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Topic: 0.999. (Read 105934 times) 

pythagoras
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Re: 0.999.
« Reply #50 on: Aug 12^{th}, 2002, 1:17pm » 

infinity + 1? As said earlier, infinity needs to be thought of differently than numbers. It is generally accepted that ... = infinity  1 = infinity = infinity + 1 = ... I can't quite follow NoYes's logic, but when we "reach" the limit we have "reached" infinity. Neither of these can actually be done, and since infinity is really the same as infinity + 1, there is no real problem. A.M. There are bigger consequences of treating infinity as a number. Say S = 1 + 2 + 4 + 8 + ... S = 1 + 2(1 + 2 + 4 + ...) S = 1 + 2S S = 1 S = 1


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heywood
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Re: 0.999.
« Reply #51 on: Aug 12^{th}, 2002, 1:40pm » 

Heres another example/riddle dealing with infinite sums that I find quite amusing. Suppose you have a perfect superball that bounces half its height on every bounce. Disregard all friction and other sources of energy loss (i.e. if the ball starts at 1 meter then the next bounce it will reach EXACTLY 1/2 a meter.) Now the question is, how long does it take the ball to stop bouncing? This was a great problem from an early calculus class I had.


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HaPpY
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Re: 0.999.
« Reply #52 on: Aug 12^{th}, 2002, 2:22pm » 

as n approaches infinity where .999... n is the number of digits past the decimal the value approaches 1. therefore if infinity is "reached" the value becomes one.


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Paul Hsieh
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Re: 0.999.
« Reply #53 on: Aug 12^{th}, 2002, 4:25pm » 

Grumble ... 1. You cannot "set n = infinity" for the simple reason that "infinity" is not a number. Its a concept. 2. One cannot so easily just "take the limit as n goes to infinity" unless you have established each step very carefully. It turns out in this trivial case, there are no issues and doing so presents no problems, but it is in a sense a circular argument. I.e., can't do it until you've established a convergence, which is really part of what we are trying to do in the first place. I did not mean that: (1a^{n}) = (1a) * (1+a+a^{2}+...+a^{n1}) Implies that: 0.9999 ... = 9/10*(1.1111 ...) These are actually two seperate and independent facts that are just true on their face. I tie the two together in later steps in my analysis.


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Mongolian_Beef
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Re: 0.999.
« Reply #54 on: Aug 13^{th}, 2002, 10:02pm » 

I think the issue really is our system and the conversions that it entails. The point is that the way our system functions, as many of you seem to be arguing in its context, .9999999 repeating is equivalent to 1. Arguing over it also appears pointless because regardless of opinion the variance between .99999999 repeating and 1 is negligible in any real world context.


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chris mallinson
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Re: 0.999.
« Reply #55 on: Aug 22^{nd}, 2002, 9:41pm » 

It has been said here that .99999....... multiplied by 10 is 9.999999999.... .99999....... plus 9 is also 9.999999999.... Is there something to be said for the fact that the latter will theoretically approach the number 10 faster than the former? chris


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Jason Short
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Re: 0.999.
« Reply #56 on: Aug 22^{nd}, 2002, 11:42pm » 

The notation I've seen for 0.999... (infinitely repeating) is to use 0.9 with a line over the 9. Instead, I'll put the line under the 9 so it becomes 0.9. If we switched from base 10 to base 20, then we could call our digits {0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I,J}. Thus "J" is the digit "19". This may be comparable to the # digit spoken of earlier: the average of 0.9_{10} and 1 is 0.J_{20}. So the number 0.9_{10} could be compared to 0.J_{20}, and we would be tempted to think that the latter was larger. But we are implicitly assuming here that they are different; in actuality both are simply equal to 1. Here is a related topic: a paradox I once heard. Imagine a game of chance in which a coin is flipped indefinitely until eventually it lands tails. You then count the number of times it landed heads (n), and the person playing the game gets 2^{n} dollars. The question then becomes: how much money does a player expect to win playing this game? Logically, we would think this is ((1/2)*1+(1/4)*2+(1/8)*4+...) since there is (for instance) a 1/2 chance that heads will be flipped 0 times, and if this happens the player will win $1. But each term simplifies to 1/2, so the series becomes (1/2+1/2+1/2+...). The infinite series of 1/2's seems to sum to infinity: it seems as though the payoff for this game is infinite. Yet this seems rather rediculous; in fact we've assumed that there is an "expected" payoff. The actuality is that the expected payoff is undefined. My point is that infinity isn't just a rather large number, it is a concept that has undefined value. There is no such thing as "infinity plus one" (^^). You simply cannot treat infinity as a value. Thus it is meaningless (although seemingly clever) to consider the value 0.99, and claim that it is larger than 0.9. In this case, assuming the given decimal representation has a value (which is sort of implicitly true, since decimal notation is simply an abstraction we made up to represent values), the best we can do is treat it as an infinite series and reduce that series to find the value. This has been done several times elsewhere: the value is 1. By contrast, the value 9 is "infinite", and certainly does not have a value. It would be quite meaningless to compare it to, for instance, 8, and try to say that it is "larger" since 9>8. 9>8 does not prevent the fact that 88>9.


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wowbagger
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Re: 0.999.
« Reply #57 on: Aug 24^{th}, 2002, 4:16pm » 

on Jul 30^{th}, 2002, 1:52am, Kozo Morimoto wrote: [...] So can you start with any finite number and keep dividing it by 10 and add 0.9 to it, and if you repeat the process to infinity you get 1? [...] 
 As far as I followed the discussion, nobody replied to this specific question yet, so here's my proof that yes, you get 1: What you do is calculate a sequence of numbers generated by the recursion formula a_{n+1} = a_n / 10 + 0.9, starting with some arbitrary a_0 (e.g. 0.8) If you "repeat the process to infinity", you take the limit of both sides of the recursion equation as n goes to infinity. If the limit of the sequence a_n exists  let's call it A  you can substitute a_n and a_{n+1} by A, since being elements of the same sequence, they must have the same limit A. So we have A = A / 10 + 0.9 9 / 10 A = 0.9 A = 1 Note that this result doesn't depend on your initial value a_0, it works for all real a_0. If you start with a_0 > 1, you will approach the limit from above, however. In the same way one can show that a_{n+1} = (a_n + 2 / a_n) / 2 goes to sqrt(2), i.e. A*A = 2 in this case. Well, I don't know whether this is of much help as some will probably feel uncomfortable with the substitution of the limit A, but I didn't want to leave the question unanswered. Kozo: I really respect your being sceptical. BTW, I'm convinced (not to say I know) the infinite series does sum up to 1. The bit about the "#" was really weird, though. Contributions of yours to other threads I read are also stimulating  keep it up!

« Last Edit: Aug 24^{th}, 2002, 4:41pm by wowbagger » 
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pythagoras
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Re: 0.999.
« Reply #58 on: Aug 25^{th}, 2002, 3:39pm » 

on Aug 22^{nd}, 2002, 11:42pm, Jason Short wrote:By contrast, the value 9 is "infinite", and certainly does not have a value. 
 Well, actually this has been argued against by people I know. It's kind of nonsensical to imagine that the 9's repeat to the right in this case, so say they repeat to the left. I'll denote something repeats to the left in parentheses like this: (9)9. Now consider (0)0  (0)1, that is: ...000000 ...000001 Let's try to do this like we don't know what a negative number is. Since 1 is greater than 0 we have to borrow a 1 from, uh, somewhere... (An imaginary 1 in front of all the zeroes?) to get: ...000000 ...000001  ...xxxxx9 and since we borrowed our theoretical subtraction will end up: ...000000 ...000001  ...999999 so (0)0  (0)1 = (9)9. That's odd. There is actually a concept called nadics that deals with this sort of thing, and this is just an example in the 10adics.


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justin m
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Re: 0.999.
« Reply #59 on: Aug 28^{th}, 2002, 2:40pm » 

just like the limit of [f(x+h)  f(x)] / [xh] as h>0 is a "pretty good approximation of the rate of change of f its not just "pretty good", if you evaluate the limit, this IS the rate of change of f, its a simplet hing called the derivative. just like the infinite geometric series 9/10 + 9/100 + ... is not just a "pretty good" approximation, when you evaluate the limit you get an exact number, 1 additionally, 1.0000...001 is mathmatical nonsense, because you cant terminate a decimal that, by definition with the "...", is nonterminating. perhaps a college math course or two will help you swallow your pride on Jul 29^{th}, 2002, 6:40am, Kozo Morimoto wrote: 0.9 is pretty close to 1, an approximation of 1. You add 0.09 to make it 0.99 and its even closer to 1 than 0.9 but still not 1. Repeat to infinity. You get ever closer and closer to 1 from the low side, but you never get there? Using the ideas given above, does it mean that 1.000 ... 001 with infinite zeroes between the two 1s make it equal to 1? So Does it mean that 0.999... = 1.000...001 ? How about 0.999...998 = 0.999... = 1.000...001? 



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wowbagger
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Re: 0.999.
« Reply #60 on: Aug 29^{th}, 2002, 2:27am » 

on Aug 28^{th}, 2002, 2:40pm, justin m wrote:just like the limit of [f(x+h)  f(x)] / [xh] as h>0 is a "pretty good approximation of the rate of change of f 
 Of course, the correct formula is f'(x) = lim_{h>0} ( f(x+h)  f(x) ) / h Perhaps people writing illiterate posts shouldn't encourage others to take courses?


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Kozo Morimoto
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Re: 0.999.
« Reply #61 on: Aug 31^{st}, 2002, 11:32pm » 

OK, here is another thought exercise  I haven't work it all the way thru but... Imagine a number system with 20 digits A to T, but still in base 10 so that A maps to 0, C maps to 1 etc until S maps to 9 and T maps to something between 9 and 10. (B maps to somewhere in between 0 and 1, D maps between 1 and 2 etc) So when you have 0.9 in normal system, that would be same as A.S, however, there is a number A.T which is bigger than 0.9 and less than 1.0. So 0.999... would map to A.SSS.... but there is a number A.TTT... which is bigger than 0.999... but less than 1.0?


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Pietro K.C.
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Re: 0.999.
« Reply #62 on: Sep 1^{st}, 2002, 11:57pm » 

Sorry to insist on this point, but it seems necessary for the argument I'm going to give regarding Kozo's question. The set of characters "0.999..." DOES NOT MEAN adding the numbers 0.9 + 0.09 + etc, it is not the result of some infinite process, nor any other complicated nonsense like that. For the simple reason that an infinite process (be it addition or whatever) does not yield a result. Srowen and aadash have explained it best. The set of characters "0.999..." means the limit that the sequence (0.9 ; 0.99 ; 0.999 ; etc) tends to, in the sense of Cauchy, which is the following: a sequence (a(n)) = (a1, a2, etc) tends to a limit L if, for every real number r, there exists a natural number N such that a(n)  L < r for all n > N. The definition of the particular sequence we are dealing with does involve addition, but the concept of a limit does not depend on it; the concept of infinite addition is inconsistent. Come on, what is the point in knowing complicated integrals if you don't have a grasp of the beautiful underlying structure? No one denies that no finitenumbered term of the sequence equals or exceeds 1, but to say that, or to attempt proofs just by adding the succesive terms, is to miss entirely the meaning of the expression. Now, all of you have excellent problemsolving skills, as can be seen by your posts. Surely no one wil have trouble proving that the number 1 is a limit of the sequence (0.9 ; 0.99 ; etc) using the definition above. Hint: the difference between 1 and the nth term of this sequence is 10 taken to the (n)th power. (As an exercise, if you haven't already done it, you may also prove that, if a sequence of real numbers has a limit, it is unique) Hence the question "0.999... = 1" is resolved, and the answer is affirmative. About Kozo's alternative number system: the fact that a number T is less than 10 does not mean that 0.TTT... is less than 1. For instance, 9.9. If we take 0.TT to mean T/10 + T/100, then 0.TT = 0.99 + 0.099 = 1.089. Since 0.TTT... > 0.TT, it must also be greater than 1.089, which is greater than 1. However, my spidersense (and the previous posts) tells me that this isn't enough of an argument for Kozo. He may argue that there is some number not expressible in our decimal system for which the properties 9 < T < 10 and 0.TTT... < 1 simultaneousy hold. I intend to show that no such number exists. Never mind the number system, and suppose that 9 < T < 10. I take it we can represent ANY number by the letter T. Now construct the sequence S = (0.T ; 0.TT ; etc) by means of sums with progressively more numerous terms, in the same manner as we constructed the infamous "0.999...". By the expression "0.TTT..." we shall then mean the limit of this sequence, if indeed it does approach a limit. To prove that the limit exists, we could resort to the axiom of completeness of the set of real numbers, or to the axiom of existence of a supremum of an increasing bounded sequence (which is equivalent, cf any introductory text on real analysis, of which there are many  if you would like the proof, email me). But, since it obviously suffices to show a number that satisfies the limit criterion, and this is a simple case, we shall do so. I say that T/9 is the limit of the sequence S. To see this, consider the difference T/9  (the nth term of S); a simple proof by induction shows that this difference is T/(9*10^n) for all natural n. Clearly, for any number T, this difference can be made as small as we please by increasing n; in Cauchy's words, for all r > 0, there is an N such that T/(9*10^n) is less than r for all n > N. Hence, T/9 satisfies the definition of a limit of the sequence S = (0.T ; 0.TT; etc). And therefore 0.TTT... = T/9. It is not a coincidence that 0.TTT... "=" T*0.111... = T*1/9 = T/9. The multiplicative property of limits can be used to rigorously remove the sarcastic quotation marks around the equal sign. Now, you may notice that T/9 > 1 for all T > 9, which was the original hypothesis. So no number "not expressible in our decimal number system" exists that satisfies both T > 9 and 0.TTT... < 1 at once. That is the basic flaw in the # argument. As a side point, I would like to remark that there exists no numbers that cannot be arbitrarely well approximated by decimal expansions (prove this); so any eventual 0.###... would have a representation involving only the digits 0 through 9. If we were to make it greater than 0.999... but less than 1, the first digit would have to be 9; but also the second; and so forth, so that a number greater than 0.999... and less than 1 cannot exist. But this type of argument alone cannot make up for the rigor of the preeding proof. I'm sorry if I wrote too much, it's just that I would like to clear this up once and for all, so that such smart people (I was impressed by many others of Kozo's ingenious posts) would not waste their valuable time on matters of definition.


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Xanthos
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Re: 0.999.
« Reply #63 on: Sep 2^{nd}, 2002, 5:42pm » 

What about using the sum to infinity? If we take 0.999.... as 0.9+0.09+0.009 etc. then it is a geometric sequence. Therefore if we use the sum to infinity formula it should give us what 0.999.... is equal to. S_{inf}= a/(1r) In this case the starting number (a) is 9 and the common ratio (r) is 0.1. Substitute and you get 0.9/0.9, which is equal to 1. I might have missed something and it might have been said before but I thought I'd weigh in with that. Xanthos


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Kozo Morimoto
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Re: 0.999.
« Reply #64 on: Sep 2^{nd}, 2002, 7:49pm » 

0.T maps to 19/20 = 0.95 which is > 0.9 and < 1 0.TT maps to 19/20 + 19/400 = 0.9975 which is > 0.99 and < 1 0.TTT maps to 19/20 + 19/400 + 19/8000 = 0.999875 which is > 0.999 and < 1 so on. so wouldn't 0.999... < 0.TTT... < 1 ?


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NickH
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Re: 0.999.
« Reply #65 on: Sep 2^{nd}, 2002, 11:27pm » 

"so wouldn't 0.999... < 0.TTT... < 1 ?" No. In the limit, both are equal to 1. What you have written as 0.TTT... is simply the base 20 equivalent of 0.999.... We can also consider 0.111... (base 2), 0.222... (base 3),.... In all cases the sum to infinity of the geometrical progression is 1. Nick


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S. Owen
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Re: 0.999.
« Reply #66 on: Sep 3^{rd}, 2002, 5:41am » 

How about this, define two sequences: A_{i} = 0.999...9 (there are i 9s) B_{i} = 0.TTT...T (there are i Ts) So the sequences are: {0.9, 0.99, 0.999, ...} {0.T, 0.TT, 0.TTT, ...} It is true that A_{n} < B_{n} < 1 for any n. However, the limit of both sequences is still 1.


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James Fingas
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Re: 0.999.
« Reply #67 on: Sep 3^{rd}, 2002, 11:44am » 

This is a killer thread! Has anyone done epsilondelta proofs? This is the most basic (consistent) way of saying that a given limit exists and converges to the specified point. The way you use them in this case is to say: The series 0.9 + 0.09 + 0.009 + ... converges to 1 iff for every epsilon around 1, there exists an M so that by summing up the first M terms of the series, the sum is within epsilon of 1. Basically, an epsilondelta proof challenges you to find a differenceany differencebetween the series 0.9 + 0.09 + 0.009 + ... and the number 1. It says that any purported "difference" that you find will vanish after I sum enough terms. Since you cannot find a difference, then the numbers MUST be the same. To prove this, I show how to get M from epsilon. This leads to the following argument: 1) Assume that the series 0.9 + 0.09 + 0.009 + ... sums to a real number a. By the ratio test, the series converges to a single value. 2) Now we assume that a != 1. 3) This means that a1 has a nonzero value. Let's take half the absolute value of a1, and call it epsilon. 4) But if we take M = 1  log10( epsilon ), and sum the first M terms of the series, we will find that it is closer to 1 than epsilon. This value of epsilon is therefore rejected. Since this holds for any epsilon, there are NO NUMBERS between 1 and a. 5) But this contradicts our assertion that a1 != 0. Consequently, it must be true that a=1. If there is no difference between a and 1, then I argue that they must be the same number.


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Pietro K.C.
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Re: 0.999.
« Reply #68 on: Sep 3^{rd}, 2002, 2:39pm » 

Yes, I did do a "deltaepsilon" proof, even though there is no delta when dealing with sequences, but as I thought that epsilon was an ugly name for a variable, when written out whole, I used the letter r instead. And in keeping with the spirit of this site, I didn't spell out all the details, such as constructing the number N such that limit  sum up to the nth term < epsilon for all n > N. Maybe I do deserve not being read for posting such a huge text. But I repeat: arguments of the type "the series sums to..." will not convince Kozo, and he is right. These arguments are incosistent, and his rebuttals go to the heart of this incosistency. Kozo, I don't know if I misinterpreted your previous construction or if your last post is a new question,but anyways. You can apply the same type of findthedifferenceasafunctionofn proof I gave in my LARGE post to show that "0.TTT..." = limit (0.T ; 0.TT; etc) (now in base 20) = T/19. For T = 19, this limit is 1. Same as my proof for 0.999... in base 10. Let me give YOU a puzzle now. Let the sequences S1, S2 be defined by: S1(n) = 1  (1/n) S2(n) = 1  (2/n) and let the numbers A, B be their limits as n>oo, respectively. Would you say that A < B, just because each term in S1 is less than the corresponding term in S2? That is not keeping with the definition of a limit. The limit of a sequence need not be included in that sequence, as the above examples demonstrate. Clearly, A = lim S1 = 1, and B = lim S2 = 1, but 1 is not a term in either sequence. The same phenomenon occurs in the definition of 0.999... and 0.TTT...; these symbols indicate limits of sequences, and are not actually terms, and so the inequality does not have to hold. The key here is that "a(n) < b(n) for all n" does NOT imply that "lim a(n) < lim b(n)". Hope that helps.


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Kozo Morimoto
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Re: 0.999.
« Reply #69 on: Sep 4^{th}, 2002, 5:21am » 

I've come to the conclusion that this has come to a point of definition/symmantics. I googled and came up with: http://www.math.toronto.edu/mathnet/answers/infinity.html which I think covers this discussion to my satisfaction in an easily accessible form. The definition I was using is covered by point 1 on the referred website, and in this context, I was right. The definition that everyone else here was using is covered by point 2 on the referred website, and in that context, everyone was correct. In this definition, convergence and infinity is used interchangeably but I was treating them as 2 different 'things'. It certainly has been an interesting discussion though!


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Pietro K.C.
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Re: 0.999.
« Reply #70 on: Sep 4^{th}, 2002, 12:08pm » 

I looked at the site, but I still don't get how point #1 validates your point of view. It says infinity doesn't exist as an element of a "number system", but very few of us proposed that it did. I really didn't understand what you said. I still think that a string of symbols such as 0.999... requires the concept of convergence to make sense, and is meaningless in the "number system" context alone.


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James Fingas
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Kozo was right!
« Reply #71 on: Sep 5^{th}, 2002, 11:43am » 

It turns out that Kozo was right all along. There is a difference between 0.9999... and 1. At least the floor function says so


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Dustin
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Re: 0.999.
« Reply #72 on: Sep 6^{th}, 2002, 5:58pm » 

.99999... < 1 if someone wrote .99999... on a piece of paper for there entire lives, and every last decended of theirs did the same thing, it would still never reach 1. Someone also earlier said that if 1/3 = .3333... 2/3 = .6666... then technically 3/3 would = .9999.... However since we know that 3/3 = 1 then .9999... must also = 1 This idea is flawed because of the fact that the representation of both 1/3 and 2/3 is also flawed. 1/3 does not = exactly .333... and 2/3 does not = exactly .666... The reason math labels it as that is because with out math system the closes representation we can get of 1/3 is .333... but that does not make it = to .333... and hence 3/3 is not = .999.... and therefor .999.... is not = to 1


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Jeremiah Smith
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Re: 0.999.
« Reply #73 on: Sep 6^{th}, 2002, 7:05pm » 

on Sep 6^{th}, 2002, 5:58pm, Dustin wrote:.99999... < 1 if someone wrote .99999... on a piece of paper for there entire lives, and every last decended of theirs did the same thing, it would still never reach 1. Someone also earlier said that if 1/3 = .3333... 2/3 = .6666... then technically 3/3 would = .9999.... However since we know that 3/3 = 1 then .9999... must also = 1 This idea is flawed because of the fact that the representation of both 1/3 and 2/3 is also flawed. 1/3 does not = exactly .333... and 2/3 does not = exactly .666... The reason math labels it as that is because with out math system the closes representation we can get of 1/3 is .333... but that does not make it = to .333... and hence 3/3 is not = .999.... and therefor .999.... is not = to 1 
 Did you miss the entire thread, somehow? Read > Comprehend > Post


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Dustin
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Re: 0.999.
« Reply #74 on: Sep 6^{th}, 2002, 7:58pm » 

I read the first group of posts and then skipped over the rest because of the fact that the same things were being said.


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