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Math Help - Proof 0 < 1 in an ordered field

  1. #1
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    Proof 0 < 1 in an ordered field

    Hi,
    I know you are supposed to prove that 0 < 1 by contradiction, but i'm not sure how to do it exactly using only the ordered field axioms.
    Thanks
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  2. #2
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    Quote Originally Posted by tbyou87 View Post
    Hi,
    I know you are supposed to prove that 0 < 1 by contradiction, but i'm not sure how to do it exactly using only the ordered field axioms.
    Thanks
    Theorem: If F is an ordered field, then the square of any non-zero element is positive.

    Proof: Either x>0 or x<0, by trichtonomy theorem. If x>0 then x*x=x^2>0 by closure property.
    If x<0 then -x>0 trichtonomy property but then, (-x)*(-x)=x^2>0 by closure property.
    Q.E.D.

    Corollary:In an ordered field (note it cannot be trivial). 1>0

    Proof:
    1=1^2 use above result.
    Q.E.D.

    ----
    Note, this prooves the complex numbers cannot be ordered because 1^2>0 bet yet i^2=-1<0.
    Contradiction.
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  3. #3
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    1 > 0

    Assume that the above theorem is false. We know that 1 cannot be equal to 0 and by trichotomy, the only option left is : 1 < 0.

    If that is indeed true, then -1 > 0. Then if we square the expression

    (-1)^2 > 0 and we will get 1 > 0. This contradicts the assumption that 0 < 1.

    Hope that helps.
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