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This is an old revision of this page, as edited by Huon (talk | contribs) at 15:07, 24 January 2006 (→The debate resumes - Problems with Rasmus's Proof). The present address (URL) is a permanent link to this revision, which may differ significantly from the current revision.

Because of their length, the previous discussions on this page have been archived.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)