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Supremum Proof of .999 = 1

  1. Jul 24, 2011 #1
    I know .999... = 1. I'm just arguing against this method of proof.

    A common proof I see that [itex].999 \ldots = 1[/itex] is that [itex]sup\{.9, .99, .999, \ldots \} = 1[/itex], but this is only true if you assume [itex].999 \ldots \ge 1[/itex]. If you assume, as most argue, that [itex].999 \ldots < 1[/itex], then [itex]sup \{.9, .99, .999, \ldots \} = .999 \ldots < 1[/itex]. Of course, by assuming [itex].999 \ldots < 1[/itex], you get the nonsense expected at the end of a proof by contradiction, but you still have to proof that [itex].999 \ldots < 1[/itex] is nonsense by proving [itex].999 \ldots = 1[/itex]. Therefore, the supremum method is useless.

    Is my logic correct?
     
    Last edited: Jul 24, 2011
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  3. Jul 24, 2011 #2

    Hurkyl

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    I'm not entirely sure. I think no, because you seem to be confusing the proof sketch with the actual proof. Computing that supremum is the idea of the proof. The actual proof consists of actually carrying out the relatively easy computation. (assuming it's being presented in a context where calculation is expected to be so obvious/trivial it can be emitted)
     
  4. Jul 24, 2011 #3
    Okay we just had a first look at sups, but I thought I understood it to mean : sup [-1,1) = 1
    for example.
    So, why should sup{.9999...}=1 only if .999....>=1? shouldn't it require lessthan/equal 1?
     
  5. Jul 24, 2011 #4
    Assume 1 > .999... Obviously, 1 is greater than any number in {.9, .99, .999, ...}, but so is .999... So, since the supremum is the lowest upper bound, and 1 and .999... are bother upper bounds, .999... is the sup or lub.

    sup is the lowest upper bound, so if a is the lowest upper bound, and b is an upper bound, then b >= a. This is a direct consequence of the definition of sup.
     
  6. Jul 24, 2011 #5

    lavinia

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    .9999... is meaningful as a limit. The partial sequences .999... of a finite number of 9's are a Cauchy sequence.The difference 1 - ,9999... of 1 with n 9's is .00000...1 with n+1 leading zeros.
     
  7. Jul 24, 2011 #6
    Yeah, I've seen the Cauchy sequence proof, too. I see no flawed assumption. But do you see what I mean about the Supremum method?
     
  8. Jul 26, 2011 #7

    Char. Limit

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    I personally prefer the infinite series proof in my signature.
     
  9. Jul 26, 2011 #8
    I don't see a signature. But, I know which one you're referring to.
     
  10. Jul 26, 2011 #9

    Fredrik

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    You can't prove that 0.999...=1 without first defining what the left-hand side means. The standard definition is [tex]0.999\dots=\sum_{n=1}^\infty \frac{9}{10^n},[/tex] but [tex]0.999\dots=\sup\{0.9,0.99,\dots\}[/tex] is a perfectly acceptable alternative definition. What definition are you using?
     
  11. Jul 26, 2011 #10
    It hinges on the definition of limit.
     
  12. Jul 26, 2011 #11

    Char. Limit

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    Unless I make .999...=1 an axiom, I'd have to hinge it on something. I think the definition of a limit is sufficiently stable.
     
  13. Jul 26, 2011 #12
    I don't see a signature either.
     
  14. Jul 26, 2011 #13

    Fredrik

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    What's your point? Every proof of the result 0.999...=1 hinges on the definition of 0.999... and the definitions of the terms used in that definition, and so on, all the way down to primitives (the terms left undefined). The primitives are usually taken to be the concept of "set" and "membership" (what it means for a set to be a member of a set).

    By the way, all the acceptable definitions of 0.999... are equivalent to simply stating that 0.999...=1. Each proof tells us precisely that the definition of 0.999... it relies on is equivalent to 0.999...=1.

    The point of such a proof isn't to find out if 0.999...=1 is "really" true, but to confirm that the definition we chose does the job we intended it to do. A definition of 0.999... that can be used to show that 0.999...≠1 would simply be dismissed.
     
    Last edited: Jul 26, 2011
  15. Jul 26, 2011 #14
    @TylerH:

    So, you just define .999... to be sup{.9, .99, ...} and then show that 1 is also the sup of this set. I think this is they way the proof you mentioned is supposed to work.
     
  16. Jul 27, 2011 #15
    Okay, now that makes sense. It would force .999... = 1 by the transitive property of equality.
     
  17. Aug 8, 2011 #16
    0.111...... * 9 = 0.999......
    but 1/9 = 0.111......
    so 0.999..... = 0.111..... * 9 = (1/9) * 9 = 1
     
  18. Aug 8, 2011 #17
    when I I have first know the recurvesing numbers i used to use this method

    x=0.999.....
    10x=9.999.....
    9x=9
    x=9/9=1


    But you can consider the number as a geometric series
     
  19. Aug 8, 2011 #18

    Char. Limit

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    And indeed, this is what you're doing. Just in a more veiled manner.
     
  20. Aug 8, 2011 #19
    The only problem is that when you multiply .99... by anything, you have to have a definition for what that means. Of course, we think of it as just moving the decimal point over, and this is a good way to give a convincing argument to a layman that .999... is 1, but to do it rigoursly, I think you need to do the series expansion or the supremum thing or something else.
     
  21. Aug 8, 2011 #20

    Hurkyl

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    What was the point of this? What bearing did it have to do with anything in the thread or even the title?

    I don't see any point in reopening the thread, especially in this fashion. Thread closed.
     
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