TITLE: A field that is an ordered field in two distinct ways
QUESTION [9 upvotes]: Question: Explain the construction below (taken directly from Counter Examples in Analysis): 
An ordered field is a field $F$ that contains a subset $P$ such that 
  $P$ is closed with respect to addition and multiplication and exactly one of the following are true:
  $$x \in P; x = 0; -x \in P$$
The set $F$ of all numbers of the form $r + s\sqrt{2}$, where $r$ and $s$ are rational and the operations of addition and multiplication are those of the real number system R of which $F$ is a subset, is an ordered field. 
  Let $P$, defined above, be the set of all members of $F$ that are positive members of R. A second way in which $F$ is an ordered field is provided by the subset $B$ defined:
  $$ r+s\sqrt{2} \in B \iff r - s\sqrt{2} \in P.$$

Based on how I've read it these few questions in particular stand out to me.

Q1. Isn't the inverse of addition destroyed by selecting only the positive members of $P$?
Q2. Can we say that this construction is effectively showing that there exists two automorphisms for the field extension ${\mathbf Q}[\sqrt{2}], {\mathbf Q}\rightarrow {\mathbf R}$? (${\mathbf Q}$ being the rationals)
Q3. What are the practical consequences for having these two distinct ways? If there are many, can you sum up their theme? (Read: why is this a counter example except to the rationals and irrationals, or why is this counter example important?)

Note: I'm new to fields, if this is an easy/poorly asked question let me know. I'll delete it and ask it again after I've read more, although I would be thankful for a hint in the right direction.

REPLY [2 votes]: Another answer to Q3: What practical consequences are there to having these two ways.
The real numbers have exactly one order which respects the field operations (i.e., one subset P as in the definition you guys are using).  The complex numbers have NO orders which respect the field operations (i.e., you can not order the complex numbers in such a way that 

$a>b \implies a+c\geq b+c$,  
$a>0 $ and $ b>c \implies ab\geq ac$

for all $a,b,c$ complex numbers, see 'ordered field' wiki or google 'real field').  
So: the field $Q(\sqrt{2})$ described above, where elements look like $a+b\sqrt{2}$, has (at least!) two real orderings, the reals numbers have exactly one, and the complex numbers have none.  It turns out that fields that have this second property are 'very nice' for studying abstractly, and these kinds of fields are called real closed.
By showing that $Q(\sqrt{2})$ has at least two orders, we see that it can't be real closed.