b for all m. i.e it is an upper bound. Now if it is an upper bound it must at the very least be equal or greater than a=1. Then in step 2, you erroneously proceed to say: Also let b=0.999... Problem is that you have already assumed it is at least 1. So you have contradicted yourself. No need to go any further. 158.35.225.231 16:32, 30 January 2006 (UTC)

Concerning Rasmus' proof: Do you agree that all finite sums sum (i=1 to m) 9/10^i are less than one? I assume you do; if not, please correct me. Then why should 0.9999... share any properties with the finite sums sum (i=1 to m) 9/10^i? Especially why should it be a contradiction that all those finite sums are less than 1 and 0.9999... is not? 0.9999... obviously is no finite sum. Besides, I cannot see where Rasmus uses 0.9999...=1. He proves it, but in the proof he does not use it.

Now on to my own proof. I stated that b is an upper bound for the a_m's and that b=0.9999... You say that's a contradiction. Do you imply 0.9999... is not an upper bound for the a_m's? That there is an m for which 0.9999... with infinitely many nines is less than 0.9999...9 with m nines? The difference 0.9999...-0.999...9 = 0.000...09999... with m zeroes after the decimal separator followed by infinitely many nines does not look negative to me.

You seem to have gotten my proof backwards. First of all I claimed that 0.9999... is an upper bound, more or less "by looking at it", because the notion that a number with infinitely many nines should be less than one with only finitely many nines seems rather strange. It was the one property of 0.9999... I assumed. Then, I indeed reason that since 0.9999... is an upper bound, it must be equal or greater than 1.

By the way, I would only get a contradiction by claiming both b=0.9999... and b>=1 (as you said I did) if I knew that 0.9999...<1. I don't know that; else I would not set out to prove 0.9999...=1. Thus, your criticism is in effect one of circular reasoning (assuming 0.9999...>=1 in order to show it), not of self-contradiction. Of course, if you were right, my proof would still be faulty. Yours, Huon 17:57, 30 January 2006 (UTC)

Blah, blah, blah...

Quack, quack, quack... Revolver 22:27, 26 January 2006 (UTC)