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B Solve this math?

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  1. Nov 28, 2017 #1
    Let f(n) be of the form F(1) = 0.999999 F(2) = 0.9999999 F(3) = 0.99999999 Etc

    For arbitrary N >= 1

    Let g(n) = f(n)^2

    G(1) and g(2) end in 8

    By linear relation all g(n) end in 8

    Assume 0.999... = 1

    0.999... X 0.999... must end in 8 per (4)

    1 x 1 does not end in 8

    By (5),(6),(7), contradiction

    Therefore 0.999... = 1 is a contradiction
     
  2. jcsd
  3. Nov 28, 2017 #2

    Mark44

    Staff: Mentor

    No. How are you getting 8 as the final digit?
    Let's start in a more organized fashion.
    ##f_1 = 0.9##
    ##f_2 = 0.99##
    ##f_3 = 0.999##
    Then ##f_1^2 = 0.81##
    ##f_2^2 = 0.9801##
    ##f_3^2 = 0.99801##
    and so on. The sequence ##\{f_1, f_2, f_3, \dots \}## is approaching 1, which is easy to prove,
    and it can easily be shown that the sequence ##\{f_1^2, 4_2^2, f_3^3, \dots \}## also has 1 as its limit.
    No
    No.
    Thread closed.
     
    Last edited: Nov 28, 2017
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