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Doubts about 0.999 being equal to 1

  1. Oct 24, 2006 #1
    If anyone has any further doubts about 0.999... being equal to 1, please direct your attention to today's featured article on http://en.wikipedia.org/.
     
  2. jcsd
  3. Oct 25, 2006 #2
    There are only 500 threads about this...
     
  4. Oct 25, 2006 #3

    arildno

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    Eek! Aargh! Grumble, shriek, not another thread on this!!:grumpy: :grumpy: :grumpy:

    Hung, drawn and quartered is too good for you! :grumpy: :grumpy: :grumpy:
     
  5. Oct 25, 2006 #4
    Really? It became featured article only yesterday.
     
  6. Oct 26, 2006 #5

    matt grime

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    500 hundred threads *here*, all by people arguing the point.
     
  7. Oct 26, 2006 #6

    uart

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    I did think the following quote from the linked article was interesting though. :)
     
  8. Oct 26, 2006 #7
    i honestly don't see how anybody can reject the equality.
     
  9. Oct 26, 2006 #8

    HallsofIvy

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    All that is required is a firm belief that a "number" is exactly the same as the "numeral" used to represent it. Since "1" and "0.9999...." are different it follows immediately that they are different numbers! Just like 0.5 and 1/2are different numbers!

    Actually not too many people will deny that 0.5= 1/2 but people who deny 1= 0.9999... are likely to deny 1/3= 0.3333....!
     
  10. Oct 28, 2006 #9

    StatusX

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    It's too bad everyone here is too well-informed, and there's no one left to argue the other side. It's kind of fun seeing what they can come up with when backed into a corner. Just for fun, here's how I would lay out the argument:

    1. [tex] 0.999... = \sum_{n=1}^\infty \frac{9}{10^n}[/tex]
    (by definition)

    2.[tex] \sum_{n=1}^\infty \frac{9}{10^n} = \lim_{N \rightarrow \infty} \sum_{n=1}^N \frac{9}{10^n} [/tex]
    (by definition)

    3. [tex] \sum_{n=1}^N \frac{9}{10^n} = \frac{(9/10) - 9/(10^{N+1}) }{1-(1/10)} = 1-{\left( \frac{1}{10} \right) }^N [/tex]
    (easy to show with algebra, and not many people would deny it because there aren't any infinities)

    4. Let [itex]\epsilon>0[/itex]. Then there is some N with [itex](1/10)^n<\epsilon[/itex] for all n>N.
    (again, pretty intuitive, doesn't involve infinites yet)

    5. [tex]\lim_{N \rightarrow \infty} {\left( \frac{1}{10} \right) }^N =0 [/tex]
    (by (4) and the definition of a limit)

    6. [tex]\lim_{N \rightarrow \infty}1- {\left( \frac{1}{10} \right) }^N =1 [/tex]
    (by (5) and the continuity of subtraction)

    7. Therefore, by (1),(2),(3),(6), and the transitivity of equality
    [tex]0.999... = 1 [/tex]

    Now, if they want to deny the conclusion, they have to pick a premise to deny. (1) or (2) would just be disagreeing with definitions everyone else uses, and there's no point in arguing about something like that. (3) and (7) are pretty undeniable. (4) is a little tricky, but again, there are no infinities, so I don't think it would be too hard to convince people of. (6) is actually the most technical line, but I think it jives with intuition, so I don't think it would be a problem.

    That leaves us with (5). Again, debating this would just be disagreeing with a definition. The essence of the problem is that people have a preconceived notion of what a limit is, as some sort of process, but this doesn't agree with the epsilon delta definition (in fact, it doesn't really make sense at all). If they can understand this definition, then I don't see how they could both accept all the definitions used above and still deny that 0.999...=1.
     
    Last edited: Oct 28, 2006
  11. Oct 28, 2006 #10

    HallsofIvy

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    The usual "objection" is that 0.999... is not the LIMIT of the sum, but the sum itelf, which the typical objector does not realize IS a limit itself.
     
  12. Oct 28, 2006 #11

    HallsofIvy

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    The usual "objection" is that 0.999... is not the LIMIT of the sum, but the sum itelf, which the typical objector does not realize IS a limit itself. That's what I meant when I said it is a confusion of "number" with "numeral".
     
  13. Oct 28, 2006 #12
    why don't people disagree with 1.000...=1. seems almost the same as .999...=1 to me.
     
    Last edited: Oct 29, 2006
  14. Oct 29, 2006 #13

    HallsofIvy

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    Because "zeroes don't count".
     
  15. Oct 29, 2006 #14

    Hurkyl

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    Okay, guys. A lack of people arguing [itex]0.\bar{9} \neq 1[/itex] isn't a bad thing, and we don't need to be picking up the slack. :tongue:
     
  16. Oct 30, 2006 #15
    Let c = 0.9...

    10c = 9.9...
    10c - c = 9
    9c = 9
    c = 9/9


    "If you don't know where you are going, any road will get you there."
    -- Lewis Carroll
     
  17. Oct 30, 2006 #16

    HallsofIvy

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    To accept that you have to accept that ordinary arithmetic operations work on infinite sequences as well as finite. That's true but no more obvious than that 0.999...= 1 to begin with.

    And having had the last say, I'm locking this silly thread!
     
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