I'm reading Rudin's real analysis where he defines what an order on a set $S$ is:
An order on $S$ is a relation, denoted by $<$, with the following two properties:
1) If $x,y \in S$, then one and only one of the statements: $x = y, x < y, y < x$ is true.
2) If $x,y,z \in S$ and $x < y$ and $y < z$ then $x < z$
Later, he defines the concept ordered field:
An ordered field is a field $F$ which is also an ordered set, such that:
1) If $x,y,z \in F$ and $y < z$ , then $x + y < x + z$
2) If $x,y \in F$ and $0 < x, 0 <y$, then $0 <xy$
In all the proofs that follow, he never mentions that he means that $a < b$ is the usual order relation (for example $2.3 < 3$), but all the properties of this usual order relation can be proven from the definitions he provided.
So my question is:
Does this mean that this relation is unique, given the properties in the definition? Can I assume that this is the order relation that I'm familiar with, or is there another relation that can be defined on such a field that satisfies those properties (so in fact I'm asking whether this relation is unique).
More concrete, I'm interested where the field is either $\mathbb{Q}$ or $\mathbb{R}$
There is an alternative (and equivalent) route to define an order on a field.
Actually it is enough to state that a set $P\subset F$ exists such that:
Then the relation $<$ can be defined as: $$x<y\iff y-x\in P$$
If you start with the other definition then $P$ is exactly the set $\{x\in F\mid 0<x\}$.
Here letter "P" associates with "positive".
In this light your question can be reworded as: is $P$ necessarily unique in this?
If we are dealing with field $\mathbb R$ (or $\mathbb Q$) then the answer is yes.
It starts with proving that $1\in P$.
If we assume that $1\notin P$ then we arrive at $1=(-1)(-1)\in P$ and a contradiction is found.
Then $\{1,2=1+1,\dots\}\subseteq P$.
Then (after properly defining) $\mathbb Q_+\subseteq P$.
And finally (after properly defining) $\mathbb R_+= P$.