# Proof using Ordered Field Axioms

• Apr 7th 2007, 04:44 PM
tbyou87
Proof 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
• Apr 7th 2007, 05:51 PM
ThePerfectHacker
Quote:

Originally Posted by tbyou87
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

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.