Another proof using Ordered Field Axioms

Let a, b element R. Show that if a <= b1 for every b1> b, then a <= b.

I think the triangle inequality may be helpful in this but i'm not sure. I really don't know where to begin here.

Thanks

Quote:

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Originally Posted by tbyou87

Let a, b element R. Show that if a <= b1 for every b1> b, then a <= b.

I think the triangle inequality may be helpful in this but i'm not sure. I really don't know where to begin here.

Thanks

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Theorem: If R is an ordered field, then for any positive element x there shall exist another positive element y such that y<x.

Proof of Problem:
Assume by contradiction that a>b.
Then, a=b+c where c is positive, i.e. c=b-a.

Choose b1=b+d where 0<d<c as in above theorem.

Then,

b1=b+d<b+c=a and b+d=b1>b.
A contradiction.
Thus, a<=b.
Q.E.D.