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Proving that the Archimedean axiom is true

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that the Archimedean axiom O5 follows from the Least Upper Bound Property O6, together with the other axioms for the reals.

    2. Relevant equations
    O5 = [if a,b > 0, then there is a positive integer n such that b<a+a+a+...+a (n summands)] or [if a,b > 0, then b < na or b/a < n]

    O6 = if A is any nonempty subset of R that is bounded above, then there is a least upper bound for A.

    3. The attempt at a solution
    My teacher told us to do this as a proof by contradiction so that's the format I'll be doing.

    Suppose the Archimedean axiom is false towards a proof by contradiction. Therefore, there exists some a,b > 0 such that b [itex]\geq[/itex] na, or b/a [itex]\geq[/itex] n.
    Then the set, say N, is bounded above by b/a and so sup(N) exists. Write sup(N) = S.

    And then I can't figure out how to finish this proof.
     
    Last edited: Sep 13, 2011
  2. jcsd
  3. Sep 13, 2011 #2

    micromass

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    That [itex]S=\sup(\mathbb{N})[/itex] means that there is a natural number n that is close to S. But then n+1>S...

    Try to formalize this.
     
  4. Sep 13, 2011 #3
    If n+1>S, then there exists a natural number not bounded above by S.
    This is a contradiction as the set N is the set of whole positive integers and adding 1 would not exclude any n previously in the set N.

    Is that right?
     
  5. Sep 13, 2011 #4

    micromass

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    Yes, that is correct. How would you choose n though?
     
  6. Sep 13, 2011 #5
    I would choose n to be close to S, or S-1<n<S.

    So for any n bounded above by S but greater than S-1, if n+1>S, then there exists a natural number not bounded above by S.

    And so on, and so on. Is that specific enough?
     
  7. Sep 13, 2011 #6

    micromass

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    Yes, but you need to state why such an n exists. You probably know it, but I want to make sure.
     
  8. Sep 13, 2011 #7
    Oh. Um, it exists because S is the supremum of the set?
     
  9. Sep 13, 2011 #8

    micromass

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    Yes, do you understand why?
     
  10. Sep 13, 2011 #9
    S is the supremum of the set because the set is bounded above by b/a, which is what sup(N[/]) is defined as at the beginning of the proof (b/a ≥ n).
     
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