Ordered field $F$

Question

If $F$ is an ordered field and $a\in F$ and $b\in F$, then how can I show that $a-b\in F$ ?

$F$ has a nonempty subset $P$ such that;

1. $a,b\in P$ ⇒ $a+b,ab\in P$

2. $F=P \cup \{0 \} \cup -P$

3. $P$,$\{0\}$,$-P$ are mutually disjoint

$a\leq b$ iff $b-a\in P$ or $a=b$

Answer 1

I suspect that your 9 axioms for a set and two operations, which I denote by $+$ (addition) and $\cdot$ (multiplication), to be a field are the following:

1. Closure [$a,b \in F \implies a\cdot b, a+b \in F$]
2. Associativity [$(a\cdot b) \cdot c = a \cdot (b \cdot c), (a + b) + c = a + (b + c) \quad \forall \;a,b,c \in F$]
3. Commutativity [$a \cdot b = b \cdot a, \; a + b = b + a$]
4. Additive identity [$\exists 0\in F \;s.t. a + 0 = a \quad \forall a \in F$]
5. Multiplicative identity [$\exists 1 \in F \; s.t. a \cdot 1 = a \quad \forall 0 \neq a \in F$]
6. Additive inverse [$\forall a \in F, \exists b =: -a \; s.t. a + (-a) = (-a) + a = 0$]
7. Multiplicative inverse [$\forall a\neq 0 \in F, \exists b =: \frac{1}{a} \; s.t. a \cdot \frac{1}{a} = \frac{1}{a} \cdot a = 1$]
8. Distributivity [$a(b+c) = ab + ac, \; (a+b)c = ac + bc$]
9. Non-triviality [$1 \neq 0$]

Or at least the 9 axioms will be more or less equivalent (maybe not including (9.) - that's not so important). The idea is that additive inverses are denoted by negative numbers. Thus the additive inverse of $a$ is $-a$. And we know the field is closed under addition, thus $b + (-a) \in F$ is guaranteed. This is completely independent of your ordering.