In his Principles of Mathematical Analysis, Rudin defines ordered fields as follows:
An ordered field is a field $F$ which is also an order set, such that
(i) $x+y<x+z$ if $x, y, z\in F$ and $y<z$,
(ii) $xy>0$ if $x\in F$, $y\in F$, $x>0$, and $y>0$.
In the above definition, is it necessary to explicitly require for the order on $F$ to satisfy these two conditions? In other words, is it possible to have a field with an order such that (i) and (ii) do not hold?
Rudin defines orders as follows:
Let $S$ be a set. An order on $S$ is a relation, denoted by $<$, with the following two properties:
(iii)If $x\in S$ and $y\in S$ then one and only one of the statements $$x<y, ~~x=y,~~ y<x$$ is true.
(iv) If $x, y, z\in S$, if $x<y$ and $y<z$, then $x<z$.
Best Answer
Here is a simple example to demonstrate that a field together with an ordering on the underlying set is not automatically an ordered field.
The field is $\Bbb R$ with usual addition and multiplication.
The ordering is as follows. Let $\ll$ denote the following relation: $x \ll y$ if $x$ and $y$ are both rational and $x<y$ by the usual order, or if $x$ and $y$ are both irrational and $x<y$ by the usual order, or if $x$ is rational and $y$ is irrational.
In this order, we have both $\pi \gg 0$ and $-\pi \gg 0$, but their sum is not $\gg 0$.