Talk:0.999.../Archive 1

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Contents

Keep in mind your knowledge of mathematics

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)

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

Using the proofs in this article, your question can be reduced to 2/2. This is 1. silsor 21:20, 7 November 2005 (UTC)

Wrong. It cannot be reduced to 2/2 - this is the point. —Preceding unsigned comment added by 192.67.48.22 (talk • contribs) 2005 November 9

Well, you've convinced me! silsor 03:59, 10 November 2005 (UTC)

Sarcasm is the lowest form of wit. 1.999... is not equal to 2 for the same reasons 0.999... is not equal to 1. —Preceding unsigned comment added by 71.248.131.9 (talk • contribs) 2005 November 10

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\dots999
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)

This is also wrong. It would only be correct to say:

  sup(partial sums(0.999...)) = 1

but it is incorrect to write Lt x<n> = 0.999...

It is incorrect to write:

  sup(partial sums(3.14...)) = pi

because we do not know the full extent of pi. What are the *final* parts of pi? Where does it end? What exactly is the value of pi? We don't know!! By definition a supremum is the least upper bound and if it exists, then we can find it by the completeness principle. Consequently, not all numbers can be expressed as the supremum of the partial sums of decimal expansions. —Preceding unsigned comment added by 68.238.110.56 (talk • contribs) 14:56, 22 October 2005 (UTC)

Okay, so if sup(partial sums(3.14...)) is not pi, what is it? According to the completeness principle, it exists, and hence, according to you, we can "find it". Eric119 01:53, 23 October 2005 (UTC)

No. For you to say it exists, means you already know its value. You don't. Pi can only be known as an approximation in a some radix system. According to the completeness principle, it exists only if it is a number whose full extent is known. We don't know the full extent of pi just like we don't know the full extent of 0.999... —Preceding unsigned comment added by 68.238.110.56 (talk • contribs) 13:16, 23 October 2005 (UTC)

And now you're adding the completeness principle to the list of concepts you're redefining or misunderstanding. The correct statement of the completeness property of the real number system is this: if a non-empty set of real numbers has an upper bound, it has a least upper bound. Nothing to do with "extent". Given the rest of the discussion, including your recent rant against "PHds" (I am not one.), I will not further participate in this discussion, having deemed it fruitless. Eric119 03:48, 25 October 2005 (UTC)

If you read the posts carefully, you would have noticed I was talking about this right from the start. My response was clear: you cannot find the *exact* value of pi. I used extent instead because it explains it better. You are as open-minded as the rest - your mind has already been made up and even if you did understand, you would refuse to accept truth. Suit yourself. 68.238.110.56

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! —Preceding unsigned comment added by 192.67.48.22 (talk • contribs) 18:42, 21 October 2005 (UTC)

It is not correct to say that pi has a least upper bound; rather, pi is the least upper bound of the set of finite truncations of its decimal expansion, or of the set of circumferences or regular polygons inscribed in a circle of diameter 1, or of any of various other sets. The things that have least upper bounds are non-empty bounded sets of numbers. pi is not a set of numbers; pi is a number. (And in the paragraph above, when the anonymous poster uses the word "liars", and in the subsequent rhetoric, you begin to see that that poster is a crackpot or a troll (and I suspect the latter because of the claims to be a retired supermodel).) Michael Hardy 23:05, 21 October 2005 (UTC)

So you say pi is the least upper bound? Interesting. But pi cannot be computed completely since it is infinitely represented. How then can it be the l.u.b ? You can say 1 is the l.u.b of 0.999... but you do not know exactly what pi is, so this is evidently non-sense. Please don't respond that you can calculate it to as many places as you like. Everyone knows this. Rather tell me if you are able to calculate pi completely. Numbers are defined in real analysis as Cauchy sequences that are in fact sets in many respects (sets can be formed out of the partial sums) and thus capable of having least upper bounds. In fact, the completeness principle states this clearly. It is so easy to call someone a crackpot and/or a troll when you have shown how ignorant you are and can't respond in any sensible way. You became sarcastic and then expected the poster to remain polite? Oh forgive me my math god for I have sinned grieviously against your sarcasm and disdain. Provide *proof* and sensible rebuttals, not rhetoric and sarcasm. Finally, just because you say something, does not make it true or divine law. You are wrong and you have not answered any rebuttal satisfactorily. Again, prove to me that the sum of 9/10 + 9/100 + 9/1000 is 1. Be careful! Don't write nonsense. Think about every word you use. You are not even half as smart as you think and you do not know everything as well as you think. —Preceding unsigned comment added by 68.238.104.130 (talk • contribs) 01:00, 22 October 2005 (UTC)

There is no danger that any mathematician reading this will conclude that I am the one who is ignorant. No, I did not "become sarcastic and expect the poster to remain polite"; rather, I responded to the poster's abuse.
It is not correct that real numbers are defined as Cauchy sequences, but rather, one of the way of characterizing them is a equivalence classes of Cauchy sequences. And yes, pi is indeed the least upper bound of the set of all finite truncations of its decimal expansion, and it is also the least upper bound of the set of all perimeters of polygons inscribed in a circle of diamter 1. Michael Hardy 18:24, 22 October 2005 (UTC)

Same thing, different words. It is perfectly correct the way I stated it. No mathematician who *knows* what he is talking about will agree with you. And no, pi is *indeed* not the l.u.b of the set of all finite truncations of its decimal expansion. A l.u.b must be completely defined. pi is not. As for sarcasm and abuse, I think you may have a bad memory in this regard too. Why don't you reread the posts and see who started to be disdainful and nasty? —Preceding unsigned comment added by 68.238.110.56 (talk • contribs) 21:49, 22 October 2005 (UTC)

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)
Wrong, 0.333... is an indeterminate number just like 0.999... In your example you are dealing with fractions that are well defined. 0.333..., 0.666..., 0.999... are not well defined numbers. (y/3)x3 proves nothing because all it means is a value of 'y'. y can be anything you want in this example. —Preceding unsigned comment added by 68.238.109.228 (talk • contribs) 00:53, 2005 October 31

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
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

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. --24.126.30.46 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 ? —Preceding unsigned comment added by 192.67.48.22 (talk • contribs) 16:50, 20 October 2005 (UTC)

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 ! —Preceding unsigned comment added by 68.238.104.130 (talk • contribs) 21:30, 21 October 2005 (UTC)


Hey guys, isn't the controversy _really_ just one of definition? If you define 0.999... as being equal to the limit of 0.999... as you let the number of nines "go towards" infinity, then 0.999... = 1. But people should be free to think of 0.999... as simply 0.999 followed by an infinite number of nines. They're hampering their own communication with the rest of the world of mathematics, since they've actually started to use a non-standard notation. I think mathematicians should respect the difference between things that are true by definition, and things with a deeper meaning (yeah, I know, if such a meaning exists is in itself a matter of discussion). —Preceding unsigned comment added by 194.192.22.33 (talk • contribs) 08:35, 24 October 2005 (UTC)

No. It is *never* true that 0.999... = 1. The confusion is really about the difference between:

   "An infinite sum"  -  impossible to calculate                      (A)

and

   "The limit of the partial sums of a sequence"                      (B)

Common misconceptions amongst PHds:

-) They can't tell the difference between the above two.
-) They will insist that (B) is equal to the number (Michael Hardy is an example) - a
   fallacy since *none* of the partial sums *ever* reaches the limit (supremum).
   Hardy will insist these are by *definition* the same thing. A common tactic by those
   who lack understanding is to use the phrase *by definition*. 
-) They feel comfortable with real analysis which is fraught with non-sense and inaccuracies.
   They do not fully understand many concepts but will do their utmost to convince others
   of their standpoint because they are arrogant and believe their viewpoints are probably
   more accurate.
-) They will perpetrate non-truths (same as *lies* but then I risk being called a troll)
   to suit themselves and save face, not to mention the 'rigour' of mathematics. There is no
   rigour except what humans perceive as rigour.
-) They feel comfortable with contradictions and confusion. The first universities were
   established by churches and the idea was to get rid of superstition and nonsense.
   Today the universities are controlled by a different kind of clergy - the ignorant 
   PHds.

   sup(partial sums(0.999...))  =  1

   limit(infinite sum(0.999...))  IS NOT equal 1  (In fact it never reaches 1)

Unfortunately, it is not one of definition. Finally, no one can just make up a definition - it has to be logical and must be justifiable. 0.999... has always meant the sum of 9/10 + 9/100 + 9/1000 + ... —Preceding unsigned comment added by 68.238.110.56 (talk • contribs) 11:01, 24 October 2005 (UTC)

You are wrong about (A). You can find the limit of an infinite sum. It's an infinite sum you cannot calculate. —Preceding unsigned comment added by 192.67.48.22 (talk • contribs) 18:17, 24 October 2005 (UTC)

That's correct. I typed this up too quickly. Aside from this error, the rest seems okay. Not that it matters much as far as most are concerned: they don't see a difference between the two. 68.238.110.56

Could you please provide proof that shows the two are not equal? I'd think that if all experts on the matter agree that they do, that the burden is on you to provide solid evidence against their equality. --BradBeattie 13:09, 25 October 2005 (UTC)

Look at the section titled Explanation in this page. He seems to have made a sufficiently solid case in my opinion. There is a clear difference between 'infinite sum' and 'limit of an infinite sum'. He challenged Michael Hardy to prove that 9/10 + 9/100 + 9/1000 + ... = 1. So far, no proof has been forthcoming. In fact, Hardy will not prove this because it is evidently not true. Way I see it, 1 has a finite representation in base 10. There is no need to approximate it. 1/3, 1/7 and many other finite numbers have no finite representation in base 10. Whether you like it or not, any calculations involving infinitely represented numbers in base 10 have always been approximations. The completeness principle does not apply to infinitely represented numbers. Every proof in favour of 0.999... = 1 is false. 0.999... is an indeterminate number that is very close to 1 but is never equal to 1. Although the difference between the two is infinitesimal, there is a difference. 0.333..., 0.999... and any infinitely represented number in any base does not make sense. It can only be used as an approximation. --Unsigned by user at 192.67.48.22

But, as was previously pointed out in this discussion, numbers only look like approximations in certain bases. I think we'd both agree that pi is irrational in base 10, but pi is represented as 10 in base pi and is therefore an integer in that base. Semantics aside, all of these symbols we use comprise the language of mathematics. They represent the idea of a number. The symbols 1/2, 0.5, 0.1 in base 2, 0.4999... represent the same value. It's important not to confuse the denotation with the connotation. --BradBeattie 15:55, 25 October 2005 (UTC)
Please. There is no such thing as "irrational in base 10" as opposed to irrational in some other base. An irrational number is simply one that cannot be written as a ratio of two integers. Bases have nothing to do with the definition. Michael Hardy 21:08, 25 October 2005 (UTC)

No. You are sorely mistaken. Pi is not 10 in base pi because pi is not completely defined. The only bases that are permissible are those that are defined. In fact, pi cannot be fully denoted in any base. It is an irrational number. 0.999... in base 10 is also an irrational number. pi is an approximation in any well defined base you like. 192.67.48.22

Actually, you can have bases of complex (or at least irrational) numbers. Take for example phinary. --24.126.30.46 07:36, 27 October 2005 (UTC)

What good is a number system in which representation is not unique? It leads only to chaos and confusion. Just look at the confusion caused when someone tries to have 0.999... = 1. This statement is false whichever way you look at it. -- 192.67.48.22

rewrite

Most students will first encounter this topic below the college level. For them, a proof using limits and Cauchy sequences is likely to be unhelpful. Others will revisit this topic at a more sophisticated level. For them, the arguments given were sorely lacking in substance. The completely rewritten article separates these two audiences, and offers proofs at both levels. Given the history of edits to this page, it would not surprise me if some other editor soon trashes it; but I hope this more honest acknowledgement of the importance of foundations makes that less likely. Enjoy. --KSmrqT 05:11, 27 October 2005 (UTC)

Nonsense. There is no proof at undergraduate or college level that 0.999... = 1. —Preceding unsigned comment added by 68.238.102.180 (talk • contribs) 11:05, 2005 October 27
I'd like to see you prove that ... Charles Matthews 11:47, 29 October 2005 (UTC)
The very same proof that is used against it, proves it. Just read the above post. There is a difference between 'infinite sum' and 'limit of an infinite sum'. sup(sum of(0.999...)) = 1 but 0.999... is not equal to 1. If you want to go to such lengths to define the number 1, then you first have to define 0 and 9 and what a fraction is. Then you have to define operations on these numbers. In fact, it is the stupidity of academics that has blinded them: by trying to carefully construct the real numbers, they have introduced circular definitions everywhere. What came first, numbers or their definition? Can you use numbers to define numbers? I can understand trying to use set theory, e.g. the cardinality of a set could be interpreted as a number. However, it becomes slightly more complicated to define parts of a number such as fractions. 1 is well defined, i.e. 1 x 1 unit. 10 is well defined, 1 x 10 + 1 x 1 unit. 1/2 is well defined, 5/10. All arithmetic operations work only on finite numbers. The dimensions of numbers such as 0.333..., 0.999..., pi, e, etc are infinite. Whatever 0.999... is, it cannot be comprehended. We can only state that it's infinite sum has a limit that it never reaches. Why do you want to define numbers as the limit of a number that immediately preceeds it? The difference between the two is infinitesimal. You would first have to accept infinitesimals. Real analysis rejects infinitesimals but then uses the same to prove a lot of things. Problem is: it is fraught with inaccuracies and contradictions. Yes, mathematics today is no more rigorous than it was in the time of Archimedes. —Preceding unsigned comment added by 68.238.102.180 (talk • contribs) 14:33, 2005 October 29