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Topic: 0.999. (Read 125647 times) 

kenny
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Posts: 12


Re: 0.999.
« Reply #175 on: Jan 6^{th}, 2003, 12:07pm » 

Wow, there are a lot of posts to this thread. Here's my answer. By saying a number "X = 0.999...", what I mean is "X is the limit of the sequence 0.9, 0.99, 0.999, ..." This can also be expressed as: x_i = 1  10 ^ (i). X is limit as i approaches infinity. Clearly, the x_i get closer and closer to 1 as i approaches infinity. Also, it is clear that for any positive number epsilon there is an N such that all x_i for i > N are closer to 1 than epsilon. (Try N =  log(epsilon)  + 1 (log base 10)). Thus, *by the definition of limit* 1 is a limit of the sequence. It is an interesting fact about sequences that if two numbers are both limits of the sequence, then they must be equal. This has been proven. Thus, X = 1. The only way to argue that X might not be 1 is to use a nonstandard definition of a mathematical limit. Or else to have a different definition of what 0.999... means. I believe my definition above is standard.  Ken


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jim rodriguez
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Re: 0.999.
« Reply #176 on: Jan 28^{th}, 2003, 8:20pm » 

hey math is great and everything, and some of this stuff is kind of interseting, but take it easy guys. heres the answer: 1/3 does not equal 4!!!! whoever said that is a moron! your pal jim Rodriguez cruz


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poo
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Re: 0.999.
« Reply #177 on: Jan 28^{th}, 2003, 8:22pm » 

MATH LEAGUE RULES!!!!!!!!!!!!!!!!!!!! jim


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The math guru
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0.999repeated < 1
« Reply #178 on: Jan 29^{th}, 2003, 5:45pm » 

no matter how you slice it, so to speak, 0.9999999999999999999999999999999999999999999999999999999999999999999999 999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999 etc is less than one, no matter how close it gets to being one.


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S. Owen
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Re: 0.999.
« Reply #179 on: Jan 29^{th}, 2003, 5:48pm » 

No matter how many 9s you type, you will also not reach 0.999... (an infinite number of 9s), so this reasoning does not pertain to the riddle.

« Last Edit: Feb 25^{th}, 2003, 5:07pm by S. Owen » 
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Mira
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Re: 0.999.
« Reply #180 on: Jan 29^{th}, 2003, 8:16pm » 

No matter how long it goes on for 0.9999... will never equal 1, it'll always be 0.0000...0001 away from it. therefore 0.9999...<1


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redPEPPER
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Posts: 160


Re: 0.999.
« Reply #181 on: Jan 30^{th}, 2003, 1:34am » 

No matter how many 9s you add to 0.999... it will never equal 1 as long as you add a finite number of 9. If you have an infinite number of 9 (that's the number 0.99~ we're dealing with here) then you can't have a number 0.00...01 to add to it, because you can't put a 1 at the end of an infinite string of zeroes. An infinite string has no end. An infinite string is not the same as an arbitrarily long string.


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Senor Incognito
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Re: 0.999.
« Reply #182 on: Feb 2^{nd}, 2003, 9:00pm » 

Standard proof: 0.999' = x x10 x10 9.999' = 10x x x 9 = 9x x = 1


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Ryan Sullivan
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Re: 0.999.
« Reply #183 on: Feb 25^{th}, 2003, 4:00pm » 

It's foolish to believe that .99999999999999999(inf) = 1. Using an integer system, and the theories some have provided... is there any number between 8 and 9? No. Therefore, they are equal. It's blatantly untrue, and .99999999inf is NEVER, and will NEVER be 1.


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Icarus
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Re: 0.999.
« Reply #184 on: Feb 25^{th}, 2003, 4:26pm » 

Mr. Sullivan  The difference between 8 and 9 is 1. What is the difference between 0.999... and 1? As well, what is (0.999... + 1)/2 ? Surely this number must lie between 0.999... and 1, where you say no number is (I say it too, but only because nothing is between a number and itself). The oft repeated demonstration, whose last occurence immediately precedes your post, already shows that you would have us abandon the rules which everyone knows for manipulating decimal expressions, just so you can demand 0.999... and 1 are different numbers. Do you have any clue just how many properties of real numbers you would have us give up? Goodbye the zeroproduct property! So long to the fundamental theorems of algebra and calculus! Adieu to the existance of multiplicative inverses! But, we gain this: If two numbers aren't written out the same as decimals, then they must be different! That certainly is an improvement over all those silly useful theorems and properties isn't it?

« Last Edit: Feb 25^{th}, 2003, 4:44pm by Icarus » 
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S. Owen
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Re: 0.999.
« Reply #185 on: Feb 25^{th}, 2003, 5:18pm » 

on Feb 25^{th}, 2003, 4:00pm, Ryan Sullivan wrote:It's foolish to believe that .99999999999999999(inf) = 1. Using an integer system, and the theories some have provided... is there any number between 8 and 9? No. Therefore, they are equal. It's blatantly untrue, and .99999999inf is NEVER, and will NEVER be 1. 
 But of course, the original riddle deals with reals, not integers, so this "counterexample" does not apply to the given reasoning. I'd like to repeat... If 0.999... < 1, then you must be able to find at least one real number that is between them. When you can find that, I'll believe you.


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otter
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Re: 0.999...
« Reply #186 on: Feb 28^{th}, 2003, 11:57am » 

on Jan 30^{th}, 2003, 1:34am, redPEPPER wrote:No matter how many 9s you add to 0.999... it will never equal 1 as long as you add a finite number of 9. If you have an infinite number of 9 (that's the number 0.99~ we're dealing with here) then you can't have a number 0.00...01 to add to it, because you can't put a 1 at the end of an infinite string of zeroes. An infinite string has no end. An infinite string is not the same as an arbitrarily long string. 
 Perhaps a restatement of redPEPPER's message is the best answer. Definition: 0.999... = the sum of the infinite series 9/10 + 9/100 + 9/1000 ... (i.e. an infinite number of 9's) Let us assume that 0.999... and 1 are different. Then there must be some number x equal to 10.999... The value of x must be zero. Any value of x other than zero must be the result of an operation on a pair of finite numbers (as redPEPPER indicates, you can't put a 1 at the end of an infinite number of zeroes) which invalidates the original assumption (remember we defined 0.999... to be an infinite series) and thus the assumption is false. Therefore 0.999... and 1 are, in fact, the same. QED


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aero_guy
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Re: 0.999.
« Reply #187 on: Feb 28^{th}, 2003, 5:53pm » 

Have you all noticed that there is a critical length to these posts of about two pages? After that people don't feel like reading the entire thread before adding whatever "insight" they have come up with. Of course, by the time the thread is that long odds are their idea was already hashed over. Or, in the case of an 8 page thread like this, hashed over, rejected, reasserted, derided, lauded, thrown out with the garbage, dragged back in, and then finally buried in the back yard. Is this currently the longest thread going?


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wowbagger
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Re: 0.999.
« Reply #188 on: Mar 3^{rd}, 2003, 3:34am » 

on Feb 28^{th}, 2003, 5:53pm, aero_guy wrote:Have you all noticed that there is a critical length to these posts of about two pages? After that people don't feel like reading the entire thread before adding whatever "insight" they have come up with. 
 Well, I do  if I have the time, that is. Quote:Is this currently the longest thread going? 
 Yes, indeed!


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udippel
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Re: 0.999.
« Reply #189 on: Mar 3^{rd}, 2003, 8:58am » 

okay; so why not simply freeze it ?? It seems everything has been said; any possible explanation, any possible wrong explanation, any possible correct explanation, anything at all. Can William not  like an auctioneer  321(sound of a hammer)closed ??


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Jeremiah Smith
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Re: 0.999.
« Reply #190 on: Mar 3^{rd}, 2003, 9:37am » 

Because if he closes it, it will eventually drop off the front of the forums list since it won't get updated anymore. And since new people are always coming, we would, eventually, get a "help me with .99999... = 1 probleme!!!!!!!!!111" thread, and this would all start again.

« Last Edit: Mar 3^{rd}, 2003, 10:09am by Jeremiah Smith » 
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Icarus
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Re: 0.999.
« Reply #191 on: Mar 3^{rd}, 2003, 5:10pm » 

on Mar 3^{rd}, 2003, 9:37am, Jeremiah Smith wrote:Because if he closes it, it will eventually drop off the front of the forums list since it won't get updated anymore. And since new people are always coming, we would, eventually, get a "help me with .99999... = 1 probleme!!!!!!!!!111" thread, and this would all start again. 
 Maybe that would be a good thing. Most people who have never come across this aspect of decimal notation before have quite normal confusions on the answer. But when they look at this humongous thread, it seems to confuse them even more. Witness the post from Jim Rodriguez! (I went looking, but never figured out why he thought someone said 1/3 = 4 ) Perhaps a fresh discussion, without the interference of those of us with towering egos , would allow them to thresh it out in their own minds!


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David Ryan
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Re: 0.999.
« Reply #192 on: Mar 4^{th}, 2003, 10:20am » 

3 things. 1) 1.000...01 does not exist. The ... implies and infinite number of zeroes, hence, no end. the .01 implies an end, and so the very concept is paradoxical. 2) .###... cannot be represented as .95, because that is 2 digits to represent one. hence, .# + .0# does not equal .95 + .095. 3) Infinity is not a number, but a concept.


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aero_guy
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Re: 0.999.
« Reply #193 on: Mar 4^{th}, 2003, 11:07am » 

Will could make it a sticky locked topic. That way it is closed but stays on page one. As to inifinity being a concept not a number, don't get these guys started on the different types of infinity!


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Alex Wright
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Re: 0.999.
« Reply #194 on: Mar 4^{th}, 2003, 1:55pm » 

My grade 10 IB math class explored this one day in class. My Teacher showed us this proof: 9.999...=10 9.999...X=10X multiply both by 10: 99.999...X=100X Subtract original values: 99.999...X=100X 9.999...X=10X you end up with: 90X=90X Thus: X=X :. 9.999... = 10 exactly.


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Kozo Morimoto
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Posts: 114


Re: 0.999.
« Reply #195 on: Mar 4^{th}, 2003, 3:29pm » 

on Mar 4^{th}, 2003, 10:20am, David Ryan wrote:1) 1.000...01 does not exist. The ... implies and infinite number of zeroes, hence, no end. the .01 implies an end, and so the very concept is paradoxical. 
 Why can't it exist? Why is it paradoxical? I posed this question on this thread before but people just ignroed it. If you have a skipping rope and you hold one end in the left hand and the other end in the right hand, and then you get your friend to take the middle of the rope and start walking away from you. At infinity, don't you get 1.000...0001? Both ends are bound yet its infinite in the middle? OR, at infinity, you end up with 2 ropes and not 1?


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S. Owen
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Re: 0.999.
« Reply #196 on: Mar 4^{th}, 2003, 3:53pm » 

I don't totally see the analogy with the jumprope, but I would note that the friend will always be a finite distance away, and is never "at infinity"... "infinity" is a place that the friend can go towards, but can never reach. I think the trick is that we all well understand what "1.0001" means, and can easily extend our reasoning to understand "1.000...01 (a million 0's omitted)", so it's natural to extend this and think we know what "1.000...01 (an infinite number of 0's omitted)" means. The latter looks reasonable enough, but what is it really denoting? It seems to denote a finite and infinite number of 0's at the same time. That's the real issue  not the math at all, but what the notation means. You can probably only reasonably interpret "1.000...001 (infinite 0's)" as "the value that it gets closer to as the number of 0's increases"  it's a limit, and the limit is 1.


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Icarus
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Re: 0.999.
« Reply #197 on: Mar 4^{th}, 2003, 4:17pm » 

on Mar 4^{th}, 2003, 10:20am, David Ryan wrote:3 things. 1) 1.000...01 does not exist. The ... implies and infinite number of zeroes, hence, no end. the .01 implies an end, and so the very concept is paradoxical. 2) .###... cannot be represented as .95, because that is 2 digits to represent one. hence, .# + .0# does not equal .95 + .095. 3) Infinity is not a number, but a concept. 
 3 replies: 1) Actually, 1.00...01, with the ... representing infinitely many zeros is not paradoxical, and can be defined (if you were to search through the posts, you will find one where I did so). However, depending on how you define it, you either get yet another representation for the number 1, so why bother? Or you get some other real number, but have to give up any sensible rules for manipulating decimals. Or, you get something that is not a REAL number, but is part a larger set of "numbers". In any case, they have no bearing on the question of whether 0.999... = 1. 2) You can define a decimaltype representation for real numbers which includes an 11th digit # = to .95 (or equal to any other number you choose, including one of the other digits). The value of any such pseudodecimal expression is defined the same way: if x = .d_{1}d_{2}d_{3}... where the d_{i} are the digits of the notation, then x = [sum]_{i=1}^{[supinfty]} d_{i}10^{i} What you get by doing so is multiple pseudodecimal expressions for every nonzero real number, instead of the 1 or 2 supplied by ordinary decimals. 3) Infinity can be a number (look again at my posts). Just not a REAL number, or a COMPLEX number. on Mar 4^{th}, 2003, 11:07am, aero_guy wrote:As to inifinity being a concept not a number, don't get these guys started on the different types of infinity! 
 Too late! The mad mathematician strikes again! on Mar 4^{th}, 2003, 1:55pm, Alex Wright wrote:My grade 10 IB math class explored this one day in class. My Teacher showed us this proof: 9.999...=10 9.999...X=10X multiply both by 10: 99.999...X=100X Subtract original values: 99.999...X=100X 9.999...X=10X you end up with: 90X=90X Thus: X=X :. 9.999... = 10 exactly. 
 ACK! I very deeply hope that this is a case of you not understanding what your teacher was saying, because the only alternative is that your teacher very deeply deserves to be fired, and is currently destroying any hope of mathematical understanding in all of his or her students! What you have given here proves nothing whatsoever. You start off by assuming the result you intend to prove. You then prove a trivial result, and because of it believe your assumption was true?? I could use the same techniques to prove any statement you want: 5=21 Proof: assume 5=21. Then 5x0=21x0 so 0=0 true. Therefore 5=21! The actual proof you are aiming at has been reproduced on this thread many times. The last one, by Senor Incognito, is here, but a better exposition of it is the one here by SPAZ.

« Last Edit: Aug 19^{th}, 2003, 8:55pm by Icarus » 
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Icarus
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Re: 0.999.
« Reply #198 on: Mar 4^{th}, 2003, 5:27pm » 

on Mar 4^{th}, 2003, 3:29pm, Kozo Morimoto wrote:Why can't it exist? Why is it paradoxical? I posed this question on this thread before but people just ignored it. 
 As I said to Mr. Ryan, I did not ignore it. Yes, you can define expressions of the form 1.000...01. But S.Owen has it right  the real crux of the matter is: what do you mean by such an expression? Just because you can define 1.000...01 does not mean there is real number out there that it represents. If you claim it is a real number, then YOU have to define what number it is. The definition for ordinary decimals is as I gave in the previous post, but that definition does not apply here. You never did this. In your posts you merely assumed that if you could conceive of such a thing, then it must be a real number, but you never explained how. If you were to try to come up with a mathematically valid definition for which real number it is, you would discover that one of the following two situations held for your definition: 1) That the transinfinite decimals have no effect on the value of the number: 1.000...01 = 1.000...9329489024 = 1 2) That the basic rules for manipulating decimals do not apply in any reasonable fashion to your decimals. The other possibility is that you accept transinfinite decimals as defining new numbers, which are not REAL. Then you get all sorts of fascinating behaviour, but you still discover one thing: since you are not allowed to redefine the current decimals, even with such numbers, 0.999... = 1. In this extended set of numbers, 1  0.000...1 = 0.999...9 != 0.999... (All the ... here represent an infinite repetition of the preceding digit. Part of what you lose with this extension of the REALs are the algebraic properties of inequalities, so yes 1  0.000...1 > 1). Alas, no matter how you look at it, the argument falls up short of disproving 0.999... = 1.


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Janet
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Re: 0.999.
« Reply #199 on: Mar 4^{th}, 2003, 5:55pm » 

If 0.999...=1 because there is no space in between the two numbers, would 1.999...=2, or 4.5999...=4.6? What about 4999.999...=5000 Or is "0.999...=1" a special case, possibly a loophole of our number system?


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