Talk:0.999...

Keep in mind your knowledge of mathematics

I do not want to sound rude by saying this, but please do not try to refute this (universally accepted among mathematicians) proof unless you have some background in mathematics that goes beyond your mere intuition. Consider the following proof that I propose be substituted for the one in this article: assume first that 0.999... and 1 are distinct. Between any two distinct real numbers lies another real number (infinitely many, in fact). Since there is no real number between 0.999... and 1, we have reached a contradiction and our original assumption must be false. Therefore, 0.999... and 1 are not distinct (i.e. they are equal in value). Q.E.D.. Shutranm 01:27, 7 May 2005 (UTC)

I don't mean to be rude but I think that you should not comment unless you know what you are talking about.

1 + 0.999...  = 1.999...

But what is 1.999... divided by 2 equal to ? Please do not tell me the result is 0.999... because you cannot compute this quotient. It involves a quantity that is *infinitely* represented. All our arithmetic works only on *finite* numbers or approximations used for numbers we know are finite in base ten. Thus you cannot state that the quotient is 0.999... and then conclude that no number exists between 0.999... and 1, therefore these must be the same number. Please prove to me that 1.999... divided by 2 is 0.999... and then tell me the completeness axiom is the reason. Otherwise please do not post anything more - you will simply show how narrow-minded you and most mathematicians really are! — Preceding unsigned comment added by 192.67.48.22 (talk • contribs) 17:53, 17 October 2005 (UTC)

The disagreement of Santa Claus

"I do not want to sound rude by saying this, but please do not try to refute this (universally accepted among mathematicians) proof unless you have some background in mathematics that goes beyond your mere intuition."

-Why/What are you so afraid of? Santa Claus

"Consider the following proof that I propose be substituted for the one in this article: assume first that 0.999... and 1 are distinct. Between any two distinct real numbers lies another real number (infinitely many, in fact). Since there is no real number between 0.999... and 1, we have reached a contradiction and our original assumption must be false."

Firstly 0,999... is not a number at all, it's a function. Secondly what natural number is between natural numbers 5 and 6? Do you even understand what is line segment? -Santa Claus

Point 1: "A number is an abstract entity used originally to describe quantity." So 0.999... is a number. As per the proof in the article, it describes the quantity 1. Furthermore, 0.999... is not a function. --BradBeattie 17:02, 15 Jun 2005 (UTC)

Using three dots makes it (0.999...) a function -Santa Claus

And I quote "a function is a relation, such that each element of a set (the domain) is associated with a unique element of another (possibly the same) set (the codomain, not to be confused with the range)." Doesn't sound like 0.999... is a function. Sounds more like a number. Besides, the "..." is just shorthand. Furthermore, we have a proof that 0.999... = 1 in this article. It relies on a series converging. If there's a problem with any particular step in that proof, please reveal it. As it stands, we have a solid proof that runs counter to what you're claiming. --BradBeattie 20:19, 15 Jun 2005 (UTC)

You should tell me how should understand in unambiguous manner these three dots. As far as I am concerned that "0.999..." ends to third (from left to right)dot. (in Europe decimals are separated from whole numbers by "," ...expressed 0,999 not 0.999) -Santa Claus

Please read the first proof provided on this article. It unambiguously defines $0.999\ldots =-9+9\times \sum _{k=0}^{\infty }\left({\frac {1}{10}}\right)^{k}$--BradBeattie

Above seems to me a function of k (0.999... expression does not include k) $F(k)=-9+9\times \sum _{k=0}^{\infty }\left({\frac {1}{10}}\right)^{k}=9\times \sum _{k=1}^{\infty }\left({\frac {1}{10}}\right)^{k}$ -Santa Claus

You don't set k; it is declared in the summation. If you chose k=3 in your example above, how does the sigma work? As it reads currently, it doesn't make sense. --BradBeattie 13:47, 16 Jun 2005 (UTC)

Point 2: There is no natural number greater than 5 and less than 6. --BradBeattie

Natural number 5 is less than 6 and 6 is greater than 5, true? - Santa Claus

True, but you're picking 2 seperate values here. You initially asked for one quantity that is greater than 5, less than 6 and within the natural number set. No such number satisfies those properties. --BradBeattie

Okay, true then. So could tell me any real number (X) differentially less than 1 (and greater than 0) so that there is no other real number (B) "between" X and 1. Define X, when 0<X<B<1 that there is no B? -Santa Claus

Yeah. Thought you might be heading in that direction, which is why I chose my wording carefully in my last point 2. Thing is, 0.999... is the same value. They're different representations, but they express the same value. Besides, you're trying to make an argument by analogy between natural numbers and real numbers, but the two have very different properties. --BradBeattie 21:15, 15 Jun 2005 (UTC)

e.g. 999 is in ENGLISH figure, tally, count, SWEDEN tal, siffra) GERMAN Zahl, FRENCH chiffre), FINNISH OFFICIALLY luku. One of 0,1,2,3,4,5,6,7,8,9 is in ENGLISH OFFICIALLY digit MIDDLE ENGLISH nombre, GERMAN digiTAL, Ziffer, (SWEDEN OFFICIALLY siffra, UNOFFICIALLY numret, nummer, (SELDOM) numro, FRENCH un numéro, LATIN numerus, (OLD LATIN) NUMERVS, FINNISH numero. You may plot your decimal "numbers" in Cartesian co-ordinates (X,Y), X-axis expresses "whole" numbers, Y-axis expresses desimals like (7,91), but no point presents (0,999...). "999..." impression does not tell where in the Y-axis "999..." ends. Did you think "0.9990999099909990999..." when you wrote (0.999...)? Do you think that "10000/9999" is number? Do you think that "0,333..." is number too? Do you think that "...666" is number too? Do you think that "...66,6*15=...333,0" is number too? Do you think that "1000/999/666" is number too? Do you think that "1000/666/999" is "number" too? Do you think that "1500" is number too? Do you think that "1000/999/999" is "number" too? Do you think that number "one" 1 is not divided (undivided) and not multiplied (unmultiplied) (Greek Atom). Do you know what means in France le numéro atomique (le nombre atomique, le nombre ordinal)? Did you know that multiplication is a making long division? Do not tell to a tennisplayer that you feel "love" for him or her (he or she may think that love is 0, nothing)... Santa Claus 22 Jun 2005

Santa Claus, as much as I'd love to continue this discussion, my initial proof remains. We could argue indefinately about semantics of math, but it won't change the proof displayed in the article. If you find a flaw in that I'll continue this discussion. At this point, I'm tired of arguing something I've already proven. --BradBeattie 13:52, 16 Jun 2005 (UTC)

BradBeattie, I've read through this "Santa Claus" person's arguments, and they show that he either doesn't know what people are talking about, or pretends not to. I don't think he's being serious here, he is very probably a troll. He is best ignored. — JIP | Talk 12:54, 22 Jun 2005 (UTC)

Either way, I'm through trying to convince here. We have an elegant proof that reduces the problem to a convergent series. We have 3 alternate proofs that illustrate the point further. If someone finds a flaw in the initial proof, I'll certainly take a second look at it, but for now I think we're done here. Cheers. --BradBeattie 13:24, 22 Jun 2005 (UTC)

Creation of this entry

I created this page in response to two threads I saw and the confusion that arose. Figured it was something worth noting. --BradBeattie 18:58, 6 May 2005 (UTC)

I think you are right. I submitted it first for deletion because the title looked a bit misleading. This is not a series of nines, the series is if you wish of

${\frac {9}{10^{n}}}$

Cheers, Oleg Alexandrov 19:01, 6 May 2005 (UTC)

True, the title was a little slap-dash. Thanks for the improvement. --BradBeattie 19:03, 6 May 2005 (UTC)

Alternative proofs

I'm not convinced this is the most intuitive proof available ... wouldn't the x = 0.999..., 10x-x = 9.999... - 0.999... = 9x = 9, x = 1 proof be more appropriate? Just a thought Mallocks 23:23, 6 May 2005 (UTC)

The second one on the Dr. Math page now I look at it Mallocks 23:24, 6 May 2005 (UTC)

The problem with that argument is that it doesn't bring convergence to the table. The summation of the infinite series is key to understanding this equation. The 10x-x proof could be construed as handwaving by those that don't already understand it. --BradBeattie 23:35, 6 May 2005 (UTC)

While I suppose that is true, this page is not understandable for people who haven't studied mathematics on a high level (me included). For them, the 10x-x proof is more understandable. --Pidgeot (t) (c) (e) 12:00, 8 May 2005 (UTC)

I don't know. I mean, it might convince some people, but to others it might seem like handwaving. In the proof currently given, it reduces the problem to the summation of a series (clearly leading into limits). The other proof is somewhat indirect and less likely to convince. We could put both of them, I suppose. --BradBeattie 14:13, 8 May 2005 (UTC)

I agree that we can include the 10x-x proof in a section called say "Alternative proof" or "Another proof", after the series proof. It certainly adds value to the article. Oleg Alexandrov 14:41, 8 May 2005 (UTC)

And by the way, one should write

== Alternative proofs ==

instead of

 == Alternative Proofs ==

Oleg Alexandrov 14:41, 8 May 2005 (UTC)

Ah, right. You mentioned this before but I forgot. I'll try to keep it in mind. Thanks! Just trying to clean this page up at the moment. --BradBeattie 14:43, 8 May 2005 (UTC)

Can we just move the alternate proofs to the talk page? As it is, the main page could easily be cluttered with dozens of proofs. While they all might be valid, we really only need one proof. --BradBeattie 19:49, 19 May 2005 (UTC)

Subpages

I created Proof that 0.999... equals 1/Alternate proofs to help keep the main page brief, but also allow us to post as many proofs as we'd like. --BradBeattie 15:39, 8 May 2005 (UTC)

Seems that some user didn't like the alternate proof page idea. Thoughts on this? --BradBeattie 15:56, 8 May 2005 (UTC)

Subpages aren't allowed in the main article namespace (see Wikipedia:Subpages), you can put it on a different page, but I'd just put them on the main page, I think. People don't have to scroll down if they don't want to. Maybe we could create a suppage of this talk page and put all the proofs there while discussing which should be included. --Mike C | talk 16:13, 8 May 2005 (UTC)

No, subpages are not good. BradBeattie, what do you think? If we agree that that subpage is unnecessary, then BradBeattie, you will need to ask it to be speedy deleted. That is as simple as going to that page, writing in the following:

{{D}}

and a convincing explaination. It has to be you to ask that, because you created that page. Oleg Alexandrov 16:49, 8 May 2005 (UTC)

Seems to have been taken care of. :) --BradBeattie 18:34, 8 May 2005 (UTC)

New section: Arguments against

I think the page needs a section on common misconceptions and flaws in the reasoning of these misconceptions. If the abracadabra section on the talk page says anything, it's that some people don't really get this. BradBeattie

But the new section should probably be not too long. (The set of all common misconceptions of people is by several orders of magnitude larger than the set of all knowledge. :) Oleg Alexandrov 22:24, 6 May 2005 (UTC)

I very much agree. Maybe something as short as the top 3 misconceptions. No doubt our anonymous friend here is providing us with detail. The links I found these arguments taking place on should also provide a starting point. I'll look through later on and try to put a short summary together. --BradBeattie 22:30, 6 May 2005 (UTC)

I think this is fruitless. If you define the recurring decimal as a limit, and believe in geometric series, you are done. WP should consist of true facts; but doesn't in general need to convince people of themm if they insist on being wrong. There is no alternate POV to be represented. Other pages that are just magnets for discussion threads have in the end been dealt with in a similar way. Charles Matthews 18:01, 8 May 2005 (UTC)

Merger with recurring decimal

I think this page should just be merged into recurring decimal. Charles Matthews 16:17, 8 May 2005 (UTC)

infinity divided by infinity

Your "proof" lies on two notions. First there is that $\infty +1=\infty$. You can you say $XXX...=\infty$ But Your second notion ${\frac {\infty }{\infty }}=1$ is totally nonsense. -Santa Claus 21:02 At Northpole Time.

That is nonsense: There is no mention of ∞/∞ in this article. Michael Hardy 21:22, 8 May 2005 (UTC)

Infinite number of finite things

I originally wrote this, but it got deleted.

I learned in my topology classes that an infinite number of finite things is a completely different beast from an infinite thing. There are an infinite number of numbers like 0.9, 0.99, 0.999, 0.9999, etc. in the domain [0, 1[. All of these numbers have a finite number of nines, despite the fact that there's an infinite number of the numbers themselves. In contrast, in 0.999..., there's an infinite number of nines, and therefore it's not in [0, 1[. — JIP | Talk 15:30, 8 May 2005 (UTC)

Doesn't etc. (et cetera) stand for "..."? (anonymous comment from User:213.216.199.18)

That's not what I am talking about. I might have been unclear in my message. Sorry about that. Here is an attempt at a more helpful way of putting it:

This is an infinite number of finite-length numbers.
0.9 (one nine)
0.99 (two nines)
0.999 (three nines)
0.9999 (four nines)
0.99999 (five nines)
etc. (or ... if you prefer)

This is an infinite-length number.
0.999... (infinitely many nines)

Although there are an infinite number of the things (put in rows) in the first example, they're all in [0, 1[. In contrast, the second example is not in [0, 1[. — JIP | Talk 07:51, 9 May 2005 (UTC)

What means "infinite-length"? -Santa Claus At 1:06 Northpole Time

You need to take a calculus course, to learn what limit is. Then, a good exercise would be to prove that if one considers the sequence

$x_{n}=0.999\dots 999$

where the number of 9's is n, then,

$\lim _{n\to \infty }x_{n}=1.$

If you don't understand what limit is, no offence, but it is pointless to have this discussion to start with. Oleg Alexandrov 22:29, 9 May 2005 (UTC)

I feel what means "limit" (i.e. finitus) when I drop myself from the sky and then hit the ground or when I hit my face on the wall. It's a kind of frustrating to type such a dialect (mutation) of Latin (e.g. "English") which contains so much/many synonyms and yet they are expressed visually different combinations of symbols (not to mention about the enunciation, pronunciation, vocalization i.e. making a sound based on symbols (phonemes, letter, symbols of alphabet). Infinite (endless, boundlessness) "is" something you can't actually observe. I was frontier (border etc.) guard (observer) during the military service and I sure can tell you that there truly is very really fine (infinitesimal) line between the border zone and the non-border zone :) And yet they keep claiming that you need two infinitesimal points (amorphous dots) to draw a line (segment) between them. -Santa Claus

Your statement is FALSE, because...

0=1

9=10

99=100

999=1000

9999=10000

999...=1000...

IS FALSE also.-Santa Claus

"This is an infinite number of finite-length numbers" What a jargon. Do I need to say more(?) ;D

My reversal

Today I reverted some new content added to this article by anonimous user 130.230.47.130. My comment was "reverted ignorant contribution". I must say that I was wrong, actually that is meaningful text. But looks kind of long. Would anybody take a look at that contribution [1] and see if anything can be included in the article? Thanks. Oleg Alexandrov 18:47, 13 Jun 2005 (UTC)

about math

Explanation

The key step to understanding this proof is to recognize that the following infinite geometric series is convergent:

$\sum _{k=0}^{\infty }\left({\frac {1}{10}}\right)^{k}={\frac {1}{1-{\frac {1}{10}}}}={\frac {10}{9}}$

Yep, that's true. Check the article page for how that's multiplied by 9 and then 9 is subtracted from that. It all works out. --BradBeattie 13:18, 16 Jun 2005 (UTC)

Timeout! It's time for a lesson on high school limits for Mr. Hardy and those who agree with BradBeattie the masculist.

Your above formula is incorrect. The infinite sum is not equal to 10/9. I hate to break this to you but you cannot compute the above infinite sum or any other infinite sum. That's what Mr. Hardy would have taught you (wrongly of course). So where does the expression a/(1-r) come from? Well, the sum of a series in which there is a common ratio r, a first term 'a' and n terms is:

      (a - ar^n)/(1-r)                                        (A)

If we let n go to infinity, all we can do is speculate about what happens to (A). So we break up (A) as follows:

      a/(1-r)  -  ar^n/(1-r)                                  (B)

We can see that (B) has a maximum (clues: upper bound, completeness princple and *limit*) value if the second term is zero. Now does this ever happen? NO!!! However, as n gets infinitely large, the second term gets closer and closer to zero. In other words you can make the second term as close as you want to zero BUTTTT you cannot make it zero. Why? Coz if you do, you are assuming that the infinite sum is equal to a/(1-r) and that is clearly *wrong*. You cannot discard the second term from (B) and innocently claim the infinite sum is a/(1-r)!

So, 10/9 is the value that your sum will never reach. What does this mean? It means that it has an upper bound or a limit. This is another reason why not all numbers can be expressed as a decimal expansion. Take pi for example. If this were true that all numbers can be expressed as a decimal expansion, then by default pi must have a least upper bound (sup). Uh oh! Pi does not have a supremum. Everything falls apart here: is the completeness principle still true? Yes. It's just the teachers who claim every number can be expressed as a decimal expansion that are liars. Be very careful when you are taught - do not only hear what you want, but what is actually spoken. Beware of teachers like Mr. Hardy. No wonder so many MIT grads don't know the difference between 0.999... and 1!

Disagree with merging to invalid proof

Unlike all the items listed in that article, this one has valid proofs. Not appropriate there. Oleg Alexandrov 16:24, 23 July 2005 (UTC)

I disagree as well. I'm reverting the change. --BradBeattie 13:58, 24 July 2005 (UTC)

Another proof... Maybe?

You see;

(y/3)x3=y

y=1

1/3=0.3333... 0.3333... x3 =0.9999...

then 0.999... equals 1

correct? --MrBird 19:01, 27 September 2005 (UTC)

yes, this is correct, but this is just another form of the 1/9 = 0.1..., 2/9 = 0.2... 9/9 = 0.999, 9/9 = 1, which is already there :) Mallocks 19:13, 27 September 2005 (UTC)

Yeah, you're right.--MrBird 15:03, 29 September 2005 (UTC)

Suggestion

I feel that this article should be marked as "under debate". There is a difference between the terms "converges to" and "equals". --192.233.129.254 23:08, 10 October 2005 (UTC)

That's only an issue with one proof though, the article itself is sound. What is the difference, by the by? Mallocks 23:18, 10 October 2005 (UTC)

How else would you define the result of an infinite decimal expansion? Eric119 02:43, 11 October 2005 (UTC)

The number represented by any infinite decimal expansion is equal to the real number that the sequence of its finite initial parts converges to. Therefore the objection expressed above is not cogent. Michael Hardy 02:56, 11 October 2005 (UTC)

So what you are saying Mr. Hardy is that we can only perform approximate calculations with any numbers represented *infinitely* in the decimal system - this is true. So which is a better representation of 1 - 0.999... or 1 ? I find it completely ridiculous to state that 1 = 0.999... It is *not* equal to 0.999... just as pi is not equal to 3.14..., just as 1/3 is not equal to 0.333.... 0.999... is an indeterminate quantity. 0.333... is an indeterminate quantity. It is *impossible* to represent 1/3, 2/3, 4/9, etc. in the decimal system except to approximate the same. There is no *known method* in any mathematics to compute the sum of an *infinite* number of terms. How is it that you are *interpreting* the *limit of a sum* to be the *actual* sum? What does

 Lt    (a - ar^n)/(1-r)  
 n->Inf

actually say? For 0.999... it says that no matter how many terms we sum in this sequence, the sum will

• never exceed* 1! It is *not* saying the sum is 1. Although the completeness axiom is true, a result of it that is used to show a contradiction here is untrue because arithmetic can only be performed on *finite* numbers. I believe this article is non-sense and should be deleted. It is not encyclopedic. — Preceding unsigned comment added by 192.67.48.22 (talk • contribs) 17:41, 17 October 2005 (UTC)

Sorry, you're wrong. It does not mean that the sum will never exceed 1. Since all terms are positive, it could be taken to mean that 1 is the smallest number that it will never exceed. But since some terms in some series are negative, it is better to define it thus: you can make it as close as you want to 1 by using a large enough number of terms (how large is large enough depends on how close you want to make it), and 1 is the only number of which that is true.

To say that "the is no *known method* in any mathematics to compute the sum of an *infinite* number of terms" is either to say that you failed to learn such methods in secondary school when your classmates learned them, or to say that you reject that on some philosophical ground, about which you should be explicit instead of pretending that the mathematics is what you're writing about. Michael Hardy 23:30, 17 October 2005 (UTC)

Thanks, but I don't need anyone to defend me. Your argument "can make it as close as I want" makes no sense whatsoever. I can make anything I like as close as I want to 1. Here are some examples:

  0.9
  0.99
  0.999
  0.9999
  0.99999

Are all these numbers the same as 1? You are telling me that if I sum the terms of the sequence 0.999... I will end up with 1 and I am telling you that I do not believe you. There is no way I will ever reach 1 even if I continue to sum the terms forever. In order for this to happen, a *carry* would have to occur at some point in the sum. It is clearly evident that a carry is *impossible* therefore the sum can never be 1. To write what you have written only shows that you and many of your classmates were not listening when these things were taught to you in high school. The completeness principle is *not* violated in this instance because there is no number between 0.999... and 1. 1 is a rational number but 0.999... is not a rational number, thus the completeness principle still stands and your arguments are false! 0.333 is an approximation for 1/3. 3.14 is ane approximation for pi. 2.718 is an approximation for e. 0.999 is an approximation for 1. BUT 0.333 is not a 1/3, 3.14 is not pi, 2.718 is not e and yes, 0.999 is not 1 ! 0.333... , 3.14..., 2.718..., 0.999... are all irrational numbers that are close approximations of other rational numbers. — Preceding unsigned comment added by 192.67.48.22 (talk • contribs) 13:54, 18 October 2005 (UTC)

---

How can he be wrong? He is referring to sequences with positive sums. What are you talking about? He is correct in stating it will never exceed 1. In fact it will never reach 1 either. As for computing infinite sums, I also, know of no formula. All I learned in high school and university is how to compute the limit of an infinite sum. The two are quite different. Finally, the fact that you can take the sum as close as you want to 1 does not mean it is equal to 1. It means that you can take it as close as you like but you will never reach 1. Just sit down and start adding up the terms and I guarantee you that you will sum until your last breath and still you will not have reached 1. Someone can continue to sum after you and he too will die summing the terms because the sum will always be less than 1. Philosophical grounds - hmmm? No, I think he is just using simple high school math. — Preceding unsigned comment added by 68.238.99.105 (talk • contribs) 00:12, 18 October 2005 (UTC)

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

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

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

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

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

Your logic is 'impeccable': would it really equal 3? You appear to be very confused. It cannot equal whatever you like. It will never equal, nor exceed 1 - that is the assertion. I am talking about the "limit of an infinite sum", not "limit of the sequence of finite sums". The formula he quoted is used in determining whether an infinite sum has an upper bound. There is no assertion that it is equal to this upper bound. Your assertion is plain wrong: there is a very easy way to check yourself - start adding up the terms and I can gaurantee you, you will always have a sum that is less than 1. Please don't tell me you are dealing with a finite sum because then your assertion that the infinite sum is 1 is absolute nonsense! You may be confusing yourself with the fact that the terms are getting closer and closer to zero (Cauchy sequence). This does not mean that any term will ever be zero. — Preceding unsigned comment added by 68.238.97.2 (talk • contribs) 10:49, 18 October 2005 (UTC)

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

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

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

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

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

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

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

I think you know exactly what I mean when I talk about Cauchy sequence: the distance between the terms is getting closer to zero. That's the definition of Cauchy sequence, i.e. Lt d(p,q) = 0 as min(p,q) approaches infinity. You write: "The value of an infinite sum IS the limit of the sequence of finite partial sums." This is not true. An infinite sum IS *indeterminate*. Get that? The formula used in this article to prove that 0.999... equals 1 proves exactly the *opposite*, i.e. that 0.999... does not equal 1. All the formula shows is that the limit of any of the partial sums of this sequence is 1. To say this is the infinite sum, shows extreme ignorance. The limit of any partial sum is *not* the infinite sum. You taught at Mit? So what, do you think I ought to bow down and be intimidated? You are wrong. You have been taught wrong too. This article is non-sense. The fact that you can write what you do, displays a fundamental lack of understanding. This non-truth of 0.999... = 1 has taken hold because most people don't understand that 0.999... is *not* a rational number. Neither is 0.333... a rational number. Partial sums from these sequences are used to approximate 1 and 1/3 respectively. I can understand using 0.333 to approximate 1/3 in base 10 (only because it can't be represented finitely in base 10) but cannot understand why 0.999 should be used to approximate 1 which has an *exact* representation. Don't you think it's about time you started thinking for yourself? I know how you will respond: You will say a rational number is any number that can be expressed as a/b where a and b are integers (b not 0). This definition of rational number is part of the problem. If a number cannot be represented *finitely* in a base that is well defined, then the number is not rational. Pi, e, sqrt(2), etc are irrational because there is no well defined base in which these can be represented finitely. You can't suggest that Pi, e, etc be respresented in their own base since these numbers cannot be completely determined. i.e. pi is not equal to 1.0 in base pi because the extent of pi is unknown. Similarly for e or any other irrational number. 15H51 18 October 2005 — Preceding unsigned comment added by 68.238.97.2 (talk • contribs) 21:07, 18 October 2005 (UTC)

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

Not entirely true. You start with 0.333... and then you try to show that it can be expressed as a/b. Or you start with 0.999..., 3.14... or some other representation and then try to show it can be expressed as a/b. A number is rational if it can be expressed in the form a/b (b <>0) with a,b integers. I maintain this is insufficient, you also need to add that the representation must be *finite* in some radix form. If indeed 0.999... is rational (it's not), then so is pi since pi can be expressed in the form a/b (i.e. 3 + 1/10 + 4/100 + ...) But of course pi is not rational because there is no number system besides pi in which pi can be expressed finitely in radix form. In base pi, pi is *rational*, i.e. pi = 10 (i.e pi + 0 units). 0.999... cannot be expressed as a/b in any number system. Please don't tell me it's 1 or 1/1 - this assumes that it is equal to 1. You need to think really hard about this. 68.238.97.2

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

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

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

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

You have told me nothing I did not know in your first and second points. In fact, if you read my posts, you would see that I said this. Your third point is false and I pointed this out in my previous response. You cannot prove that 0.999... = 1 because you do not know the difference betwen the limit of an infinite sum and an infinite sum itself. In fact, 0.999... 'is not' equal to 1. You do not understand the formula used to show that the sum of an infinite sequence is bounded from above. Maybe you should sit down and think about it again? You have been unable to refute anything I have said and you have not even tried to understand it. If I am incorrect in stating that finite representation is 'required' for the definition of a rational number, then pi, e and sqrt(2) are all rational seeing these are the sum of their respective expansions. Frankly it has nothing to with semantics, only simple logic that even a ex-professor from MIT can't see or won't see?. 68.238.97.2

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

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

Talk about a lot of 'words'.... Could your rebuttal possibly be a little more abstract. You know you are wrong and just can't admit it. ... sour grapes? 68.238.97.2

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

Wrong again. Retired supermodel. More profitable and unlike teaching/(child minding), no fake power-trips: the runway is a 'real' power-trip. But don't quit your day job. If your posted photo is recent, I can't tell you won't make it. Sorry, don't mean to be rude, just realistic. 68.238.97.2

Did you mean 'can tell he won't make it' ? :-) He is probably still figuring out how to make 0.999... add up to 1.

Yes, that should have read: "I can tell you won't make it." I see he has not responded to your rebuttal. Instead he chooses to be sarcastic and rude to a lady. Frankly, he skirts the rebuttals and tries to be cunning and humorous. 68.238.97.2

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

I am sorry but I have no idea who Paris Hilton is? Is he a mathematician? Wait, I can goolge it. Hopefully there aren't too many with that name. Now I know you love chatting but I really wish you would give this subject more thought. I can answer any questions you might have. You ought to be grateful for this. 68.238.97.2

Do Not Feed The Troll (and a comment)

What's the point of having the -9 and 9? I find it demeaning, superfluous, and do to my overly analytical nature, confusing. The summation converges to 1 on it's own, why bother with adding the 9? Unless of course you're trying to make a parallel to the dumbed down version, which is neither necessary or has been noted. --? — Preceding unsigned comment added by 24.126.30.46 (talk • contribs) 03:50, 16 October 2005 (UTC)

The 9 and -9 are put in so that you have the 9 x sum etc. bit, which is neccesary for the proof. Mallocks 20:30, 17 October 2005 (UTC)

Not convinced. Removed the -9. --Tob 15:29, 20 October 2005 (UTC)

10x - x

As I see it, there is no fault or handwaving in the $10x-x$ proof. It all depends on what machinery you have available. When writing things such as $0.333\dots$, most people assume the existence of some set of reals and then define $0.333\dots$ as the limit of $0.3,\,0.33,\,0.333,\,\dots$ in that given set of reals. But then elementary calculus tells you that multiples and sums of converging sequences are again converging and the obvious rules, i.e. $2\cdot 0.333\dots =0.666\dots$ apply.

After taking everything into account which necessarily has to be known or accepted to be able to 'ask' the question whether $0.999\dots =1$, a useful proof from my point of view is

$1=3\cdot {\frac {1}{3}}=3\cdot 0.333\dots =0.999\dots ,$

since most readers probably believe the lemma

${\frac {1}{3}}=0.333\dots ,$

and those who actually know how to prove it can easily see how to modify that proof for one to the original question.

Of course this is biased since I am speaking from personal experience; this is how I convinced myself at the age where I asked that question for the first time.

I object to the 'Property of real numbers' proof, though. It is correct that there are infinitely many real numbers between any given two distinct ones. But why is there no number between $0.999\dots$ and $1$? --Tob 15:29, 20 October 2005 (UTC)

There is no real/rational number between 0.999... and 1 because 0.999... is not a rational number. If pi had a least upper bound say x, would there be a number between x and pi ?

I will answer this question for you. In fact, pi must have a least upper bound (even though we cannot calculate it) for it is *finite*. circumference = diameter * pi. The circumference of any circle as we know is finite. Unlike converging Cauchy sequences in which we can state a least upper bound exists and we can usually find it, pi has a least upper bound that we cannot find. In reality, pi (like e and sqrt(2)) are numbers that cannot be represented finitely in any number system we know - these numbers can only be approximated. They are mysterious and exotic numbers because they do not reveal their full extent. Well, I am personifying them but it's just for effect. There are not infinitely many numbers between two real numbers. This is in fact a complete contradiction and once again exposes the weaknesses and errors of real analysis. On the one hand its supporters will claim there are no infinitesimals and on the other hand they will make a statement to the effect that there are infinitely many numbers between any two real numbers. The sad truth is that mathematics has been hijacked by a lot of idiots with PHds in mathematics. Almost 99% of dissertations I have read are not even worth the paper they are written on. Yet these fools were granted PHds. The result? Millions of idiotlets believing that 0.999... = 1 !