According to wikipedia:
A prepositive cone or preordering of a field $\mathbb{F}$ is a subset $P \subset \mathbb{F}$ that has the following properties:
▪For $x$ and $y$ in $P$, both $x + y$ and $xy$ are in $P$.
▪If $x$ is in $\mathbb{F}$, then $x^2$ is in $P$.
▪The element $−1$ is not in $P$.If in addition, the set $\mathbb{F}$ is the union of $P$ and $−P$, we call $P$ a positive cone of $\mathbb{F}$. The non-zero elements of $P$ are called the positive elements of $\mathbb{F}$.
Now, my question is if it is posible to have two different positive cones of $\mathbb{R}$. Intuitively I feel the answer should be no, but I tried to derive a contradiction by assuming there exist two different positive cones and I didn't succed.
On the other hand, it seems we can define precisely the "usual" positive cone of the reals if they were constructed from the rationals using Cauchy sequences, Dedekind cuts or any other approach, in such a way it contains all non-negative rational numbers and none of the negative ones. But can we define (maybe for this we would need an unconuntable amount of instructions) a subset $P$ of $\mathbb{R}$ which contains some negative rational numbers but not all of them and which is a positive cone?
By definition, a prepositive cone $P\subset\mathbb{R}$ must contain all nonnegative real numbers, since any nonnegative real number is a square (here I mean "nonnegative" in the usual sense). So it suffices to show it cannot contain a negative number. But if $x\in P$ is negative, then $-1/x$ is positive and hence is also in $P$. It follows that $x\cdot(-1/x)=-1\in P$, which is a contradiction.