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Topic: 0.999. (Read 123637 times) 

Kozo Morimoto
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Re: 0.999.
« Reply #75 on: Sep 6^{th}, 2002, 8:07pm » 

Ok, how about this... You have a set A = {0.9, 0.99, 0.999, ...} which is inifinite in size. You also have a set W = {1, 2, 3, 4, 5...} which is the set of whole numbers and its size is also infinite. You can see that there is a 1to1 mapping between set A and set W, where number of 9s in an element of A is an element in W eg 0.9 > 1, 0.9999 > 4 etc etc So if 0.999.... = 1, then 1 must be an element in set A. If so, what does 1 in set A map to in set W? Since it doesn't, set A must no include 1, therefore 0.999... <> 1


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Pietro K.C.
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Re: 0.999.
« Reply #76 on: Sep 6^{th}, 2002, 11:59pm » 

No, you're still hanging on to the notion that a sequence's limit must also be a term in that sequence. You have successfully proved that 1 is not in set A, and then used the fact that 0.999... is in that set to conclude that they are different numbers. There is a tiny flaw, and to uncover it I now pose the same question: what number does 0.999... map to? Surely not infinity! You yourself referred us to a website that said infinity does not exist in the context of a "number system". No, let's work backwards. What is an arbitrary member of set W? A positive integer x. A FINITE positive integer x. What element of A is mapped onto that FINITE positive integer? The number 0.999...9, with exactly x 9's after the decimal point. And that is surely not equal to 0.999... repeated indefinitely. So 0.999... is not mapped onto any member of W, so it is itself not in A. At this point I would like to encourage anyone who would object to the above argument by saying that infinity IS included in the set W (the integers) to pick up a book on the subject, study it, and stop waving their intuitions around like they're foolproof methods of ascertaining mathematical truth. If the book is too boring, at least pick up one on the history of mathematics, and read about the nineteenth and twentieth centuries. Anyways. If I were to use your kind of reasoning, Kozo, I could easily construct an example where 1 IS included in the set A. For instance, construct the set B as: B = {1/1, 1/2, 1/3, etc}. There is also a onetoone mapping between A and B. If we assume that 0.999... is in A, we must also assume that 0 is in set B. The situation is exactly the same. If you tell me that 0 is not in B because, no matter how large n is, 1/n is not zero, then I reply that, no matter how many 9's we put after the decimal point, 0.999...9 (terminating) is not 0.999... (repeating). If, however, you accept that 0 is in B, suppose that 0.999... = 1, and that it is in A. What does 1 map to in B? Why, 0, of course! Because the mapping is onetoone, and any terminating decimal with n 9's maps to 1/n > 0. Voilą, the contradiction is gone. The point is, for the same reason you don't believe that "infinity" is in the integers, you mustn't assume that a sequence's limit is in that sequence. I gave examples of this in my previous posts. And 0.999... indicates precisely the limit of set (sequence) A. It makes no sense to argue about inequalities satisfied by elements of A, because 0.999... is NOT in A. Please, read the posts of S. Owen, Aadash and myself a bit more carefully  the main ideas are there. And, if you come up with another of these "thought experiments" (VERY ingenious  you're like a modern Zeno ), try to solve them yourself using the concepts we presented you. Chances are, you'll succeed. P.S.  Dustin, don't you think that "infinitely good approximation" is something of an oxymoron? Maybe you should think a bit more about the subject, because it is not easy  it took mankind about 3000 years to formalize it properly.


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Kozo Morimoto
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Re: 0.999.
« Reply #77 on: Sep 7^{th}, 2002, 12:52am » 

But 0.999... is in set A, otherwise size of set A would be finite. How can size of set A be infinite, but still not include 0.999...?


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Pietro K.C.
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Re: 0.999.
« Reply #78 on: Sep 7^{th}, 2002, 9:04am » 

In exactly the same way that {1, 2, 3, etc} can be infinite in size without including anything like "infinity". I ask you again, if 0.999... is in A, what does it map to in W? With that question you went to the heart of the matter; a sequence (not a set, mind you) of real numbers like (a1, a2, a3, etc) is in reality a function f : N > R, from the natural numbers (positive integers) to the reals. When we list the terms of the sequence, what we are saying is that f(1) = a1, f(2) = a2, f(3) = a3, etc. Therefore, a number x is in the sequence IF AND ONLY IF there is some positive integer n such that f(n) = x. Since there is no positive integer n satisfying f(n) = 0 in my previously defined sequence B, then 0 must not be in B  the same goes for 0.999... and sequence A. The concept of infinity, and things like infinite sets whose elements are all finite, is very often counterintuitive. From a nonformal point of view, what makes the set of natural numbers infinite is that, given any one of them, you can always find a bigger number by adding 1 to it. You may argue that, if we imagine this process carried on indefinitely, as it must be, "infinity" will be included in the set of naturals. But it will not. By adding 1 to a number, you are simply showing that there is a contradiction in assuming that N is finite. The argument is an abbreviation of "suppose N is finite. Pick its largest element. Add 1 to it. The new element is a natural number, but it is larger than the others. Therefore it was not included among the finite list of numbers. But that list was arbitrary, so N must have an infinite number of elements". If you do not assume that N is finite, then you cannot necessarily pick its largest element  which is another thing the argument says, that there are arbitrarily large naturals. Do you see this? That, after you stop assuming that N is finite, you have nothing to assure you that addition CAN be carried on indefinitely? It cannot. BORING PROOF Addition, too, is a function, one that maps two naturals on to a third one. What a + b = c means is that +(a,b) = c, where + is a function. However, the value A = 1+ 1+ 1 + ... = +(1, +(1, +(1, ...) ) ) is not defined. Because, if defined, it is either a natural number, or it is not. And we stated that + takes pairs of naturals as arguments. So, if A is not a natural, A = +(1, A) is not defined. If, however, it is a natural, it must satisfy +(1, A) = +(0, A), so 1 = 0 (by the properties of function +). So A must not be a natural, and therefore not anything.  I think you can see how subtle this is. The modern point of view is that a set A is infinite if and only if it can be put in onetoone correspondence with a part of itself. For instance, the naturals. The function f(n) = n + 1 is a onetoone mapping from {1, 2, 3, etc} to {2, 3, 4, etc}, that is, from N to N  {1}. Hence N is infinite. Now I define set A to include ONLY numbers of the form 0.999...9, with FINITELY many 9's, and ALL such numbers. I am deliberately excluding 0.999... from A. If I define f(0.999...9) = 0.999...99, that is, if given a number in A I map it to another which has exactly one more decimal place, I establish a onetoone correspondence between A = {0.9, 0.99, etc} and {0.99, 0.999, etc} = A  {0.9}. So A is infinite too, and includes nothing but finitely representable numbers. I refer you to texts on modern set theory (read about Georg Cantor's work) for more examples on how infinite sets can be confusing. The following page has one of the most basic examples: http://users.pipeline.com.au/owen/infinity.html

« Last Edit: Sep 9^{th}, 2002, 10:57am by Pietro K.C. » 
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Pietro K.C.
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Re: 0.999.
« Reply #79 on: Sep 7^{th}, 2002, 8:23pm » 

Thought of something today... suppose set A is infinite, and does contain 0.999... . However many 9's there are, that is still just one element. If we remove it, set A will no longer contain 0.999..., but will still be infinite. How about that?


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Jason
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Re: 0.999.
« Reply #80 on: Oct 1^{st}, 2002, 3:24pm » 

If .9999... = 1, then (.9999...)^(infinity) should equal (1)^ infinity. However, any number less than 1, no matter how close, will eventually approach 0, not 1. The number 1 raised to 'n' as n > infinity will always be an absolute 1. Therefore, .9999... < 1.


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S. Owen
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Re: 0.999.
« Reply #81 on: Oct 1^{st}, 2002, 3:41pm » 

on Oct 1^{st}, 2002, 3:24pm, Jason wrote:However, any number less than 1, no matter how close, will eventually approach 0, not 1. 
 It sounds like you are presupposing that 0.999... < 1!


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Icarus
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Re: 0.999.
« Reply #83 on: Oct 10^{th}, 2002, 7:49pm » 

Let's back up here. The first tool ever devised for handling when two numbers calculated in two different ways are equal was Eudoxus' Method Of Exhaustion. It works like this: suppose you want to know if x = A, but you don't have a way of expressing what x is exactly, only a means of approximating x as closely as you like. (Eudoxus invented it to handle such things as the ratio of the circumference of a circle to its diameter. But it applies here. We can approximate x=0.999... as closely as we want with a terminating sequence of 9s.) Eudoxus reasoned that if for every number B < A, you could show that x > B, then x could not be less than A, otherwise x would have to be less than itself!. Similarly, if for every number C > A, you can show that C > x, then x cannot be greater than A. If x is not less than A, or greater than A, then x IS A. So first, can we truly approximate x with a terminating sequence of 9s? Yes. I will not prove it here, but if you doubt, ask and I will supply one. So given any [epsilon]>0, there is a number y=0.99...9 (the dots represent a finite set of 9s here) so that xy < [epsilon] Second let C be an arbitrary number > 1. let [epsilon] = C1, so [epsilon]>0. Then there is a y=0.99...9 so that xy<C1. In particular, xy<C1. But 1>y, so x1<xy<C1 Add 1 to get x < C. Since C was arbitrary, x is strictly less than every number greater than 1. Third let B be an arbitrary number < 1. This time we will be more restrictive with [epsilon]: [epsilon]=(1B)/2. We must also note some thing else, just like we can pick a y so that xy<[epsilon], we can also pick a y=0.99...9 so that 1y<[epsilon]. (Again, if you don't see this, ask and I will supply the proof). Further, we can pick one y that will do for both (the more 9s you add, the closer you get to both x and to 1, so pick a y with enough 9s to satisfy both conditions). This gives: y  x < xy < [epsilon] 1  y < [epsilon] Adding gives (1  y) + (y  x) < 2[epsilon] = 2(1  B)/2 1  x < 1  B B < x Since B was arbitrary, x is strictly greater than every number less than 1. x=0.999... is greater than every number less than 1, and less than every number greater than 1. There is no other choice: it has to be 1. As if this post were not long enough already, I have some comments. Some of you will see this as an [epsilon]e[delta] proof in disguise. I answer, No: [epsilon][delta] proofs are Method of Exhaustion proofs in disguise. Eudoxus was first. Cauchy borrowed from him. A few might also see another proof in that 0.99...9 is shown (actually only stated, but it can be shown) to approach both x and 1 as 9s are added. I answer, Not really. That proof requires a proof of uniqueness of limits, which brings us back to the above. I offer this proof not because it is better, or more accurate than previous ones, but because it involves the least number of concepts of any I have seen. It basically depends only on the following: That 0.99... actually is a real number. (If anyone has questioned this, I missed it, but it is something that should be addressed to be completely thorough.) That all real numbers are comparable: i.e. given any real numbers a and b, one of the following must be true: a < b, a = b, a > b That < interacts with + and  according to the wellknown rules. And some basic properties of decimal notation, used in the parts I skipped. No, I'm not done yet: It is possible to define numbers that are greater than 0.99...9 but less than 1. It is done in NonStandard Analysis all the time. But these are not Real numbers. (I am using "Real" here as the name of a set of numbers, not as opposed to "fake" numbers. All numbers are only concepts, and no concept possesses greater reality than any other.) These new "infinitesimal" numbers obey most, but not all, of the standard laws of arithmetic. They cannot be expressed as decimal numbers, and they do not negate the proof above. But they do show the fallacy of simply stating that there can't be any numbers between 0.99... and 1. If you would like to learn more about Nonstandard analysis, infinite numbers, or nadic numbers (someone mentioned them above, or I might have gone on about them myself  they're cool!) look on google. For infinite numbers, I suggest searching for "Transfinite numbers".

« Last Edit: Aug 19^{th}, 2003, 7:27pm by Icarus » 
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Jeremy
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Re: 0.999.
« Reply #84 on: Oct 11^{th}, 2002, 11:44pm » 

... right... how about this. what does .999...  1 = ? Either A) 0 (in which case 1 = .999...) B) something other than 0... i'm thinking most people would say "0.00001", or "0.0000000001"... basically the number of 0's will equal the number of 9's that you took away from the original one. well it's easy to figure out the number of 0's with small values of .999, for example .9  1 = .01 .9999  1 = .00001 ok... well .999... by definition has an infinite 9's after it. well infinity by definition does not stop... it just goes on and on and on and on. if it were to ever stop it just wouldn't go on for infinity. s you'll never get to that one at the end... just lots and lots of 0's... so you'de never reach that ...01. Well so B can't be the case, so lets all go back to A. ok i suck for not using any math to try to prove this, and instead using intuition... but i think the best bet was .999... = x .999... * 10 = x * 10 10x = 9.999... 10x  x = 9.999...  .999... 9x = 9 x = 1 but you know what? everything i just said has already been said before. yeesh why is this topic so long? think we'll ever get to intinity posts?

« Last Edit: Oct 12^{th}, 2002, 12:29am by Jeremy » 
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FenderStratFatMan
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Actually
« Reply #85 on: Oct 16^{th}, 2002, 6:30am » 

I really don't see were you are coming from saying that _____ _____ .99999 = 1 because if .99999 then there is no way that 1 = .99999. Obviously a number isn't = to another number unless it is written in a form of the number. And I sure as heck can't see how .99999 is any way of writing 1.


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TimMann
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Re: 0.999.
« Reply #86 on: Oct 16^{th}, 2002, 9:21am » 

By that reasoning 1/2 is not equal to 2/4 because it's not written in the same form. I.e., that's not an argument at all, just a random wrong statement. Hmm, I sound crabby. Got to stop posting before breakfast.


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Jeremiah Smith
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Re: Actually
« Reply #87 on: Oct 16^{th}, 2002, 10:05am » 

on Oct 16^{th}, 2002, 6:30am, FenderStratFatMan wrote:And I sure as heck can't see how .99999... is any way of writing 1. 
 Perhaps, then, you should consider reading the thread first, as my fellow boardmates have explained it very well, in about twenty different ways. I'm crabby, too.


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GRAND_ADMRL_THUORN
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Re: 0.999.
« Reply #88 on: Oct 29^{th}, 2002, 8:30pm » 

(MY SECOND POST, YAAAAAA! ) unfortunatly, it seems that just about everyone in here is wrong, im surprised most of you intelligent individuals didnt see the answer from the begining .9999..... is GREATER than 1!!! no computer can show you .999.... actually written out you cant actually put the true .999.... on paper its completely impossible but showing 1 is easy and takes up infinitly less space on paper or hard drive/ram so in actuality .999.... is infinitly larger than 1 1x=.999.... (lol, i really wanted to add something to this wonderful thread, even if it was only a joke )


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Icarus
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Re: 0.999.
« Reply #89 on: Oct 29^{th}, 2002, 9:09pm » 

0.99... 1 Which did you say was bigger?


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Marc Nielsen
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Re: 0.999.
« Reply #90 on: Nov 4^{th}, 2002, 9:41am » 

How long can you guys keep arguing wether 0.999....=1 I think that it was pretty clear in one of the first postings. And even if the 1/3+1/3+1/3 = 1 argument doesn't convince you, then there's a much easier way to mathematically prove it. how did you learn to 'round up' (I don't know if that is the English term  I'm not thinking of a ranch ) a decimal figure Notice that the figure 1 isn't 1,00000... Anyways... Only the last two statements are true. That's a fact. I don't know if it's mentioned anywhere in this thread  I'm not about to read all four pages And I'm not going to reveal it right away... You try 'n figure it out (pun intended) regards Marc Nielsen


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Marc Nielsen
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Ooops...
« Reply #91 on: Nov 4^{th}, 2002, 10:54am » 

'Why O Why, didn't I read all 4 pages ' Sorry 'bout that  didn't see the last couple of posts... regards Marc


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Guess Who
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Re: 0.999.
« Reply #92 on: Nov 5^{th}, 2002, 10:37pm » 

let n = a real number, lets not deal with imaginaries yet. if n = real number that is NOT 0 n / 0 would not exist ( n divided by 0 is undefined ) yet, any n / m where m is a number that APPROACHES 0 is positive / negative infinity by definition, as the only undefined rational expression is when the divisor is 0. as any number divided by a very small number becomes a very very large number, and if this number approaches 0, then n / m would approach positive or negative infinity. > means approach E means from the set of R means reals so {mm > 0, m E R} does NOT equal 0 because as a divisor, one produces [approaching infinity] and the other is undefined and since there are 2 different "quotients" (infinity and the crossed out 0 denoting undefinition), the divisors therefore cannot be the same number so if a number that approaches zero is not zero, how can a number that is approaching one be one? to dispel the theory that 1/3 is 0.33333...., do 1 / 3 by long division, there will always be a remainder of 3 that doesnt belong anywhere. Sure, it is really small, and you can divide it into another decimal place, but THE REMAINDER IS ALWAYS THERE; THERE IS NO WAY OF ELIMINATING THIS REMAINDER COMPLETELY. It will always lag behind the 3 in the smallest place. The remainder is the essential difference between 1 / 3 and 0.33333333..... WHICH NEEDS A REMAINDER OF 0.000000........3 before it can = (1/3) , so 1/3 actually does NOT equal 0.33333......., as there will always be something missing one place further down


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Guess Who
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Re: 0.999.
« Reply #93 on: Nov 5^{th}, 2002, 10:44pm » 

oh, and get it through your heads: infinity DOES EXIST in the real world, so you cant ignore it and disregard the 1 or whatever digits that does happen after an " infinity " of 0's, as the 1 would be getting smaller more quickly than you can put 0's on. i mean, hey, theres an infinity of numbers between every single real number, so there is a concept of grasping beyond infinity. ( if 1 < n < 1.0001, then n can be an infinite number of numbers )


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Pietro K.C.
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Re: 0.999.
« Reply #94 on: Nov 6^{th}, 2002, 4:41am » 

AAAAAAARGHHHHHH!!!! I really REALLY try to be nice to the best of my abilities in answering people's doubts and wrong conclusions, because I myself have made more than my fair share of mistakes. But because of the "get it through your heads" comment, which is VERY snobbish, I will go against custom this time. So my answer is this. I won't even try to argue with you, since you have clearly not read a SINGLE post of this thread, and would therefore not have the patience to read the lengthy exposition necessary to clarify your misconceptions. Further, you obviously have no idea about what an expression such as "0.999..." actually MEANS, and are basing your arguments on things which have nothing to do with the question. Enlightening though it was that you pointed out that "infinity exists" (oooh... how clever...), that sort of infinity has no bearing whatsoever on the matter at hand, and I must also disagree that it exists in the "real world" (which is even more irrelevant). Now if you REALLY want to argue about 0.999..., go read some of Kozo Morimoto's posts, which at least are interesting. So there you have it. A post of mine on just four hours' sleep and with a project due in half an hour.


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richi
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Re: 0.999.
« Reply #95 on: Nov 6^{th}, 2002, 5:00am » 

I must say I am surprised by the vast amounts of incorrect proof that ".999...(rep indef) = 1 .999...(rep indef) is a number that approaches 1 as the number of significant figures (9's) approaches infinity, but as the number of significant figures never reachs infinity, .999...(rep indef) never reaches 1.


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S. Owen
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Re: 0.999.
« Reply #96 on: Nov 6^{th}, 2002, 5:41am » 

on Nov 6^{th}, 2002, 5:00am, richi wrote:I must say I am surprised by the vast amounts of incorrect proof that ".999...(rep indef) = 1 .999...(rep indef) is a number that approaches 1 as the number of significant figures (9's) approaches infinity, but as the number of significant figures never reachs infinity, .999...(rep indef) never reaches 1. 
 The point is that "0.999..." does not equal any decimal like "0.999...9" that has a finite number of significant digits. The notation can easily fool one into thinking about "0.999..." as just a special kind of "0.999...9", one with "infinite" digits, and thus something that must be less than 1. That's the point of the riddle  this seemingly reasonable reasoning is wrong. "0.999..." can only mean "the value that "0.999...9" approaches, but never reaches, as the number of digits grows to infinity;" this is 1. The true issue here is not what "0.999..." equals, but what it really means to begin with. Once that is clear  and I think the meaning of "0.999..." can only be what I reiterated above  the value of "0.999..." is unquestionably 1.


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richi
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Re: 0.999.
« Reply #97 on: Nov 6^{th}, 2002, 5:35pm » 

It seems to me that the point has been blatantly missed. To say "is equal to" is not an approximation. The induction is: as the number of significant figures of (lets call it x) increases, x approaches 1. Because, by definition, we can never reach infinity, x will never reach 1.


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S. Owen
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Re: 0.999.
« Reply #98 on: Nov 6^{th}, 2002, 5:47pm » 

"0.999..." is not an approximation of anything. It is, exactly, the limit as the number of significant figures increases, and you rightly agree that this limit is 1. To say that "0.999..." equals 1 is not to say that any number of significant figures ever gets us to 1. It is, again, the limit. As you write more 9s, you never get to "0.999..." (an infinite number of 9s). The value never gets to 1. The key point is that these are saying the same thing! If you still aren't convinced, try this. Let the value of "0.999..." be x. You say that x < 1. But surely you see that for any x < 1, I can create a value y = "0.999...9" with enough 9s so that y > x. So if I reach y with after adding a bunch of 9s, how can adding more 9s get me to x, if x < y?


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Icarus
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Re: 0.999.
« Reply #99 on: Nov 6^{th}, 2002, 8:21pm » 

The definition of decimal notation is this. A decimal sequence {d_{i}}_{i=N}^{[subinfty]} for some integer N [le] 0, is a sequence of digits d_{i} from the set {09}, commonly written (in the USA and some other places  others use a , or a raised . instead of a lowered .) as d_{N}d_{N+1}...d_{1}d_{0}.d_{1}d_{ 2}. . . (I have reversed the signs on the index because it makes the rest of the notation more conventional, though it is a bit confusing here.) This sequence by definition represents the unique real number L such that L = lim_{k[to][subinfty]} [sum]_{i=N}^{k} d_{i}10^{i} Note that L is not any of the partial sums. L is the limit. Now the definition of the limit of a sequence: the limit L of a sequence {a_{i}} is the unique number L satisfying this condition: for every [epsilon]>0, there is an N such that for every i > N, a_{i}  L < [epsilon]. If you don't believe the number L is unique, consult any good calculus text. They all contain the proof that it is. (If the one you picked doesn't, then rest assured that it is NOT a good calculus text!) Or try to prove it yourself. The proof is not hard. Some things to notice here. 1) There is no such decimal notation as 0.9999...(infinitely many 9s)...9 decimal notations are a "shorthand" for decimal sequences, which are in turn functions from the set of all integers [ge] N for some N [le] 0 into the set {0,1,2,3,4,5,6,7,8,9}. There is no integer to map to that last 9. (If you disagree with this, go look up the definition of "Integer".) You are free to define such a notation (say, use ordinals instead of integers for your index). But then it is also up to YOU to define what real number it represents! You are also to free to introduce new numbers to be represented by your new notation. Just don't mistake these for real numbers. (There are many ways of defining the real numbers, but they are all equivalent to this: The set of Real numbers is the smallest topologicallycomplete ordered field.) 2) 0.999... is not the same as 0.999...(finitely many 9s)...9, no matter how many 9s are in that finite sequence. 0.999... is a written representation of the infinite sequence {d_{i}}_{i=0}^{[subinfty]} with d_{0}=0 and d_{i}=9 for all i>0. This sequence is by definition of decimal notation a representation of the number L = [sum]_{i=1}^{[supinfty]} 9*10^{i}. It has been proven many times in this forum that L is the real number 1. 3) A limit is NOT an approximation. The elements of the sequence are the approximations. The limit is the number that is being approximated. This is a fundamental difference that is behind the confusion several respondents have shown. By definition, the decimal sequence 0.999... represents the limit, not any of the approximations. 4) If you don't agree with the results of a proof, don't assume that it is wrong and you are right. If you can't find a fault in its reasoning, you better check YOUR reasoning a little more carefully. 5) To take a limit does not require you to do an infinite amount of anything. If you will look again at the definition of the limit of a sequence, you will notice that there is no mention of infinity anywhere in it (other than saying that this is what the phrase "limit as i goes to [infty]" means).

« Last Edit: Aug 19^{th}, 2003, 7:20pm by Icarus » 
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