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Topic: 0.999... (Read 31970 times) 

Vincent Lascaux
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Hi, I'm from France. We actually use semi colon to separate numbers in coordinate, sets... For example, we would say that the coordinate of a point are (0,5;0,6), where you would write (0.5,0.6). Actually, I'm so used to see (0.5,0.6) on computers that its odd to see (0,5;0,6), but on paper it looks ok


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Jetru
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Wow, great discussion. .999... IS 1 .999... IS THE SAME AS 1 .999... = 1 Alright tell me a number between .999... and 1. Oops. No such number. You can't say .999... with an EXTRA 9 because .999... already has infinite nines. Even if you put an extra nine, it's the same number. 1.9 =.1 1.99=0.01 therefore, 1.99... =0.0000...1 there is NO such number as '0.0000...1'. You DO NOT even get to the terminating 1, because there are INFINITE 0's. Therefore, 0.0000...1 IS 0. Anyway, .999... is NOT a real number, because there are infinite nines.Infinite itself is not 'real'.


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towr
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Re: 0.999...
« Reply #52 on: Mar 29^{th}, 2005, 6:44am » 
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on Mar 29^{th}, 2005, 5:44am, Jetru wrote:Anyway, .999... is NOT a real number, because there are infinite nines.Infinite itself is not 'real'. 
 It is a real number, because it is 1, and one is a real number. Every number equal to a real number is real. Altough you might maintain that nothing is real, and thus not 1 nor 0.999... nor anything. But aside from that it is, and infinity is just as real as anything else, especially in mathematics. Heck, even imaginary numbers are real in mathematics


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Grimbal
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Re: 0.999...
« Reply #53 on: Mar 30^{th}, 2005, 3:56am » 
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on Mar 29^{th}, 2005, 5:44am, Jetru wrote: 1.9 =.1 1.99=0.01 therefore, 1.99... =0.0000...1 there is NO such number as '0.0000...1'. You DO NOT even get to the terminating 1, because there are INFINITE 0's. 
 But then, where did the extra 1 go? It's not because it is too far to see that it does not exist. PS: it's a joke. Of course the diff becomes infinitely small, which is zero.


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Sjoerd Job Postmus
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Re: 0.999...
« Reply #54 on: Mar 30^{th}, 2005, 4:26am » 
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If I were to describe this as an infinite sum, I'd use: Probably not too much help... Just in case it might give you something to work with. You've probably thought about it already, I guess... anyway


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Icarus
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Re: 0.999...
« Reply #55 on: Mar 30^{th}, 2005, 2:56pm » 
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It's in the proofs post at the start (with 10^{n} instead of 0.1^{n}). But you can't read it right now because the mathematical symbolry is currently disabled. Hopefully William will be able to get the symbolry back up sometime soon.


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Jetru
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I think this thread should be locked. The first few posts make it clear that .999...=1. If people don't understand it, then they should learn maths, then think about it again.


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The_Godfather
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Just a question, but if there is a contradiction or problem in math with many solutions based on opinion, views, or moral standpoints, what makes that a riddle?


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Icarus
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Re: 0.999...
« Reply #58 on: Mar 31^{st}, 2005, 6:35pm » 
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1) There is no contradiction here. 2) There is only one solution. Its sole truth is unaffected by opinion, view, or moral standpoints. It is a direct consequence of the definition. 3) It is a truth that many people have trouble finding or understanding, because their past experience has led them to make false assumptions. 4) Point (3) is the most common basis for riddles. Most "riddles" rely on leading the person to make false assumptions. All those riddles you've recently posted in the easy forumare like this. 5) This forum and site does not restrict itself to "riddles" as you might think of them, but allows more general problems with some aspect that makes them interesting.


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ricecakeboy
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Re: 0.999...
« Reply #59 on: Apr 27^{th}, 2005, 12:19am » 
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why is everyone arguing over this ... this is the way u proof repeating decimals anyways its like arguing over why 0.333.... = 1/3 0.666... = 1/6 0.999... = 1/1 i dont see why 0.999... is so special to have over 20 pages http://en.wikipedia.org/wiki/Recurring_decimal


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towr
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Re: 0.999...
« Reply #60 on: Apr 27^{th}, 2005, 1:46am » 
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You're right, it's terribly easy. Yet many people simply can't believe it, regardless of the number of excellent proofs they're confronted with. (btw you mistakenly typed , 0.666... = 1/6, instead of 2/3 )


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Deedlit
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Re: 0.999...
« Reply #61 on: Apr 27^{th}, 2005, 8:00pm » 
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It was pointed out in a math FAQ somewhere that the "trivial" proofs of .99999... = 1 hid some important issues  namely, what an infinite decimal really means, and how you can justify using the same operations and theorems on them as you do on finite decimals. Of course, we really mean that .a_{1}a_{2}... = [sum] a_{i} 10^{i}, where an infinite sum is defined as the limit of partial sums, etc. Then, showing that the things we do with infinite decimals are justified requires a bit of work. Since showing 1/3 = [sum] 3 * 10^{i} isn't any easier than showing 1 = [sum] 9 * 10^{i}, the above proof doesn't really shorten anything. Similarly for the 10x  x = 9 approach  you're subtracting two infinite sums, and you need to know that's justified when the two sums are absolutely convergent. So the easiest rigorous proof is probably just straghtforwardly showing the limit is 1. On the other hand, it's easier to give the naive proofs to doubters, if it will shut them up! Odds are they will tune out if you try to explain limits to them.


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SMQ
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Re: 0.999...
« Reply #62 on: Apr 28^{th}, 2005, 5:32am » 
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on Apr 27^{th}, 2005, 8:00pm, Deedlit wrote:Since showing 1/3 = [sum] 3 * 10^{i} isn't any easier than showing 1 = [sum] 9 * 10^{i}, the above proof doesn't really shorten anything.. 
 Maybe not in full rigor, but demonstrating the reverse  that ^{1}/_{3} = [sum]3*10^{i}  is relatively simple through long division. In fact, couldn't you construct a rigorous inductive proof based on the long division algorithm that, while less elegant than the delta proof given somewhere above, would be intuitively easier to grasp by someone unfamiliar with infinite series? SMQ


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Icarus
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Re: 0.999...
« Reply #63 on: Apr 28^{th}, 2005, 6:37pm » 
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It might be easier to understand, but it still would require the same delta style argument in the end. On of the proofs given in my summary (suggested by Sir Col) is essentially exactly this sort of longdivision based proof. But there is a conceptual leap between saying that 1 = 3(.33...33) + .00...01 and saying that 1 = 3(.333...) that most people who are bothered by this problem are unable to overcome.


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Deedlit
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Re: 0.999...
« Reply #64 on: Apr 28^{th}, 2005, 8:04pm » 
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As Icarus says, technically you need the epsilondelta proof as that's how limits are defined; still, it's commonplace to just assert sequences go to zero when they obviously do, so perhaps we can be a little casual here. But it seems that the rigorous justification of the long division will justify 1/1 = .99999... just as easily (instead of requiring the remainders to be less than 10^{k}, make them less than or equal to 10^{k}). So there's still no advantage to proving 1/3 = .333... first. Of course, a lot of people will probably see one as rock solid and the other as suspicious, so again it may well be a more convincing argument.


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Icarus
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Re: 0.999...
« Reply #65 on: Apr 29^{th}, 2005, 3:11pm » 
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Unfortunately, the problem people have had is exactly where you are suggesting we be casual. These people either do not accept that the sequence goes to zero (because they do not understand the concept of a sequence, having never been exposed to it before). Or else, they do not understand the relationship between the various finite decimals 0.999...9 and the infinite 0.999..., so the fact that the sequence (10.999...9) goes to zero is immaterial to the value of 0.999... to their mind. This is why you can't really convince someone of this until they have a good understanding of the notation. And once they do have it, the value of 0.999... is trivial.


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SMQ
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Re: 0.999...
« Reply #66 on: Apr 30^{th}, 2005, 7:21am » 
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Which is why I thought going through 1/3 = .333... might be useful: Most people, because of their experience with long division, seem ready to accept that 1/3 = .333... with a neverending series of 3's. Many people can, from there, just multiply both sides by 3 to get 3/3 = 1 = .999... , but these people aren't the one's we're concerned with (and that step did indeed hide an important detail of the multiplication) I was thinking that with a little reflection, at least some of the remaining people might be able to agree with the mathematical conjecture that 1/3 = .333... = .3 + .03 + .003 + ... = sum from i = 1 to infinity of 3*10^{i}. Now, strictly mathematically, 3*1/3 = 3/3 = 1 = 3*(sum from i = 1 to [infty] of 3*10^{i}) = sum from i = 1 to [infty] of 3*3*10^{i} = sum from i = 1 to [infty] of 9*10^{i} = .999... But maybe those who would accept that line of reasoning are already convinced by other demonstrations. SMQ

« Last Edit: Feb 26^{th}, 2006, 2:30pm by Icarus » 
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Whiteshadows
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you people will believe anything these days won't you? you are all narrow minded and anything that is put on your plate, you will eat it. even if it is poop which is everyone who believes .9999... is equal to 1. so you narrowminded guys think this.... 1/3=.3333... 2/3=.6666... 3/3=.9999... and since 3/3=1, therefore 1=.99999... that is the most ridiculous bull crap i have ever heard. we use .3333 etc. as a decimal because there is no real decimal to describe thirds. it is impossible to describe thirds in decimals becease 10/3 just doesn't end. saying .9999 is equal to 1 is like saying 9=10 or 99=100 or 1000=999. IT JUST DOESN'T WORK THAT WAY YOU IDIOTS. you guys are freaking stupid to believe that .9999=1. it doesn't. 1 is only equal to .9999999 if you add .0000001 to it. which we are not so sal;fjdsl;fjkdsa;flkj. man you guys are so stupid that if your brains were dynamite then there wouldn't be enough to even blow your nose. that is if you think .9999... is equal to 1. phew! i am glad i got that out of my system. good luck trying to deny me. you can go ahead and break a couple of sweats trying to disprove me if you want but you might as well give up. Blackshadows out.


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Icarus
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Re: 0.999...
« Reply #68 on: Oct 2^{nd}, 2005, 7:51pm » 
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From the Misconceptions list: on May 1^{st}, 2004, 7:20pm, Icarus wrote: You know, it really funny when someone like you comes around pronouncing how smart you are and insulting every one else, while at the same time spouting utter nonsense in your actual reply. At least the guy whose post I linked to as an example knew what he was talking about. You are completely clueless! For the record, since it is not clear that you even know what it is we are talking about: By 0.999..., we do not mean 0.999 or 0.999999 or the like. The three dots indicate that the sequence of 9s continues forever. This is more often denoted by placing a bar over the final 9, but that convention is not available in this forum. So we use an ellipsis (the three dots) instead to indicate the same thing. Because the 9s continue on forever, the behavior of 0.999... differs significantly from that of 0.999 or 0.999999 or any finite number of decimal places. The difference is why 0.999... = 1 while all the finite versions fall short. I am not going to try and prove it to you. If William ever gets the math symbolry repaired, then you can read several proofs in the second post of this thread. But you would still reject out of hand the ones simple enough for you to have some hope of following (as you already have one of them). So instead, if you are still in school, I strongly suggest asking your math teacher. If you are not in school, I suggest calling the math department of your nearest university. Ask them, and they will tell you that 0.999... = 1.


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Grimbal
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Re: 0.999...
« Reply #69 on: Oct 3^{rd}, 2005, 1:00am » 
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on Oct 2^{nd}, 2005, 6:28pm, Whiteshadows wrote: lol! The kind of post that tells nothing about the subject at hand, but a lot about the poster.


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Whiteshadows
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Please explain to me where i said that i was a whole lot smarter than you? that i am positive i didn't say. and i was just not caring enough for the triple period because there is no such thing as an infinite number of nines. something always ends somewhere, except for pi. who taught you math anyway icarus? your mom? your two year old sister? your threats are so scary i almost peed my pants. or died from laughter of the stupidity of your comments. i will let that tingle in your brain MR. Always right. and, since i am done ridiculing you, it doesn't even matter how many nines there are, that just means the difference is that much smaller even though it may be an extremely small difference, there is still a difference. and for grimbal, what in the world does me saying,"whatever" have to do with anything? just wondering is all. blackshadows out.


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towr
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Re: 0.999...
« Reply #71 on: Oct 6^{th}, 2005, 4:16am » 
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on Oct 5^{th}, 2005, 9:08pm, Whiteshadows wrote:and i was just not caring enough for the triple period because there is no such thing as an infinite number of nines. 
 I think you are confusing reality and mathematics. In mathematics there is such a thing as an infinite number of nines following a decimal point. In reality, well, arguably; but noone will write it down. Quote:something always ends somewhere, except for pi. 
 And how about the average of 1 and pi? Or simply 1+pi, or 2+pi etc Rather peculiar if there was only one such number wouldn't it? There are in fact infinitely many (and not just ones you get by taking the average of pi with 1,2,3,.. either; try sqrt(2) for example) Quote:who taught you math anyway icarus? your mom? your two year old sister? 
 He has a degree from University, as well as several decades experience. But he's not one to brag about his credentials. Quote:it doesn't even matter how many nines there are, that just means the difference is that much smaller even though it may be an extremely small difference, there is still a difference. 
 An infinitely small difference is not a difference. If 0.999... is smaller than 1, then the average of 0.999... and 1 must lie in between the two, and be distinct from both. I'll personally nominate you for the nobel prize for mathematics if you tell me what number that is

« Last Edit: Oct 6^{th}, 2005, 4:28am by towr » 
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BNC
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Re: 0.999...
« Reply #72 on: Oct 6^{th}, 2005, 7:01am » 
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on Oct 5^{th}, 2005, 9:08pm, Whiteshadows wrote: who taught you math anyway icarus? 
 LOL... Asking this of Icarus of all people...


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Icarus
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Re: 0.999...
« Reply #73 on: Oct 6^{th}, 2005, 3:34pm » 
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on Oct 5^{th}, 2005, 9:08pm, Whiteshadows wrote: Please explain to me where i said that i was a whole lot smarter than you? 
 Okay! How about here: on Oct 2^{nd}, 2005, 6:28pm, Whiteshadows wrote:man you guys are so stupid that if your brains were dynamite then there wouldn't be enough to even blow your nose 
 on Oct 5^{th}, 2005, 9:08pm, Whiteshadows wrote:and i was just not caring enough for the triple period because there is no such thing as an infinite number of nines. something always ends somewhere, except for pi. 
 As towr has indicated, didn't it ever occur to you that pi might not be such a special number (at least in that aspect)? It is quite true that you cannot write down an infinite number of digits, not even for pi. However the concept of decimal expressions containing an infinite number of digits has long been a part of mathematics. Indeed, mathematicians generally consider all decimal expressions to contain an infinite number of digits, even if they are not written down (in the case of terminating decimals, they are considered to end in repeating zeros). For repeating decimals we even have a notation that denotes exactly this: overlining the block of digits that repeats indefinitely. These expressions are welldefined, meaning that there is a unique real number associated with each (but this does not mean that each real number has a unique decimal expression ). on Oct 5^{th}, 2005, 9:08pm, Whiteshadows wrote:who taught you math anyway icarus? 
 I earned my PhD in mathematics from Wichita State University in 1991. on Oct 5^{th}, 2005, 9:08pm, Whiteshadows wrote:your threats are so scary i almost peed my pants. 
 I am sorry to hear about your problems with incontinence, but you will have to point out where I made any threats. Rereading my post, I find nothing that even slightly resembles a threat. Unless you count my telling you that if you would just ask anyone you know who, unlike you, is educated in mathematics, they would confirm that 0.999... = 1. I strongly suggest you try it, just as soon as you put on a fresh diaper. on Oct 5^{th}, 2005, 9:08pm, Whiteshadows wrote:i will let that tingle in your brain MR. Always right. 
 Not always, but definitely so here. on Oct 5^{th}, 2005, 9:08pm, Whiteshadows wrote:it doesn't even matter how many nines there are, that just means the difference is that much smaller even though it may be an extremely small difference, there is still a difference. 
 And what nonnegative numbers are so small that they are less than any positive number? Let x = 1  0.999..., and you can show that for any h > 0, h is also greater than x. In particular, let n be an integer > 1/h (that such an n exists is guaranteed by the Archimedean principle). Then 10^{n} > n > 1/h, so h > 1/10^{n} = 10^{n}. Therefore 1  10^{n} > 1  h. But 1  10^{n} = 0.999...(n1 9s total)...9998 < 0.999...(infinite 9s)... . So 0.999... > 1  10^{n} > 1  h. This gives us that x = 1  0.999... < 10^{n} < h So the difference is between 1 and 0.999... is smaller than any number above zero. It cannot be anything other than zero itself. That is one of the many proofs that 1 = 0.999...

« Last Edit: Oct 6^{th}, 2005, 3:37pm by Icarus » 
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rmsgrey
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Re: 0.999...
« Reply #74 on: Oct 7^{th}, 2005, 12:52pm » 
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on Oct 2^{nd}, 2005, 6:28pm, Whiteshadows wrote:good luck trying to deny me. you can go ahead and break a couple of sweats trying to disprove me if you want but you might as well give up. 
 I deny you. No that that's out of the way, could I just ask which part of your post is meant to constitute a proof of anything for us to try and disprove? You make a number of unsupported assertions, such as: on Oct 2^{nd}, 2005, 6:28pm, Whiteshadows wrote:IT JUST DOESN'T WORK THAT WAY YOU IDIOTS. 
 but never offer an actual proof. If necessary, Icarus could produce an answer to any question of "why is that assertion true" for any of his assertions in proving that 0.999...=1  ultimately, being reduced to "because that's just the way we choose to assume things to be"  at which point, you're free to take exception with any of the axioms that reflect several thousand years of mathematical thought (in fact, nonEuclidean geometry requires you to discard an axiom that was used for geometry for a couple of thousand years) or choose to replace any of the definitions or terminology or notation used that you don't like, though you should really explain where you choose to use nonstandard terms, and then you shouldn't be surprised if your results don't apply to the standard terminology... Can you provide any proof that "it just doesn't work that way" or are we to take it as an additional axiom  in which case I should point out that it's inconsistent with the usual sets of axioms (and would enable us to prove that you are the Pope)


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