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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- Apr 9th 2007, 02:35 PMtbyou87Proof 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 - Apr 10th 2007, 07:15 PMThePerfectHacker
**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.

Contradiction. - Apr 10th 2007, 08:24 PMWintaker99
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.