Talk:0.999.../Arguments/Archive 3

message: Thinking about what Jerr and Maelin write above, it occurs to me that some people might have something like this structure in mind:  Suppose we consider formal strings of decimal digits (but with no negative numbers; this will be important) and define addition and multiplication on them by looking at sums and products of their initial segments, and taking the limit in the topology of pointwise convergence. That is, to get the nth digit of a+b, you find long enough initial segments of a and b that the nth digit of the sums of those initial segments, or any longer ones, will always be the same.
It's not instantly obvious that these operations are well-defined, but in fact they are; this isn't hard to see. Call one of these numbers "terminating" if it eventually reaches an infinite string of zeroes. Then the sum and product of a and b are terminating if and only if both a and b are terminating; this follows from monotonicity of addition and multiplication. Now it's easy to see that the digits of the sum/product of a and b are precisely the digits of their usual sum/product, disambiguating between ...000... and ...999... by the rule in the last sentence. (This doesn't work if you allow negative numbers.)
The usual laws of commutativity, associativity, and distributivity hold (see this by noting that the answer is terminating if and only if all inputs are terminating).  However most nonzero numbers have no multiplicative inverse (none of them have an additive inverse), and neither the additive nor the multiplicative cancellation law holds. Another inconvenience is the structure's topology, which is that of the Cantor space, not so good for calculus.
Still, this is a well-defined structure in which affirmations like Jerr's make some sort of sense, and maybe that ought to be acknowledged. Is there a standard name for it? --Trovatore 07:16, 15 October 2006 (UTC)
replies: text: Sounds to me like the commutative semiring of "decimal numbers" from 0.999...#Breaking subtraction. Richman doesn't seem to be aware of a name for it, nor for that matter any prior research on it at all. Of course, he may have simply failed to look hard enough. Melchoir 07:28, 15 October 2006 (UTC)
thread_title: A structure for the skeptics

message: 1 = 1
0.999... = 0.999...
:1 + 2 = 1 + 2
:4 - 1 = 4 - 1
:Therefore, 4 - 1 ≠ 1 + 2 Supadawg (talk • contribs) 02:44, 25 October 2006 (UTC)
replies: text: Was that a response to unsigned or to Supadawg? In any case, technically unsigned has only made true statements, but has implied the conclusion 0.999... ≠ 1, which is just as fallacious as Supadawg's sarcasting conclusion 4 - 1 ≠ 1 + 2. -- Meni Rosenfeld (talk) 16:54, 27 October 2006 (UTC)
replies: text: Oh. I get it now. I thought that whole thing was stated by Supadawg. Now I see the satire.Xparasite9 20:00, 30 October 2006 (UTC)
thread_title:  1 = 1 

message: I just probably find another proof of my theory that .999... isn't equal to 1.
I was reading the recent modifications of the article and I that is says:
lim n->inf   1/(10^n)=0   but that isn't true because lim->inf (1/10^n)*10^n = 1
and 0 multiplied by any number doesn't equals 1....
also in my thinking i also got to this .333 =/= 1/3 <--> .999 =/= 1
That means .333 isn't equal to to 1/3 only and only if .999 isn't equal to 1.
You can't prove that .999... = 1 with with the afirmation that .333... is 1/3 
because one affirmation depends of the other =P. Even if you can threat .333 as
1/3 because the difference is minimum the difference is still in there.
Well that's just what i think... but i'm sure that what i said about 10^n is right.
I hope you can understand my english.
--Rensato 04:06, 25 October 2006 (UTC)
replies: text: Sorry, but this is a little off. Lim n->inf 1/(10^n) and Lim n->inf (1/10^n)*10^n are completely different functions.  Because Lim n->inf (1/10^n)*10^n = Lim n->inf 1.  The limit of a constant is the constant itself, so Lim n->inf (1/10^n)*10^n = Lim n->inf 1 = 1.  However, Lim n->inf 1/(10^n) is totally different.  The denominator contains an exponentially increasing function, so as n tends to infinity, 10^n gets gigantic.  So the denominator get very large, which means 1/10^n gets very small.  So Lim n->inf 1/(10^n) approaches 0 as n approaches infinity.  Tachyon01 05:23, 25 October 2006 (UTC)
thread_title:  Probably another proof 

message: Limits, such as shown at limit of a sequence and limit of a function, work as a value is approached, but not at the value. Why is it, then, that the limit proof used in the article can be applied at .999...? -- kenb215 talk 05:18, 25 October 2006 (UTC)
replies: text: Because 0.999… is defined as the limit of the relevant sequence. You can't evaluate a sequence at infinity, so we don't even try to do so. Melchoir 05:57, 25 October 2006 (UTC)
thread_title:  The limit proof 

message: Imagine that all the numbers that exist, starting from zero, were placed one after the other in order of ascending value. Surely the number that would appear one place below 1 in that list would be 0.999... And since 0.999... appears prior to 1 in a list of ascendding value it must be less than 1. Why am I wrong? 82.35.70.17 07:06, 25 October 2006 (UTC)
replies: text: Well, unfortunately, the first step is wrong. The real numbers can't be listed in ascending order. In fact, they can't be listed at all! See Cardinality of the continuum. The impossibility of an ascending list is easier to  see than the general statement, though: if you have any two adjacent elements on the list, where does their average go? Melchoir 07:10, 25 October 2006 (UTC)

text: That list is nondecreasing, to be sure, but it isn't strictly increasing, because two of its items are equal. Melchoir 07:22, 25 October 2006 (UTC)

text: There isn't anything "just below" 1 or any other real number. You can't shoehorn the continuum into either a list indexed by the integers, or a single collpsed point. It's a much more interesting structure than either of those. And, in fact, there are ordered sets even more interesting than the continuum; see Long line (topology) for example. Mathematical structures are often surprisingly rich, to the point where easy methods of investigation, like listing, are inadequate. Melchoir 07:34, 25 October 2006 (UTC)
thread_title:  Why am I wrong? 

message: :Moved from main talk page. --Brad Beattie (talk) 12:23, 25 October 2006 (UTC)
replies: text: These criticisms are addressed in the Skepticism section. The number 0.999... is not a process. It is not a function. It does not have a limit, it does not move anywhere. It is just a single point on the real line. That point is coincident with 1, i.e. they are equal. Yes, 0.9, 0.99, 0.999, any such number with a finite number of nines, no matter how many, is less than 1. But 0.999... is not any of them. It has an INFINITE number of nines. It's true that when you write 0.9, that number is not equal to 1, and no matter how many nines you append, the number is never equal to 1. But it's never equal to 0.999..., either. 0.999... is the limit of the sequence {0.9, 0.99, 0.999, ...} and that limit is, as I'm sure you agree, 1. Maelin 06:44, 25 October 2006 (UTC)
replies: text:  You say "it does not have a limit". Then, you say "0.999... is the limit of the sequence {0.9, 0.99, 0.999, ...} and that limit is, as I'm sure you agree, 1". You just contradicted yourself. Which is it? Does it have a limit or not? I agree with your second statement. 0.999... does have a limit. The limit is 1. But that's not what the article says. I read the Skepticism section. The students SHOULD be skeptical. The article is wrong, and based on a misunderstanding of limits. Math is an exact science. The article is non-rigorous, an example of math misused. By the way, "It is just a single point on the real line" is just more circular reasoning. Dagoldman 07:17, 25 October 2006 (UTC)
replies: text: Look more closely. 0.999... is a limit. It doesn't have a limit. "To be" and "to have" are really basic verbs; I should think we all understand the difference between them? Melchoir 07:27, 25 October 2006 (UTC)
replies: text: Yes, you said 0.999... "is" the limit. I misread your comment (it would never enter my head that someone could say such an outlandish statement), and you did not contradict yourself. But where do you get the idea that 0.999... IS a limit? Isn't this just more circular reasoning? If you assume that 0.999... IS a limit, then of course it is equal to 1. But 0.999... is NOT a limit. It is an infinitely repeating decimal. It HAS a limit. You can get as close as you want to 1, but you'll never get there. Dagoldman 07:45, 25 October 2006 (UTC)
replies: text: Look, definitions are not optional. 0.999… is a limit. If you want to use the phrase "0.999…" to mean something inconsistent with its given definition, this is of course your personal right, but it will interfere with your ability to communicate with the rest of humanity. I can choose to call the color of lava "blue" and then claim "The sky is not blue". Now is that statement really a valuable insight? Melchoir 08:05, 25 October 2006 (UTC)
replies: text:  The argument about colors applies to you just as much as it does to me. You haven't added anything. Dagoldman 18:26, 25 October 2006 (UTC)

text: That's an interesting point. By what I was thinking 1 would come out as "an infinity of infinitesimals" with .999... coming out as "infinity -1 of infinitesimals", but as "infinity -1" is still infinity they'd be the same. Interesting, it kind of relates to the conceptual difficulty of thinking that a number minus one equals itself as that wouldn't work in real numbers. (Getting us back to square 1)--T. Anthony 22:03, 25 October 2006 (UTC)

text:  A number can't *have* a limit. Only expressions with variables can have a limit. The number .999... (or better yet .\bar{9}) is not a number with an arbitrary number of 9's more than 3, it is a number with infinite 9s. Fresheneesz 08:13, 25 October 2006 (UTC)
replies: text: The conceptualisation of the number however, requires a thorough and rigourous definition and understanding of the concept of infinity, which is itself only well explained using limits IMO. Ansell 08:50, 25 October 2006 (UTC)

text: Just about every professional mathematician, historical and contemporary, you find will disagree with you Dagoldman, maybe your mathematical abilities surpass theirs though.  Brentt 08:33, 25 October 2006 (UTC)
replies: text: Maybe the professional mathematicians do disagree. But yes, I think they are wrong. I am not a professional mathematician. However, in my defense, I never made less than 100 on a math test in high school or college (sometimes there was extra credit). And that includes advanced calculus and number theory at Yale. I didn't go into math because the mathematicians I met seemed strange. Maybe that was a mistake, because I do love math. Anyway, back to the problem, I think a source of much of the misunderstanding here is the notation. 0.999... seems a terrible (not useful) notation to me. It's a representation of a concept. But you can't do anything with it. It leads to people multiplying and subtracting "0.999... numbers", which is not demonstrated to be valid. And it leads to bizarre statements such as "0.999... is a limit". 0.999... is a representation of an infinitely repeating decimal. This infinitely repeating decimal is properly defined, manipulated, and notated as a geometric series, which is what it really is. Forget about 0.999... notation for a second. Would anyone disagree that the series for this infinitely repeating decimal has a limit of 1? I think limits (and calculus) are very difficult for most people to understand on a fundamental level. I just happened to see this as the "article of the day". I have no idea how many high-level mathematicians would contribute to this article. But maybe zero, because the article is so obviously non-rigorous with it's inclusion of the circular-reasoning "fraction proof" that it doesn't even pass the smell test. The "fraction proof" assumes 0.333... is exactly equal to 1/3 to "prove" that 0.999... is exactly equal to 1. The "proof" under "Infinite series and sequences" is also obviously circular reasoning. It assumes that an infinitesimally small amount is equal to zero to "prove" that 0.999... is equal to 1. The "algebra proof" is not circular reasoning, but is also bogus, because the 0.999... notation is not a number that you can multiply by 10 and and then subtract. Try adding 0.999... plus 0.999... one or more times and see what happens. It doesn't work. Where do one or more 8's (9 + 9) go? Infinitesimally small differences cannot be just swept under the rug. They exist, just as much as large differences. A way to deal with the infinitesimally small is to define a limit, which is my whole point. 0.999... is as close to 1 as you want, but is infinitesimally different from 1. Three bogus proofs. Where's the beef? Dagoldman 09:34, 25 October 2006 (UTC)
replies: text: This is not a mathematical question, but I'm curious about your thinking. Is 0.999… a real number or not? Melchoir 10:30, 25 October 2006 (UTC)
replies: text: I would agree 0.999... is a real, irrational number. I think most mathematicians would agree that the concept of irrational numbers and exactly where to put them on the number line is not easy for to grasp. You probably have read how irrational numbers horrified and mystified Pythagoras. The diagonal of a 1 x 1 square has an exact length. Yet it's not a rational number, so how can it have an exact length? This was unacceptable to Pythagoras. Is it any easier for us to grasp? A possible argument for "0.999... equals 1" might be: \sqrt{2} ends up "exactly equal" to some "diagonal exact length", just like 0.999... ends up exactly equal to 1. But what would the diagonal "exact length" be? Nobody knows. This is a fundamental difference between 0.999... and \sqrt{2}. 0.999 is either exactly 1 or has a limit of 1. \sqrt{2} seems to be just \sqrt{2}. You can't say it equals any other number, or has a limit. So I don't think the other irrational numbers get us any closer to resolving the 0.999... issue. Dagoldman 11:30, 25 October 2006 (UTC)
replies: text: Do you believe that there are non-zero, infinitesimal, real numbers? Melchoir 11:47, 25 October 2006 (UTC)
replies: text: Do you? What's your point? Dagoldman 18:27, 25 October 2006 (UTC)
replies: text: Given your intelligence, I would love to read what you come up with for the value and properties of 0.999...².  That's my favorite conundrum for those who argue that 0.999... ≠ 1.  Is its square equal to itself or not?  You say 0.999... is irrational, so it is a real number -- does it have the same properties all real numbers must have?  --dreish~talk 18:26, 26 October 2006 (UTC)

text: Here is where the discussion of infinitesimals is going. the field of real numbers is an Archimedean field -- it does not contain infinitesimals. If .999... is a real number, then 1-.999... is also a real number, and cannot be an infinitesimal. So either there is a real number between .999... and 1 or .999... is equal to 1. As for .999... as a limit, what would you say is the limit, as n approaches infinity, of the summation from 1 to n of 9*.1^n?--Epee1221 05:05, 28 October 2006 (UTC)
thread_title:  This article is just wrong 

message: :Moved from main talk page. --Brad Beattie (talk) 12:27, 25 October 2006 (UTC)
replies: text: It is not clear what you mean by "reaching" infinity, but yes, we can discuss infinite quantities as long as we properly define our terms. And no, 0.999... = 1 follows rigourously from the definitions - nothing "graphical" about that. -- Meni Rosenfeld (talk) 13:49, 25 October 2006 (UTC)
replies: text:  Saying that 0.999... = 1 sounds to me to be exactly the same as writing 1 - 0.1+∞ = 1, or if you like better 0.1+∞ = 0. My only problem in this is simply the use of +∞ as a real number, which it isn't. That's all. That's the reason why I was insisting on the fact that infinity cannot be reached, we can only tend towards it. Metropolitan 15:43, 25 October 2006 (UTC)

text: +∞ is definitely not a real number. If that's what you mean by "unreacheable", then we agree. I don't see why you would call it that way, though - not being real does not make it somehow "illegal". The important thing is to pay extra attention to what we wish to say about it. -- Meni Rosenfeld (talk) 16:27, 25 October 2006 (UTC)

text: If infinity and infinitesimals aren't real numbers then isn't an infinite string of nines simply an impossibility? So saying .999.... equals 1 is essentially a way of saying .99... doesn't exist, you should simply say 1.--T. Anthony 21:48, 25 October 2006 (UTC)
replies: text: No. Note that "real number" is a technical term with a specific meaning, and just because infinity isn't a real number doesn't mean it's outside of mathematical reasoning. Melchoir 21:55, 25 October 2006 (UTC)
thread_title:  Can infinity be reached ? 

message: :Moved from main talk page. --Brad Beattie (talk) 13:45, 25 October 2006 (UTC)
replies: text: We're not 'doing arithmetic' with infinite series. We are doing arithmetic with one single real number. It is the limit of an infinite series, and such things are perfectly well defined - in fact, in one construction of the real numbers, numbers are defined as the limits of infinite series. By the way, 0.999...*10 = 10 is true. In fact 0.999...*x = x for every x in the real numbers. Maelin 13:53, 25 October 2006 (UTC)
thread_title:  Infinity 

message: All of the proofs that have been presented, I have noticed, seem to work off a finite representation generated by rounding or truncation of an infinite decimal, and extended by assumption the original decimal. For example, the first proof assumes that the final digit (should there be one) in the recurring decimal for \frac{1}{3} is a 3; for all we know, it could, in fact, be a 4, which, when multiplied by 3, returns the number to 1 again.
The most common error I have noticed, and this is one which I have even caught myself on, although only once, is the decimal-to-fraction algorithm (which is taught in many schools in Australia and, I assume, around the world). Because of the repetition of the singular digit, it is assumed that there is a rational representation for it; the result generally ends up formulating the assumption that the two numbers are alike.
The only proof, at least that I have noticed, that disproves the assumption, is the Egyption fraction-esque representation (it uses a similar structure to Egyptian fractions); this form explicitly shows the inequality of the recurring decimal:
\frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{9}{10000} + \cdots
With each new fraction of the form \frac{9}{10^n} \ (n \in \mathbb{N}, n \geq 1 \mbox{ to clarify}), the difference between it and 1 lowers to \frac{1}{10^n}. Because the decimal is recurring, i.e. it has no terminating digit, then it can be assumed that n extends to \infty, making the difference \frac{1}{10^{ \infty }} (as far as I know, it has not be either proven or disproven that \frac{1}{x^{ \infty }} = 0 (x \in \mathbb{Z} \backslash \{ 0 \}), an infinitesimally small number, and an irrational real, making 0. \dot{9} irrational as well.
Saying that, like I said, it is solely an irrational real that has been given a rational (let alone integral) 'synonym' through rounding, and it should be accepted that the equality is untrue.
(I have also come to notice that \frac{0. \dot{9}}{2} = 0. \dot{4} \dot{9} \neq 0.5. Can someone explain that?) --JB Adder | Talk 15:18, 25 October 2006 (UTC)
replies: text: No. There does not exist a real number, rational or irrational, that has anything to do with 0.999…, except the real number 1. Melchoir 15:37, 25 October 2006 (UTC)

text: JB-adder--  your arithmetic is wrong--  \frac{0. \dot{9}}{2} = 0.4 \dot{9}=0.5. --Alecmconroy 16:33, 25 October 2006 (UTC)

text: There is no last digit. You seem to say "maybe there is, maybe there isn't" and that's wrong. Also, ∞ is not a real number, and in this context (as opposed to other contexts) we have not defined arithmetic operations on it, so we cannot say that the difference is 1/10∞. We're not rounding anything, we're following definitions and theorems rigorously. -- Meni Rosenfeld (talk) 17:08, 25 October 2006 (UTC)
thread_title:  A simple, but major, error caused by rounding 

message: I probably should read the article but am I right that since 0.999...=1 anything else with X999... equals (X+1). For example. 3.9999... = 4 | 1.5334999...=1.5335 ? Of course, I'm not particularly sure if you'd ever come across these in reality but I guess the concept remains the same Nil Einne 16:02, 25 October 2006 (UTC)
replies: text: Yes; see 0.999...#Generalizations. Melchoir 16:16, 25 October 2006 (UTC)
thread_title:  Perhaps a stupid question 

message: So does that mean that:
#1.111... is also equal to 1?
#1.999... equal to 2? —The preceding unsigned comment was added by Ejh102 (talk • contribs) 15:57, 25 Oct 2006  (UTC)
:1; No, 1.111... is equal to 1+\frac{1}{9} 
:2: Yes →AzaToth 16:07, 25 October 2006 (UTC)
replies: text: First one, eh say what? Why will 1.111...=1? This doesn't make sense to me. Second one, well I have the same question see above Nil Einne 16:04, 25 October 2006 (UTC)

text: No, 1.111... doesn't equal 1, because base "1" isn't the lowest digit-- 0 is.  Note, however, that 1.000... DOES in fact equal 1.   Or another way to show you--  the difference between 1.111... and 1 is at LEAST .1 ( ie, 1.111...-1>.1).  In contrast, if you try to find a positivenumber that is DEFINTELY smaller than the difference betweeen 1 and .999..., you won't be able to find one.   --Alecmconroy 16:26, 25 October 2006 (UTC)
thread_title:  Random question... 

message: What is x, such that x-1=0.99999...?
--Rlwest 13:03, 25 October 2006 (UTC)
replies: text: x = 1, since 0.9999... = 1 -- The Anome 13:05, 25 October 2006 (UTC)

text: There is no such thing as non-exact numbers (though there may be non-exact numerical data in applications). If asked, "what is x such that x-1 = 444 / 148", is there something wrong with replying, "444 / 148 = 3, therefore x = 1/3"? 0.999... is just equal to 1, there is no way around it - the same way 444 / 148 is just equal to 3. -- Meni Rosenfeld (talk) 14:02, 25 October 2006 (UTC)
replies: text: Well put Meni.  Just to be clear, there is no need perform mathematical operations on .999... without substituting 1 because we have proved that .999... = 1!  That's the beauty of proofs! -bobby 14:07, 25 October 2006 (UTC)
thread_title:  A Naive Question 

message: I've seen a lot of arguments back and forth on this subject over and over again, so let's do what our teachers told us to do: show your work. I.e. prove your answer in a format that even the lamest of morons could follow the numbers.
What we know
------------
Most high school algebra teachers have taught their class the method to find the fractional form of a repeating decimal. And they show the proof through the most common method, taught to us in preschool: reverse the problem to check your answer. So with that in mind, a small lesson:
One of the most fun fractions to play with is 1/7, which equals 0.142857...  Now the way to determine the fractional value of a repeating infinite decimal is to use 9's. Take the section of the sequence that repeats, and place it over a series of nines with an equal number of digits.
Ex. 142857/999999
Notice there are six digits on top and bottom. This it the fraction used to determine the value of 0.142857... Repeating. Reduce the fraction (you all do remember how to reduce, right? If not, break out a calculator and divide the bottom by the top, I promise it will always come out to an integer) and you are left with 1/7, our original fraction. Now check your work by reversing the problem. Multiply the fraction 1/7 by 142857 to arrive at the fraction 142857/999999, and likewise divide one by seven (I advise a calculator, unless you like long division that continues into infinity) and there you have it, reversal-proof that 0.142857... repeating equals 1/7.
If you care to try it with a few other fractions with prime denominators to get used to the concept go ahead and do so.
Where we go from there
----------------------
Now we come to the wonderful 0.9... Repeating. We repeat the process, taking the sequence, which is 9 by itself, and place it over a series of nines with an equal number of digits, i.e. 9/9. Do I need to go on?
:Unfortunately, this is not really a proof, just an example. It's a nice thing to think about, but not really what we need here. -- Meni Rosenfeld (talk) 14:03, 25 October 2006 (UTC)
thread_title:  Show your work 

message: Do we want to answer this objection more explicitly? 
:(Dispute. If you cannot make the assumption that 1 = 0.9.... then you also cannot make the assumption that 1/3 = 0.3...., in both cases it is true that the limit is equal to 1 and 1/3 respectively, however the arguement that 0.3... is strictly less than 1/3 is as valid as the argument that 0.9... is strinctly less than 1)
It may be enough to observe than 0.9999... is not the result of long division, so the cases are distinct. Septentrionalis 16:37, 25 October 2006 (UTC)
replies: text: If you wrote a computer program that prints real numbers in decimal, and, when given a "1", it starts to print a bunch of nines instead, virtually everyone reading this would conclude this program has a bug.  If the mathematicians burst in and explain that, "well, you know, this program, as long as it runs forever, is actually quite correct" would be dismissed just as easily as someone who suggested that, by golly, you can indeed unload your groceries into the kitchen, one molecule at a time!  This is ultimately the source of the difficulty with this business.  People know what is "right", Occam's razor is a very deep computational element in the human brain.  Probably the mathematicians know this, since this particular result has essentially no use beyond the novelty factor.  mdf 18:14, 25 October 2006 (UTC)
replies: text: Well actually computers don't and shouldn't deal with infinity and mathematicians don't deny this and I'm pretty sure nearly every single one would agree a computer program giving the result as 0.999.... instead of 1 is buggy. Just because they mean the same thing doesn't mean you should use them interchagably. Nil Einne 18:39, 26 October 2006 (UTC)
thread_title:  Cogent objection? 

message: Isn't it possible to prove that 0.999...=1 by use of open sets?  In other words if S=(0,1) is an open set then 0.999... is not a member of S since (0.999...,1) is not an open set.  However, 0.999... is a limit point of S so it is in the closure of S (since it is a limit point) which shows that 0.999...=1. Unmasked 19:02, 25 October 2006 (UTC)
replies: text: Well, reasoning with (0.999...,1) before determining what 0.999... is seems dangerous. Melchoir 20:16, 25 October 2006 (UTC)
replies: text: OK I suppose what I'm saying is that it should be possible to prove that 0.999... is not in S with this line of thinking.  I'm not sure why you moved this to the Argument's section as I'm just wondering if this is another method to prove 0.999...=1. Unmasked 20:38, 25 October 2006 (UTC)
replies: text: Well, this page isn't just for cranks; any mathematical discussion that doesn't concern the article can get dragged here. Anyway, sure, you can prove that 0.999... is not in S, but at that point, since 0.999... <= 1, you've proved 0.999... = 1 already! Melchoir 20:43, 25 October 2006 (UTC)
thread_title:  Open set proof? 

message: lets assume 1 = 0.999...
if x = 1, and x is then multiplied by 2, and then 1 is subtracted, x will still equal 1
if this process is repeated infinite times, x will still equal 1.
now, if 1 = 0.999..., then the same result should be reached if x equaled 0.999... instead of 1
but if you multiply 0.999... by 2 and subtract 1, you get 0.999...8. this then becomes 0.999...6, then 0.999...2, and so on. Repeated infinitly, you would end up with negative infinity, rather than 1.
therefore, 1 cannot equal 0.999...
i hope that all made some sort of sense, let me know if you see anything wrong with it
Diabl0658 19:45, 25 October 2006 (UTC)
replies: text: The problem with your "proof" is that we are talking about real numbers, not surreal numbers, so the notation 0.999...8 is not allowed. →AzaToth 19:49, 25 October 2006 (UTC)

text: Yes, it's a legal equation. You can rewrite it as:
0.8 + \sum_{x=1}^n1.8(0.1^x)
→AzaToth 20:41, 25 October 2006 (UTC)
thread_title:  Proof that 1 does NOT equal 0.999...