As per this question here, Rudin's definition of an ordered set is completely equivalent to simply defining a (non-strict) total order on the set and then defining the corresponding strict total order.
My question is about whether we have the same result for ordered fields. That is, if Rudin defines an ordered field as
An ordered field is a field $\mathbb{F}$ which is also an ordered set obeys the following two axioms:
(1)For all $x,y,z$, if $y>z$ then $x+y>x+z$.
(2) For all $x,y$, if $x>0$ and $y>0$ then $xy>0$.
This definition is meant to interact with his choice to define an ordered set in terms of a strict total order. I want to confirm that this new definition interacts nicely with the alternative definition in terms of a nonstrict total order given in the link. That is, if I were to define an ordered field as
An ordered field is a field $\mathbb{F}$ which is also an ordered set obeys the following two axioms:
(1)For all $x,y,z$, if $y\geq z$ then $x+y\geq x+z$.
(2) For all $x,y$, if $x\geq 0$ and $y\geq 0$ then $xy\geq0$.
and then defined $>$ as in the link, I would recover Rudin's definition (and vice versa, if I use $<$ to define $\leq$)? I am getting lost in all the cases, and would frankly be happy even with a "yes/no" here. I do wish Rudin used the precise langugae of order theory etc. so that I don't get confused in the future, but oh well.
Best Answer
The properties
(1) $y\geq z\implies x+y\geq x+z$
(2) $x,y\geq 0\implies xy\geq0$
clearly imply
(1') $y>z\implies x+y>x+z$ (because $x+y=x+z\implies y=z$)
(2') $x,y>0\implies xy>0$ (because $xy=0\implies x=0$ or $y=0$).
Conversely,
(1') $\implies$ (1) (because $y=z\implies x+y=x+z$)
(2') $\implies$ (2) (because $x=0$ or $y=0\implies xy=0$).