Talk:0.999.../Archive 11

message: I've noticed that many pages that are subject to vandalism/controversy have an HTML comment at the beginning saying something like "The content of this article is well-established. If you plan to make a significant change, please consider discussing it on the talk page." Is it time for 0.999... to have one? Confusing Manifestation 01:49, 1 November 2006 (UTC)
replies: text: That's not a bad idea.  However, keep in mind that that won't deter some users with a seemingly religiously held conviction that the entire article is wrong, so we should do our best to divert them to the Arguments page. Supadawg (talk • contribs) 02:04, 1 November 2006 (UTC)
replies: text: That's what I thought. What about a wording like "WARNING: This article contains several proofs that 0.999... = 1. If it is your intention to try and disprove this, please see the Arguments page at http://en.wikipedia.org/wiki/Talk:0.999.../Arguments first, as most of the common objections are dealt with repeatedly there. Also, please understand that the earlier, more naive proofs are not as rigorous as the later ones as they intend to appeal to intuition, and as such may require further justification to be complete. If there is a significant flaw with a proof, however, please discuss it on the Talk page." Confusing Manifestation 02:12, 1 November 2006 (UTC)
replies: text: Done.  I hope this is a good wording. } (talk • contribs) 02:23, 1 November 2006 (UTC)
thread_title:  Time for a comment? 

message: Moved to Talk:0.999.../Arguments#Proofs need to be reexamined by Calbaer 01:30, 2 November 2006 (UTC)
thread_title:  Proofs need to be reexamined 

message: Is anyone else surprised that there exists a mathematics article in 14 languages such that none of them is German? Melchoir 19:15, 1 November 2006 (UTC)
replies: text: I am. I suggest we get someone to translate this to the language of our Vaterland so that the Germans can start arguing over the correctness of rigorous proofs. =) --King Bee 20:32, 1 November 2006 (UTC)
replies: text: Maybe Germans just get it and so are busy laughing at these silly English-speaking people and no one has bothered to translate/write and article on such a silly topic which is understood by everyone Nil Einne 10:12, 5 November 2006 (UTC)
replies: text: They actually know that, regardless of all so-called "proof" or argument, logic must dictate that 1 cannot be equal to any other number. (in my opinion - and you are entitled to it) ¢Joseph C 13:52, 26 January 2007 (UTC)
replies: text: HA! Well, just one thing though - 1 is in fact not equal to "any other" number.  The fact is that 0.999... and 1 are THE SAME number. Tparameter 15:25, 27 January 2007 (UTC)
thread_title:  Interwiki 

message: There is currently a proposal suggesting this addition to the global CSS of Wikipedia:
 span.texhtml {
  font-family: sans-serif;
 }
This would mean that inline mathematics would be displayed in sans-serif rather than serif. This proposal was shot down twice before, but it seems that the strange nature of Wikipediamocratics allow for it to be suggested a third time. Maybe you would like to take a look there and partake in the discussion, since you're all active editors in the field of mathematics. —msikma <user_talk:msikma> 06:54, 2 November 2006 (UTC)
thread_title:  Proposal to globally change inline mathematics into sans-serif 

message: One interesting thing to note is that there are other ways of lexicographically representing real numbers in a fixed range (say, [0,2]) by infinite strings of digits (or bits or other alphabets), and they all have the multiple representation "problem."  By lexicographically, I mean that x >= y if and only if x is after y in alphabetical (dictionary) order.  In any such system, there will be no string between "0hhh..." and "1000..." where 'h' is the highest digit of the alphabet, 9 in decimal.  If these two strings represent two different real numbers, their average will fail to have a represenation.  Thus they must represent the same real number.  An interesting example of an alternative representation is LCF (), in which the representation is based on continued fractions, and the alternative representations, rather than being 0.999... and 1.000, are 0 + \frac{1}{1+0} and 1 + \frac{1}{\infty}.  I'd add this, but I'm not sure how to do so without possibly causing additional confusion. Calbaer 20:58, 5 November 2006 (UTC)
thread_title: A proof of the more general concept

message: Many of the examples of students misunderstanding this concept are apocryphal and not referenced. It seems that they should be rewritten or removed. —The preceding unsigned comment was added by 209.174.60.3 (talk • contribs) 02:44, 9 November 2006  (UTC)
:Could you please be more specific? Melchoir 02:45, 9 November 2006 (UTC)
replies: text: Some of the papers are not well-attributed.  For example, a paper which I assume is "Conceptual and Formal Infinities, Educational Studies in Mathematics, 48 (2&3), 199–238" is only attributed as "Tall 2001, p. 221."  The references could use more info by whomever added them on. Calbaer 03:13, 9 November 2006 (UTC)
replies: text: The full citations are in the reference section, rather than in the footnotes. -- SCZenz 03:17, 9 November 2006 (UTC)
replies: text: There is no Tall 2001 in the reference section. Calbaer 03:30, 9 November 2006 (UTC)
replies: text: Gosh, you're right; sorry about that.  Can someone add it, please? -- SCZenz 03:38, 9 November 2006 (UTC)
replies: text: Whoops! Actually, those should simply read "Tall 2000"; I messed up my own records. I'll fix them now. Melchoir 04:01, 9 November 2006 (UTC)
thread_title: Student stories

message: This proof is intuitive, but poorly worded.  I would like to see some more refined version of this in the article.  Does anyone agree, does anyone have criticisms?
:0.999… is a real number.  If you are thinking that it is not – that’s okay; but, understand that we’re referring to two different numbers.  I am specifically referring to the real number 0.999… 
:Since 0.999… is a real number, and since 1 is a real number, then if they are different, there are an infinite number of real numbers between the two, by the properties of real numbers.
:Then we will construct a real number between 0.999… and 1 by modifying one or the other by any digit in order to construct a number between the two.
::Case 1: We will modify 1.000… by changing some digit to construct the number.  If we add any digit to any place after the decimal, i.e. change the kth digit from 0 to some other numeral – then we have constructed a number greater than 1. If we change 1 to 0, then the number we constructed is 0, which is < 0.999…  Lastly, if we change 1 to any other numeral, then the number we constructed is > 1.  Hence, we cannot modify any digit that makes up 1 to construct a number between 0.999… and 1.
::Case 2: We will modify 0.999… by changing some digit.  Since 0.999… is a decimal followed by an infinite string of 9s, then any digit on that string that we change will construct a number not > 0.999…, since 9 is the largest numeral in base 10.  Hence, there is no way to construct a number between 0.999… and 1 by changing some particular digit of 0.999…
:Conclusion: Since we cannot construct any number between 0.999… and 1, and since there are an infinite number of real numbers between any two distinct real numbers, then 0.999… is not distinct from 1; i.e. 0.999… = 1.
Flame on! Tparameter 02:44, 2 December 2006 (UTC)
replies: text: I see why that "proof" might be warranted what with all the people who (implicitly if not admittedly) believe that there's something holy about the decimal representation of numbers.  But it's a bit awkward and not at all formal.  Also, it seems like a lot of the trouble is with people who refuse to believe that there cannot be two adjacent real numbers with nothing in between or those who believe "0.999..." could not possibly represent a real number.  Better we stick to the shorter, clearer proofs. Calbaer 03:22, 2 December 2006 (UTC)

text: I like that proof. However, a lot of people who come here with misconceptions about the real numbers might not understand why there must be any number between the two. For some reason, people seem to think of the two as beads on a string, right next to & touching each other. So going into the whole background of WHY there are always an infinite number of real numbers between any two real numbers might complicate the proof beyond what you intended. Good thinking, though. Argyrios 16:19, 1 January 2007 (UTC)
thread_title: Another Proof

message: Fraction Proof assumes that \frac{1}{3}=.333\ldots , so therefore \frac{3}{3}=.999\ldots .
My understanding tells me that this is incorrect. \frac{1}{3}\ne .333\ldots.
My understanding tells me that .333\ldots < \frac{1}{3} < .333 \ldots 4.
Algebraic Proof
\begin{align}
c        &= 0.999\ldots \\
10 c     &= 9.999\ldots \\
10 c - c &= 9.999\ldots - 0.999\ldots \\
9 c      &= 9 \\
c        &= 1 
\end{align}
My understanding tells me that 9>9c>8.999 \ldots, or that 9c=9 \times .999\ldots.
The only way for the above statements of mine to be true is for 'Fractional Infinitesimals' to exist. Therefore  \frac {1}{3}=.333\ldots\frac{1}{3} and 9c=8.999\ldots(\frac{9}{10}\times.999\ldots)
My understanding tells me that the following statement is true.
\lim_{n\to\infty}\sum_{k = 1}^n\frac{9}{10^k}  = 1.\,
However, when limits are not used, the function \frac{1}{10^n} where n=\infty is no longer equivalent to 0.
\frac{1}{\infty} \ne 0; \frac{1}{\infty}=0.000\ldots 1, This supports the existance of asymptotes.
Thefore we can conclude that 1 - \frac{1}{\infty}=.999\ldots
202.150.120.237 12:02, 2 December 2006 (UTC)
replies: text: The problem is that mathematics based on your understanding (your conception of the real numbers) would not be consistent. For instance, you could easily prove that 0.999\ldots = 1-\frac{2}{\infty} instead, and after that it would not require many steps to prove that 1=2. Mathematicians have struggled with such inconsistencies, and finally resolved them with the modern conception of the real numbers, in which 0.999...=1.--Niels Ø 12:52, 2 December 2006 (UTC)
thread_title: The validity of various proofs

message: Why, oh why... Nice article, by the way – Gurch 22:10, 23 December 2006 (UTC)
replies: text: Yeah, check out Talk:0.999.../Arguments. It's pretty absurd. Larry V (talk | contribs) 22:57, 23 December 2006 (UTC)
replies: text: Indeed. Imagine how many articles could have been written if all the time spent on those pages was directed elsewhere – Gurch 23:17, 23 December 2006 (UTC)
replies: text: It seems almost every online forum I've been to has had the subject crop up, wasting everyone's time.  I'm not sure how much of it is ignorance, and how much is trolling, but having this article to direct them to (and the arguments page if they still won't give up..) saves a lot of effort.  It's kind of like the Monty Hall problem in that common sense and truth don't completely overlap. 69.85.162.224 01:39, 24 December 2006 (UTC)
replies: text: Well, at least the Monty Hall problem (which intellectually I know is proven, but emotionally seems utterly insane to me, even though it's correct) can be demonstrated and simulated and at least that generally shuts up the doubters. This one, though, has no physicality, no way to demonstrate it in other than mathematics. --jpgordon∇∆∇∆ 02:48, 24 December 2006 (UTC)
thread_title:  People actually ''argue'' about this? 

message: I made the notes into two columns to try and make it smaller. If anyone objects please revert, I don't want to step on feet. Thought I'd leave a note. (haha) Cheers,--Flying Canuck 06:57, 2 January 2007 (UTC)
replies: text: It doesn't show up as two columns in Safari. Anyone else experiencing this issue? –King Bee (talk • contribs) 16:15, 2 January 2007 (UTC)
replies: text: Works in Firefox 1.5, but not Internet Explorer 7. Supadawg (talk • contribs) 19:37, 2 January 2007 (UTC)
replies: text: It also works in Firefox 3.0a2pre (Minefield) but not in Opera 9.10 (though that doesn't even display the images in the article - maybe I need to alter some settings). Raoul 20:18, 2 January 2007 (UTC)
thread_title:  Notes into two columns  

message: 2 - .999..... = 1.000...... because the .999... goes on forever you can never get 1.000......1 idk if this continues a rational proof I am just saying. Sorry for grammar. Or lack of math vocabulary. 72.178.44.126 21:06, 3 January 2007 (UTC)
replies: text: The problem I would see is that the familiar algorithm we were all taught in grade school for subtraction begins with the rightmost digits, which obviously is not possible with 0.999....--Trystan 22:36, 3 January 2007 (UTC)

text: You are simply assuming your conclusion, just like all the other proofs here.  2 - .999..... = 1+ an infinitesimal.  The irony is that while none of the proofs here prove anything to my satisfaction, I know of one approach that optionally does seem to come out with .999=1.  However it does so in a way that I think the others here will not like.  Algr 07:04, 4 January 2007 (UTC)
replies: text: None of the proofs in the article assume their conclusion. You may have come to the conclusion that they do because you do not recognise the results on which they are built, but the proofs are all built on solid mathematical results that are derived inevitably from the axiomatic construction of the reals. If you really feel that a proof is assuming its own conclusion, please describe how you think it is doing so on the Arguments page. Maelin (Talk | Contribs) 07:49, 4 January 2007 (UTC)
thread_title:  I think a simpler proof 

message: Can we add the proof in the article that 0.999...^2=0.999... and in our number system there are only two values that when you square them they equal themselves so 0.999... must be equal to 1.
:Sorry, but I doubt this proof is a useful expansion of the article. There are a few problems:
:#You would have to know how to multiply decimal representations, and even if you know the principle, it's quite a task to show that the result really is 0.999...
:#Those who believe that there's a tiny amount missing to 1 won't be convinced anyway, since to them for 0.999...^2, there's a larger, but still infinitesimal amount missing to 1, but 0.999...^2 isn't 0.999...
:The "simple" proofs we give rely on long division and multiplication by 10, both of which seem easier than multiplication of decimal representations. --Huon 20:43, 9 January 2007 (UTC)
thread_title:  I Have a New Proof for why 0.999...=1 

message: Does 1.999... equal 2? I would think that it would but I'm just making sure.-- Jacroe |  Talk 06:41, 15 January 2007 (UTC)
replies: text: Yep. Given that 0.999... = 1, we can then see that 1.999... = 1 + 0.999... = 1 + 1 = 2 . Maelin (Talk | Contribs) 07:10, 15 January 2007 (UTC)
thread_title:  1.999...