Talk:0.999.../Archive 5

message: Dear all,
I assume all of us are well-meaning people, wanting a good result for this article. That makes me ask the following favor to anonymous contributors:
* Please make an account. It is not productive for us to deal with a person who always uses a different IP address. It is impossible for us to keep in touch with you this way. Making an account will take you five secods. Just choose an imaginary username which has nothing to do with your real name, and a password. No more. But so much gained. 
* Please sign your posts. Use four tildas for that, like this: ~~~~. I would really, really ask you to give it a try. You leave a lot of unsigned comments and nobody can tell which is what you wrote, and which is somebody else, and which is another anonymous user. 
I truly appreciate you taking your time to read all this, and do us a the small favors I asked which will take you a very small amount of time (not infinitesimal, but close :) and will make it so much more pleasant for us to have constructive discussions. Sincerely, Oleg Alexandrov (talk) 06:04, 18 December 2005 (UTC)
thread_title:  Rules of engagement 

message: An anonymous IP added a proof idea in the "Elementary" section that essentially duplicated a proof in the "Advanced section". While a good-faith edit (thanks), it doesn't seem to offer any improvement. The "squeeze play" idea is a good one, and does sometimes appear in informal explanations. The point here is that we already cover the same ground more rigorously. --KSmrqT 18:56, 10 January 2006 (UTC)
thread_title:  Explanation for removal of good-faith insertion 

message: Quack, quack, quack... Revolver 22:27, 26 January 2006 (UTC)
thread_title:  Blah, blah, blah... 

message: Rasmus starts his real argument for 0.999... = 1 with:
x=1-0.9999...<1-\sum_{i=1}^{m}\frac{9}{10^i}=\frac{1}{10^m}.
I was almost fooled by this proof until I considered what would happen if m indeed were allowed to run through infinity and one accepts that 0.999... = 1. What happens is that it results in an absurdity: 0 < 0 = 0.   Let's suppose that 0.999... = 1. Then 1-0.999... = 0 and 1-\sum_{i=1}^{m}\frac{9}{10^i}= 0 (as m approaches infinity). 
This is the evidence that this proof is blatantly incorrect. Crafty fellow that he is (or not, I don't know), Rasmus almost convinced me. However, I know that by definition 0.999... is less than 1. It is part of the architecture of the decimal system. It is not only idiotic to try to prove otherwise but also an exercise in futility.
71.248.131.234 12:12, 2 February 2006 (UTC)
replies: text: I never claimed the inequality held for "m = ∞". I claim, however, that the inequality holds for all natural numbers m. Please make sure you understand the difference - it seems to be fundamental to this discussion. Rasmus (talk) 12:23, 2 February 2006 (UTC)

text: The inequality does not hold for "m = ∞", it does not need to. At no stage does the proof rely on 1 being the limit of the sequence of partial sums. I rely on 0.999... being greater than the partial sums to justfiy the step \frac{1}{10^{m}} > x. If you do not understand that you do not understand the proof. Mikekelly 15:47, 2 February 2006 (UTC)
thread_title:  Rasmus' proof and its error 

message: The fact that 0.999... equals 1 is hardly something worth of a "proof", it follows directly from the definition of real numbers. I think the article should reflect that.
Dedekind cuts and Cauchy series are mentioned, which is good. Most proofs mentioned, however, start from some arbity point that you just have to accept. If you don't think 0.999... is 1 you might not accept that 1/3 is exactly 0.333... Same goes for the algebra proof. You might think that 10c − c = 8.9999...
If you just understand what it means when 0.999... is written, the subject becomes very clear. Therefore one should try to explain what it means, exactly, and dispel any unclear mental images such as "endless number of digits".
Briefly, "0.999..." means the limit of (0.9, 0.99, 0.999, ...) which is 1, exactly.
More generally, real numbers can be defined as cauchy series of rational numbers: x = (a1, a2, a3, ...). If the series approaches a rational number it can be identified with a rational number. x = (1, 1.0, 1.00, ...) => x = 1. The notation "0.999..." actually means the series (0, 0.9, 0.99, ...) that approaches 1. Therefore 0.999... and 1 are the same, in a real number sense that is.
Comments? Mistakes? English is not my native language so some math terms can be wrong...
194.251.240.114 17:46, 10 February 2006 (UTC)
replies: text: Actually, the main argument on the talk page and its archives seems to have been the idea that the infinite sum is somehow a separate entity from the limit of the partial sums. However, see the above section for my proof that they are equal. Confusing Manifestation 12:37, 11 February 2006 (UTC)
thread_title:  A few points to consider. 

message: How is it that you all feel comfortable saying:
    -1) sqrt[6*[1+1/4+1/9+1/16+.... ]]  equals to some number pi
and 
    -2)  2 + 1/2! + 1/3! +   equals to some number e
However, you do not feel comfortable saying:
     9/10 + 9/100 + 9/1000 +   equals to some number y   ??
No, you have to say y = 1. Well, hypocrites, do the same for pi and e then!
158.35.225.231 19:49, 10 February 2006 (UTC)
thread_title:  Levels of hypocrisy. 

message: I always see this concept and every time I see it I get hung up on a rash of bad assumptions. Because I don't do decimal math all the time, when I see .999 (rep) I don't recognize it. Actually, it's not a complex number nor magic bullet -- if you remember where it comes from:
1/3 x 1 = .333 (rep) = 1/3
1/3 x 2 = .666 (rep) = 2/3
1/3 x 3 = .999 (rep) = 3/3 = 1
1/3 x 4 = 1.333 (rep) = 4/3 = 1 1/3
1/3 x 10 = 3.333 (rep) = 10/3 = 3 1/3
I fall into the trap everyone else does of overcomplicating the entire thing. Our number .999 (rep) is nothing but the remainder created when doing math on some fractions with reoccuring decimals. If we do math in decimal format, we get a decimal answer, which we then have to convert to a integer answer. Converting .333 * 3 = 1/3 * 3 = 1 is so natural that I never even stop to figure out how to  get from the first to the last.
More examples:
   1 / 9 = .111 (rep)
 + 8 / 9 = .888 (rep)
  --------------------
   9 / 9 = .999 (rep)
   2 / 11 = .1818 (rep)
 + 9 / 11 = .8181 (rep)
  -------------------
   11/ 11 = .9999 (rep)
Remember, .999 (rep) is the initial remainder. To convert reoccuring decimals into integer values, divide by an equal number of nines:
  .999   999
  ---  = ---  = 1
  .999   999
Thus we have a remainder of 1 full item to add to our total!
This is grade school decimal math.
I understand all the confusion because I fall for it every time. Here is the biggest false assumption I have to dispel for myself.
False Concept: .333 (rep) is .3 with inifnitely repeating digits. WRONG! This is "notation": it means literally: When we try to determine the decimal equivilent of 1/3 we kept getting a remainder. Because the remainder repeats every time, we can do the math as many times as we need to. At any point we can stop and sum up the rest as a remainder and have a finite number. When we do math we always have a remainder that is equal to the repeating decimals. Thus .333 (rep) means we have as many 3s as we need to handle any math job, and when we do math we have some form of .333 (rep) left over. For example:
100,000 x .333 (rep) = 33,333.333 (rep) = 33,333 1/3
10,000,000 x .333 (rep) = 3,333,333.333 (rep) = 3,333,333 1/3
The remainder of 1/3 is given by taking the digits after we stop and finding their integer values by using the division by 9 process.
And what does .999 (rep) mean? It too is a finite number. It's different, though because it means that we have a series of 9s. At any time we can stop and we have a remainder that is some form of 1. When we add our 9s and the remainder, we get 1. The following are examples where I stopped mathing out .999 (rep) and created a finite version at completely arbitraty points:
  .999 (rep) =  .9999 + .0001 
             =  .999 + .001
             =  .99 + .01
             =  .9 + .1
             = 1.0
I don't pluck the "1" format for the remainder out of the air. I used 1 because that results from the dvision by 9s rule. If I decide to stop at say the sixth place:
  I have   .99999 remainder .000009
                    .000009     
  Which is .99999 +    --
                        9
                  
  Which is .99999 +  .00001
Summed up: all numbers have three notational formats (ways of expressing them): for 1 these are integer 1, finite decimal 1.0, and repeating decimal .999 (rep).
When you begin to differentiate between the various notations for numbers, then you understand what the proofs are saying.
208.252.26.3 21:04, 10 February 2006 (UTC)
replies: text: I like your point about notation, although the remainder stuff is pretty sketchy. Would you like to log in? Melchoir 21:19, 10 February 2006 (UTC)
thread_title:  Apples and bannas: not enough explanation in the concept