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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- April 7th 2007, 04:44 PMtbyou87Proof 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 - April 7th 2007, 05:51 PMThePerfectHacker
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.