Talk:0.999.../Arguments/Archive 1

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1 Geometry and comparing real numbers
2 Induction proof
3 An infinite sum is NOT equal to the limit of an infinite sum
4 Mathematical Induction
5 Here you go:
6 What is a rational number?
7 Farewell
8 You go wrong because you are all wrong...
9 THIS ARTICLE SHOULD STAY
10 and therefore this proof erroneous
11 You will always be challenged on this article because it is untrue.

Geometry and comparing real numbers

It does not matter what you think and it is you who appears to have difficulty grasping even the most fundamental concepts in Mathematics. Of course there is a philosophical element; mathematics started with philosophy but you appear to be ignorant of this fact. Tell me, does a finite line have infinitely many points? And now that you mentioned definitions, please tell me what in your opinion a point is. Furthermore, your ignorance shows and if I were you I would refrain from posting comments that you have not given sufficient thought.71.248.147.163 01:28, 24 February 2006 (UTC)

That's your second. Melchoir 01:43, 24 February 2006 (UTC)

I have a lot to say, but this is really getting us nowhere, so I will cease this argument. Sorry. -- Meni Rosenfeld (talk) 06:15, 24 February 2006 (UTC)

Just to give the definitions we were asked for: A line is, in the context of elementary geometry, a one-dimensional affine space. Since here we only consider objects over the field of real numbers, all lines have infinitely many points (even uncountably many). (If we instead considered some finite field, lines would have only finitely many points, but that is irrelevant for the subject at hand.)

I could not find a definition for a "finite line", unless a line segment is meant. Since the interior of a line segment can be bijectively mapped to the entire line, a line section also has infinitely many points.

Finally, according to Euclid, a point is that "which has no parts" (or something to that effect). But even Euclid did not use that definition but rather a set of axioms points shall satisfy. In analogy to my "line" definition above, I could also define a point as a zero-dimensional affine space.

I fail to see a connection between these definitions and whether 0.9999...=1 or not. Huon 11:16, 24 February 2006 (UTC)

You fail to see what a point has to do with this fact? My, oh my. It has everything to do with it. You are trying to have 0.999... which you claim is a number on the real line equal to 1. Euclid did not define a point: "Simeon estin ou meros outhen" means in Greek: "A point has no part." This so called definition is such a load of rubbish. It is meaningless and was written down by Euclid because he could not define the concept of a point without introducing self-referential terminology. In fact a point and a line are defined in terms of each other. It is completely ridiculous to talk about rigour in modern mathematics because modern mathematics is based on this garbage. It is based on concepts and notions that are not clearly defined. This is one of the reasons real analysis has failed. If you talk about a real number line, you are talking about points and if you are talking about points, one of these points represents 0.999... and another point represents 1. Now I have in the archives shown and proved conclusively that having 0.999... < 1 does not violate any mathematics (not even the archimedean property). I asked you why you compare real numbers differently and you were not able to answer, nor anyone else as for that matter. Again, I ask you this question: Why do you compare the limit of 0.999... with 1 and why do you compare the partial sum of pi with any other number? Just start by answering this last question. We can try to redefine point in a logical, non-referential (or cyclic) way later. This is extremely difficult to do. 71.248.136.114 18:00, 27 February 2006 (UTC)

Huon, I think it is better in this context to state that a point needn't be defined at all. Every mathematical theory needs several elementary notions which are not defined and receive their meaning from axioms - otherwise the definitions would inevitably be circular. -- Meni Rosenfeld (talk) 07:57, 26 February 2006 (UTC)

Meni: My answer is both yes and no. In a purely geometrical context, you are right; lines also can be characterized axiomatically. But I assumed that we wanted to consider the special case of the "real line" and points on that line. In that case, the concept of points is more limited. Yours, Huon 13:16, 26 February 2006 (UTC)

Anon: I agree Euclid's definition of a point is useless. But neither Euclid nor any mathematician since then (that I know of) made use of that definition. I also fail to remember any set of self-referential definitions of point and line (could you provide a reference?). Now in our context of 0.9999... and 1, we do not need the complete geometric baggage. For example, we do not have more than one line, and while every totally ordered set (like the set of real numbers) could be thought of as a line, that seems to be unnecessarily complicated. But I happily agree to use geometric language if you prefer.

Now you claim that 0.9999... and 1 are represented by different points on the real line. These distinct points have a non-zero distance, say x. (The one advantage of your geometric interpretation is that we can easily form sums, differences, and products with natural numbers.) You also claim that the Archimedean property holds. Thus, there is a natural number n such that n*x>1. Especially, x*10^n>x*n>1, so x is greater than 1/10^n. Then 0.9999...=1-x<1-1/(10^n)=0.9999...9 (with n nines). Please note that every operation I performed can be done geometrically (with the exception of finding n, but you agreed that the Archimedean property holds, so n exists). Somehow I doubt you are willing to accept my line of reasoning. Please be precise in stating where you disagree and give full details.

Finally you ask: "Why do you compare the limit of 0.999... with 1 and why do you compare the partial sum of pi with any other number?" I don't quite understand the question, probably because of its terminology. To me, 0.9999... does not have a limit, it is a limit (say, of the sequence (0.9, 0.99, 0.9999, ...)). Of course I compare that limit with 1, since I cannot easily compare sequences (the set of sequences is not totally ordered, at least not canonically). Now pi can also be considered as a limit (although to give a sequence is more difficult). Of course then I also compare the limit of that sequence to other real numbers if I want to get statements abut pi. If I only need approximate statements (say, whether 0.999...>0.5 or whether pi>3), then it may suffice to compare only partial sums (as I can probably find monotonous series), but pi and 0.9999... are not treated in different ways. As an aside, there is not the partial sum of pi. Any series has infinitely many partial sums. Yours, Huon 21:39, 27 February 2006 (UTC)

The reason I state that there are infinitely many numbers between 0.999... and 1 is because you can in reality only compare 'partial sums' where most numbers (numbers are defined in terms of series) are concerned. It makes no sense in some cases to compare the limits of certain series and in other cases to compare the partial sums. You state that pi can be considered as a limit even though you can't provide a sequence - this is untrue for a limit is defined in terms of a sequence. So, pi cannot be considered a limit. Furthermore, if pi cannot be found in its 'entirety', its limit cannot be known, hence you are not comparing a limit when you compare pi to any other number. 71.248.129.191 15:53, 28 February 2006 (UTC)

If numbers are defined in terms of series, then one always has to consider them as limits of those series (or, more precisely, as the limit of the sequence of partial sums). One cannot consider them as partial sums, since some partial sums may coincide for different series. Now concernung pi, there are several series. From the article pi:

$\sum_{n=0}^{\infty} \frac{4(-1)^{n}}{2n+1} = \pi$ or
$\sum_{k=0}^\infty\frac{1}{16^k}\left [ \frac {4}{8k+1} - \frac {2}{8k+4} - \frac {1}{8k+5} - \frac {1}{8k+6}\right ]=\pi$ .

The $\sum_{k=0}^\infty$ notation means, of course, the limit of the sequence of partial sums. There are lots of further sequences converging to pi. You can choose whichever you like. The partial sums of the series given above certainly differ, their limit can be shown to be the same (although that's probably non-trivial).

I still do not understand why we now discuss pi and its properties. It may be an interesting analogy, but shouldn't we be discussing 0.9999...? My point of view is clear: 0.9999... is the limit of the sequence of partial sums (0.9, 0.99, 0.999, ...), and thus obviously equal to 1. Rasmus and I even provided proofs that every real number which is greater than all of those partial sums but not greater than 1 must be equal to 1 (see here and here). By now, you have claimed 0.9999... to be a real number (although you also claimed the opposite), you claimed there is no partial sum 1-1/(10^n) which is greater than 0.9999..., and I never heard you claim 0.9999...>1. I could cite you on the first two counts if you insist. Thus, Rasmus' and my proofs hold. Your "proof", in comparison, had a fatal flaw, trying to use induction in ways it's not designed to work. If you disagree, please repeat that proof in full detail. Yours, Huon 16:54, 28 February 2006 (UTC)

You write:

"If numbers are defined in terms of series, then one always has to consider them as limits of those series (or, more precisely, as the limit of the sequence of partial sums)"

So what is the limit of the sequence of partial sums of pi? You are willing to say that 0.999... = 1 because you can find the limit of the partial sums of 0.999... but you cannot say pi = k where k is a limit of the partial sums of pi since you cannot determine k.

My proof using induction is correct. Your's and Rasmus's proof are horribly flawed. As for 0.999... being a real number - I still claim it is as real as pi but I also claim it is irrational just as pi is irrational. Look, if I reason the way you do, then I can prove that 1 is irrational. Suppose 1 = 0.999... then I cannot express 9/10+9/100+9/1000+... as a finite number of terms so that it can be written as a/b (a, b integers, b not zero), hence it must be irrational. Please don't tell me that you can express it as 1/1 because this is a limit and by your own admission ("...the limit of the sequence of partial sums...") you stated the series is the limit. The way you compare real numbers makes it ridiculous and outrageous for you to say that 0.999... = 1. Again, if this were true then you should be able to write k = pi and then tell me exactly what k is. You cannot. All you can do is approximate k. You cannot find the limit of the k series by any means. What is so hard for you to see here?71.248.129.191 23:10, 28 February 2006 (UTC)

I have given two different series converging to pi; do you mean the sequence of partial sums of one of those? Anyway, I would probably call the limit pi. Of course pi is irrational, and I cannot give a representation as a/b with a an integer, n natural. I also cannot give a finite decimal representation for pi. But why should I want to?

I also fail to see why I should not be able to express 1 as 1/1. And if 0.9999...=1, then 0.9999... inherits that representation. I don't see any connection between being able to write one limit as a quotient of non-zero integers and being unable to write another limit as such a quotient. Concerning pi, I don't need to find the limit. I know it exists (as the series given above can be shown to converge), and whenever I write "pi" I mean this limit. With 0.9999..., on the other hand, I can not only show that (0.9, 0.99, 0.999, ...) converges, but also that it converges against the rational number 1=1/1.

Simply put, if you are not willing to repeat your proof in full detail, especially those parts which were challenged the first time you proposed it, that's a bad sign. In comparison, while you continue to claim Rasmus' and my proof to be "horribly flawed", you seem to be unable to give details.

By the way, here you agreed that "There is no m for which 0.9999... is less than sum (i to m) 9/10^i." Now my line of reasoning above shows that assuming 0.9999...<1 contradicts this fact. Now I would claim this alone shows 0.9999...>=1 (that's a short version of Rasmus' proof). You probably disagree, but could you say where you see problems? Yours, Huon 00:09, 1 March 2006 (UTC)

It does not matter which series of partial sums you use for pi. What matters is that you cannot find a limit for either series. The reason you would want to find a limit is because you can do this for 0.999... and then compare it with 1. What I am telling you and you keep 'sidestepping' is that you are comparing pi and 0.999... differently. Till now you have not been able to show me how you compare pi with any other real number. First you say it is the limit, then you say it is the partial sum.... Seems to me you are very confused! I am not going to debate this with you anymore because you are clearly unable to provide any answer to the question of comparison. For the time being I suggest you forget about Rasmus's proof because I don't think you understand it. Think only about how you compare pi with other real numbers and then tell me if you compare it the same way as you do 0.999... and 1. 70.110.83.42 02:24, 1 March 2006 (UTC)

The short answer is: I do compare pi to other real numbers in the same way as I compare 0.9999... and 1. Besides, I never said pi is "the partial sum".

Let me give an example: I want to know whether pi>11/4. or not. As a more interesting example, I will use the first of the series given above, not the second. (The second happens to be monotonous, simplifying matters greatly. On the other hand, the first is so slow to converge that showing pi>3, which I would have preferred, becomes too much work without any new ideas.)

Let $s_n=\sum_{i=0}^n\frac{(-1)^i4}{2i+1}$ be the n-th partial sum. Then $\pi=\lim_{n\to\infty}s_n$ . Now $s_3 =4-\frac{4}{3}+\frac{4}{5}-\frac{4}{7} =\frac{304}{105} =\frac{1216}{420} =\frac{1155}{420}+\frac{61}{420} =\frac{11}{4}+\frac{61}{420}$ . So all I have to do is to show that pi > s_3-61/420. Now since the s_n converge to pi, there is an n_0 such that for n>=n_0 we have |pi-s_n|<1/7=60/420. Without loss of generality we may assume n_0 to be odd: n_0=2(k_0)+1 (if n_0 happens to be even, we take (n_0)+1 instead, which has the same property). Thus, |pi-s_{2(k_0)+1}|<1/7.

Lemma: For every natural number k>1, we have $s 2 k + 1 > s 2 k - 1$ .
Proof of the lemma: $s_{2k+1}=\sum_{i=0}^{2k+1}\frac{(-1)^i4}{2i+1}=\frac{4}{4k+1}-\frac{4}{4k+3}+\sum_{i=0}^{2k-1}\frac{(-1)^i4}{2i+1}=\frac{8}{(4k+1)(4k+3)}+s_{2k-1}>s_{2k-1}$ .

Now using the lemma k_0-1 times (for k=2, k=3, ..., k=k_0), we see $s_3<s_5<s_7<...<s_{2k_0+1}$ . Now finally $\pi>s_{2k_0+1}-\frac{1}{7}>s_3-\frac{1}{7}>\frac{11}{4}$ .

Note that while my computations use quite a lot of partial sums, the result I show is about the limit of those.

As a final remark, your repeated unwillingness to discuss Rasmus' proof is telling. That proof is based only on assumptions you agreed to, and it uses only computations that can definitely be performed in your chosen geometric context. I am pretty sure I understand it, and if it were horribly flawed as you claim, it should be easy to denote exactly what goes wrong. You propose to discuss ever more far-flung theories instead of the problem at hand. Do you lack arguments? Yours, Huon 11:33, 1 March 2006 (UTC)

You have just confirmed what I stated: you do not compare 0.999... and pi the same way. Who do you think you are fooling? Do you think any person reading this will agree with you? Of course you used the partial sums (as I predicted) and of course these are NOT the limit. I rest my case. 71.248.129.19 13:42, 1 March 2006 (UTC)

I beg to differ. My comparison of 0.9999... and 1 also used partial sums as a tool in the process, but what was compared in the end was the limit. That's exactly the same method as in the calculation above. Besides, I do believe most people reading this will agree with me. While mathematics is, of course, not a democracy, I am pretty certain I would win a popular vote on this. Finally, I agree that the limit is (in general) not the same as a partial sum. Now I claim 0.9999... is the limit. Do you think of it as a partial sum? If so, which one (as there are infinitely many of them)? Yours, Huon 14:11, 1 March 2006 (UTC)

As much as this discussion forces me close to the point of breaking WP:CIVIL and WP:NPA, I will continue for as long as I feel I can keep within those bounds. I will also point out that as far as any induction proofs have been suggested against the case, to my memory they have all relied on the (completely incorrect) assumption that proving S(n) for integer n proves S(infinity). If you can show me a reasonable induction proof that doesn't use that fact, directly or indirectly, I will look at it in more detail. Confusing Manifestation 12:14, 1 March 2006 (UTC)

Incidentally, because no-one got to really comment on my proof, I've put it at User:ConMan/Proof that 0.999... equals 1 (Limit proof) along with a little lemma to clarify one of the steps. Comment here or there, I don't care (but please, if you argue the conclusion, you must at least argue one of the steps that leads to it). Confusing Manifestation 12:52, 1 March 2006 (UTC)

THE REASON INFINITE 0.9 IS <> 1 IS BECAUSE!.......

N = 1/3 = O.33333333333...... This calculation results in an INFINITE stream of 0.3's But more important is

the fact that its a perfect Calculation! there are no numbers! or fractions of numbers left over! what I mean

by this is! if it was possible to look at the last .3 it would onlly be .3 there are no other numbers Hiding!

So now we want to find the true Value of INFINITE 0.9 Easy! N * 3 = 0.999999999......No numbers left over!

Or as below!

0.3333333333333333333333333333333....................... + 0.3333333333333333333333333333333....................... + 0.3333333333333333333333333333333.......................

0.9999999999999999999999999999999....................... =

A.R.B

So you show that 0.9999... = 3*N = 3*(1/3) = 1. While I agree (and while that proof is listed in the article as the "fraction proof"), I fail to see the point of your post, especially considering the headline. Yours, Huon 13:37, 17 May 2006 (UTC)

WRONG!!

1/3 = 0.3... + 1/3... = 0.3.... + 1/3... = 0.999999......... NOTHING LEFT OVER!

A.R.B

What exactly is wrong? Are you saying that 3*(1/3) is not equal to 1? Yours, Huon 13:37, 18 May 2006 (UTC)

EXACTLY! 3*(1/3) CAN ONLY EQUAL 1 IF YOU ROUND THE NUMBERS UP!

1/3 = 0.3333... THE LAST NUMBER? IF POSSIBLE MUST BE .3 SO 3 * .3 = .9 = INFINITE 0.99999 ...

A.R.B

I'm sorry to say so, but you seem not to understand fractions. More or less by definition, 1/3 is the number which, when multiplied by 3, gives 1. (In general, a/b is the number which, when multiplied by b, gives a.) This does not depend on rounding. Yours, Huon 14:13, 18 May 2006 (UTC)

IM SORRY TO SAY THAT WHEN YOU MUTIPLY 0.3 * 3 YOU GET 0.9

A.R.B

You needn't be sorry for that; I agree. But I fail to see how 0.3 (which can be written as the fraction 3/10) is related to 1/3. Surely, three thirds equal one?

By the way, I would ask you to sign your posts, using four tildes (~). Also, writing in all caps is usually considered equivalent to SHOUTING. Yours, Huon 14:08, 19 May 2006 (UTC)

Induction proof

ConMan: The induction proof is not required to show S(infinity). No induction proof ever shows S(infinity). This is something Rasmus tried to use to defend his faulty proof because Rasmus starts his proof with a result from induction. 71.248.129.19 13:44, 1 March 2006 (UTC)

You can show (by induction or otherwise) that for every natural number n the finite sum $\sum_{i=1}^n\frac{9}{10^i}$ is less than 1. That result is undisputed. But then it was claimed that $\sum_{i=1}^\infty\frac{9}{10^i}$ would also have to be less than 1. That would, in effect, be S(infinity): The very same statement which was shown for natural numbers n, where n is now replaced by infinity. Could you clarify this step?

By the way, Rasmus' proof does not use induction at all. If you claim otherwise, please specify where, in your opinion, it does. Yours, Huon 13:59, 1 March 2006 (UTC)

I don't know if you are being obstinate or just plain ignorant? How on earth do you claim to use the limit when you do not know what the limit is for pi? Now you say 0.999... is the limit - how is it that you then say 1 is the limit? 0.999... and 1 are not the same.

For the last time: "RASMUS DOES USE A RESULT OF INDUCTION" - he starts by saying that:

     0.999....  >  Sum 9/10^i (i=1 to m)

This is a result of induction. Is this true if i goes to INFINITY? OF COURSE NOT!!!!! PLEASE DO NOT dispute this anymore because you are beginning to annoy me intensely.

Now, although it is legal for me to say:

           Sum 9/10^i (i=1 to m) < 1 for every m.

It is illegal for you to say:

           0.999... > Sum 9/10^i (i=1 to m)

because you do not know what 0.999... is. However, I know what Sum 9/10^i (i=1 to m) is exactly, not approximately.

71.248.129.19 16:06, 1 March 2006 (UTC)

Concerning the limit: I can use it even if I do not know, say, the full infinite decimal representation (assuming that's what you mean by saying I "do not know what the limit for pi is"). I know enough properties of that limit (especially the property of being the limit) in order to deduce further results.

By the way, you seem to have gotten my arguments backwards: Since both 0.9999... and 1 are the limit of the sequence (0.9, 0.99, 0.999, ...), one by definition, the other by easy calculations, they are equal. After all, that's what I want to show.

Concerning Rasmus' proof: You agreed here that "There is no m for which 0.9999... is less than sum (i to m) 9/10^i." That's the same as claiming that for every m, 0.9999...>= sum (i to m) 9/10^i. Have you changed your mind? If so, why should I denote by 0.9999... a number which is less than some number 0.999...9000... with only finitely many nines?

This "greater or equal" result implies a strict inequality (since sum (i to m) 9/10^i < sum (i to m+1) 9/10^i <= 0.9999...), but that's rather unnecessary: Rasmus' proof will also work with ">=" instead of ">" at the line you mention.

If I remember correctly, Rasmus and I (and you, see the link above) agreed that 0.9999... should have that property, or it would be a mis-nomer. Now if you accept a definition of 0.9999... as, say, the limit of the sequence (0.9, 0.99, 0.999, ...), then it could indeed be proven to possess that property, and it might be possible (though not necessary) to do it by induction. As you correctly noticed, that result is wrong if I consider the limit for m tending to infinity (explicitly, we do not have 0.999.... > lim_{m->oo} Sum 9/10^i (i=1 to m); if I misunderstood your "true if i goes to INFINITY" remark, please correct me). On the other hand, you also correctly noted that "No induction proof ever shows S(infinity)." So we know Rasmus cannot show it by induction (and he does not try to), it is false, but since it he also does not need that limit result in his proof, why does its falsehood invalidate the rest of the proof? Yours, Huon 17:00, 1 March 2006 (UTC)

You can type screeds of stuff but you still don't get it: You are comparing the two numbers differently (pi and 0.999...). And yes, you do not know the limit of the pi series or sufficient enough to use it. In fact you contradict yourself here because you do not do the same thing with 0.999..., i.e. you don't just take a few partial sums, say 0.9999999 and then deduce it less than 1; no, instead you take the limit and say that it is equal to 1. Next, it's strange that you now agree no induction proof ever shows S(infinity) but you have been claiming throughout that my induction proof does not do this and hence it can't be true. Why don't you make up your mind? 71.248.129.19 20:25, 1 March 2006 (UTC)

Huon: Are you a student of Rasmus? Why don't you try thinking for yourself? 71.248.129.19 20:30, 1 March 2006 (UTC)

Concerning your induction proof: While we agree you cannot show S(infinity), I still fail to see what distinguishes the result you claim to have shown (i.e. that 0.9999...=Sum 9/10^i (i=1 to infinity) < 1) from S(infinity). Thus I say that you do have correctly shown Sum 9/10^i (i=1 to m) < 1 for every natural number m (albeit most details were missing), but not what you intended to show, and even more, the result you truly want to get cannot be shown by induction - because it is S(infinity). I hope that clears up my problems with the induction proof.

And concerning the comparison of numbers: Of course all the partial sums of 0.9999... are less than 1. And if I intended to show, say, 0.9999...>0.99995, it would suffice to use just some of them to get it done. But unfortunately even though they all are less than 1, that alone cannot be used to show that 0.9999... is less than one, for all of its partial sums are also less than 0.9999..., and when I have two numbers which are both greater than all the partial sums, then I cannot immediately conclude which of them is greater (nor whether they are different).

Maybe the pi>11/4 example given above was not that good. Obviously, any proof concerning the equality of a limit and something else cannot be done with just a finite number of partial sums (since there are series converging to different limits but having the same first few partial sums). If, for example, I intended to show that pi if defined as the limit of a sequence of partial sums is equal to, say, the area of a circle of radius 1, then I would also have to use all the partial sums, or at least an infinite number of them, just like the 0.9999...=1 case.

By the way, there is still a question open for you. Quite a bit earlier on this arguments page I showed that assuming 0.9999...<1 leads to the existence of an m such that 0.9999... is less than sum (i to m) 9/10^i, which you agreed cannot be. What do you say to that?

Finally, I am not a student of Rasmus. Yours, Huon 21:51, 1 March 2006 (UTC)

Quoting anon: "You do not know what 0.999... is". Yes we do. It is the decimal representation of $\sum_{i=1}^{\infty}\frac{9}{10^i}$ . And given that, we know lots of things about it. We know, for example, what the digit in the five billionth decimal place is (Guess what? It's 9!). We also know that, by the trichotomy, for all n, either $0.999... < \sum_{i=1}^{n}\frac{9}{10^i}, 0.999... = \sum_{i=1}^{n}\frac{9}{10^i}$ or $\sum_{i=1}^{n}\frac{9}{10^i}$ , and it is possible to prove that the first two possibilities result in ridiculous conclusions without even touching on induction. I really don't understand how you can continue to make contradictory arguments - first trying to argue your point based on manipulations of 0.999... as a real number, then saying that since we "don't know what it is" we can't actually do any maths with it - basically implying that it isn't a real number, then trying to argue its very definition which has been clearly stated on the argument page as well as in just about every archive of this completely ridiculous discussion (not to mention several points where you explicity agree with said definition). Confusing Manifestation 22:46, 1 March 2006 (UTC)

Look, if I have made contradictory statements, I am not the only one. This discussion is about whether it makes any sense at all to compare the limit of 0.999... with 1 and then set it equal to 1. It makes no sense whatsoever. I have shown that the Archimedean property is not violated by providing examples of numbers between 0.999... and 1. The problem is that you are defining an infinite sum to be equal to its limit. This is not only untrue but it makes no sense whatsoever. Now if you are going to compare 0.999... and 1 in this way, then you ought to compare pi and other real numbers in the same way. You cannot compare any real number to the limit of pi because as Houn stated, it is impossible to write down all the representation in a radix system. It is also not correct to use the partial sums of pi because you do not use the partial sums of 0.999... when comparing to 1. When we say that Lim 9/10+9/100+9/1000+... = 1, we are in fact stating explicitly that 0.999... < 1. This is true. 70.110.86.62 22:20, 2 March 2006 (UTC)

"The problem is that you are defining an infinite sum to be equal to its limit." That is the definition of an infinite sum. See Series_(mathematics)#Infinite_series. Mdwh 23:17, 2 March 2006 (UTC)

If an infinite sum is not the limit of the partial sums, then what else shall it be? Please give a definition which either makes no use of the "+..." notation (which also commonly implies some sort of limit), or define that notation as well. You were the champion of non-self-referential definitions above; here is a task worhty of your talents!

Next, the Archimedean property once more. Giving numbers between 0.9999... and 1 does not solve the problem (I also have problems with those numbers for other reasons, but that's beside the point): The difference 1-0.9999... must be either infinitesimal or zero (or negative), or we run into the problem of having 0.9999... being less than one of the partial sums Sum 9/10^i (i=1 to m). That's what I show above, you have not yet answered to that. Differences of real numbers cannot be infinitesimals (that's where I use the Archimedean property of that difference); that 1 should be less than 0.9999... is, shall we say, strange; and we all agreed that 0.9999... should be greater than any of those partial sums. Thus, only one option remains: The difference is zero.

Finally, comparing pi. In order to compare real numbers, we need not necessarily know the representation in a radix system. And using partial sums is correct if and only if they are used in a way that allows conclusion about their limit. As an example, for the pi>11/4 proof above, I explicitly chose a partial sum which was closer to pi than a given difference and then showed that that partial sum was greater than 11/7 by more than that difference. I do not claim pi>11/7 just because that partial sum was, but because the limit was so close to that partial sum that the limit could not fail to be greater than 11/7. As another example, for 0.9999..., I cannot do that trick: Showing that the difference between 0.9999... and one of the finite partial sums is less than the difference between 1 and that partial sums (which would be sufficient to imply 0.9999...<1) quickly becomes equivalent to showing 0.9999...<1 itself; nothing is gained.

By the way, I fail to see how having 0.9999...<1 would help with the problem of comparing pi to other real numbers. We still do not know the full decimal representation of pi. But imho that shouldn't be the main part of our discussion. Yours, Huon 23:32, 2 March 2006 (UTC)

Huon: There is nothing wrong with this grammar. Let me break it down for you: Why do you insist on defining an 'infinite sum' as the 'limit of an infinite sum' rather than what it is, i.e. an 'infinite sum'? Furthermore, if an infinite sum is defined as a limit, then what on earth is an infinite sum itself? I guess you and Hardy fail to see this too, eh? 70.110.83.235 17:50, 4 March 2006 (UTC)

"An infinite sum is an infinite sum"? What sort of definition is that?! An infinite sum being defined as the limit is not Huon's definition, it is the accepted mathematical definition. As for "what on earth is an infinite sum itself", the infinite sum is the limit. How many more times does this have to be repeated? Mdwh 18:32, 4 March 2006 (UTC)

And, for the umpteenth time, the "infinite sum is equal to the limit" is not just a necessary definition, but a provable fact, that I notice has *still* gone unchallenged. Frankly, until I see evidence that you've either (a) accepted that fact, or (b) legitimately refuted the proof, I'm leaving this pointless discussion behind. Confusing Manifestation 03:41, 3 March 2006 (UTC)

IM SORRY TO SAY THAT WHEN YOU MUTIPLY 0.3 * 3 YOU GET 0.9

A.R.B

An infinite sum is NOT equal to the limit of an infinite sum

Show me one reference where this has been stated. There is no such reference. Why? Because it is utter nonsense and a lot of fools have misinterpreted a result that was illustrated earlier in this archive, .i.e The sum of a geometric series to n terms. It seems to me the Wiki team searches for certain vague references and tries to use these to back up their views.

Firstly, an infinite sum is NOT POSSSIBLE. Secondly, the limit of an infinite sum DOES NOT EXIST if the partial sums do not converge. Thirdly, it is not always possible to find the limit of a converging sum (e.g. pi). Fourthly, it is a PROVABLE FACT that all we can speculate about an infinite sum is that it may or may not have a limit and in certain cases, we can calculate this limit.

And yes, ConMan, unless you agree to this, there is no point contuining this discussion.

Proof that an infinite sum is not equal to the limit of an infinite sum:

I am assuming that you agree the sum to n terms of a geometric series is:

    Sum = (a - ar^n)/(1 - r)  where a is the first term and r the common ratio.

First note that if |r| is not less than 1, the series does not converge, so there is no limit. Secondly, we are more interested in infinite sums where |r| < 1 because we can deduce information about a possible limiting value. So we work with this case.

Suppose that n approaches infinity. We can rearrange the above as follows:

     Sum = a/(1-r) + ar^n/(1-r)

This gives us a clue to what happens to the sum as n -> infinity. The first term remains unchanged, i.e. a/(1-r) is not affected by n. However, the second term ar^n/(1-r) becomes smaller and smaller so that it approaches 0 (but NEVER becomes 0!!!!!). Infinity is NOT a NUMBER. Although we CANNOT compute the infinite sum, we are now in a position to say something about a limiting value (either a LOWER bound or an UPPER bound depending on the value of a). We can say with confidence that the sum will never be less than a/(1-r) or greater than a/(1-r). But is this the sum? Of course not!! a/(1-r) is NOT the value of an infinite sum. It is a limiting value.

ConMan and Huon: I learned this in high school.It is as simple as can be. No tricks. No nonsense. It seems to me you may not have learned it properly?

Now once you understand this, we can move along to the problem of you comparing limits and partial sums for 0.999... and pi.

71.248.134.207 13:10, 3 March 2006 (UTC)

(via edit conflict) OK, fine, one more post just because this is about as close as we have come to agreeing on something since, well, since the beginning of the debate. I'll respond to these in the order I think is most appropriate.

"Secondly, the limit of an infinite sum DOES NOT EXIST if the partial sums do not converge." I agree. But in the specific case we are considering, we know that the partial sums converge so that isn't a problem.

"Thirdly, it is not always possible to find the limit of a converging sum (e.g. pi)." Not exactly. It is not always possible to find an exact decimal expansion of the limit of a converging sum (and by exact I mean know enough of it to be able to calculate the digit in an arbitrary decimal place). But in the greater scheme of things, decimal expansions aren't all they're cracked up to be. In the sense of being able to know enough facts about a particular partial sum that we can use it mathematically, it is possible to "find" the limit of a converging sum. For example, it is possible to prove that a particular partial sum (some examples given higher on the page by other people) is equal to the ratio of a circle's circumference to its diameter, or to the value of x' for which the area under 1/x between 1 and x' is equal to 1.

"Firstly, an infinite sum is NOT POSSIBLE." Again, sort of. It's certainly possible to write an infinite sum, but of course it's not something that is particularly useful to work with unless you can give it some kind of meaning. But then, the same holds true of any mathematical notation. -1 doesn't mean anything unless you provide the construction of a set that includes elements you define as 0 and 1, and then associate an operation you call addition, and finally append an element that acts as the additive inverse of 1. The symbol $\infty$ itself has no meaning unless you know that you're talking about some kind of limit - for that matter, something as simple as dy/dx is nothing more than a fraction that can be cancelled down to y/x unless you define the first principles of differentiation. And likewise, the "infinite sum" doesn't mean anything unless you give it a consistent definition. What definition can you possibly give that sum that is consistent? I have shown that it is perfectly consistent to equate the infinite sum with the partial sums. That's why all reputable sources give that definition.

"Fourthly, it is a PROVABLE FACT that all we can speculate about an infinite sum is that it may or may not have a limit and in certain cases, we can calculate this limit." Well, sure. It may or may not have a limit. And if it does, as I have said above, the most logical value to give it is that limit. And if it doesn't, then you say it's undefined. No problems there. And in certain cases, we can calculate this limit - again, yes, to any decimal precision you choose, as much as that means. Or, you can equate that limit to some other known number or expression.

As such, this is obviously where your understanding of these so-called "infinite sums" and ours differ, and yes I agree that until those understandings meet there will be no consensus - but I will feel a little better if I can at least get you to admit that defining the infinite sum to be the limit of the partial sums (where it exists of course) is both convenient and consistent, like all good definitions in mathematics. And if you can't even agree to that, I am curious to see how you interpret two special cases of infinite sums:

$\sum_{i=0}^{\infty}0$ <- what is this if it isn't zero?
Taylor (and other) series, such as $e^x = \sum_{i=0}^{\infty}\frac{x^i}{i!}$ . Confusing Manifestation 14:30, 3 March 2006 (UTC)

And next, to what you posted while I was typing this diatribe: Again, I both agree and disagree with what you say. Yes, for any integer n, that extra term is non-zero. Yes, in the limit that extra term approaches zero. And yes, you cannot just say something about the infinite case just because you can say it about the finite cases. But the point is that the infinite sum doesn't mean anything unless you define it to equal the limit. Like you said, "infinity is not a number" - meaning that while a_n > a/(1-r) for any number n, it says nothing about infinity. Hell, you could define a_inf = 1 if you wanted, but that wouldn't be a mathematically useful definition. And yet you seem to be happy to first say "such-and-such is true for all numbers n, but infinity is not a number, but I still say it's true for infinity even though it's definitely false for the infinite limit". Confusing Manifestation 14:30, 3 March 2006 (UTC)

An infinite sum is equal to the limit of the sum as the number of terms tends to infinity, as explained here, and as I said before. This is the established definition. This is not up for debate. Your personal "definition" posted below which says the infinite sum is defined as the infinite sum is a tautology - you don't say how the sum from one to infinity is defined or calculted. But even if you did propose some definition, that's beside the point, as Wikipedia is not the place to promote original research.

As for your "proof", you state that ar^n/(1-r) has a limit of zero. Whether it ever reaches zero or not is irrelevant, so your point "but NEVER becomes 0!!!!!" is irrelevant. What's important is that the limit is zero, so the limit of the sum as n tends to infinity is a/(1-r), and hence, by defintion, the limit is the infinite sum.

"Firstly", an infinite sum is possible, since a limit is possible. "Secondly", no one is claiming that an infinite sum exists if the series doesn't converge. The issue is about series which do converge. "Thirdly", just because we can't write out the limit doesn't mean it doesn't exist. "Fourthly", there is no speculation, as an infinite sum is defined as the limit, and limits can proven to exist.

Now, do you actually have a point to all this? Arguing on Wikipedia isn't going to change established mathematics. Mdwh 02:11, 4 March 2006 (UTC)

1) An infinite sum is NOT possible as it is undefined. 2) You are claiming that an infinite sum exists based on your definition involving a limit. 3) And YES, just because you can't write out the limit DOES INDEED MEAN IT DOES NOT EXIST!. 4) I think you better read what I wrote again and this time allow yourself some time to absorb the meaning. 70.110.84.242 14:52, 6 March 2006 (UTC)

1+2: That is not "my" definition, it is the accepted mathematical definition. 3: So you are saying that pi, or the square of 2, don't exist? 4: I read you quite clearly; an infinite sum does not "have a limit" - it is the limit by definition. This limit may or may not exist - but in the case of 0.9..., we can prove the limit exists. And guess what? It equals 1. Mdwh 23:53, 6 March 2006 (UTC)

ConMan: you write: "But in the greater scheme of things, decimal expansions aren't all they're cracked up to be." I am sorry but I am going to tear you apart for making stupid statements like this. What do you mean by such a statement? The decimal system is all we have. If I understand you correctly and 'radix systems aren't all they are cracked up to be', then why are you making a fuss about 0.999... = 1 for by definition 0.999... is less than 1. This is a property of comparing numbers in radix systems. In fact, we use radix systems to compare pi with other real numbers - it is the only tool we have!! There is no other way to compare numbers.

You wrote: What if the sum of 0+0+0+0+.... is not zero. This is again a very silly and bad argument. It is very easy to show that the sum can be nothing else but zero. Regarding the Tayor series, I interpret these as useful approximations to numbers such as pi, e, etc. You cannot find any limits using Taylor expansions, only approximations. Again, what meaning does 1+1+1/2!+1/3!+... have to you? It is an aproximation to the number we call e. We do not know its limit (just as we don't know the limit of pi), only that it converges to a real value. In fact, the only reason we know that numbers such as pi and e converge is because we know by their definitions they must, i.e pi = circumference/diameter and we know that both circumference and diameter are finite real quantities. Of course either the diameter or circumference will always be an irrational number or both will be irrational.

Okay, so if you both agree and disagree, then let's be reasonable: why do we need to define an infinite sum in terms of its limit when it may not even have one?! Why not call it what it is: a limit of an infinite sum? Why refer to it as an infinite sum when it is in fact not?! I have no problem with anyone saying the limit of 9/10+9/100+9/1000+... = 1. I do have a problem if you say 0.999... = 1 because to most people 0.999... = 9/10+9/100+9/1000+... In actual fact, 0.999... is always less than 1 by induction. And I have shown in the archive that this does not violate any of the mathematics including the archimedean property. And hell yes, you can always arrive at ODD RESULTS if you start off with unsound statements. We have a lot of so called 'paradoxes' which are in fact not paradoxes but nonsense because they are based on incorrect reasoning. Some examples are Xeno's paradox, Russell's paradox. These are not paradoxes. Please I don't want to hear of what Tarski did or the hotel problem or anything else related to these. Let's just stay with the topic at hand. I simply mentioned this because it illustrates how you can arrive at absurd conclusions when your foundations are shaky. 71.248.134.207 16:07, 3 March 2006 (UTC)

Concerning the reference you asked for: Granville, Smith and Longley: Elements of the differential and integral calculus, 1941 (the first edition seems to have a 1904 copyright). In Chapter XIX, it states: "A series is the indicated sum of the terms of a sequence. [...] When the number of terms is unlimited, the [...] series is called an [...] infinite series." The book continues to state that "[...] in a convergent series, the value of the series is the number u (sometimes called the sum) [...]", where u is given by $u=\lim_{n\to\infty}S_n$ , where S_n is the n-th partial sum. While I find GSL's notation a little imprecise at times, I am content to work with it. No, this reference does not explicitly state that an infinite sum is the limit of an infinite sum. But I never said it should be; I was always careful to call an infinite sum (what GSL (and, by the way, Melchoir somewhere in the archives) call "the sum of an infinite series") the limit of the partial sums.

You still have not provided definitions of your own for an infinite sum, the limit of an infintie sum and, if you intend to use it in the previously mentioned definitions, the "+..." notation. It will be extremely difficult for you to convince me of anything about those terms without giving definitions first. Yours, Huon 15:48, 3 March 2006 (UTC)

You have been picking at me about definitions. How about you defining whatever you state? You write: "While I find GSL's notation a little imprecise at times, I am content to work with it." What an irony! Is not mathematics about being precise? Let's see, if Hardy makes a statement that is illogical, does it automatically become divine law?! Look, if any of the great mathematicians made such a statement, I would reject it because it makes NO SENSE. It does not mean I am being disrespectful, it only means that I have found certain concepts or definitions to be unacceptable. Now we both know that not everything has been clearly defined. There is no such thing as modern mathematical rigour. It makes me laugh when I hear fools using this phrase. It is the same group of fools who post articles on websites containing phrases like: "We harness infinity ...", "Infinity behind the scenes ..." and you can go on and on. In many respects we are not really that more advanced than those who lived in the time of Newton. As for real analysis: it contains contradictions and should not be treated as an indisputable subject that explains all the properties of real numbers. Real numbers are in fact not well defined. Much of the theory of real analysis is useless and I am thinking of Weierstrass and Cantor especially. Please don't tell me that because it's been around for over 200 years that it is sound and proven. It is not fact by any stretch of the imagination. Finally, just keep remembering Huon, that wherever you get your references from, these people were also grappling with the same problem - that of clearly defining and understanding mathematics. It is my opinion that we build theories and whenever they fail, we start all over again and try to more sense with new theories. Nothing is set in stone. 71.248.134.207 16:26, 3 March 2006 (UTC)

The problems I have with the book are not primarily in the parts I quoted. And those problems are mostly due to that book's age. Actually, mahtematics has developed and become more rigorous, especially since Weierstraß (who is not one of my favourite mathematicians, since I find his works a bit too pedantic). But you seemed to have problems with all current mathematicians and their work; thus, I was not too unhappy with finding a rather old book.

The important definitions (of which I gave a variant just before you asked) are: An infinite series is a sequence of partial sums $s_n=\sum_{i=0}^n a_n$ . The value or sum of the series is defined as the limit of that sequence of partial sums (if it exists), following GSL. It is denoted by $\sum_{i=0}^\infty a_n:=\lim_{n\to\infty}s_n$ ; this notation usually implies convergence. The term "infinite sum" should be avoided, but if I use it, it denotes the sum of the series. Finally, the "+..." notation should also be avoided, but if it is used, it means: $a_0+a_1+a_2+...:=\sum_{i=0}^\infty a_n$ . It might have happened that by some abuse of notation I used the word "series" to refer to the sum of the series; but from the context it should usually be clear what is meant. Especially for comparison purposes I cannot use the series but only its sum, since comparision of sequences is not defined, while the sum is a real number. Those are my definitions. Do you agree to them? If not, give your own (as you have once more avoided to do).

By the way, you might want to have another look at your answer to Confusing Manifestation. I had the impression that in your zeal to express your opinion you lost command of the grammar... Yours, Huon 18:23, 3 March 2006 (UTC)

Lost command of the grammar? Sorry, you need to be more specific. No, I don't agree with your definition of an infinite series. I would define an infinite series as: $s_\infty=\sum _{i=0}^\infty a_i$ . Your first definition is that for a finite series. However, I do agree that your formula with the + notation is correct. Okay, if by grammar you mean my language - I only have this to say: life is not made for wimps. If you lose sleep over anything that's posted on the web, you may need to see a therapist. If it makes you feel any better, let me say this: I neither dislike nor hate anyone. Feel better now? BUT, I believe this article is a good reason to start a Math Intifada/Jihad. 71.248.134.207 19:09, 3 March 2006 (UTC)

A finite series would be something like $s_m=\sum _{i=0}^m a_i$ . That is, of course, not an infinite series. As I wrote, an infinite series is a sequence $(s_m)_{m\in\mathbb{N}}$ of partial sums s_m, not just one of them.

Concerning your definition, I fail to see whether you try to define $s_\infty$ as $\sum _{i=0}^\infty a_i$ or the other way round. Anyway, both of those notations were not previously defined by you (s_oo didn't even appear before, if I'm not mistaken), so please use somewhat more basic concepts for your definitions.

Finally, while you sometimes do seem to have problems with remaining civil in the heat of a discussion, I indeed meant grammar, not language; especially the sentences: "why do we need to define an infinite sum in terms of its limit when it may not even have one?! Why not call it what it is: a limit of an infinite sum? Why refer to it as an infinite sum when it is in fact not?!"

Do you really want to suggest calling an infinite sum the limit of an infinite sum instead of calling it the limit of an infinite sum? Whether I agree with you on this point or not, I had the strong impression that you intended to say something else, but didn't want to guess. Yours, Huon 21:48, 3 March 2006 (UTC)

Huon: There is nothing wrong with this grammar. Let me break it down for you: Why do you insist on defining an 'infinite sum' as the 'limit of an infinite sum' rather than what it is, i.e. an infinite sum? Furthermore, if an infinite sum is defined as a limit, then what on earth is an infinite sum itself? I guess you and Hardy fail to see this too, eh? 70.110.83.235 17:53, 4 March 2006 (UTC)

Well, this page is now coming off my watchlist so I can take some time to contribute to real articles, not arguments on talk pages that never go anywhere. But, anonymous, I will ask you one thing: When you say that 0.999... < 1 by induction, how in the hell do you manage that? The only "induction" I can think of that comes close is something along the lines of "0.9...9 < 1 for n 9s, so it is < 1 for n+1 9s, so it is < 1 for any number of 9s", which as we both agree, says nothing about the case for infinite number of 9s. For all that "proof" (the details of which are simple to provide) says, 0.999... with an infinite number of nines could very well be infinite itself, or negative, or complex, or a fish. Please, leave a full induction proof on my talk page that actually proves the infinite case without making an assertion of the above kind. I look forward to seeing it. Confusing Manifestation 12:55, 4 March 2006 (UTC)

A proof has been posted in the archives - it is very simple and needs no deep mathematics. No proof by induction ever proves the infinite case as you state. It is both impossible to prove the infinite case and it is irrelevant. 70.110.83.235 17:53, 4 March 2006 (UTC)

Would you mind to either repeat that proof in full detail or to provide a link to the relevant part of the archives (preferably the difference)?

I wonder - you claim to have proven by induction that $0.9999...=\sum_{i=1}^\infty\frac{9}{10^i}=s_\infty$ is less than 1, but at the same time you claim that the infinite case (and what but the s_oo case would be the infinite case?) cannot be proven by induction. Do I misunderstand you, or is that a self-contradiction?

Just to repeat my point of view: I agree that for any natural number m the sum $s_m=\sum_{i=1}^m\frac{9}{10^i}$ is less than 1; that can indeed be shown by induction. But that does not imply s_oo < 1, and your proof lacks justification for this final step. Yours, Huon 23:19, 4 March 2006 (UTC)

You keep harping on the case of s_oo. Induction does imply that s_oo < 1 and there is no need to show S(infinity) - in fact this cannot be done. Standard induction DOES NOT REQUIRE having to show S(infinity). For the LAST TIME: You CANNOT prove S(infinity) and it is not required. This is something that Rasmus decided to use to justify his proof that is based on a FALSE starting point. If you believe this induction proof is false, then please PROVIDE JUST ONE VALUE of m for which it is false. You CANNOT provide a COUNTER-EXAMPLE so stop beating a DEAD HORSE please!!!! 68.238.107.132 13:31, 5 March 2006 (UTC)

The one (and only) counter-example is infinity. You keep saying that while induction does not show S(infinity) (I agree), it does imply that s_oo < 1 (I disagree). I fail to see the difference between the statement S(infinity) when S(m) is "s_m < 1" on the one hand and the statement "s_oo < 1" on the other hand. So please clarify how induction implies s_oo < 1. Yours, Huon 14:40, 5 March 2006 (UTC)

Quite incredible that you say: "the one (and only) counter-example is infinity." Infinity is not a counter-example because it is NOT a number. Did you not admit this yourself? Ergo, there is no counter-example and I claim it is true. 68.238.107.132 17:51, 5 March 2006 (UTC)

Of course infinity is not a number. But I repeat: Why then should the fact that for every number m you have s_m < 1 imply s_oo < 1? I do not doubt the first fact, but the implication. Yours, Huon 18:42, 5 March 2006 (UTC)

And I repeat: this is mathematical induction. That's why if for every number m you have s_m < 1, then this implies that s_oo < 1. This is the principle of mathematical induction. If it is untrue, then mathematical induction is a load of rubbish. Do you think it's a load of rubbish? Do you understand mathematical induction? 68.238.107.132 20:34, 5 March 2006 (UTC)

No, that's not mathematical induction. Mathematical induction says: If I can show a statement depending on an integer m to hold for one m_0 and if I can show that if said statement holds for m, it also holds for m+1, then it holds for all integers greater or equal to m_0.

Thus, you can show for all natural numbers m that s_m is less than 1, but you cannot show s_oo < 1, since infinity is not an integer. Yours, Huon 21:26, 5 March 2006 (UTC)

Yes, it is. Infinity is not an integer and it does not matter either way. AGAIN: YOU DO NOT HAVE TO SHOW s_oo < 1, it is IMPLIED!!!!68.238.107.132 21:31, 5 March 2006 (UTC)

Mathematical Induction

The truth of an infinite sequence of propositions $p i$ for i=1,..., $\infty$ is established if (1) $p i$ is true, and (2) $p k$ implies $p k + 1$ for all k. This is the principle of induction.

68.238.107.132 21:29, 5 March 2006 (UTC)

I've refrained from joining in this discussion for months, but I'd just like to say one thing, which has been said many times before. If this conception of induction is valid, then we could argue as follows: Let p_i be the assertion "0.9999...999, with i nines, is less than 0.999..." For i = 1 we have 0.9 < 0.999..., which is clearly true. Now suppose p_k is true. Now, clearly, 0.9999...999, with k + 1 nines, is less than 0.999... (This doesn't require the hypothesis that p_k is true, but I included it to meet the form of an inductive argument.) Now we have shown that p_oo is true. Therefore 0.999... < 0.999... But a number cannot be less than itself. Eric119 04:12, 6 March 2006 (UTC)

You are wrong. Your proof is incorrect right from the very first step where you say 0.9 < 0.999... Why? Answer is very simple: you do not know what 0.999... is, i.e you do not know the full extent of 0.999... as you do with 1. 1 is completely defined whereas 0.999... is NOT. Rasmus's proof is not incorrect because it uses induction, it is incorrect because it uses incorrect induction. How? To say Sum 9/10^i (i=1 to m) < 0.999... for every m is in fact UNTRUE for the very same reason stated earlier: you do not know the full extent of 0.999... So if you proceed to base your proof on such a fallacy, you will obviously arrive at an incorrect conclusion as Rasmus showed. Induction assumes that you know all the dimensions of any proposition. So you have shown nothing and your statement is evidently in error. This conception of induction is perfectly valid. Naturally if your propositions contain undefined elements, you may arrive at an illogical conclusion. For months, I have stated that 0.999... is by definition less than 1. I maintain the same viewpoint till this day. I have refuted all Wiki's so-called proofs and demonstrated that 0.999... < 1 does not violate any mathematical property. Furthermore, I have challenged Wiki's team to prove that they indeed compare al real numbers in the same way - they have not been able to do this because they do not compare 0.999... and 1 in the same way as they might compare pi, e or any other irrational number. In the case of 0.999... and 1, they are comparing a limit to a sum. Radix systems were designed to facilitate comparisons. It is because of radix systems that we are able to talk about the real numbers being complete seeing we have a means to determine ordering. In actual fact, to say that radix systems are irrelevant is the most stupid thing any educated person could say. To have duplicate representation is moronic because it defeats the whole purpose of comparisons between numbers. Unfortunately most of the Wiki team that has been posting is not very bright. Sigh,... 70.110.84.242 14:04, 6 March 2006 (UTC)

Ok, I'll give an easier counterexample to your version of induction. Let proposition

p k

be "k is a natural number". Obviously 1 is a natural number, so p_1 is true. Now if k is a natural number, so is k+1;

p k

implies

p k + 1

for all k. Thus infinity should be a natural number... Yours, Huon 14:46, 6 March 2006 (UTC)

You missed the point! I just stated that your propositions must be well-defined! What is infinity Houn?!!!! I do not use infinity anywhere in my induction proof. 70.110.84.242 14:55, 6 March 2006 (UTC)

The statement heading this section reads: "The truth of an infinite sequence of propositions

p i

for i=1,..., $\infty$ is established..." So you used infinity first. Whatever you meant by infinity in that sentence, your concept of induction makes it a natural number, by my "proof".

If you claim infinity is not well-defined (I might agree), then what is that infinity in your induction concept? Yours, Huon 15:17, 6 March 2006 (UTC)

See, I am correct in stating that you do not understand the principle of induction. You also did not understand what I wrote. I do not use infinity in any of my propositions. However, mathematical induction is about an infinite sequence of propositions. These two are not the same: i.e. an infinite sequence of propositions is not the same as a proposition about infinity. Rasmus's propositions contain statements about infinity. Mine do not. There is no contradiction or falsehood in anything I wrote. 70.110.84.242 15:28, 6 March 2006 (UTC)

Let me see if I can break this down very simply using Eric's example: Let's suppose that P(1) is true (in fact it is not true beyond any reasonable doubt but be that as it may) and consider P(k) and P(k+1). The problem with these is that we must have k < infinity. Since we do not have this restriction (k < oo) and it makes no sense since infinity is not a number, Eric's proof is incorrect. As for asking me to define infinity: I cannot. The principle of mathematical induction does not attempt to define it in any other way except to state that in context it means never ending. 70.110.84.242 15:32, 6 March 2006 (UTC)

Your "sequence of propositions" has a last proposition $p_\infty$ . Now an infinite sequence usually does not have a last element (not even if it converges); it only has an element for every natural number. So what do you mean by that? By the way, you have also claimed that in the special case of 0.9999... and 1, $p_\infty$ is given by " $\sum_{i=1}^\infty \frac{9}{10^i}<1$ ". There seems to be an occurrence of infinity in this last proposition, unless you can provide a definition for an infinite sum which does not make use of infinity (such as the limit definiton I gave above). Yours, Huon 15:46, 6 March 2006 (UTC)

Where do you see a last proposition? Are you thinking? I just stated that infinity means 'never ending' in the context of the induction principle. I will not respond to the rest of your questions since these follow on this incorrect conclusion of yours. 70.110.84.242 16:04, 6 March 2006 (UTC)

Your notation "1,..., $\infty$ " suggests that the index set for your propositions is $\mathbb{N}\cup\{\infty\}$ , not just $\mathbb{N}$ . Then the proposition corresponding to oo would be the last, in a sense.

You did claim to have proven $\sum_{i=1}^\infty \frac{9}{10^i}<1$ . How did you arrive at that result, if the sequence of propositions $\sum_{i=1}^m\frac{9}{10^i}<1$ shown by induction does not include that last one? Yours, Huon 16:13, 6 March 2006 (UTC)

P(infinity) in this context means 'never ending' because there is no last proposition. What else can it mean since infinity is not a number? $p_\infty$ does not mean a proposition about infinity, it means that there is an infinite number of propositions. It may be a bit confusing if you take it out of context. So okay, $p_\infty$ in this case does not mean the last proposition. It means an infinite number of propositions. (70.110.84.242 16:04, 6 March 2006 (UTC)

Induction is irrelevant. For the series 0.9, 0.99, 0.999, ... to have a limit of 1, it isn't necessary for it to ever equal 1, just that it remains within any given neighbourhood of 1. See Limit of a sequence. Mdwh 00:09, 7 March 2006 (UTC)

Here you go:

The following definition should clear it up:

The truth of an infinite sequence of propositions $p i$ for i=1,... is established if (1) $p i$ is true, and (2) $p k$ implies $p k + 1$ for all k. This is the principle of induction.

Happy now? You cannot consider all the propositions - this is not what induction principle is about. In fact, you never consider a last proposition - this is absurd! In fact if you could consider a last proposition, you would not need mathematical induction, would you?

Are you a school teacher or a professor Huon? 70.110.84.242 16:18, 6 March 2006 (UTC)

Yes, now I'm happy with your concept of induction; it's in essence what I wrote here, although you worded it quite a lot better. But I still fail to see how you can show $\sum_{i=1}^\infty\frac{9}{10^i}<1$ by induction. I do agree you can show $\sum_{i=1}^m\frac{9}{10^i}<1$ for all natural numbers m, but that's not the same, is it? Yours, Huon 16:32, 6 March 2006 (UTC)

What are your qualifications, if you're going to start asking us ours? Mdwh 00:04, 7 March 2006 (UTC)

You can show that $\sum_{i=1}^\infty\frac{9}{10^i}<1$ is true for all natural numbers. You don't need to consider infinity (infinity is not a natural number and we can't show anything about it) - the principle of mathematical induction has never demonstrated P(infinity). If this could be done, then there would be no need for mathematical induction. In short, it is the same as showing $\sum_{i=1}^m\frac{9}{10^i}<1$ for this is what induction demonstrates. In considering $\sum_{m=1}^\infty\frac{9}{10^m}<1$ , there are no propositions about infinity. We are making a deduction about the LHS being less than the RHS which is well-defined. We don't know what the LHS evaluates to. So we test it by the process of induction and when we cannot find an m for which it is untrue, we deduce that it must be true for all m. 70.110.84.242 17:05, 6 March 2006 (UTC)

Unless you use a highly non-standard notation, the statement " $\sum_{i=1}^\infty\frac{9}{10^i}<1$ " does not depend on any natural number (the i is just a summation index). So what do you mean when you say that statement is true "for all natural numbers"?

Let me try a similar proof. For all natural numbers m, $\sum_{i=1}^m\frac{9}{10^i}$ is a rational number whose denominator (when fully reduced) is a power of 10. Now let's have a look at the statement " $\sum_{m=1}^\infty\frac{9}{10^m}$ is a rational number whose denominator (when fully reduced) is a power of 10". We make a deduction about the infinite sum being an element of the set of rational number whose denominator is a power of 10, which is well-defined. We may not know what the infnite sum evaluates to. So we test it by the process of induction and when we cannot find an m for which it is untrue, we deduce that it must be true for all m. That "proof" uses exactly the same methods as yours. So 0.9999... is a rational number, and its denominator is a power of 10. Yours, Huon 18:14, 6 March 2006 (UTC)

So, I don't see any problem with this proof. Yes, I would agree that the denominator will be a power of 10. You are saying the denominator is not a power of 10 and thus the induction conclusion is wrong? 70.110.84.242 19:18, 6 March 2006 (UTC)

Here you claimed 0.9999... to be irrational. Now it is "proven" to be rational, and what is more, it is "proven" to have a finite decimal representation. That sounds strange, and it is still a counterexample to both your demand of unique decimal representations (since it has both a finite one and an infinite one) and to the use of decimal representations in ordering numbers - by comparison of digits, 0.9999... should be larger than any number less than 1 with only a finite amount of non-zero digits. Yours, Huon 19:49, 6 March 2006 (UTC)

You are quoting me completely out of context. I did not say any of what you claim in the above paragraph. I stated that the denominator will be a power of 10 but I don't know what the dimensions of the denominator are or even if the denominator is a real number. I still claim that 0.999... is irrational because there is no representation for it such that a/b = 0.999... There is no contradiction here and nothing strange about it. You may argue that since the denominator is a power of 10 it must be rational. However, not only can you NOT determine the denominator but you CANNOT determine the numerator also. It follows that 0.999... must be irrational. Furthermore, you are basing your incorrect conclusions on your idea that you can actually determine 9/10+9/100+9/1000+... when in actual fact you CANNOT. Your induction only tells you the denominator must be some power of 10, it does not tell you that it can be determined and you have by no means shown that you can determine the denominator. Try as hard as you like, you cannot refute the validity of mathematical induction. 70.110.84.242 20:13, 6 March 2006 (UTC)

You are missing the relevant point: In the proof you didn't see any problems with, I explicitly "showed" $\sum_{m=1}^\infty\frac{9}{10^m}$ to be a rational number. I also showed something about its denominator (whatever that should be if the infinite sum were not a rational number), but rationality was part of the bargain. I challenge you: Find an m for which $\sum_{i=1}^m\frac{9}{10^i}$ is not an element of the set of rationals. You cannot provide a counterexamle! Do you now see why I have problems with your method of using induction? Yours, Huon 21:12, 6 March 2006 (UTC)

You did not show anywhere that sum 9/10^m (m=1 to infinity) is a rational number. All you showed is that its denominator is a power of 10. I will have to think about the second part of your comment. 70.110.84.242 23:10, 6 March 2006 (UTC)

Here is the answer: mathematical induction works only if your propositions are true. In your example, you are setting out to prove that a rational number can be expressed as an infinite sum, i.e. 9/10+9/100+9/1000+... This is a false proposition and the fact that one cannot find an m so that a finite sum of terms is irrational only proves this fact. This also illustrates that 0.333..., 0.666... and any recurring radix representation is in fact irrational since it cannot be expressed as a finite sum of the powers of x/10^i or more commonly in the form of a/b with b not 0. 70.110.84.242 23:40, 6 March 2006 (UTC)

If mathematical induction only works if I know beforehand that my statements are true, then it's worthless - why should I prove what I already know to be true? It's just the opposite: If I can use induction (i.e. show that the first proposition is true (0.9 is rational) and that the truth of the k-th proposition implies the truth of the (k+1)-th (when $s_k:=\sum_{i=1}^k\frac{9}{10^i}$ is rational, so is $\sum_{i=1}^{k+1}\frac{9}{10^i}=s_k+\frac{9}{10^{k+1}}$ ), my propositions are true for all natural numbers k. And indeed for all natural numbers k, the finite sum s_k is rational. That's what we agreed induction to do. Now you made the extension of saying that whatever can be shown (by induction) to hold for all of $\sum_{i=1}^k\frac{9}{10^i}$ must also hold for $\sum_{i=1}^\infty\frac{9}{10^i}$ - that's the strange part.

Secondly, I was not trying to show that a finite sum of terms is irrational, but that an infinite sum of terms is rational. As the methods I employed are almost literally those used in your 0.9999...<1 proof (only changing "is less than 1" into "is an element of the set of rationals") , you should have severe difficulties to discredit my "proof" without renouncing your own.

Thirdly, you seem to claim there are real numbers without any decimal representation. Usually, every real number is said to have a decimal representation; do you really want to introduce numbers which do not have even one decimal representation? Yours, Huon 10:12, 7 March 2006 (UTC)

You are correct in some of the things you say. For example, you cannot know beforehand that your induction is true. However, you can know that your propositions are true or false. In the case of your rational example, it is easy to see how this is true for any finite number of terms. Introducing an infinite number of terms automatically asserts that you can perform the process of infinite summation - herein lies the problem: you cannot. Thus your induction result will be incorrect. The induction conclusion must follow on true propositions. In my example, I never assume anything about an infinite sum - not even in my propositions. And this is how induction is intended to work. Regarding your other point: I do claim that many real numbers do not have an exact decimal representation. Can you represent 1/3, 2/7, 1/6, pi, e, etc exactly in the decimal system? Answer is no. You can represent 1/3, 2/7, 1/6 exactly in some radix system? Yes. See, the definition of a rational number is a bit sketchy if you don't consider this fact too. i.e. a rational number is a number that can be represented exactly in some radix system. This is a more rigourous definition in terms of radix systems and it's less confusing. And unlike the author of the problem article, radix systems have everything to do with this anomaly and the question of rationality. In fact, what does 1/3 mean? What does a/b mean? It is completely meaningless unless we can somehow quantify it or define it by means of a measuring system, in this case a radix system and in particular the decimal system. This leads us to the next important activity of being able to compare numbers in the form of a/b. Once again, we can only do it by using a measuring system. In some cases it is easy to do by inspection: for example, which is larger - 718/719 or 7/8 ? Answer is 718/719. But this is not always easy to see. What if I used 629/719 and 7/8 - would you be able to tell me readily which is larger? Answer is no. You would have to use a measuring system. 68.238.104.216 14:35, 7 March 2006 (UTC)

I have determined that 7/8 > 629/719 without looking at their decimal expansions. 629*8 = 5032 and 719*7 = 5033. Whether or not I used a "measuring system" is anybody's guess. 1/3 is precisely the quantity such that (1/3)*3 = 1. 2/7 is precisely the quantity such that (2/7)*7 = 2. 718/719 is precisely the quantity such that (718/719)*719 = 718. We don't need to consider radix systems at all to define (and compare) rational numbers. The assertion that this is "more rigourous" is laughable. Eric119 16:53, 7 March 2006 (UTC)

How would you compare pi/3 and 1.13 without radix representation? 198.54.202.195 15:10, 9 March 2006 (UTC)

There are many ways. Here's a relatively simple one: It can be shown that Pi is smaller than any number of the form:

$4\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-...+\frac{1}{4n+1}\right)$

So for example:

$\frac{\pi}{3} < \frac{1}{3}4\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}\right) = \frac{4}{3}\frac{315-105+63-45+35}{315}=$

$\frac{4 * 263}{3 * 315}=\frac{1052}{945}=\frac{105200}{94500}<\frac{106785}{94500}=\frac{113*945}{100*945}=\frac{113}{100}=1.13$

So pi/3 < 1.13. The idea is that you just have to compare pi with some rational number using math, and then it's just a matter of operations on integers - which do not require a radix system. -- Meni Rosenfeld (talk) 15:29, 9 March 2006 (UTC)

Imagine having to use one of the many pi series and formulas every time you need to compare pi? This is why we approximate pi using a radix system and then it is easier for us to compare it with other numbers. Now let's think if we were using a number like the orginal poster described, i.e. 3.14......6 with 6 billion digits represented by the dots. Would you still want to use the method you suggested? I think not. In fact, I think he has a point in saying that we do not compare pi the same way as we compare 0.999.... Why don't you compare 0.999... and 1 in the way as you compared pi/3 with 1.13? My last post cheers 198.54.202.195 20:21, 9 March 2006 (UTC)

In practice, it is convenient to use decimal expansions to compare numbers, and there are theorems that it works the way you'd expect, in the absence of an infinite string of 9s. Melchoir 01:36, 10 March 2006 (UTC)

Two point should be emphasized:

What is convenient or not is irrelevant. What matters is what can be done mathematically. And all these comparisons you mention can be done mathematically. It is not convenient to solve quartic polynomial equations, but it can be done. Quintics, on the other hand, can't (using radicals). This is the kind of distinction that interests us.
I am comparing 0.999... and 1 the same way I'm comparing Pi/3 and 1.13. Just like I can prove mathematically that Pi/3 < 1052/945 < 1.13, I can prove mathematically that 0.999... = 1, and that has been done many times in this discussion. -- Meni Rosenfeld (talk) 10:11, 10 March 2006 (UTC)

Actually, there is a slight difference in the methods used to compare pi/3 with 1.13 and 0.9999... with 1. The reason for this difference is that in the first case, the numbers are different, while in the second, they are equal. The method of comparing pi/3 with 1.13 is more akin to the method one would use to compare, say, 0.9999... with 1.01, while the method used for comparing 0.9999...with 1 is more akin to what we would have to do to compare, say, pi given as the limit of some sequence with the circumference of a circle of diameter 1. Yours, Huon 10:26, 10 March 2006 (UTC)

What is a rational number?

If k is a rational number, then k can be expressed as a/b where a and b are integers and b CANNOT be 0. So, if 0.999... is a rational number, then there exists integers a and b such that a/b = 0.999... No such a and b exist, thus 0.999... cannot be rational. 70.110.84.242 20:25, 6 March 2006 (UTC)

Yes it does. a = b = 1. If you disagree, then there must be a number between 0.9... and 1 (an infinite number, in fact), so please give me one. Mdwh 00:03, 7 March 2006 (UTC)

No it does not. a=b=1 => a/b = 1 and not a/b = 0.999... Here is a number: Let A=99, then p = A/100+A/10000+A/1000000+... and there are infinitely many more. I leave the rest as an exercise for you. 70.110.84.242 00:33, 7 March 2006 (UTC)

Do you want to imply that (90/100+9/100)+(90/100²+9/100²)+... is not equal to 90/100+9/100+90/100²+9/100²+...? That seems to defy the law of associativity. Yours, Huon 10:18, 7 March 2006 (UTC)

I do not imply this at all and it has nothing to do with the law of associativity. The following two numbers are different:

  9/10+9/100+9/1000+....
  A/100 + A/10000 + A/1000000 + ....         A = 99

It is easy to verify by checking the partial sums:

  A/100 > 9/10                       <=>   .99 > .9
  A/100 + A/10000 > 9/10 + 9/100     <=>   .9999 > .99

In fact you cannot simply add the terms as you like because in the case of irrational numbers, there is no pattern so it is important to compare the partial sums with the correct order of the terms. 68.238.104.216 14:15, 7 March 2006 (UTC)

Farewell

I think you will all be happy to read this: I will no longer be continuing this discussion. Hope somebody else with some brains will be able to pick up where I left off. ANON/BRAVE-IP/Whatever...

You go wrong because you are all wrong...

(2 - 0.999...)^[1/(1-0.999...)] = ?

If 0.999... is equal to 1, then the above expression is meaningless. However, as long as 0.999... < 1, the above expression is equal to e (2.718...).

This has nothing to do with 'quixotic' quests or preconceived ideas or cranks. I suppose Meni Rosenfeld or some other Wiki puppet will say the expression is meaningless.

Wikipedia has two choices:

1) Stay with the nonsense of user KSmrq

or

2) Get rid of this false article.

71.248.135.27 19:59, 17 April 2006 (UTC)

It is true that $\lim_{n\to\infty}\left((1+\frac{1}{10^n})^{10^n}\right)=e$ . But what you claim seems to be $\left(\lim_{n\to\infty}(1+\frac{1}{10^n})\right)^{\lim_{n\to\infty}10^n}=e$ , and that is indeed wrong, or at least ill-defined. If you disagree, it would be extremely helpful if you were able to give precise definitions for 0.9999... and e, and a proof why e should be equal to the expression given above. Yours, Huon 08:38, 18 April 2006 (UTC)

I claimed that: Limit n-> oo { 2 - [1 - 1/(10^n)] } ^ (10^n) = e.

This is accurate and just a different form of the common definition. If indeed:

    Lim n-> oo  [ 1 - 1/(10^n)] = 1, then the above does not produce e.

If however, we say Lim n-> oo [ 1 - 1/(10^n)] = 0.999..., then there is no problem since 0.999... < 1. Of course there is still a problem with misconception about an infinite sum being equal to the limit of its partial sums. There is no clear statement of this in any of the respected mathematics archives. 71.248.138.107

<sarcasm>Of course, proven facts known by every mathematician in the world have no weight when compared to some rants made by an anonymous person with no desire to use any kind of accuracy in his arguments, but only to annoy people. We should obviously delete this article, because the aforementioned anonymous person lacks the mathematical understanding required to comprehend its contents. </sarcasm> -- Meni Rosenfeld (talk) 15:35, 18 April 2006 (UTC)

Sarcasm becomes you Rosenfeld - it is the lowest form of wit. You were proved wrong in your previous argument regarding your attempt to show that pi and other real numbers are compared the same way as 0.999... and 1. You were challenged to use the same argument you presented in your defense to show how pi and another real number are compared - you could not because it is untrue that pi, e and any other irrational number is compared like any other real number. So you would be wiser to listen rather than spew out rubbish you have been taught. 71.248.138.107 16:21, 27 April 2006 (UTC)

Well, non sarcasm way. I can point you why are you wrong. Let x=0.9999... , y=1.0000 then you consider

(2 - x)^[1/(y-x)]

now, you state

If x=y, the above expression makes no sense (well it does not, but 1/(2-2) doesn't either, but that doesn't mean 2 is not equal to 2)
If x not equal to y, then above expression equal to e. Yes, and if 2=3, then above expression is also equal to 3. I mean, any affirmation of the form "if P then Q" is valid when P is false. You're actually not proving this entry is false. All you're saying

If the entry were false, the limit would be e.

Then you switch to limits. So let's do it the proper way. Let a_n = 0.99999 with n 9's. The claim on the article is that limit a_n = 1, and it's clear that a_n is NOT 1 for any finite n.

Let's go through your argument now

I claimed that: Limit n-> oo { 2 - [1 - 1/(10^n)] } ^ (10^n) = e.

which rewritten is

$(1- 1/(10^n))^{(10^n)} \to e$

which indeed has the same limit as

$\lim_{t\to\infty} (1-1/t)^t = e$

But your argument fails to deliver for the same reasons that

$\lim_{x->2}\frac{x^2-4}{x-2} =4$

yet you cannot just "substitute" x=2 on the quotient. The reason is that limit is all about values that are close but not exactly the given point (in the later, values close but not equal to 2 and in the former values close to 0.999... but not exactly it). In each cases you cannot substitute x=0.999999... or x=2 but that doesn't contradicts the fact the limit exists. -- ( drini's page ☎ ) 16:54, 27 April 2006 (UTC)

All your points are valid except for the very last one. What is wrong with:

Lim (t=1 to infty) (1-1/(10^t))^(10^t) = e (*)

I am not disputing the limit but stating that if

 0.999... =  Limit (t=1 to infinity) 1-1/(10^t)

and

$\lim_{t\to\infty} 1-1/(10^t) = 1$

then (*) is absurd since Limit (t->oo) (1+t)^t = (1+t)(1+t)....

which by the limit product law is the product of the limits. i.e.

  Lim  f(x).g(x)  = Lim f(x) * Lim g(x)

Finally, I am challenging this contributor (Rosenfeld) because he failed to prove that all real numbers are compared the same way. For example, pi and another real number is not compared the same way as 0.999... and another real number because there is confusion about what an infinite sum and the limit of an infinite sum really is. Perhaps you too ought to read this entire thread before you make up your mind. 70.110.87.245 21:28, 27 April 2006 (UTC)

(*) has nothing to do with Lim f(x)g(x) = Lim f(x) * Lim g(x)

Why not? Because, as t (or x or whatever) increases, the number of factors increases as well. And while every single factor converges to 1, the given limit, where the number of factors becomes ever larger, need not (and does not) converge to 1. Let me give a related example for clarification: $\lim_{n\to\infty}\prod_{i=1}^{n}\frac{n+i}{n+i-1}$ . Again every single factor (n+i)/(n+i-1) converges to 1 (whatever i may be, as long as n+i gets large), but the product of the n factors is always equal to 2 (due to cancellation); thus, so is its limit.

Finally, considering the comparison of real numbers again: Let us begin be recalling the definition of a real number. We could use Dedekind cuts, but personally I prefer Cauchy sequences. So a real number is an equivalence class of Cauchy sequences of rational numbers, where two sequences (a_n), (b_n) are considered to be equivalent if the Cauchy sequence (a_n-b_n) of rationals converges to 0 (this is convergence in the set of rationals, no reals here). Then how do I "compare" real numbers? I may ask questions like:

Are two numbers equal?
Is one number less than another?

Let us consider now the case of 0.9999... and 1 as an example. The real number 1 also is an equivalence class of cauchy sequences - the equivalence class containing the constant sequence (1, 1, 1, ...). On the other hand, 0.9999... is the class containing the sequence (0.9, 0.99, 0.999, ...). In order to show that the equivalence classes coincide, I need to show that the sequence (1-0.9, 1-0.99, 1-0.999, ...) converges to 0 (which is rather obvious).

If I am to compare pi to some other number, I again need some Cauchy sequence of rationals which is contained in the equivalence class I call "pi". Examples of such sequences (or, more precisely, of series whose sequences of partial sums are of this type) are given above; to show that the two series I gave in my 28 February edit indeed give equivalent sequences of partial sums probably is non-trivial. The other number pi shall be compared to is also given by some equivalence class of Cauchy sequences. Again I have to choose a representative of the equivalence class, form the difference series, and decide whether this series converges to 0 or whether its elements eventually will all be positive (or negative) without convergence to 0. Depending on that result, I have shown pi and the other number to be equal, or I have shown one of them to be less than the other.

This is, in principle, the method employed to compare real numbers - no matter whether they are 0.9999... and 1 or pi and 3 (or even 1 and 2). Of course, this complicated method is usually simplified - for example, by identifying the equivalence classes of the constant sequences with the rational numbers, or by using propositions relating the limit of a sequence of reals to the elements of that sequence.

Now where do you see differences between comparing 0.9999... to 1 and other comparisons of reals? Being rather sorry for this long rant, Huon 10:13, 28 April 2006 (UTC)

There is no cancellation possible in the example I stated and thus:

      Lim f(x)g(x)k(x).....  = Lim f(x) * Lim g(x) * Lim k(x) * ...

So the number e would not exist if 0.999... = 1.

Now your Cauchy example is irrelevant but I'll address it quickly:

 If we compare numbers using convergence on differences in
 Cauchy sequences, then this is not the same as comparing *limits*.
 All this process does, is tell one that the differences converge,
 it says nothing about the numbers being equal.

An equivalence class is not about equal numbers, rather it's about numbers that have similar properties. You have incorrectly applied this knowledge in your example. You cannot say that because an equivalence exists in terms of differences converging that the numbers are equal. Just take a look at 0, 5, 10, 15 - one can build an equivalence class for these under certain conditions, however they are by no means *equal*. In fact, reflexivity, transitivity and symmetry imply equivalence. 71.248.139.36 14:21, 28 April 2006 (UTC)

This is exactly the point. You really should read Construction of real numbers - then you'll see that a real number is defined as an equivalence class of cauchy sequences of rational numbers (I assume we have an agreement about what a rational number is - if not, we can clarify it). There are of course other (essentially identical) definitions, but we'll use that one. The moment something is defined, our intuition about what is right or wrong is irrelevant - What matters is the definition. So your statement, "an equivalence class is not about equal numbers", is simply wrong here - if a real number is defined as an equivalence class, then if two objects are the same equivalence class, they are the same real number. That's really all there is to it.

If I take your example, suppose for every integer n I define n* = the equivalence class of n under congruence with modulus 5, then I'll get 0* = 5*. So, while 0 and 5 are different integers, 0* and 5* are the same equivalence class - the same object. So if I have something like 37* and I want to compare it to the other objects, I will say that 37 is congruent to 2 mod 5, therefore 37 is in 2*, therefore 37*=2*. The same for real numbers - if I show that two numbers are representatives of the same equivalence class - that is, the same real number - then these 2 real numbers are equal.

And here comes the crucial point - you want very badly a mathematical structure where 0.999... < 1. And it just happens that, as has been shown numerous times, the structure known as the real numbers, plentiful definitions of which can be found at Construction of real numbers, and which is very commonly used in mathematics - is just not up to this demand. That doesn't mean there aren't other structures which can satisfy it. I have at least one such structure in mind, but among other things, in this particular structure, when you have 2 different objects you can't always say that one is greater than the other (though, for example, you can say that 0.999... <1). Also, there are structures like hyperreal numbers and surreal numbers, where I'm not sure objects such as 0.999... are defined, but they do contain infinitesimals (surely you realize that 1 - 0.999..., if not 0, is an infinitesimal?), so they may better represent what you believe a "number" should be. -- Meni Rosenfeld (talk) 17:07, 28 April 2006 (UTC)

Really? There was no such thing as real number, hyperreal number, etc when the foundations of calculus and mathematics was laid. In fact, 'real analysis' did not even exist. This is the point, not what you have been ranting about. Real analysis has errors and contradictions. This is but just one of these errors.... There are many more. As for wanting very badly a structure that accomodates 0.999... < 1 - this is nonsense. There is no such thing as an infinitesimal. If 0.999... is not less than 1, the reals have holes but we do not treat the reals as if they have holes. They are treated as a complete field. Having 0.999... < 1 does not contradict the archimedean principle in any way. There are infinitely many numbers between 0.999... and 1. You cannot have a number system that carries blatant contradictions such as 0.999... = 1 in one case and the existence of the natural number e in the other case. Finally, you do not compare all real numbers in the same way. Please don't talk about Frechet's identity of indiscernibles - it does not make sense because a point does not make sense and has never been properly defined. Cauchy sequences support Frechet's metric space theory. In fact a Cauchy sequence is defined in terms of the distance metric. Frechet was an idiot in my opinion. 71.248.132.250 13:34, 29 April 2006 (UTC)

My example does not in principle rely on cancellation. It just serves to show that in general, Lim f(x)g(x)k(x)... and Lim f(x) * Lim g(x) * Lim k(x) * ... do not coincide. Let us, as another example, take the sequence $a_n=\prod_{i=1}^n2^{1/n}$ . There is no cancellation involved this time; the limit of 2^(1/n) is 1, but still a_n is equal to 2 for all n; so is the limit.

Concerning equivalence classes, Cauchy sequences and real numbers, I have nothing to add to Meni Rosenfeld's explanations. Yours, Huon 17:32, 28 April 2006 (UTC)

Your example is not the same as Lim f(x) * Lim g(x) * Lim k(x) * .... It is incredible how you continually fail to get your facts right Huon. How is 1/2 the same as f(x) or g(x) ? The two examples are completely different. 1/2 is not dependent on anything. It is constant. 1/2^n is not the same as the above definition. Can you write 1/2^n as Lim f(x) * Lim g(x) * Lim k(x) *... ? 71.248.132.250 13:34, 29 April 2006 (UTC)

Pardon me for saying this, but your last post strengthens my belief that you understand very well that 0.999... = 1, but that you, for some reason, enjoy the discomfort caused to people by arguing to the contrary. I'm sure you understand why. But never mind that. I have a question: 0.999... is some number, right? Denote it by c. You say c < 1. So it must be that 1 - c > 0. So there is definitely a number $\frac{1}{1-c}$ . Surely this is a positive number, and surely it is greater than any integer. So how exactly does the archimedean principle hold in such circumstances? And another thing - a point is (in some contexts) undefined, the same way sets are undefined. Sets derive their meaning from axioms - so do points in synthetic approaches to geometry. -- Meni Rosenfeld (talk) 14:35, 29 April 2006 (UTC)

I know very well that the limit of 9/10+9/100+9/1000+... = 1. As for a number between 0.999... and 1, there are infinitely many. Just check the archive - there are examples of these. 71.248.151.133

OK, even more details. You said above: "[...] then (*) is absurd since Limit (t->oo) (1+t)^t = (1+t)(1+t)...." There is a typo here; you probably meant (1+1/t)^t. Then t obviously is a natural number; else (1+t)(1+t)... (or (1+1/t)(1+1/t)...) would not be well-defined. Now for t a natural number, we have $(1+1/t)^t=\prod_{i=1}^t(1+1/t)$ . Now your argument about the limit of a product being the product of the limits seems to me to be that $\lim_{t\to\infty}(1+1/t)^t=\prod_{i=1}^\infty\lim_{t\to\infty}(1+1/t)$ , and the right hand side is equal to 1. If you did not mean that, I probably misunderstood you completely; please clarify what I cited above.

Now I gave a rather analogous example: Obviously the limit of $\prod_{i=1}^t2^{1/t}$ is 2, but the product of the limits of the factors 2^(1/t) is equal to 1. So in effect, what you call f(x), g(x), k(x) would here be f(t)=g(t)=k(t)=2^(1/t), and the limit of f(t)g(t)k(t)... is not the same as the product lim f(t)*lim g(t)*lim k(t)*.... This should be a counter-example to transpose limits with products when the number of factors is not constant as we approach the limit. If you still disagree, please help me and clarify where the f(x), g(x) and so on appear in your remarks about the representation of e as the limit of (1+1/t)^t. (As a general remark, if you believe I misunderstood you, it might be helpful to give a more detailed account of what you meant instead of critizising my examples which probably will be besides the point anyway.) Yours, Huon 15:22, 29 April 2006 (UTC)

1. (1+t)(1+t)... (or (1+1/t)(1+1/t)...) is defined for both natural and real numbers but in

this context we are talking about natural numbers only.

2. As for your analogous example, it is incorrect because the limit is not

  equal to 1:

  Lim [t=1, oo] Product 2^(1/t)  = Lim [2 * 2^(1/2) * 2^(1/3) * 2^(1/4) * ...]

  the above limit is not equal to 1. How did you arrive at 1?

  Perhaps what you meant is:

  Lim [t=1, oo] Product 2^(-t) = Lim [ 1/2 * 1/4 * 1/8 * ...]

  But even in this case

  Lim [t=1, oo] Product 2^(-t) = Lim 1/2 * Lim 1/4 * Lim 1/8 * ...]

  and this limit is also not 1.

71.248.151.133 18:23, 30 April 2006 (UTC)

1. We were talking of (1+1/t)^t=(1+1/t)(1+1/t)(1+1/t)..., i.e. t factors. Then let us have a look at, say, t=pi. What should pi factors be? The left side (with the t-th power) can be defined for all real numbers (that makes implicit use of the exponential function); the right side presents difficulties...

2. You do not form the limit in the correct way. If you tried to do it for e the way you do it for my product, saying that

Lim [t=1, oo] Product 1+1/t = Lim [2 * (1+1/2) * (1+1/3) * (1+1/4)* ...],

you would conclude that e is not finite. Instead, you must keep track of the correct number of factors: You successively get 2=2, 2^(1/2)*2^(1/2)=2, 2^(1/3)*2^(1/3)*2^(1/3)=2, ... Thus, the limit of the product is equal to 2 (as the product is equal to 2 for every natural number t). This is indeed not equal to the product of the limit of the factors: 2^(1/t) tends to 1, thus Lim 2^(1/t) * Lim 2^(1/t) * Lim 2^(1/t) * ... = 1*1*1*... is equal to 1. That's just what I am saying: When the number of factors becomes infinite, limits and products are no longer transposable. Yours, Huon 22:33, 30 April 2006 (UTC)

I think you are getting confused: In the case of e you do not have:

    (1+1/2)*(1+1/3)* .... etc.

t is the same value throughout. It would not equal to e if t changed as you think. That is entirely a different product. What you need is:

    (1+1/2)*(1+1/2)

    (1+1/3)*(1+1/3)*(1+1/3)

   etc.

And yes, these limits and products are transposable which is contrary to what you think. Are you still a student Huon? 70.110.93.169 01:52, 2 May 2006 (UTC)

Of course you are right in the case of e. And my example above is just the same; again t is the same value throughout. Why did you insist on that being different?

And do you really claim that $\prod_{i=1}^\infty (\lim_{n\to\infty}f(n)) = \lim_{n\to\infty}(\prod_{i=1}^n f(n))$ ? e itself is a counterexample (for e is not equal to 1). I repeat: When the number of factors becomes infinite, products and limits can no longer be transposed. Several counterexamles have been given. I can easily produce more, for any variety of that statement I can think of. My suggestion: Try to give a rigorous proof, with full details, based on precise definitions, and you will see where it fails. Yours, Huon 09:13, 2 May 2006 (UTC)

I did not insist on anything being different. I think you are very confused. So you agree that the t's are all the same. Then for your example:

Lim [t=1, oo] Product 2^1/t = Lim 2   [t =1]

Lim [t=1, oo] Product 2^1/t = Lim sqrt(2)*sqrt(2) = 2   [t =2]

  Check transposition:    Lim sqrt(2) * Lim sqrt (2) = 2

Lim [t=1, oo] Product 2^1/t = Lim 2^3*  2^3 * 2^3 = 2   [t =3]

  Check transposition:    Lim 2^3 * Lim 2^3 * Lim 2^3 = 2

Where do you see a problem with transposition? You can open any elementary calculus book and find proof for this. I am sorry I don't have time to do it here.

e is not a counterexample. I am using e as an example to show that 0.999... must be less than 1 since this limit is used in the limit for e. 71.248.136.178 22:30, 2 May 2006 (UTC)

I'm sorry, but I still disagree. Let's just have a look at

Lim [t=1, oo] Product 2^1/t = Lim 2   [t =1]

How can there be a limit for t tending to infinity if you next claim t to be equal to 1? Why is there even a limit on the right side of the equation? Again, if we tried to have a look at e the way you now interpret my example, we would get:

e = Lim [t=1, oo] Product (1+1/t) = Lim 2   [t =1]

This is obviously wrong, especially since for [t =2] you would instead get e=9/4. Thus, your interpretation of my example must be wrong as well. It should be completely analogous to e. If you were able to provide a reference to an "elementary calculus book" to support your view, I would be grateful. Don't worry, it needn't be a nineteeth century book... Yours, Huon 09:12, 3 May 2006 (UTC)

I do not claim t = 1. What are you talking about?! There is a limit on the RHS because you were insisting that transposition is not possible. In fact it is, and I demonstrated it with your example. Yes, you would get e=9/4 and as t approaches infinity, you will get pretty close to the real value of e (2.718...). As for a reference, look at:

http://math.stanford.edu/~aschultz/w06/math19/coursenotes_and_handouts/coursenotes_060123.pdf

It shows the limit law for products (Stanford is considered a respectable university...). But there are many others so I don't know what your problem is to be quite frank. As for a nineteenth century reference, yes, you would need one to show me where it states that an infinite sum is defined as its limit. This very example of e shows exactly how absurd and ridiculous your article really is. You have got it wrong because you don't know what you are talking about. It makes no sense to define an infinite sum as its limit because if this were true, then e should equal to 1, but e does not equal to 1. See, we are really not interested in what the actual limit is (only that it exists) and in most cases we cannot find it. e.g. e, pi and any other irrational number. We care only to know that the series representing these are bounded above. 71.248.142.82 21:49, 3 May 2006 (UTC)

First of all, thank you for providing a reference. I suppose you meant the fourth part of the theorem on page two, which states: If $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ ) exist, we have $\lim_{x\to a}(fg)(x)=\lim_{x\to a}f(x)\lim_{x\to a}g(x)$ . That's true, but there are one and a half reason why that does not cover our case. The half reason is that that theorem is about limits of functions, not limits of sequences. That's not so big a problem; an analogous statement is true for limits of sequences. The other reason why that does not apply to our case is crucial: That theorem is about two factors. But for, say, e=lim (1+1/n)^n the number of factors is not constant; it tends to infinity as n tends to infinity. Thus, the Stanford theorem proves lim ((1+1/n)²)= (lim (1+1/n))², but it cannot be used to show anything like e=(lim (1+1/n))^(whatever).

By the way, if you consider Stanford as a respectable university and believe their publications to be true, have a look at this. It is the official solution to a mid-term exam (not just some student's notes). The answer to, say, problem 7 (page 10) shows that to Stanford, the sum of a series is equal to the limit of the sequence of the partial sums, or even more precisely, that for -1<a<1, $\sum_{i=0}^\infty b*a^n=b+ba+ba^2+ba^3+...=\frac{b}{1-a}$ . Thus, to Stanford, 0.9999... = 0.9+0.9/10+0.9/100+0.9/1000+... = 0.9/(1-1/10)) = 0.9/0.9 = 1.

Next, concerning my example, you wrote:

Lim [t=1, oo] Product 2^1/t = Lim 2   [t =1]

Now you said you did not claim t=1. Then what exactly did you mean by that [t =1]? Was there not t set to be equal to 1 on the right side?

Also, obviously e is not equal to 2. Neither is it equal to 9/4, and even more obviously, it is not equal to both at once. Would you mind clarifying your comments to that effect?

Finally, concerning language. I prefer not to use the term "infinite sum". That has never been defined properly. Instead, we should speak (as, once more, Stanford does) of series and the sum of a series. Then the sum of a series is (by definition) equal to the limit of the sequence of partial sums. But never did I (or anybody but you yourself) claim that "an infinite sum is equal to its limit". I suppose you mean by an infinite sum what I would call a series, and by "its limit" the limit of the sequence of partial sums?

To summarize:

The Stanford theorem does not cover our case.
Other Stanford publications support my point of view (and this article).
e is not equal to 9/4; I still do not understand your notation.
I would be glad not to hear of an "infinite sum being equal to its limit" again without some definitions.

Yours, Huon 10:31, 4 May 2006 (UTC)

You really can ramble on and on, can't you? As for Stanford, I said it is considered to be respectable, I did not say I respect it or any other institution. I used the reference because you asked for it. It is a valid reference contrary to what you have stated. I am not going to argue this juvenile point with you. You need to go back to school if you can't accept it.

It is true that the definition can be extended to as many terms as you like.

It says: lim [f(x)+g(x)] = lim f(x) * lim g(x)

This is easily extended to:

lim [f(x)*g(x)*h(x)] = lim f(x) * lim g(x) * lim h(x)

Now lim (1+1/t)^3 = lim(1+1/t)*lim(1+1/t)*lim(1+1/t)

The above is an approximation of e when t = 3, just as in the previous example, 9/4 is an approximation of e when t = 2. So, to better aproximate the value of e, one calculates more terms on the right hand side. In fact, the value that is calculated converges to the value we approximate as e.

Now a problem arises if we perform some rearrangment as follows:

lim (2 - 0.999...)^(1/0.999...)

lim (2 - [1-(1/10^t)])^(10^t)

lim (2 - [1-(1/10^t)])^(10^t)
    =Lim (2 - [1-(1/10^t)])*Lim(2 - [1-(1/10^t)])*...*Lim(2 - [1-(1/10^t)])

We know that as t->oo, the LHS becomes e. However, looking at the RHS, if 
0.999... = 1, then we have the RHS evaluating to 1 which is clearly *not* e!

BTW: Please don't write screeds of irrelevant rubbish, you are not impressing
me. 71.248.152.55 23:11, 4 May 2006 (UTC)

Let us focus on just one question for the moment: When you state

lim (2 - [1-(1/10^t)])^(10^t)
    =Lim (2 - [1-(1/10^t)])*Lim(2 - [1-(1/10^t)])*...*Lim(2 - [1-(1/10^t)])

how many factors are there on the right hand side? BTW: Please don't make ad hominem attacks, you are not impressing me either. Yours, Huon 23:37, 4 May 2006 (UTC)

There are 't' factors. 71.248.132.13 00:45, 6 May 2006 (UTC)

The left hand side does not depend on t (as it is a limit for t tending to infinity). And yet you say the number of factors on the right hand side depends on t? How is this number of factors connected to the rest of the formula? Yours, Huon 14:03, 6 May 2006 (UTC)

Wrong. Both sides depend on the value of t. The rhs is always an approximation. 68.238.107.103 16:55, 11 May 2006 (UTC)

So the left hand side depends on t? Then obviously your definition of a limit is different from mine (and that of, say, Wikipedia). It does not even matter whether you take the limit of a function or (as I believe you should) the limit of a sequence. Would you please provide your definition of a limit, and explain why it depends on the variable that tends to infinity? Thanks, Huon 22:37, 11 May 2006 (UTC)

THIS ARTICLE SHOULD STAY

under the category "Fun Mathematical Teasers". We could add an article such as "Proof that -1 = 1". Stuff like this is fun and is great for provoking thought. (For those of you that haven't seen the proof of -1 = 1, here it is)

Given:

$X=1 \,$

Solution…

$1=X^2 \,$

$0=X^2-1 \,$

$0=(X+1)\cdot(X-1) \,$

$\frac{0}{(X-1)}=(X+1) \,$

$0=X+1 \,$

$-(X)=1 \,$

…Substitute 1 in for X…

$-(1)=1\,$

…Therefore…

$-1=1 \,$

Many people don't catch that $\frac{0}{(X-1)}=(X+1) \,$ is an illegal operation.

Since X = 1, (X-1) =0; you are therefore saying $\frac{0}{0}=(X+1) \,$

Macwiki 00:00, 7 May 2006 (UTC)

Actually, there is an article on invalid proofs containing this example. Huon 02:23, 7 May 2006 (UTC)

Thanks for the info Macwiki 03:03, 7 May 2006 (UTC)

Can we merge this article and the invalid proofs one? Macwiki 03:31, 7 May 2006 (UTC)

No. Melchoir 03:35, 7 May 2006 (UTC)

I'm not sure what your point is - your "proof" is flawed (as you point out yourself); the proof that 0.9recurring equals 1 however is valid. Mdwh 00:14, 7 May 2006 (UTC)

Looks like Macwiki thought that 0.999... is not really 1, and that this article gives invalid proofs for the equality. Not too sure what gave him that idea, but it only shows why care should be taken when presenting invalid proofs such as this -1=1 thing. -- Meni Rosenfeld (talk) 12:15, 8 May 2006 (UTC)

$0.\overline{3} \ne \frac1{3}$ and therefore this proof erroneous

Are you sure this proof is valid? I beg to differ.

The article incorrectly assumes that $0.\overline{3} = \frac{1}{3}$ .

$0.\overline{3} \approx \frac{1}{3}$ , but not equal to. $0.\overline{3}$ is a "decimal approximation" (not exactness) of 1/3.

$0.\overline{3} < \frac1{3}$

$\lim_{n \to \infty} \sum_{i=1}^n \frac3{10^n}=\frac1{3}$ and is the point exactly after $0.\overline{3}$

All this article does is prove that $0.\overline{3} \ne \frac1{3}$ .

If $0.\overline{3}$ equaled $\frac1{3}$ , then $3 \cdot 0.\overline{3}$ would equal $1 \,$ and not 0.999999999…

Macwiki 03:03, 7 May 2006 (UTC)

No, 0.3recurring equals 1/3. Rigourous proofs that 0.9recurring equals 1 (which can be used for 0.3recurring equals 1/3) are given under "Advanced proofs". You agree yourself that the limit is equal to 1/3 - well, 0.3recurring equals that limit by definition. See decimal expansion. Mdwh 03:30, 7 May 2006 (UTC)

We're going to have to agree to disagree on this. My understanding is that the limit in this case has the same function as an asymtode. (It gets closer, but never reaches, like the kiss property. Concerning "You agree yourself that the limit is equal to 1/3 - well, 0.3recurring equals that limit"—I probably rushed it, I was trying to find a way to explain it mathematically. I do however, back up all other arguments I listed Macwiki 03:58, 7 May 2006 (UTC)

Your understanding is wrong. The sequence 0.3, 0.33, 0.333, etc, has the property you explain, but the actual completed infinite expression $0.\overline{3}$ is exactly equal to 1/3.

Just the same, you have correctly identified a gap in the 1/3-based proof; this proof assumes that there exists a decimal expression that, when multiplied by 3, gives exactly 1. This is a step in need of proof. The proof's not hard, but if you leave it out then the argument is incomplete. --Trovatore 05:38, 7 May 2006 (UTC)

No. I would say your understanding is incorrect. There is no such thing as the actual completed infinite expression ! Where do you get off saying this?! What is a completed infinite expression? And once again I believe you have a Phd too. Oh well, I rest my case.... 68.238.107.103 16:58, 11 May 2006 (UTC)

OK, I think I understand you. Also, given this passage from the article–"Another kind of proof adapts to any repeating decimal. When a fraction in decimal notation is multiplied by 10, the digits do not change but the decimal separator moves one place to the right. Thus 10 × 0.9999… equals 9.9999…, which is 9 more than the original number. Subtracting the smaller number from the larger can proceed digit by digit; the result is 9 − 9, which is 0, in each of the digits after the decimal separator. But trailing zeros do not change a number, so the difference is exactly 9. The final step uses algebra. Let the decimal number in question, 0.9999…, be called c. Then 10c − c = 9. This is the same as 9c = 9. Dividing both sides by 9 completes the proof: c = 1."- If we assume that $0.\overline{3} = \sum_{n=1}^\infty \frac3{10^n}$ , does this proof divide $\frac{\infty}{\infty}$ because $n \to \infty$ in the summation? Macwiki 00:07, 8 May 2006 (UTC)

I'm not sure I understand what you are saying - when you say "It gets closer", what is "It"? The limit doesn't get closer - it is a fixed value, like an asymptote. $0.\overline{3}$ is, by definition, this limit. Mdwh 15:29, 7 May 2006 (UTC)

Sorry, 'IT' is the value of $0.\overline{3}$ . The more digits you add, the closer you get to 1/3 (for example: .333 is closer to 1/3 than .33). Even if $0.\overline{3}$ went on for $\infty$ digits, (which by definition it does) it would get closer and closer to 1/3, but never make it. So if you were to graph it using "digits used" by "value" you would get a curve that never reaches 1/3, but gets forever closer.Macwiki 23:33, 7 May 2006 (UTC)

Mdwh has explained this twice now. $0.\overline{3}$ is defined to be a limit. It does not have any number of digits, and it is not possible to "add" digits to it. It does not contain a sequence. It does not imply a curve. It does not move closer to anything. It does not somehow fail to "reach" its "value". It does not have anything to do with infinity, and its definition does not involve infinity. If you "graph it", you graph just one point: one third. Melchoir 23:57, 7 May 2006 (UTC)

Then isn't $\frac1{3} \cdot 3 = 1$ ? Macwiki 00:07, 8 May 2006 (UTC)

That is a true equation, yes. Melchoir 00:08, 8 May 2006 (UTC)

OK, I think I get it. Because there is no distance between 1 and $.\overline{9}$ , they occupy the same space and therefore must be the same thing. And $.\overline{9} \ne .\overline{9}8$ because $.\overline{9}8$ cannot exist because you can't put anything at the end of $\infty$ , thus keeping any number from equaling all numbers. Is that right? Macwiki 00:35, 8 May 2006 (UTC)

Yes, that's a good way to think about it! Your intuition about no distance can actually be formalized into a foundation for the entire real number system; see Construction of real numbers#Construction from Cauchy sequences. As for $.\overline{9}8$ , yes, most people would consider it undefined, since you can't construct a decimal expansion with all 9s and then an 8. But if you want to be liberal about it, then you might define it to be the limit of the sequence (.8, .98, .998, .9998, ...), which is 1 after all. So in some sense, $.\overline{9} =.\overline{9}8$ ! Melchoir 00:46, 8 May 2006 (UTC)

Well now, you certainly can have a string of infinitely many nines, followed by an eight. It just wouldn't be the decimal representation of a real number. I think it might be a decimal representation of a surreal number, though; namely, 1+8·10^−ω. I haven't really checked that last point so take it with a grain of salt. --Trovatore 15:45, 8 May 2006 (UTC)

I recall from Geometry class that Geometry is divided in 2 categories: discrete and something else (it's been a while). Does the validity of this proof depend on which geometry you use, or is it the same in all cases? Do I even recall in romotely the same area? Macwiki

I think the other was Euclidian geometry.Macwiki

I fail to see any connection between this proof and geometry. After all, we do not really speak of point sets, lines and such, but of numbers and sequences. One can speak of numbers in geometric terms (i.e. the line of real numbers, the complex plane and so on), but using a discrete geometry seems strange: Neither the reals nor the rationals are well-suited to it (unlike, say, the integers). If one were to give the set of reals the discrete topology, that would indeed change the validity of the proof, since then the sequence (0.9, 0.99, 0.999, ...) no longer converges (as only sequences which become constant converge), and 0.9999... ceases to be defined in any meaningful way. But that should be considered as highly non-standard (and rather useless, too). Yours, Huon 19:01, 8 May 2006 (UTC)

You will always be challenged on this article because it is untrue.

The result indeed does apply for all values of n, i.e. 0.999... is strictly less than 1. There is no counter-example. The archimedean principle is not violated. All the so-called 'proofs' use faulty concepts from real analysis (mostly based on Cauchy sequences) to show the 'equality'. Real analysis is an attempt to explain the real numbers. Real numbers were around long before real analysis was invented. Why are you and the entire wiki team so arrogant that you cannot even begin to see this? Show me one encyclopedic reference as you claimed that backs up your statement for equality. This article is not only counter-intuitive, it is in fact real nonsense and no valid proof exists anywhere that demonstrates this equality or ever has existed. This so-called surprise is a result of gross miunderstanding and misinterpretation of the limit of the series 9/10+9/100+9/1000+.... In fact, there is no stable definition of the sum of an infinite series. Prof. Hardy and others will claim the infinite sum is equal to the limit. If this is true, then why do you not compare all real numbers by their limits? 70.110.86.207 15:59, 12 May 2006 (UTC)

At the risk of violating WP:DNFT: We do "compare real numbers by their limits". Real numbers are the limits. Septentrionalis 18:31, 12 May 2006 (UTC)

Actually you do not compare real numbers by their limits. Please show me how you compare pi and 3.14 and then show me how you compare 0.999... with 1. Meni Rosenfield attempted this and ended up putting his foot in his mouth. Real numbers are not limits. Real analysis attempts to define real numbers through limits but fails misereably. 70.110.86.207 20:16, 12 May 2006 (UTC)

This discussion belongs to the arguments page. Please continue there. Thanks, Huon 20:38, 12 May 2006 (UTC)

I've moved this section from the article discussion page to the arguments page.--Trystan 20:42, 12 May 2006 (UTC)

Thanks. Here we can see some of the answers to anon's complaints. The very first topic concerns comparison of numbers. I explicitly gave a comparison of pi (defined as the limit of a certain sequence) to 2.75 (which would, as an equivalence class of Cauchy sequences, be represented by the constant sequence (11/4, 11/4, 11/4, ...)). Comparing pi to 3.14=314/100 would involve more work along the same lines. The comparison of 0.9999... to 1 also involves taking partial sums and deducing something about the limit. Indeed the methods of comparison are not completely analogous, but the difference lies not in some strange properties of either 0.9999..., 1, 11/4 or pi, but in the fact that for pi and 11/4, the numbers to be compared are different, while for 0.9999... and 1, they are the same. Usually showing two numbers to be different is easier than showing them to be equal. Comparison of pi and 3.14 should, thus, not be likened to comparison of 0.9999... and 1, but, say, to comparison of 0.9999... and 0.99=1-1/100.

Finally, there are two points in anon's "challenge" I cannot let go unanswered. Firstly, I doubt real numbers were around for a longer time than Cauchy sequences and Dedekind cuts. Please provide a reference - any reference! - where real numbers are properly defined without making use of either. Please don't give a text where real numbers are just assumed to be known intuitively, but some proper definition. Now if you could find a reference that predates Cauchy sequences and Dedekind cuts, that would be a bonus (and ist should rule out any attempt at giving an axiomatic definition, which should be possible though difficult).

Secondly, the Archimedean property states that for every positive real number x, there is a natural number n such that 1/n<x. Applying this to x=1-0.9999..., which is a positive real number if 0.9999... is a real number less than 1, would yield a natural number n such that 10^(-n) < 1/n < 1-0.9999..., and thus 0.9999... < 1-10^(-n) = 0.999...9. Thus, there is some natural number n such that $\sum_{i=1}^n\frac{9}{10^i}=0.999...9000... > 0.999...9999...$ . Can we all agree that this would be a consequence of the Archimedean property? If not, please be precise when pointing out where you disagree. Yours, Huon 21:42, 12 May 2006 (UTC)

You have an incredible ability to ramble on and on about irrelevant information. You have not shown anything Huon, only that you do not know what you are talking about. Again and again, you ask for a reference but you have not been able to provide even one reference showing that an infinite sum is equal to its limit. Furthermore, you are unable to compare all real numbers by their limits. The archive shows clearly how the Archimedean property is not violated - that you cannot understand it, does not mean it is incorrect. Go back and study it - you might get it someday. By the way, I do not need to provide any reference for stating that real numbers were around long before real analysis - Isaac Newton and many other mathematicians knew absolutely nothing about real analysis. References are as good as those who compile the articles. Just because some Phd wrote an article does not make it an indisputable fact. I can see you are a product of some university. Let me tell you why they taught you to provide references: See, many stupid people attend university and to prevent them from writing a ton of rubbish, the faculty would have them 'research' the material so that they could perhaps check if they were somehow on the right track. In other words, references were required for idiots. There is no safety simply because a lot of people start to agree on the same thing. It does not make something right by fiat. Let me ask you this question: Suppose you are asked to research some topic in which there are no references, what would you submit for your research paper then? In fact, you would not be able to submit anything because no one else is saying the same thing as you!! Thus you are a fraud if you do and your work is summarily dismissed. Seems like you need to have a peer review - okay, so what if your peers are all idiots or inferior to you intellectually? 71.248.141.160 13:04, 14 May 2006 (UTC)

If you want to discuss mathematics, I will gladly comply. But this last "ramble about irrelevant information" is clearly off-topic and shows a distinct lack of mathematical arguments on your part. Besides, you also show you either don't read or utterly ignore my posts, for example this one and this one concerning the "infinite sum equal to its limit" question. Finally, concerning peer review: Even if you happen to be smarter than all other mathematicians and able to conceive bright sparkling new theorems nobody else ever thought of, you should better be able to give a complete proof, and one even your dim-witted colleagues can understand.

I will not write again until we discuss mathematics once more. Yours, Huon 15:35, 14 May 2006 (UTC)

"Finally, concerning language. I prefer not to use the term "infinite sum". That has never been defined properly. Instead, we should speak of series and the sum of a series. Then the sum of a series is (by definition) equal to the limit of the sequence of partial sums. But never did I (or anybody but you yourself) claim that "an infinite sum is equal to its limit". I suppose you mean by an infinite sum what I would call a series, and by "its limit" the limit of the sequence of partial sums?"

First, it's good that you finally agree that "infinite sum" has never been properly defined. Secondly, the sum of a series is not equal to the limit of the sequence of partial sums. It is an actual sum. In the case of infinite series, it cannot be computed, only approximated. Thirdly, Prof. Hardy has clearly stated that "an infinite sum is equal to its limit" contrary to what you state.

In answer to your last question: 1. By series I mean a finite or infinite sum of terms. This is how it was understood in Newton's time, not as a Cauchy sequence. 2. By limit I mean the maximum value (provided the terms are greater than zero) that cannot be reached or exceeded no matter how many terms you sum.

Finally, Newton understood that all infinite series could only be approximated. In fact he was the father of finite differences. These are the forerunners of the derivative. 71.248.149.130 00:45, 18 May 2006 (UTC)

Thanks for providing some definitions. Unfortunately, firstly, I fail to understand your definition of a series in the infinite case. You state it is "a finite or infinite sum of terms". We just agreed that "infinite sum" is not defined. Then what shall an "infinite sum of terms" be?

Secondly, mathematics has progressed since Newton's times. Especially, as, for example, the article on history of calculus claims, Newton and his contemporaries had an intuitive understanding of limits and such, but no precise definitions as we have today. But intuition can err. And since Newton has written quite a lot, I don't really feel like looking for any comments by him that might be of importance here. If you insist that Newton's writings are to be considered, I would ask you to provide a more precise reference.

Thirdly, I just browsed the archives, but failed to find any remark by Hardy that "an infinite sum is equal to its limit". Here, too, some precise reference (either a link to the relevant difference or just the date of Hardy's comment) would be welcome.

I believe the main difference between our points of view is that I don't recognize the object you call an "infinite sum of terms". Yours, Huon 09:54, 18 May 2006 (UTC)

1. An infinite sum of terms is one in which there is no last term. 2. Mathematics may have progressed a little since Newton, but understanding has retrogressed. 3. If Hardy's comment regarding an infinite sum is no longer on Wikipedia, it is because a Wiki puppet/sysop has removed it to save Hardy from personal disgrace. How convenient for Hardy! It may not have been in this article. Perhaps it was in another Wiki article. If Hardy had any "balls", he would admit openly he said this. 4. The main difference between our points of view is not that you fail to recognize what "an infinite sum of terms" is, but that you do not understand even when it is explained to you. 71.248.151.19 18:02, 19 May 2006 (UTC)

1. I am still unconvinced by your definition of an infinite sum of terms. Is such an infnite sum of terms a real number, or just a formal expression (similar to, say, a formal power series)? If it is the former, then surely there are problems of convergence which should be adressed. It would also be interesting to give some sort of connection between the infinite sum of terms and the finite partial sums of only some of those terms. If all terms to be added are positive, then it should probably be larger than a finite sum of only some of those terms? That alone would be enough to show that it cannot be less than what you call the limit (that's what I prove here). On the other hand, if this infinite sum of terms were just a formal object, then it could not easily be compared to a real number; so that seems to be the wrong concept.

In effect, you define how I can write an infinite sum of terms (by indicating that there is no last summand), but not what it is, mathematically speaking. Adding two real numbers is well-defined; adding infinitely many is not.

2. Let's agree to disagree about the advance of mathematical understanding. I will only point out that we can solve more problems (of physics, for example) today than Newton and his contemporaries could.

3. This sounds as if the Cabal is out to get you... Yours, Huon 21:06, 19 May 2006 (UTC)

Categories: Talk archives

Talk:0.999.../Arguments/Archive 1

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Contents

Geometry and comparing real numbers

Induction proof

An infinite sum is NOT equal to the limit of an infinite sum

Mathematical Induction

Here you go:

What is a rational number?

Farewell

You go wrong because you are all wrong...

THIS ARTICLE SHOULD STAY

$0.\overline{3} \ne \frac1{3}$ and therefore this proof erroneous

You will always be challenged on this article because it is untrue.

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