A field $\mathbb{F}$ is an ordered field if $\exists P\subseteq\mathbb{F}$ such that: 1. $\forall a,b\in P$, $a+b,ab\in P$. 2. Only one of the following is true $\forall a\in\mathbb{F}$, 1) $a\in P$ 2) $a=0$ 3) $-a\in P$.
I know that in $\mathbb{R}$, $\mathbb{R}_{>0}$ satisfies such properties. Is it unique? Are there any other subsets of $\mathbb{R}$ that satisfy the above properties? I know how to prove all positive rational numbers are in such $P$ (if $-p/q\in P$, then $-p=(-p/q)+...+(-p/q)\in P$. $p^2\in P$, then $0=p^2-p-...-p\in P$, which can't be true). However, I think it is possible for a negative transcendental number to be contained in such $P$
I claim there is a unique positive cone (that's how this set $P$ is called) on $\mathbb{R}$.
Note that all cones $P$ on a field $\mathbb{F}$ contain $\{y^2 : y\in \mathbb{F}\setminus \{0\}\}$.
Proof: If $y\in\mathbb{F}\setminus\{0\}$ then either $y\in P$ so $y^2\in P$, or $-y\in P$ so $(-y)^2 = y^2\in P$. $\square$
Moreover, in the case of $\mathbb{R}$ we have that above set is already a positive cone.
Note the maximality property of cones: If $P_1, P_2$ are cones and $P_1\subseteq P_2$ then $P_1 = P_2$.
Proof: Let $y\in P_2\setminus P_1$. Then $-y\in P_1\subseteq P_2$. But this is impossible. $\square$
More generally, all real-closed fields (which $\mathbb{R}$ is a part of) have unique order in which the positive cone consists of non-zero squares.
Here's an exercise: Let $\mathbb{F}$ be a subfield of $\mathbb{R}$ and consider all algebraic elements over $\mathbb{F}$ in $\mathbb{R}$. Say the thus obtained subfield of $\mathbb{R}$ is $\mathbb{F}_0$. Then $\mathbb{F}_0$ also has a unique ordering (in fact, is real-closed). For example, the field of (real) algebraic numbers has a unique order which makes it an ordered field (here, is obtained by letting $\mathbb{F} = \mathbb{Q}$).
On the other hand, extending $\mathbb{Q}$ by a transcendental $t\in \mathbb{R}$ leads to distinct ordered fields for all distinct transcendentals $t$, while they're all isomorphic to the field $\mathbb{Q}(x)$. Thus this is an example of a field with continuum many different possible orders.