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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 $a=b$

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By definition of a field, $a, b \in F$, one has $a - b \in F$. Is there another question after that? – William Jun 7 '12 at 19:19
I'm missing something - forget the order, if $a,b\in F$, then $a-b\in F$ automatically by the field axioms. – Jason DeVito Jun 7 '12 at 19:19
By notational convention $a - b = a + (-b)$ where $-b$ is the additive inverse of $b$ (i.e., that (unique) element $c$ such that $b + c = 0$). – arjafi Jun 7 '12 at 19:26
Field axioms: 1) has additive inverse. 2) closed under addition. So take "a" and add to it "b inverse" – The Chaz 2.0 Jun 7 '12 at 19:27
I don't understand why there are so many downvotes here. The question is clearly in the first line, and the lines after it are the user's attempt to answer the question for himself. – Chris Taylor Jun 8 '12 at 11:41
up vote 4 down vote accepted

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.

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