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B Finitism and .999...=1 question

  1. Oct 3, 2017 #1
    I had a discussion with someone about does .999...=1. She used finitism and I don't know enough about it to reply. I read a little about finitism but I don't know what to make of it, it just seems wrong
    Can someone recommend a site with a good rebuttal
     
  2. jcsd
  3. Oct 3, 2017 #2

    mathman

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  4. Oct 3, 2017 #3

    fresh_42

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    The shortest version: ##0,999\ldots = 1##.
    The short version: The left hand side is a different representation than on the right. So first of all, the left hand side has to be explained. If it is a limit, then it equals ##1##. If it is something else, then what?
    The long version: https://www.physicsforums.com/threads/0-999-0.919086/
     
  5. Oct 3, 2017 #4

    Marc Rindermann

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    ##0.99999..... = 0.\bar9= 3 \times \frac{1}{3} = \frac{3}{3} = 1##

    I don't know anything about finitism. And I also don't know why one needs to be all philosophical about the question.

    As I see it the problem just lies in the way numbers are represented in the decimal number system. Nobody would ever argue that ##\frac{3}{3} = 1##. If we used a number system base 60 like on our clocks where ##\frac{1}{3}## would be something like 20 minutes, the question whether 60 minutes is one hour wouldn't come up either.
     
    Last edited: Oct 3, 2017
  6. Oct 3, 2017 #5

    TeethWhitener

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    Just to pile on, apparently Finitism sees some academic discussion occasionally, but unless your friend is a research logician, she probably doesn't know as much about it as she thinks she does. The type of finitism that asserts ##0.999... \neq 1## also must assert some weird propositions, like ##N+1\ngtr N## for sufficiently large ##N##.
     
  7. Oct 3, 2017 #6
    I suppose one particular way to see this equality can be to see it as a geometric series (probably already mentioned in thread linked above). For example, set:
    a = initial term = 9/10
    r = multiplication factor = 1/10

    Because r<1 we know that the following sum converges:
    a+ar+ar2+ar3+ar4......

    S=a/(1-r)=1

    Interestingly, a similar equality will seemingly follow no matter number-base system we use.

    What is more interesting would be to prove the converse: That is, the only source of ambiguity in decimal representation can occur between:
    ---- decimal representations with infinite strings of 9's
    ---- decimal representations with infinite strings of 0's
    I think it might be a bit tedious to prove but it could be done by a more careful analysis.

    ===============

    Also I think probably the term ultrafinitism might be more in line with regards to question in OP.
     
    Last edited: Oct 3, 2017
  8. Oct 3, 2017 #7

    Mark44

    Staff: Mentor

    Actually, I don't think it would be hard to prove at all. Any decimal fraction (i.e., base-10) that has an expression of the form ##.d_1d_2 \dots n999\dots## can also be represented as ##.d_1d_2 \dots m000\dots##, where m = n + 1. Of course, I'm not proving anything here, but for any given decimal fraction in the first form, you could prove that it is equal to the second form, using the same geometric series argument that you gave.

    The ambiguity of these two decimal fractions has to do with 9 being the largest decimal digit. If we were dealing with base-2 fractions, 1 is the largest binary digit, so a binary fraction of the form ##.d_1d_2\dots n111\dots## would be the same as ##d_1d_2\dots m000\dots##, again with m = n + 1, and this could be proved in the same manner. In base-3, we have the ambiguity in trinary fractions of the form ##.d_1d_2 \dots n222\dots##, as 2 is the largest base-3 digit.
     
  9. Oct 4, 2017 #8
    Yeah, basically I think the most illustrative case would be of:
    1=0.999....
    I think apart from the equality of these (as is mentioned frequently), it is also important to show that 0.9999... is the only other decimal representation for 1**. I don't think it should be quite difficult (but certainly illustrative). Once one shows that it would be reasonably clear that it can be extended as a more general result.

    Furthermore, one would also want to show that apart from certain kind of decimal representations(terminating ones and ones with infinite string of 9's), all other decimal representations uniquely describe a real number.

    ** Obviously assuming the convention that unnecessary 0's towards the end are cut-off. For example, 1.000 is counted as same decimal representation as 1.
     
    Last edited: Oct 4, 2017
  10. Oct 5, 2017 #9

    Math_QED

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    Well, how does she define 0,999... in her view?
     
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