Uniqueness of set of Positive real numbers using order axioms While constructing the real numbers, we come across the order axioms, used to define the positive real numbers, i.e there exists a set $P$ such that the properties of trichotomy, closure under addition and multiplication hold. 
The axioms guarantee the existence of such a set. Will such a set be unique as well? I've tried proving this, and it should be trivial, but I can't seem to get it.
 A: I assume "satisfies trichotomy" means "every real is positive, negative (= its additive inverse is positive), or zero.
Yes, if you have a reasonable set of axioms, you can show that the set of positive reals is uniquely defined. Specifically, you can prove that the set of positive reals is exactly $$\{x: \mbox{ for some $y\not=0$, $x=y\cdot y$}\}.$$ Note that this can be used to define the order relation: $x_1<x_2$ iff for some $y$, we have $y\cdot y+x_1=x_2$. So in fact, "$<$" does not need to be included as a primitive symbol.
Note: the list of axioms you have given so far are not enough - indeed, the field $\mathbb{Q}(\pi)$ satisfies them, but does not have a unique set of positive elements (since it has an automorphism swapping $\pi$ and $-\pi$). So I think you're missing an axiom or two . . .
How do you prove this?
Well, first suppose $x=y\cdot y$ for some $y$. Then


*

*Can you show that $x=(-y)\cdot (-y)$ as well?

*What does that tell you about $x$ (think about whether $y$ is positive or negative)?
This shows that the set I've defined is a subset of the positive reals. Now we want to show that it contains the positive reals. To do this, you need to argue that for any real $r$, either $r$ or $-r$ has a square root. You can't do that with only the axioms you've listed. I think you have another axiom or two lying around . . .
A: For general orderable fields, $P$ isn't unique — all you can say is that it must contain all sums of nonzero squares.
To see that $P$ must contain squares, note that if $x \neq 0$, then you have two cases:


*

*$x \in P$, and thus $x^2 \in P$

*$-x \in P$, and thus $(-x)^2 \in P$


Since $x^2 = (-x)^2$, either way we conclude $x^2 \in P$.
I believe there is a theorem that says that if neither $y$ nor $-y$ is a sum of squares, then there is a choice for $P$ containing $y$, and another choice for $P$ containing $-y$.

The reals are a special case, since the set of all nonzero squares already satisfies trichotomy, so $P$ can't contain anything else, thus $P$ is unique.
