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Topic: 0.999... (Read 33909 times) 

Icarus
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Re: 0.999...
« Reply #25 on: Nov 11^{th}, 2004, 3:48pm » 
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on Nov 10^{th}, 2004, 5:19am, Three Hands wrote:If 0.999... is the same as 1, then it presumably means that there can be no difference of 0.000...1. So, if you subtract 0.000...1 from 0.999... (resulting in 0.999...8 ), then you still have 1. 
 If you will note in my post right above yours, I state Quote:Note that in the notations above the 3 dots represents the digits repeating to infinity. As such, there is no such decimal notation as "0.000...1", for it would require an infinite number of 0s to occur before the 1. If you check the definition of decimal notation given above, you will see that it does not include such transinfinite decimals. 
 Your whole argument simply ignores this problem, pretending that 0.000...1 exists! Let me try to be clear here: THERE IS NO SUCH REAL NUMBER AS 0.000...1 OR 0.999...9 OR 0.999...8! The definition of decimal notation simply doesn't extend to these transinfinite decimal positions. If you want to talk about such "numbers", YOU have to define them first. And even if you do define them, they have no bearing on the question of whether 0.999... = 1, as these two notations already have definitions, which are not changed by defining additional ones. By those standard definitions, both refer to the same real number. Quote:This process could then be repeated, eventually resulting in 0=1 
 Actually, this process cannot be carried out even once (unless you can come up with some sort of reasonable definition for your new notations  and even then, it is NOT going to be repeatable in such a way as to get to 0). You seem to think that 0.999... has some "last 9" that we don't bother to write down. This is most definitely not the case. 0.999... has an infinite number of 9s  one nine in order for each of the natural numbers. A last "9" would mean the same as a "highest natural number". But there is not a highest natural number. If you give me a proposed highest, I can always beat it by adding 1. So also 0.999... does not have a last 9 for us to decrement to get 0.999...8. Quote:Presumably people do not want this to be the case. Hence I would argue that the difference of 0.000...1 between 0.999 and 1 is important in that respect, and that 0.999 is less than 1 But then, someone has probably already pointed this out, and the discussion suggested that they are some ignorant person who doesn't fully understand the intricacies of infinity, etc. But, hey, I never claimed to be a mathemawhatsit... 
 Ignorance itself is not a problem. Only when ignorance is willful is it something to be ashamed of. To believe wrong concepts is a common condition of all people. To hold to these concepts without bothering to examine the arguments against them (when such arguments have been presented), or to reject such arguments simply because you don't understand them, is unexcusable. And to presume you know the tenor of a conversation you haven't have bothered to follow is rude.

« Last Edit: Nov 12^{th}, 2004, 4:02pm by Icarus » 
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Three Hands
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Re: 0.999...
« Reply #26 on: Nov 12^{th}, 2004, 4:50am » 
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OK, fair cop, I should have thought through what I was implying a bit more before posting  especially given philosophers seem to enjoy questioning existence so much to begin with . I'll try and avoid making stupid comments in future (but almost certainly fail...)


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h7
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 The idea is that hyperreal numbers extend the reals, including in "infinitely small" numbers. That's it! we say: 1  0.999... is one of these infinitely small numbers, rather than 0! If they're hyperreals between 0.999 and 1 does that mean 0.999 still equals 1? (I don't know how to express hyperreals or complex notions so please bear with me) Suppose we add a new set of numbers to extend the decimals in which (to show the new set of numbers I'll add a colon) : 0.9 : 9 which does not equal 0.99 suppose :1 + :9 = 0.00...1 the numbers after the colon will be something that is smaller then 0.000....1 (And by this I mean the smallest thing we can represent using our number system) Something similar of how time works. 1:42:51:75:35 letting 1 be intergers, 42 to be decimals, 51 to be these new sets of smaller numbers, and so on and so on. So we'll have a number like 0.999... : # which is smaller then 1 but greater then 0.999 # = any natural number I'm really getting a headache thinking about such infinitely small things... like this. I don't think I made my post clearly enough, so if anyone has any questions (or more probable corrections/retorts in which proves that this is all babble and nonsense), please post.


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Sir Col
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Re: 0.999...
« Reply #28 on: Nov 14^{th}, 2004, 3:02am » 
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There is nothing wrong with asking genuine and original questions. But, as you said, you can always expect to be challenged. This is the arena of mathematics: everything we commit outwardly is open to refutation, including the points I make in this post. It was only during the 19th and 20th century when mathematicians were evicted from their ivory towers of certainties that we began to make such significant advancement. Now, any credible mathematician would not refuse to have premises questioned. Having challenged the perfect harmony of balance, where 0.999...=1, with your question, I would ask you to clearly define this "new set of numbers to extend the decimals". For example, I could say that x is a number that exists beyond the normal decimals, but resides between 0.999... and 1. Therefore 0.999...[ne]1. The argument fails because x cannot be defined in terms of the decimals, hence it cannot validly be used to draw conclusions based on the decimals. To say that :1+:9=0.000...1 doesn't help, because 0.000...1 is not defined in the normal decimal system anyway. Can you express your new system in terms of standard decimal notation? Or can you show how normal decimals operate in your extended system? If not, it makes no sense to combine the two distinct systems in your argument.


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Sir Col
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Re: 0.999...
« Reply #30 on: Nov 14^{th}, 2004, 2:58pm » 
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Hyperreals are an extension of the real number system and deal with infinitesimals. A positive hyperreal is defined as being greater than zero but less than every positive real number. Consequently, the decimal system, which describes real numbers, simply cannot describe the complete set of hyperreals. The only infinitesimal that real numbers can describe is zero. If you want to learn more about hyperreals, I can strongly recommend, Elementary Calculus: An Approach Using Infinitesimals, written by H. Jerome Keisler. It deals comprehensively with the entire system and you can download it from here: http://www.math.wisc.edu/%7Ekeisler/calc.html


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Icarus
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Re: 0.999...
« Reply #31 on: Nov 15^{th}, 2004, 8:10pm » 
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on Nov 12^{th}, 2004, 4:50am, Three Hands wrote:OK, fair cop, I should have thought through what I was implying a bit more before posting  especially given philosophers seem to enjoy questioning existence so much to begin with . I'll try and avoid making stupid comments in future (but almost certainly fail...) 
 That's alright. I wouldn't want you raising the bar higher than I could meet anyway! on Nov 13^{th}, 2004, 6:08pm, h7 wrote:If they're hyperreals between 0.999... and 1 does that mean 0.999... still equals 1? 
 (I added the dots to the quote, since without them you just have a trivially obvious fact: 0.999 < 1. This may seem pointless since we both know what you meant, but there are some who read this thread who have a hard time differentiating between 0.999 and 0.999.... I'd rather not make it harder for them by confusing the notation even more.) Since the hyperreals extend the real numbers, 0.999... is defined in the hyperreals exactly as it is in the Reals. I.e. 0.999... = 1 as hyperreal numbers, too. Therefore, there are no hyperreal numbers between 0.999... and 1, as no other number can come between 1 and itself. In order to have a number between 0.999... and 1, you would have to redefine what 0.999... means in the hyperreals. But even if you do redefine it, it has no bearing on the question of this thread, which is what 0.999... equals as a real number under the ordinary definition. Quote:So we'll have a number like 0.999... : # which is smaller then 1 but greater then 0.999... 
 You can define 0.999...:# as a hyperreal much as you have suggested. The key is: during addition or multiplication, when carries reach the colon they are discarded, as they are considered "infinitely smaller" than any of the digits to the left of the colon. But this changes nothing. You simply have that 0.999...:# = 1:# > 1. Your assumption that anything you tag onto the end of 0.999... is going to result in something less than 1 is false.


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rmsgrey
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Re: 0.999...
« Reply #32 on: Nov 22^{nd}, 2004, 6:35am » 
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Just to annoy Icarus... 0.999...<1[.000...] provided you use a lexicographic ordering. One reason so many people have difficulty with 0.999...=1 is that, except for the very rare cases where you compare the two representations of a terminating decimal, the lexicographic order (padded with leading 0s to align the decimal points) is the same as the usual order on the reals, and generally people judge the order of two numbers first by looking at the number of digits (to the left of the decimal point) or number of leading 0's (to the right of the decimal point) or by comparing the exponent of 10 (in the same way), then, in case of a tie, looking at the lexicographic order. This is fine except when (padded) lexicographic order differs from the usual order on the reals...


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Icarus
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Re: 0.999...
« Reply #33 on: Nov 22^{nd}, 2004, 4:45pm » 
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As abstract decimal expressions 0.999... < 1.000... lexigraphically. But not as real numbers, since they are in fact the same real number. It is not quite accurate to say that the real numbers do not follow lexigraphical order. After all, if x.xxx... [le] y.yyy... lexigraphically as abstract decimal expressions, then x.xxx... [le] y.yyy... as real numbers as well. The only difference is that sometimes x.xxx... < y.yyy... lexigraphically, but x.xxx... = y.yyy... as real numbers. The general rule is, x.xxx... and y.yyy... represent the same real number if and only if they are adjacent lexigraphically (i.e. no other decimal expression lies between them). So another way of defining the real numbers (not a very natural one) is to take the set of all decimal expressions, and mod out the relation (x is adjacent to y).


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ThudnBlunder
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Re: 0.999...
« Reply #34 on: Dec 17^{th}, 2004, 9:02am » 
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Consider the following (generally accepted) 'proof' that 0.9999999... = 1 Let 0.9999999... = x Multiply both sides by 10 9.999999... = 10x Therefore 9x = 9 x = 1 But 1) We cannot assume that subtraction applies to infinite decimals in the same way that we know it applies to finite decimals. (Hence the original question of whether 0.9999999... = 1). 2) Point 1) above implies that we cannot assume that addition applies to infinite decimals in the same the same way that we know it applies to finite decimals. 3) Given that multiplication is no more than repeated addition, we cannot therefore assume that multipication applies to infinite decimals in the same the same way that we know it applies to finite decimals. 4) Therefore the above 'proof' is begging the question (assuming to be true that which is to be proved). 5) Therefore the above 'proof' is bogus. Given that a proof is either valid or it isn't, why is the above 'proof' so often quoted as being acceptable? (Even in this thread.)

« Last Edit: Dec 17^{th}, 2004, 9:19am by ThudnBlunder » 
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towr
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Re: 0.999...
« Reply #35 on: Dec 17^{th}, 2004, 11:04am » 
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Quote: 1) We cannot assume that subtraction applies to infinite decimals in the same way that we know it applies to finite decimals. 
 I disagree, we can assume it. 'infinite decimals' are welldefined, as much as any real number, and so are the operations on them. Or would you want to object to 10*1/31/3=3 just because someone wrote it as 3.333...  0.333... = 3 ? We could always write 0.999.. as [sum]_{i=1}[supinfty]9*10^{i}, but it's still the same thing. 10*[sum]_{i=1}[supinfty]9*10^{i} = [sum]_{i=1}[supinfty]9*10^{1i} = [sum]_{i=0}[supinfty]9*10^{i} [sum]_{i=0}[supinfty]9*10^{i}  [sum]_{i=1}[supinfty]9*10^{i} = 9 * 10^{0}=9 so [sum]_{i=1}[supinfty]9*10^{i} = 1

« Last Edit: Dec 17^{th}, 2004, 11:06am by towr » 
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ThudnBlunder
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Re: 0.999...
« Reply #36 on: Dec 17^{th}, 2004, 12:04pm » 
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Quote:I disagree, we can assume it. 'infinite decimals' are welldefined, as much as any real number, and so are the operations on them. 
 1) Are you saying that 'infinite decimals' are welldefined. Therefore 0.9999999... = 1? and/or 2) Are you claiming that the proffered 'proof' that 0.9999999... = 1 is acceptable?

« Last Edit: Dec 17^{th}, 2004, 12:10pm by ThudnBlunder » 
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Sir Col
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Re: 0.999...
« Reply #37 on: Dec 17^{th}, 2004, 1:36pm » 
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I don't think anyone with less than a graduate level understanding of mathematics, in presenting that particular proof, knows what is really going on. However, what makes it such a popular proof is that it appeals to the intuition and confirms what is mathematically true, albeit with somewhat naive assumptions; that is, the heart of the proof requires a sophisticated appreciation of arithmetic with infinite decimals. Although a novice could begin to grasp what would happen if you added 0.111... to 0.222..., they would be hard pushed to explain what happens with the "carry" when 0.666... is added to 0.444... After all, their understanding of adding is aligning numbers and adding from right to left. So where do you start with infinite sums? Obviously to mathematicians, arithmetic involving infinite decimals is clearly defined. Even if the definition circumvents the question of what is actually going on. Which human can honestly grasp the infinite or the infinitesimal. As I say to my students, "Quite literally, the infinite is too big to understand and the infinitesimal is too small." The real question is, does presenting a proof, which may make use of a higher concept, become invalid if the writer hasn't yet grasped the higher concept? For example, some of my students could prove a series of neat number theory results that make use of the fundamental theorem of arithmetic. Just because none of them have seen a proof for the FTA, does that make their proofs any less valid? There is an excellent book by A. Gardiner called, Understanding Infinity: The Mathematics of Infinite Processes, which asks the big question, "Why does calculus work?" It has chapters dedicated to the exploration of real numbers, infinite decimals, recurring decimals, and even recurring nines! http://store.yahoo.com/doverpublications/048642538x.html


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Icarus
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Re: 0.999...
« Reply #38 on: Dec 17^{th}, 2004, 6:28pm » 
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on Dec 17^{th}, 2004, 12:04pm, THUDandBLUNDER wrote: 1) Are you saying that 'infinite decimals' are welldefined. Therefore 0.9999999... = 1? and/or 2) Are you claiming that the proffered 'proof' that 0.9999999... = 1 is acceptable? 
 I think he is saying both, and if he isn't, then I am.


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Sir Col
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Re: 0.999...
« Reply #39 on: Dec 18^{th}, 2004, 3:24am » 
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On a more general note... I am reminded of when I extended one of my classes work on arithmetic into different bases and whilst working in base 2 they explored decimal fractions. Having previously looked at recurring decimals in base 10 and the result, 0.999...=1, the class were a little divided on whether or not they "believed" it. You can imagine their reaction when they converted 0.111..._{2} into base 10; that is, 0.111..._{2}=1/2+1/4+1/8+...=1. I suspect that I won across most of the remainging stubborn doubters; at least those who are bright enough to appreciate the significance of the digit (b1) in base b.


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towr
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Re: 0.999...
« Reply #40 on: Dec 20^{th}, 2004, 12:57am » 
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on Dec 17^{th}, 2004, 12:04pm, THUDandBLUNDER wrote:1) Are you saying that 'infinite decimals' are welldefined. Therefore 0.9999999... = 1? and/or 2) Are you claiming that the proffered 'proof' that 0.9999999... = 1 is acceptable? 
 Indeed, as Icarus suspected, I am saying both. To be honest, I probably wouldn't have if Icarus hadn't convinced me of it earlier in the discussion

« Last Edit: Dec 20^{th}, 2004, 12:58am by towr » 
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Re: 0.999...
« Reply #41 on: Dec 28^{th}, 2004, 5:38pm » 
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Well, some people might believe that 1/3=0.333... because they can work it out using long division (after "giving up" and "getting the picture"). So maybe they'll believe that 1=0.999... after using long division to divide 2 by 2, say. Since 2/2=1, they could write, 0.99.. 2)2.000000 0 20 18 20 18 2.. If they can't comprehend that, then they should probably be questioning the fact that 1/3=0.333... as well.

« Last Edit: Dec 28^{th}, 2004, 6:10pm by kellys » 
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igabo jj
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Sorry Icarus, my math needs some help. For your first proof in calculus, could you explain how you got the explicit equation from the geometric series? I'm a bit lost as to why the numerator is 1  10^(N  1). I'm not sure where the 1 in the exponent comes from.


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Icarus
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Re: 0.999...
« Reply #43 on: Jan 12^{th}, 2005, 3:18pm » 
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Let y = [sum]_{[subn]=1}^{[smiley=supcn.gif]} x[supn] = x + x[sup2] + x[sup3] + ... + x[smiley=supcn.gif] 1 + y = 1 + x + ... + x[smiley=supcn.gif] (1  x)(1 + y) = 1 + x + x[sup2] + ... + x[smiley=supcn.gif] x  x[sup2]  ...  x[smiley=supcn.gif]  x[smiley=supcn.gif][supplus][sup1] = 1  x[smiley=supcn.gif][supplus][sup1] So y = (1  x[smiley=supcn.gif][supplus][sup1])/(1  x)  1 Substitute x = 10[supminus][sup1], and you have the result.


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Qrimson Fury
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If only the question would have been worded like the Title, then it would actually be a riddle, i.e.: Which is true? .999...<1 .999...=1 .999...>1 The answer would obviously be: .999...>1 since 0.999.999.999.999 and so on > 1 Of course the leading "0." gets dropped just like if you wrote 015.


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Re: 0.999...
« Reply #45 on: Jan 20^{th}, 2005, 9:04am » 
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on Jan 20^{th}, 2005, 8:37am, Qrimson Fury wrote:If only the question would have been worded like the Title, then it would actually be a riddle, i.e.: Which is true? .999...<1 .999...=1 .999...>1 The answer would obviously be: .999...>1 since 0.999.999.999.999 and so on > 1 Of course the leading "0." gets dropped just like if you wrote 015. 
 The convention usualy used on this forum is for '.' to denote a decimal point rather than a thousands separator  in those cases where a thousands separator is used, it's usually a ',' so: 9999999.9999 or 9,999,999.9999


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Qrimson Fury
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"." may not be the commonly used convention for separating thousands, but it is an accepted convention. Am I missing something? Where exactly does the "riddle" state that we are dealing with Real numbers? I would say that (if my above solution is accepted) all three possibilities have been proven. .999... = 1 was proven several ways, and I believe the following is an agreement by Icarus that .999... < 1 on Nov 22^{nd}, 2004, 4:45pm, Icarus wrote:As abstract decimal expressions 0.999... < 1.000... lexigraphically. 
 A couple questions: 1) Can you use the 1.000... 0.999... 0.000... = 0 proof for this since, when subtracting, don't you have to start at the rightmost digit? How can you start at the rightmost digit when there isn't one? 2) Doesn't the 1=0.999...9^{1} go against what is given in the question: 0.999... has infinitely many 9s. Doesn't this mean that EVERY digit after the decimal point is a 9? 9^{1} isn't a 9. Therefore it doesn't fit the question. 3) Using the 1=0.999...9^{1} proof, does this mean that there can be a "last" digit? Thus, looking at only the last digit of .999... which, if my 2nd question is correct, would be a 9, using the 10x proof, when you multiply .999... by 10, wouldn't the last 9 shift right one place and you would be subtracting the last 9 of x from a 0 in 10x 10x = 9.999...90 x = 0.999...99 9x = 8.999...1?


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Icarus
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Re: 0.999...
« Reply #47 on: Jan 21^{st}, 2005, 6:02pm » 
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on Jan 21^{st}, 2005, 1:37am, Qrimson Fury wrote:"." may not be the commonly used convention for separating thousands, but it is an accepted convention. 
 Where you are, maybe. In the USA, it is not accepted at all. Quote:Am I missing something? Where exactly does the "riddle" state that we are dealing with Real numbers? 
 Because it never occurred to William that the riddle could be given any other interpretation than as real numbers. The convention concerning decimal notation representing real numbers is so universal that it is necessarily assumed unless specifically said otherwise. I.e., in order for 0.999... to mean anything other than a real number, you must actually say that it can. Since William did not, it refers to Real numbers. Quote:I would say that (if my above solution is accepted) all three possibilities have been proven. .999... = 1 was proven several ways, and I believe the following is an agreement by Icarus that .999... < 1 
 If you are allowed to come up with any meaning for 0.999..., then you can prove anything you want about it. But there is no point. Your result is meaningless except in the particular context of the definition you used, and the only context of only import is for 0.999... is the real numbers. In the example you gave, I specifically pointed out the context: Lexigraphical sequences of ordered characters. This context tells me nothing about the context of the question itself: decimal expressions for real numbers. Which was the point I was making in that post. Quote:A couple questions: 1) Can you use the 1.000... 0.999... 0.000... = 0 proof for this since, when subtracting, don't you have to start at the rightmost digit? How can you start at the rightmost digit when there isn't one? 
 No, you do not have to start at the rightmost digit. You can start at the left digit, and move right. Occasionally, you have to back up a bit to borrow, but whenever you have a nonzero digit in the result, you will never have to go back any further than it again, so you are still guaranteed to have unchanging digits eventually in any spot. Quote:2) Doesn't the 1=0.999...9^{1} go against what is given in the question: 0.999... has infinitely many 9s. Doesn't this mean that EVERY digit after the decimal point is a 9? 9^{1} isn't a 9. Therefore it doesn't fit the question. 3) Using the 1=0.999...9^{1} proof, does this mean that there can be a "last" digit? Thus, looking at only the last digit of .999... which, if my 2nd question is correct, would be a 9, using the 10x proof, when you multiply .999... by 10, wouldn't the last 9 shift right one place and you would be subtracting the last 9 of x from a 0 in 10x 10x = 9.999...90 x = 0.999...99 9x = 8.999...1? 
 In general, an ellipsis, "...", indicates a continuance of whatever precedes it. Usually, this continuance is finite in length. Only in mathematics does the possibility of an infinite continuance come up. But even in mathematics, the ellipsis does not always mean an infinite continuance. Usually the meaning is determined through context. In the context of decimal expressions, an expression of the form 0.999... usually means infinite continuation. But an expression of the form 0.999...9 does not, because in any decimal, any digit has only a finite number of predecessors. In this case the ellipsis represents only a finite continuation. This is why, when discussing the concept of infinitely many digits before another digit in earlier posts, I was always careful to establish the context of infinite continuation when giving these expressions (which are NOT decimal expressions). Unless specifically said otherwise, 0.999...9, and similar expressions, represent only a finite number of digits, while 0.999... represents an infinite number of digits.


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Qrimson Fury
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Since this is a site for riddles, wouldn't it be better to word the problem along the lines of finding a case where you can prove all three possibilities? As it is written, it's just a simple mathematical proof, and one that I would think many who frequent this site saw back in grade school when first taught the 10^{n}x method for finding the fractional equivalent of a repeating decimal. My 999.999.999... answer was actually in response to someone in the original thread that challenged someone to come up with showing .999...>1. My answer was the best I could come up with. I do live in the USA. I saw this notation in an old Mathematical Recreations book several years ago, but unfortunately lost it. I did a web search under 1.000.000, and it looks like it might be a common convention in Europe. I've never been to Europe, so I have no firsthand knowledge whether this is the case. I don't believe I made up a new meaning for .999... in order to prove anything.


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Sir Col
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Re: 0.999...
« Reply #49 on: Jan 22^{nd}, 2005, 3:56am » 
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Please don't feel got at, Qrimson Fury. Your reinterpretation was a nice idea, and you're absolutely correct about the comma convention. It is only in English speaking countries that the dot is used as a decimal point and the comma is used as a separator. Certainly in continental Europe, and pretty much the rest of the world, the dot is used as a separator and the comma is used as a decimal point. You can check these out by going to Control Panel, Regional and Language Options. If you change your location you will see examples of how money, time, date, and number is written in the box below. I found this interesting page: http://www.geocities.com/Broadway/1906/cultr14.html One thing I am not sure about, and I would really like this to be confirmed or refuted, but I believe that they use a semicolon in coordinates.


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