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Topic: 0.999. (Read 125646 times) 

matt
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Re: 0.999.
« Reply #250 on: Sep 8^{th}, 2003, 10:15am » 

It's pretty funny to watch you all argue over something like this. The simple answer is that 0.999... < 1. I learned that from my calculus teacher when I decided to argue the point. 0.999... != 1, but the math terminology indicates that it is "infinitely close to" 1. So that's like saying "it isn't, but it may as well be." I can tell my "authoritiative" answer doesn't thrill you. You want proof that they're at least not equal, though? Easy as pie. An irrational number is defined as "any number that can't be represented by the ratio of two integers." So I ask, what ratio of integers would result in 0.9999....? It's not 3/3... that results in 1. So, 0.999.... is an irrational number. So why are so many people trying to say that an irrational number can be equal to a rational one? If they were the same, they'd have to be within the same set. if x member of R and x=y then y member of R. Well, 1 member of REAL, 1 = 0.999...., 0.999 member of REAL? It doesn't work... it can't work. Although, it might as well work, because for all practical intents and purposes, they're the same.


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ThudnBlunder
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Re: 0.999.
« Reply #251 on: Sep 8^{th}, 2003, 10:22am » 

Quote:So I ask, what ratio of integers would result in 0.9999....? It's not 3/3... that results in 1. So, 0.999.... is an irrational number. 



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matt
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Re: 0.999.
« Reply #252 on: Sep 8^{th}, 2003, 10:38am » 

Icarus posted this (though with better formatting): Sorry, but it ain't so. By definition 0.999... = limitn {0.999...( n9s)}. Each element of the sequence is a real number. By the definition of limits, the limit of a sequence of real numbers is a real number. So 0.999... is real, not hyperreal. I have to take exception to this.... you're stating that 0.9... is the limit of 0.9! Maybe my limited math experience is showing here... but that's a bit of infinite recursion, and that's simply not possible. I really don't know what you're trying to say... but 1/3 != 0.3... because 0.3... is the base10 approximation of a number that is unrepresentable with that notation. 0.3... is by its very nature something that doesn't really exist (in the sense of a concrete quantifiable number). But, the limit of 0.3....... n3's as n>INF is definitely 1/3. As the limit of 0.9..... (n 9's) as n>INF is definitely 1. But it's not 0.9... (n9's). That's self defeating and stupid. Granted I've never heard of hyperreals, and I've only progressed through Calculus 2. But I think I've got a pretty firm grip on things. Last I checked, putting 0.9... and saying "It really should be the limit of 0.9..." doesn't change the original question. And if someone told you there's an assumed limit in "0.9..." you should get your money back for that course. Of course, I'm one of those idiots who doesn't see a major problem with 0.0....1, so I must be braindead for thinking that an alternative representation for the calculus h constant is valid. Go back to your Calculus fundamentals... it's the first thing they teach you. h is used like any other variable, and you don't balk at it there. Why is it a problem when people who haven't been exposed to it refer to it with improper terminology?


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Ulkesh
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Re: 0.999.
« Reply #253 on: Sep 8^{th}, 2003, 10:43am » 

I think I'll have to agree with matt here. I mean, when something is so intuitively obvious it seems pointless to use something as inaccurate at maths to justify a 'correct' answer. After all, Godel's incompleteness theorem quite clearly shows that maths is incomplete and therefore wrong about more or less everything. While we're on the subject, I'd just like to voice my support of the longheld view of a flat Earth. Since Physics is basically applied maths, and since I've rigorously proven that maths in inherently wrong, the misguided fools who believe in building their models of physical phenomena purely upon empirical evidence are clearly not using their own intuition and seeing what is obvious. If the earth were round we'd clearly fall off. The idea of 'gravity' is therefore fallacious, QED.


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S. Owen
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Re: 0.999.
« Reply #254 on: Sep 8^{th}, 2003, 10:52am » 

on Sep 8^{th}, 2003, 10:15am, matt wrote:An irrational number is defined as "any number that can't be represented by the ratio of two integers." So I ask, what ratio of integers would result in 0.9999....? It's not 3/3... that results in 1. So, 0.999.... is an irrational number. 
 To expand on T&B's smiley face  your argument is plainly circular. You say that 3/3 != 0.999... because 3/3 == 1. But that is only a valid argument if you assume that 1 != 0.999... ! This sort of argument doesn't show that anything is irrational anyway. For example, 3/3 != 0.8, but that does not mean that 0.8 is irrational. You would need to show that there are no integers a and b such that a/b = 0.8, not that some particular a and b do not satisfy a/b = 0.8. First step is to decide what you think "0.999..." even means. Then you can decide what it equals and does not equal.


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matt
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Re: 0.999.
« Reply #255 on: Sep 8^{th}, 2003, 10:59am » 

somebody said this, but I'm on the preview page and can't see his/her name (sorry): To expand on T&B's smiley face  your argument is plainly circular. You say that 3/3 != 0.999... because 3/3 == 1. But that is only a valid argument if you assume that 1 != 0.999... ! yeah... my mistake. I didn't mean to put the word "so" there or imply any sort of causality. I meant to prevent anyone with replying that 3/3 directly gives 0.9..., as that would be to assume the proof (like I ended up doing by mistake). I was just typing faster than I was thinking. I still challenge anybody to come up with a ratio that is == 0.9... that doesn't include some sort of unquantifiable expression (such as (INF1)/INF == 0.9...). And I still won't concede that 0.9... really means the limit of 0.9... because that's just as circular as me!


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wowbagger
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Re: 0.999.
« Reply #256 on: Sep 8^{th}, 2003, 11:28am » 

on Sep 8^{th}, 2003, 10:59am, matt wrote:And I still won't concede that 0.9... really means the limit of 0.9... because that's just as circular as me! 
 Hm, if it doesn't mean the limit to you, what does it mean? Do you see this expression as representing a real number? And if so, which one? As Icarus already pointed out, you can invent many notations, but you have to say what number it represents. Standard mathematics defines such nonterminating decimals als the limit of an infinite process  how do you define the value of your 0.999...? How is the definition via the limit circular? It has the form x = lim_{n[to][infty]} [sum]_{k=1}^{n} 9*10^{k}. There's no x in the right hand side, neither explicitly nor hidden.

« Last Edit: Sep 8^{th}, 2003, 11:32am by wowbagger » 
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S. Owen
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Re: 0.999.
« Reply #257 on: Sep 8^{th}, 2003, 12:25pm » 

on Sep 8^{th}, 2003, 10:59am, matt wrote:I still challenge anybody to come up with a ratio that is == 0.9... that doesn't include some sort of unquantifiable expression (such as (INF1)/INF == 0.9...). And I still won't concede that 0.9... really means the limit of 0.9... because that's just as circular as me! 
 Yes, I think this is the real heart of this question  what does "0.999..." even mean? We all immediately understand what the string "0.999" signifies, but although it looks similar, "0.999..." is not conventional notation for a decimal. So this is what we need to decide. I think probably the only reasonable interpretation is to simply extrapolate from what decimal notation means; so 0.999... = 0.9 + 0.09 + 0.009 + .... And this is 1. As to giving you a ratio that equals 0.999..., well, 3/3 is such a ratio. But for you to decide whether that is true, you need to decide what 0.999... equals, which is original question anyway. So, I don't think the rational/irrational line of argument gets anywhere before exploring what 0.999... even is.


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Sir Col
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Re: 0.999.
« Reply #258 on: Sep 8^{th}, 2003, 1:10pm » 

I like Matt's challenge: "So I ask, what ratio of integers would result in 0.9999....?" That's a very original slant on the discussion. Perhaps 0.999... is an illegal expression in the first place? I think this, in part, relates to the denumerability thread that I started in the General/Whatever area, here. Maybe, as Matt says, and I'm probably showing my ignorance again here, the decimal representation is incomplete (as it can never be completed) and is not the same as the rational representation. However, I think that S. Owen made some excellent points, as the heart of the question is, "What does 0.999... mean?" I agree with him, the only sensible interpretation is to extrapolate the infinite expansion of the decimal: 0.9+0.09+0.009+... Confused! I really must stop reading this thread. By the way, Icarus, thank you for your amazing summary of all those pages.

« Last Edit: Sep 8^{th}, 2003, 1:26pm by Sir Col » 
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towr
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Re: 0.999.
« Reply #259 on: Sep 8^{th}, 2003, 1:29pm » 

Let's make a decimal expansion of 1/1, using a longdevision 1/ 1.000... \ 1.000... 1  0.0 .0  .0 (etc) now let's do it again, but differently, because we like to be contrary 1/ 1.000... \ 0.999... 1  1.0 .9  .10 .09  .01 (etc) So clearly 0.999... is a rational result of dividing 1 by 1

« Last Edit: Sep 8^{th}, 2003, 1:30pm by towr » 
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Icarus
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Re: 0.999.
« Reply #260 on: Sep 8^{th}, 2003, 6:50pm » 

on Sep 8^{th}, 2003, 10:15am, matt wrote:It's pretty funny to watch you all argue over something like this. The simple answer is that 0.999... < 1. I learned that from my calculus teacher when I decided to argue the point. 
 Yet another supposed math teacher who does not know the subject they are trying to teach. My father had a math teacher (everything teacher actually  it was oneroom school) whose approach to solving math problems was to flail around at random until she managed to produce the answer in her book. He told me that she regularly included the problem number in calculating her "solutions". The point is: just because a teacher tells you something, that doesn't mean it's true. Quote:0.999... != 1, but the math terminology indicates that it is "infinitely close to" 1. So that's like saying "it isn't, but it may as well be." 
 "infinitely close" is not an math terminology. It is a phrase used solely by those who do not understand what they are talking about, or by those who do know in leading into the truth: "infinitely close" means zero distance between them; ie., they are equal. Quote:I can tell my "authoritiative" answer doesn't thrill you. 
 "Authoritative" ?? ?? I seem to have missed your qualifications as an authority here. I don't like saying this, because it sounds like I'm sticking my nose in the air, and demanding that you take things on my sayso. This not true. But several comments you have made indicate you believe that we are not well educated in mathematics. I have a Ph.D. in the subject. Several others here have degrees. Most if not all of the posters since your first post are much further advanced in the subject of mathematics than you are. This does not mean you should take any of our word on this, but it does mean that if you find yourself in disagreement  you would be wiser to question your own understanding first, instead of assuming that we are all wrong and you are right. By all means, bring up your questions. Ask why this isn't so. Demand explanations of statements you don't understand or sound false. But don't assume that everyone else doesn't know what they are talking about. on Sep 8^{th}, 2003, 10:59am, matt wrote:somebody said this, but I'm on the preview page and can't see his/her name (sorry): 
 I usually open up a second session. This allows me explore in it while creating my post in the first. I can even quote from multiple posts by doing a quote in the second and copying it back into the first. Of course, if you don't have some form of broadband, that may not be practical. Quote:yeah... my mistake. I didn't mean to put the word "so" there or imply any sort of causality. I meant to prevent anyone with replying that 3/3 directly gives 0.9..., as that would be to assume the proof (like I ended up doing by mistake). I was just typing faster than I was thinking. 
 No  I never do that! But you are wrong. You can directly get 0.999... from 3/3, as Towr has demonstrated. Quote:I still challenge anybody to come up with a ratio that is == 0.9... that doesn't include some sort of unquantifiable expression (such as (INF1)/INF == 0.9...). 
 Such an expression is not acceptable, but again this misses the point. You always get 0.999... from a/a expressions, because 0.999... = 1. [/quote] Quote:And I still won't concede that 0.9... really means the limit of 0.9... because that's just as circular as me! 
 I am not clear as to what you think is being said here. There is nothing even remotely circular about it. Either you do not have a good understanding of the concept of a limit of a sequence, or you have completely misunderstood the notation. The definition of 0.999... (where the dots here means that infinitely many 9s follow), is that 0.999... = lim_{n[to][subinfty]} A_{n} Where {A_{n}} is the sequence A_{1} = 0.9 = 9/10 A_{2} = 0.99 = 99/100 A_{3} = 0.999 = 999/1000 ... Why do you think this definition is circular? Are you under the mistaken impression that 0.999... is itself in the sequence? It isn't. Each of the sequence elements only has a finite number of 9s. Sometimes we tend to browbeat too much, but I would honestly like to discuss this with you  and even more with that calc teacher! Anyone teaching mathematics really needs to know better than that! Backing up to something else... on Sep 8^{th}, 2003, 10:38am, matt wrote: I really don't know what you're trying to say... but 1/3 != 0.3... because 0.3... is the base10 approximation of a number that is unrepresentable with that notation. 
 I addressed this as the first misconception, and what I said there is true. 0.333... is NOT an approximation. All finite length strings of threes are approximations of 1/3, with the approximations improving as the string length increases. By definition, 0.333... is the limit of these approximations. That limit is exactly 1/3. Quote: 0.3... is by its very nature something that doesn't really exist (in the sense of a concrete quantifiable number). 
 How so? 1/3 is quite quantifiable, and it is what 0.333... means, by the definition. I assume what you are refering to here is the inability to ever write down an infinite number of 3s. This is misconception number (4) in my list. I don't have to write down infinitely many threes to represent the IDEA of it. You yourself have written down exactly that idea many times. To claim it is impossible would be disingenuous. Quote: But, the limit of 0.3....... n3's as n>INF is definitely 1/3. As the limit of 0.9..... (n 9's) as n>INF is definitely 1. But it's not 0.9... (n9's). That's self defeating and stupid. 
 Was that last "(n9's)" a mistype? No one ever said limit was 0.9... (n9's). Quote:Granted I've never heard of hyperreals, and I've only progressed through Calculus 2. 
 hyperreals were a chimera someone threw out. This whole situation is defined in the real numbers. No "superset" of numbers is going to change the meaning in the real numbers. Quote:But I think I've got a pretty firm grip on things. 
 Sorry, but no. Quote:Last I checked, putting 0.9... and saying "It really should be the limit of 0.9..." doesn't change the original question. 
 When you write down "0.9", do you say "it really should be 9/10". If so, you've got a problem. Because you need to change that to "(1+1+1+1+1+1+1+1+1)/(1+1+1+1+1+1+1+1+1+1)". But no, you then need to exchange that for the definitions of "+", "/", "()", and of course "1". Nicolas Bourbaki in the formalism he developed in "The Elements of Set Theory" estimated that to express 1 in the primative elements of his set theory would require several tens of thousands of symbols. I hope you have a lot of time, as you will need to write it down 19 times! "+" is worse yet, and "/" will be a real killer. All notations are defined as shorthand representations of something else (with the exceptions of a few primatives, which are defined in terms of how they relate to each other). "0.999..." is a defined notation which stands for the concept of infinitely repeating 9s, which in turn stands for the limit of the real numbers represented by finite numbers of 9s, as the number of nines increases without bound. An examination of the formal definitions given above, or from any reference you can find, will show that these are grounded in solid finite terms, without any shaky handwaving. Quote: And if someone told you there's an assumed limit in "0.9..." you should get your money back for that course. 
 SOMEBODY should, anyway! Quote: Of course, I'm one of those idiots who doesn't see a major problem with 0.0....1, so I must be braindead for thinking that an alternative representation for the calculus h constant is valid. 
 [smiley=h.gif] is a real variable. Ie. just like every other real variable, it's values are all real numbers. The problem with 0.0...1 (with the dots representing infinitely many zeros) is that no one has defined it. 0.999... is a defined expression. 0.0...1 is not. Before you can start talking about 0.0...1, you need to define what number it represents. This can be done, but you either end up with something other than real numbers (so [smiley=h.gif] does not take them on as values in the standard real calculus), or else you just end up with another expression for 0 (so [smiley=h.gif] is not it either, since in the application you are refering to  definition of the derivative  [smiley=h.gif] is not allowed to be 0), or you get something that does not behave at all like other decimals and so is worthless for any sort of calculation. In particular, you are comparing 0.0...1 to a variable expression. All decimals refer to a single number. But you are arguing this new expression should be regarded as a variable (which can take on many values)?


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Icarus
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Re: 0.999.
« Reply #261 on: Sep 8^{th}, 2003, 6:51pm » 

on Sep 8^{th}, 2003, 1:10pm, Sir Col wrote:Maybe, as Matt says, and I'm probably showing my ignorance again here, the decimal representation is incomplete (as it can never be completed) and is not the same as the rational representation. 
 ?? Would you mind clarifying? What do you mean by "incomplete". How is it that "it can never be completed"? The only interpretation I can find for this is yet another instance of the "infinite process" fallacy. That infinite decimals require infinite processes. Conceptually, there is such a thing. But in the actual mathematics, there is not. We define the outcomes of our conceptual infinite processes in a finite fashion. If we cannot find such a finite definition, then the idea remains undefined. That is not the case here. Quote:However, I think that S. Owen made some excellent points, as the heart of the question is, "What does 0.999... mean?" I agree with him, the only sensible interpretation is to extrapolate the infinite expansion of the decimal: 0.9+0.09+0.009+... 
 Yes, when trying to determine what something means, the definition is a REALLY GOOD place to start!


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Smart Guy
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Re: 0.999.
« Reply #262 on: Oct 3^{rd}, 2003, 4:07am » 

on Jul 27^{th}, 2002, 6:32am, Kozo Morimoto wrote:"A nonmath solution: What number is between .9999... and 1? See? There isn't one. Hence: .9999... = 1" Of course there is, we just can't express it with the Arabic numeral that we use. Lets create a new digit # which is bigger than 9 but less than 10. Then .999999... < .####..... < 1 so there is a number between .999... and 1, but we just can't express it with the digits that we NORMALLY use. Just because we can't express it doesn't mean that it isn't there. 
 The answer is (0.9999......1)


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Sir Col
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Re: 0.999.
« Reply #263 on: Oct 3^{rd}, 2003, 6:02am » 

I'm afraid not, Smart Guy. You 'answer' terminates, by virtue of having a 1 at the end. In which case consider these two numbers: 0.999...1 0.999...9 Which is bigger? Now consider these two: 0.999...9 0.999...99 As soon as you terminate the decimal, it ceases to be 0.999... (recurring). And Icarus, I must object to your sarcasm: Quote:Yes, when trying to determine what something means, the definition is a REALLY GOOD place to start! 
 You'd almost pass for being English.


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ThudnBlunder
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Re: 0.999.
« Reply #264 on: Oct 3^{rd}, 2003, 12:24pm » 

Quote:You'd almost pass for being English. 
 ...or a scriptwriter for The Hitchhiker's Guide to the Galaxy??

« Last Edit: Oct 3^{rd}, 2003, 12:25pm by ThudnBlunder » 
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Icarus
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Re: 0.999.
« Reply #265 on: Oct 3^{rd}, 2003, 5:02pm » 

on Oct 3^{rd}, 2003, 6:02am, Sir Col wrote:You'd almost pass for being English. 
 That whirring sound you hear is my Scottish forebears spinning in their graves at the very thought!


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william wu
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Re: 0.999.
« Reply #266 on: Oct 6^{th}, 2003, 12:47pm » 

on Oct 3^{rd}, 2003, 12:24pm, THUDandBLUNDER wrote: ...or a scriptwriter for The Hitchhiker's Guide to the Galaxy?? 
 Haha I think Icarus would be perfect for such a position P.S. Thanks again Icarus for your uncanny summaries of this topic

« Last Edit: Oct 6^{th}, 2003, 12:48pm by william wu » 
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johnkil
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Re: 0.999.
« Reply #267 on: Oct 11^{th}, 2003, 1:37pm » 

Graphically, If, as x approaches some number c, the values "absolute value of R(x) as it approaches infinity, then the line x=c is a vertical asymptote of the graph of R with an asymptote of x=1, c=0.9999999... (ad infinitum) c+0.x would approach 1 but NEVER intersect. therefor 0.999999... < 1 !


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ThudnBlunder
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Re: 0.999.
« Reply #268 on: Oct 11^{th}, 2003, 1:49pm » 

Quote:c+0.x would approach 1 but NEVER intersect. therefor 0.999999... < 1 ! 
 Graphically, it intersects at infinity. And how many 9's does 0.999999... contain?? An infinite number of 9's. Therefore 0.999999... == 1 Duh.

« Last Edit: Oct 11^{th}, 2003, 1:51pm by ThudnBlunder » 
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S. Owen
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Re: 0.999.
« Reply #269 on: Oct 11^{th}, 2003, 2:14pm » 

Yes, the argument that 0.999... < 1 because graphically the sequence of points (1, 0.9), (2, 0.99), (3, 0.999) approaches y=1 asymptotically is incorrect. However I don't so much like the counterargument that it "intersects at infinity", since it just begs another question  what does that mean? It also perpetuates this way of analyzing the problem  where does the sequence intersect y=1? It doesn't, but the more important point is what about this scenario corresponds to 0.999...? The best point to make that 0.999... actually corresponds to that line which the sequence approaches, and does not correspond to the sequence of points. johnkill: if you're not convinced, again I would like to trot out the fast counterargument: Let 0.999... = x. If x < 1, then consider the number y = (x+1)/2. x < y < 1, but what would the decimal representation of y be? 0.999...?


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Icarus
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Re: 0.999.
« Reply #270 on: Oct 11^{th}, 2003, 6:35pm » 

I lumped this argument (it has been made before) in with some similar arguments as misconception (2). I suppose I should have listed it separately. But it boils down to the same thing: 0.999... is the limit, not the sequence or any member of the sequence. The limit is not the approximations, but the thing that is being approximated. Therefore, as S.Owen has well said: 0.999... is the asymptote, not the approaching curve! However, concerning your counter demonstration, S. Owen, I reiterate my disgreement to its worth. The reason that some people are convinced by this simple argument is that they do not understand the subject well enough to spot what needs to be demonstrated to make it rigorous. For this to be a rigorous demonstration, you must first show that every real number has a decimal notation, and that to be greater than 0.999... and 1 requires digits strictly between 9 below and (1)0 above. The latter is wellknown if not wellunderstood. The former requires a much deeper knowledge of real numbers than people who are confused on this issue have. Just as you do not like T&B's argument, I do not like this one that relies for its simplicity on the ignorance of the person hearing it. Both arguments are valid, but only with enough sophistication on the part of the listener that 0.999... = 1 should no longer be an issue for them.


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random0
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Re: 0.999.
« Reply #271 on: Nov 14^{th}, 2003, 2:53pm » 

I suppose this whole argument is about base 10?


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Icarus
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Re: 0.999.
« Reply #272 on: Nov 14^{th}, 2003, 3:57pm » 

Yes, the question refers to base 10 only. Though a similar thing occurs in any base: 0.111..._{2} = 1 0.222..._{3} = 1 0.FFF..._{16} = 1


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Sir Col
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Re: 0.999.
« Reply #273 on: Nov 14^{th}, 2003, 4:27pm » 

Boy it takes a long time to post a message in this thread. Any chance of starting a sequel thread and placing Icarus' excellent summary at the top, and maybe a link to this old thread? It's funny, I was teaching one of my younger classes (1213 year olds) how to write fractions as decimals (manual division to establish cyclic patterns), and then looking at how we write decimals as fractions. A natural question, when looking at decimals being written as fractions, is whether or not all decimals can be converted to fractions. As we'd established that all fractions recur or terminate they wondered if there are any other types of decmial. Once we wrote down a few examples (I've avoided the terminology with them for the moment), they decided to stick to recurring or terminating decimal types. This is where it got really interesting. After me showing them how to write recurring decimals as fractions (using the algebraic approach), they wondered if it was reasonable to suppose that, just because all fractions will recur or terminate, all recurring decimals necessarily come from fractions. When I presented them with a few practice questions: 0.666..., 0.0555..., 0.232323..., 0.999..., you can imagine their reaction when they came across the last one. Their question, in which they were asking to be convinced that all recurring decimals necessarily come from fractions, I found quite interesting. I wonder if this is the heart of the problem for some people in understanding the equivalence of 0.999... and 1?

« Last Edit: Nov 14^{th}, 2003, 4:31pm by Sir Col » 
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TimMann
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Re: 0.999.
« Reply #274 on: Nov 14^{th}, 2003, 11:47pm » 

I don't think I understand their question. Did they mean, "is every repeating decimal equal to some fraction"? And by "fraction", did they mean "rational number", or did they think 0.999... = 1 is an exception because 1 "isn't a fraction", or because 9/9 (if you showed them that method of converting a repeating decimal to a fraction) isn't a proper fraction?

« Last Edit: Nov 14^{th}, 2003, 11:50pm by TimMann » 
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