Definition of an ordered ring - isn't it luck which "half" we get as positive? EDIT I think "I want to prove $D^+$ is unique" might be better as a question, somehow this definition filters the negative from positive. (think of the graph z=xy, there is some notion of positive fundamental to the ring you plot the graph on, not because of the numbers you use)
I've just been given the following definition:
An integral domain $D$ is said to be an ordered integral domain if $D$ has a subset, $D^+$, where the following hold:  


*

*If $a,b\in D^+$ then $a+b\in D^+$

*If $a,b\in D^+$ then $ab\in D^+$

*$\forall a\in D$ exactly one of the following holds:

*

*$a=0$

*$a\in D^+$

*$-a\in D^+$



Now I have proved previously that $(-a)(-b)=ab$ for a ring. 
My question is if I get this set we call $D^+$ does it actually have "the positive ones" in it. 
One of my favourite proofs is "for a ring $R$, $\forall a\in R: 0a=a0=0$ - this shows that the simple definition of a ring must give you this notion of 0 (if you want the distributivity rules, you get a 0 that has this property).
Here, where does the notion of negative come from? There'll be some $x$ say, where $x^2$ is in the set that $x$ isn't in (negative x will take you to positive) but the $-x$ notion is arbitrary. $x$ and $-x$ are inverses of each other, we could call $x$ the inverse of $-x$ or $-x$ the inverse of $x$, the $-$ sign just means "the thing that when added to this without the - sign gives 0"
I can see that $a,b\in D^+\implies ab\in D^+$ might lead to something about $a^2>0$ and we've defined something as positive if $a-b\in D^+$ but I can't quite see it. 
I sense I am very close to something nicer than what I said about 0, but I can't quite see it. Bit of a non-question "what makes negative negative" but I am so close to something, I can feel it!
 A: This is a trick of language, rather than math. English language gives us a system of naming many integers satisfying $x > 0$. Thus, the most convenient way to name the integers satisfying $x < 0$ is through stating that they are the additive inverse of a number we can name: i.e. because we have named something "one", we would name its inverse "negative one".
Note that if $x < 0$, then "negative $x$" is actually a positive number.

Rings can have many different orderings. An instructive example is the ring of all numbers that can be written in the form
$$ a + b \sqrt{2} $$
where $a,b$ are integers. This has the usual ordering $\leq$ on the real numbers. It also has another ordering which I will call $\preceq$ given by
$$ a + b \sqrt{2} \preceq c + d\sqrt{2} \iff a - b \sqrt{2} \leq c - d \sqrt{2} $$
I believe that in general, in an orderable ring, only the sums of squares are "intrinsicially" positive in the sense that they are positive in every ordering of the ring. Their negations would be "intrinsically" negative. Any other nonzero element would be positive in some orderings and negative in other orderings.
(if this isn't true in general, it's at least true for some interesting class of rings I can't remember)
A: Consider the integral domain $D=\mathbb Z[X]/(X^2-2)$. 
For any injective ring homomorphism $f\colon D\to \mathbb R$ we can let $D^+=f^{-1}((0,\infty)$, i.e. we declare an element of $D$ positive iff its image under $f$ is a positve real number.
Now note that we have two obvios ring honmomorphism $D\to \mathbb R$, one given by $\overline X\mapsto \sqrt 2$, one given by $\overline X\mapsto -\sqrt 2$, hence we have two distinct notions positiveness, two distinct choices of $D^+$. This shows that positiveness is not an absolute property of all elements of $D$ (but it is for some, e.g. $1=1^2$ implies $1\in D^+$ for all choices of $D^+$).
A: It's not arbitrary because a square should be positive. In particular, $1$ should be positive. (You would be correct if you were talking about ordered abelian groups - that is, if you were just talking about addition - but the particular form that compatibility with multiplication takes breaks the symmetry.)
