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Math Help - Proof using Ordered Field Axioms

  1. #1
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    Proof using Ordered Field Axioms

    Hi,
    I have to prove that
    if 0 < a < b, then 0 < b-1 < a-1; for a,b,c element Real Numbers using the ordered field properties.
    (b-1 and a-1 means b inverse and a inverse)
    Thanks
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  2. #2
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    Quote Originally Posted by tbyou87 View Post
    Hi,
    I have to prove that
    if 0 < a < b, then 0 < b-1 < a-1; for a,b,c element Real Numbers using the ordered field properties.
    (b-1 and a-1 means b inverse and a inverse)
    Thanks
    Some definitions.

    Given a ordered ring R we defined P (the positives) to be such a subset (if it exists) such that,
    1)"Closure" a+b in P ad ab in P.
    2)"Trichtonomy" a in P or -a in P or 0 in P and exactly one of these three.

    We define a>b iff (a-b) in P.

    Thus, given,
    0<a<b we need to show, 0<1/b<1/a

    By definition, we need to show that,
    1/a-1/b in P.
    But,
    1/a-1/b =b/ab - a/ab = (b-a) * (1/ab)
    But,
    (b-a) in P.
    And 1/a and 1/b in P. Thus, 1/ab in P because of closure.
    But then,
    (b-a)*(1/ab) in P because of closure.
    Q.E.D.
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