# Thread: Proof 0 < 1 in an ordered field

1. ## 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

2. Originally Posted by tbyou87
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.

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

3. 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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