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.3333 . equals to

  1. Mar 30, 2015 #1
    I just show how .999.... = 1

    Let x = .999....
    10x = 9.999...
    10x - x = 9.999... - .999...
    9x = 9
    x=1

    So according to the above .3333.... shows as 3.

    Let x = .3333...
    10x = 3.333...
    10x - x = 3.333... - .333...
    9x = 3
    x=3

    But according to me, .333.... should equal to .4 , isn't it?

    I must be missing something simple, what is it?

    Thank you.
     
  2. jcsd
  3. Mar 30, 2015 #2

    Doug Huffman

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    Gold Member

    The fallacy of truncation.
     
  4. Mar 30, 2015 #3

    DrClaude

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    Staff: Mentor

    More like x = 1/3, which is pretty much as expected :wink:
     
  5. Mar 30, 2015 #4

    DrClaude

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    Staff: Mentor

    Why? There is an infnity of numbers between .333... and .4, 0.334 for instance. The argument for .999.... only works because there can't be any number between .999... and 1.
     
  6. Mar 30, 2015 #5
    My mistake,
    9x = 3
    so x=1/3.
    Thanks. Next number after .333.... can be .333334 or .3333334 or .3333333333334.
     
  7. Mar 30, 2015 #6

    Mark44

    Staff: Mentor

    There is no "next number" after .333... Integers have such a property, but the rationals and real numbers don't.
     
  8. Mar 30, 2015 #7
    That result comes about from truncating the repetend.

    Try keeping the repetend intact, like this:
    X = .9999…..
    X = 9 x .1111….
    10X = 10(9 x .1111….)
    10X = 90 x .1111….
    10X – X = (90 x .1111….) – (9 x .1111….)
    9X = 81 x .1111….
    X = 9 x .1111….
    X = .9999….
     
  9. Mar 30, 2015 #8

    Mark44

    Staff: Mentor

    So you have proved that if x = .999..., then x = .999...
    This is true, but not useful.
     
  10. Mar 30, 2015 #9

    Mark44

    Staff: Mentor

    The fallacy of dividing 3 by 9 and getting 3.
     
  11. Mar 30, 2015 #10
    It also comes about because it is true. This is a shaky method of "proving" it, though.
     
  12. Mar 31, 2015 #11
    Yeah, isn't the set of all integers a countable infinitely-large set of size aleph-null, whereas rationals are reals are not countable; therefore, they're considered to be of a larger infinity known as aleph-1. And then you can get even larger sets by taking the power set of aleph-1, then aleph-2, etc. I might be getting a bit off topic, but you brought up something interesting, though I'm not sure I explained it correctly.
     
  13. Mar 31, 2015 #12

    Mark44

    Staff: Mentor

    Did you mean, "whereas rationals and reals..."
    The rational numbers are countably infinite, just like the integers. The irrational numbers are uncountably infinite.
     
  14. Mar 31, 2015 #13
    Yes, I accept that it is true, (by definition) I just don’t accept that it is proven by the method of truncating the repetend.

    This is a little better, but it still won’t stand up as a rigorous proof:
    The sum of a geometric series is equal to: a(1-r^n)/(1-r)
    Where a is the first term (0.9 in this case), r is the common ratio (0.1)
    if r is less than one, as n goes to infinity the sum becomes just a/(1-r)
    In this case, Sum = 0.9/(1-0.1) =1, therefore 0.999… = 1
     
  15. Mar 31, 2015 #14
    Tom_K, a proof using a geometric series is rigorous. The "by definition" part should be applied to a decimal being defined as a geometric series. The definition of a decimal is (IIRC), ##\sum a_i10^{-i}## where ##a_i## may be equal to zero for all ##i>N## and there are finitely many values of ##i<0##. That is a definition from which 0.99...=1 can be proven.
     
  16. Mar 31, 2015 #15
    Ahhh that makes sense. Rationals and integers are countably infinite, whereas reals are not. Therefore, rationals and integers are of size aleph-null and reals are of size aleph-1. And thank you for the grammatical correction :)
     
  17. Mar 31, 2015 #16

    jbriggs444

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    Science Advisor

    Technically, it is undecidable whether the cardinality of the reals is aleph-1. That is the continuum hypothesis. The cardinality of the reals could be greater than aleph-1. It cannot be smaller.
     
  18. Mar 31, 2015 #17
    Oh, yeah. Basically, it stipulates that no set can have its size between integers and real numbers, but we don't actually know if it's true, right?
     
  19. Apr 1, 2015 #18
    As there is an infinite sequence of 9-s in the decimals, there logically can not be any other real between 0.9(9) and 1 - the density property of the reals states that there is always another real between any 2 reals, conclusion - the 2 reals are equal.
    As to what 0.3(3) is exactly equal to is answered by the axiom of continuity - it's equal to 1/3, since it's the least upper bound for the sequence and likewise there can not be another real between 0.3(3) and 1/3.

    ( I'm curious what the cardinality of reals has to do with the problem at hand )
     
    Last edited: Apr 1, 2015
  20. Apr 1, 2015 #19
    What about a number like 1/9999…… where the decimal place is infinitely far to the Right.
    Could that theoretically fit between 0.9999…… and 1.0 ?
    My only point is that just saying there can be no real number between 0.9999… and 1.0 is not a proof so much as it is a definition.
    I do accept the geometric-series proof I posted earlier that they are the same number.
     
  21. Apr 1, 2015 #20
    1/999.. most certainly does not fit between 0.9(9) and 1.
    As 9 < 99 < 999 < 9999 ... etc, then their reciprocals 1/9 > 1/99 > 1/999 ..., where 1/9 = 0.1(1) < 9*0.1(1) = 0.9(9).
    If you demand a rigorous proof, the problem should be more abstract and we should not be allowed to deal in real numbers, for real numbers have a handful of useful properties to be exploited and neither can you deny the validity of an argument that is based on the axioms and properties of real numbers.
     
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