Thread: Ordered Fields, 0<a<b

I just read a proof that says 0 < a ==> 0 < a^-1.

How do I prove 0 < a < b ==> 0 < b^-1 < a^-1?

Also, the text did not give a definition for 0 < a < b.

Would it be correct to assume 0 < a and a < b <=> 0 < a < b, similarly for 0<= a and a<=b <=> 0 <= a <= b?

Originally Posted by Noxide

I just read a proof that says 0 < a ==> 0 < a^-1.

How do I prove 0 < a < b ==> 0 < b^-1 < a^-1?

Also, the text did not give a definition for 0 < a < b.

Would it be correct to assume 0 < a and a < b <=> 0 < a < b, similarly for 0<= a and a<=b <=> 0 <= a <= b?

Of course this last definition is the correct one. For the second part recall (c.f. Rudin) that one of the ordered field axioms is that and then . Moreover, it's easy to show that . So you have that , and thus from previous discussion it follows that

My text only gave me axioms for <=
Are < and <= equivalent ... I know they're not, but can one replace the other in the ordered field axioms?

Originally Posted by Noxide

My text only gave me axioms for <=
Are < and <= equivalent ... I know they're not, but can one replace the other in the ordered field axioms?

In all cases that immediately occur to me, yes.