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The next time I hear .9~ isn't 1 I'm

  1. Jan 5, 2005 #1

    Alkatran

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    going to say "Well of course it's not 1, if it was 1, we'd write it as 1! The important thing is it is EQUAL to 1. If you say a = 2 and b = 2, b isn't a, but it does EQUAL a: they both point to the same value."

    Then I'll get a blank stare.
     
  2. jcsd
  3. Jan 5, 2005 #2
    .9...i and 1 are mathamaticly the same, but logically different. Yes, .9i and 1 are written differently, but are mathamaticly the same.
     
  4. Jan 5, 2005 #3

    matt grime

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    In what sense do you mean logically different?
     
  5. Jan 5, 2005 #4

    Zurtex

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    I'm somewhat confused, do you mean in the same way that [itex]\sin^2 x[/itex] and [itex]1 - \cos^2 x[/itex] are different?
     
  6. Jan 5, 2005 #5

    HallsofIvy

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    You have been told repeatedly that .9i has a standard meaning quite different from the way you are using it. Are you doing that intentionally?

    In any case, 0.999..., with the 9 infinitely repeating, is simply a different notation for the number 1, in precisely the same sense that 0.5 and 1/2 are different notations for the same number. I agree completely that "Yes, .9i and 1 are written differently, but are mathamaticly the same" but I don't know why you would call them "logically different".
     
    Last edited: Jan 5, 2005
  7. Jan 5, 2005 #6

    Alkatran

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    Two pointers to the same thing, yes.
     
  8. Jan 5, 2005 #7

    Zurtex

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    Sorry I should have made myself more clear, I understood your post just not Gamish's.
     
  9. Jan 6, 2005 #8
    .99999..... and 1 are still different in the realms of hyperreals. But ofcourse they are hyperreals and Non-Standard Analysis would make no sense to a common man or maybe it does, thats why he believes they are different. :biggrin:

    -- AI
     
  10. Jan 6, 2005 #9

    matt grime

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    Are they? The difference is infinitesimal and real hence it is zero. Or am I thinking of something else?
     
  11. Jan 6, 2005 #10
    Yeah they are..
    Consider them as hyperreals and their representations,
    {1,1,1,1,1,1,1,1,1,1,...................} and {0.9,0.99,0.999,0.9999,...}
    Since the terms in the first sequence are larger than every corresponding term in the second sequence, 1 > 0.9 in hyperreal.

    -- AI
    P.S -> Ofcourse my understanding of hyperreals is very minimal , considering i read most of hyperreals from this place ...
    http://mathforum.org/dr.math/faq/analysis_hyperreals.html

    So if any mistake above is purely my inability to grasp the subject properly and nothing else.
     
  12. Jan 6, 2005 #11

    matt grime

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    I too know very little about non-standard analysis. But, from what I do know, one of its features is that standard analytic results can be recovered from it. Now, you've just lost, for instance, the mean value theorem, and uniqueness of limits in R, when you say that.

    The hyper reals are an extension of the reals, and 0.99... is still 1.

    Also 1-0.999... is still real and an infinitesimal, hence it is 0.
     
  13. Jan 6, 2005 #12

    Hurkyl

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    There are some subtleties here...


    In the hyperreals, 0.999... does indeed equal one. The difference is that (when stepping back into set theory) there are a heck of a lot of more 9's in this expression.

    If you tried to directly transplant the standard 0.999... into the hyperreals (rather than doing a transfer, like you're supposed to), you'd find that it's undefined. The countable sequence 0.9, 0.99, 0.999, ... does not converge. (However, doing a transfer, so it's indexed by the hypernaturals instead of the naturals, gives a sequence that converges to 1)


    And yes, the element {0.9, 0.99, 0.999, ...} of the countable ultraproduct does map onto a hyperreal different from 1.
     
  14. Jan 6, 2005 #13

    Integral

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    What's this??? A thread on .999... =1 not started by a "non believer".~^


    I'll bet you all are just waiting for some poor sucker to step in here and say, "It can't be!"

    :biggrin:
     
  15. Jan 6, 2005 #14

    Alkatran

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    If I wanted that I would go to a non-math forum.
     
  16. Jan 6, 2005 #15
    Maybe there is a way to capitalise on this...
     
  17. Jan 6, 2005 #16
    You can't ever write the 0.9... exact. So, we use something like 0.9... with dots to say that the nine is extended to infinity... but we can't reach infinity never. In the real world, there is always an uncertainty (I'm not talking about Heisenberg this time!).

    For example, we say that [tex] \pi = 3.141592... [/tex] or [tex] e = 2.7182.... [/tex]. In theory, we use [tex]\pi, e, \hbar \ldots[/tex] but doing the Math, we can't never use the exact value, because we don't know it !

    The same thing occurs with 0.9... . It is fairly easy to demonstrate that is equal to one. For example, this one:

    1/3 × 3 = 0.333.... × 3 = 0.999....
    1/3 × 3 = 1

    So 1 = 0.999....

    Or this one. If you want to know if a periodic number is rational, you can do this:

    (1) 0.5353535353... = x
    (2) 53.53535353... = x * 100

    (2) - (1)

    53 = 99x so x = 53/99 = 0.5353535353...

    So, we can do with 0.99...

    0.999.... = x
    9.999.... = x * 10

    9 = x * 9

    x = 1

    So 1 = 0.999....

    Why we don't use 1 instead of 0.999...? Because there are not the same thing ! Both are represent the same value, but 1 is a special number. 1 is the neutral element for multiplication, and also the generator of the natural numbers. It has a lot of algebraic propierties that 0.999... don't verifies. 0.999... is not a natural number, but it is equal to one!

    This can sound strange, but occurs many times. We know that 8/2 is equal to 4 but it is written very different.
     
    Last edited: Jan 6, 2005
  18. Jan 6, 2005 #17

    matt grime

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    Hmm, do you want a list of the slightly dubious points in that post?

    You could start with "the exact value" bit, that needs to be removed.

    0.99.. has exactly the same algebraic properties as 1, that is why they are equal.
     
  19. Jan 6, 2005 #18
    So in other words, you're saying that infinitesimals are necessarily quantized. :approve:

    I actually knew this all along.
     
  20. Jan 6, 2005 #19

    matt grime

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    They aren't so I guess you were wrong all along (if e is an infinitesimal, then e^2 <e I believe, though neither of us knows the first thing about them do we?)
     
  21. Jan 6, 2005 #20
    If you genuinely want to talk about this from a mathematical point of view you should recognize that the controversy surrounding this topic is almost entirely due to the fact that mathematicians fail to include the phrase "in the limit" when talking about this idea with laymen or philosophers.

    When a mathematician says that 0.999… = 1, he or she really means that "in the limit 0.999… = 1". This is necessarily so because this is, in fact, included in the formal definition of 0.999… = 1.

    If mathematicians would simply include this little crucial tidbit of information much of the controversy surrounding this topic would fade away.

    No mathematician in his or her right mind would suggest that 0.999… = 1 outside of the concept of the limit. Actually if they did that they would be incorrect anyway, and couldn't possibly claim to be speaking for the mathematical community as a whole.

    The concept of 0.999… = 1 is formally defined on the concept of the calculus limit. Therefore when a mathematician is speaking to a non-mathematician he or she should include the phrase "in the limit" when referring to this definition. Mathematicians refrain from including this phrase when speaking with other mathematicians because any mathematician worth his salt is fully aware of the role that the calculus limit plays in this definition.

    So I really wish than mathematicians would start posting on a public forums the words "in the limit" whenever the concept of the limit is involved with a definition. It's actually more precise and correct to do this. It can also serve to get the general public more interested in learning and understanding the formal concept of a limit.
     
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