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Math Help - Another proof using Ordered Field Axioms

  1. #1
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    Another proof using Ordered Field Axioms

    Let a, b element R. Show that if a <= b1 for every b1> b, then a <= b.

    I think the triangle inequality may be helpful in this but i'm not sure. I really don't know where to begin here.

    Thanks
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  2. #2
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    Quote Originally Posted by tbyou87 View Post
    Let a, b element R. Show that if a <= b1 for every b1> b, then a <= b.

    I think the triangle inequality may be helpful in this but i'm not sure. I really don't know where to begin here.

    Thanks
    Theorem: If R is an ordered field, then for any positive element x there shall exist another positive element y such that y<x.

    Proof of Problem:
    Assume by contradiction that a>b.
    Then, a=b+c where c is positive, i.e. c=b-a.

    Choose b1=b+d where 0<d<c as in above theorem.

    Then,

    b1=b+d<b+c=a and b+d=b1>b.
    A contradiction.
    Thus, a<=b.
    Q.E.D.
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