# Ordered Fields, 0<a<b

• Jan 15th 2011, 08:04 PM
Noxide
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?
• Jan 15th 2011, 08:36 PM
Drexel28
Quote:

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 $x and $z>0$ then $xz. Moreover, it's easy to show that $x,y>0\implies xy>0$. So you have that $0, $a^{-1},b^{-1}>0$ and thus from previous discussion $a^{-1}b^{-1}>0$ it follows that $0<(a^{-1}b^{-1}}a<(a^{-1}b^{-1})b\cdots$
• Jan 15th 2011, 09:17 PM
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?
• Jan 15th 2011, 09:18 PM
Drexel28
Quote:

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.