Consequences of the Order Axioms

We work only from the field axioms of $F$ and the three defining properties of $P$ (closure under $+$, closure under $\times$, trichotomy), plus the definition $a<b \iff b-a \in P$. Throughout, we use the standard field facts $(-1)a = -a$, $-(-a)=a$, and $-(ab) = (-a)b = a(-b)$; these are pure field-axiom consequences (proved by the usual distributive-law manipulations, independent of order) and we take them as already established, since they involve no order concept.

Proof of (O1). By the trichotomy axiom applied to the element $b - a \in F$, exactly one of

holds. These are precisely the statements $a < b$, $a = b$, $b < a$ (i.e. $a>b$) respectively. Hence exactly one of $a<b$, $a=b$, $a>b$ holds. $\blacksquare$ (O1)

Proof of (O2). Suppose $a<b$ and $b<c$. By definition, $b - a \in P$ and $c - b \in P$. By closure of $P$ under addition,

Field algebra gives $(b-a)+(c-b) = c - a$. Hence $c - a \in P$, i.e., $a < c$. $\blacksquare$ (O2)

Proof of (O3). Suppose $a < b$, so $b - a \in P$. For any $c \in F$,

by field algebra (the $c$'s cancel). By definition this says $a + c < b + c$. $\blacksquare$ (O3)

Proof of (O4). By definition, $a > 0$ means $a - 0 = a \in P$. Also $-a < 0$ means $0 - (-a) = a \in P$. These are literally the same statement, $a \in P$, so they are equivalent. $\blacksquare$ (O4)

Proof of (O5). (a) Suppose $a > 0$ and $b > 0$, i.e., $a \in P$ and $b \in P$. By closure of $P$ under multiplication, $ab \in P$, i.e., $ab > 0$. (b) Suppose $a > 0$ and $b < 0$. Then $a \in P$, and by (O4), $b < 0$ means $-b \in P$. By closure under multiplication, $a(-b) \in P$. But by field algebra $a(-b) = -(ab)$, so $-(ab) \in P$, which by definition means $ab < 0$. (c) Suppose $a < 0$ and $b < 0$. By (O4), $-a \in P$ and $-b \in P$. By closure under multiplication, $(-a)(-b) \in P$. By field algebra,

using $-(ab)=a(-b)$ (field algebra) twice and the involution $-(-x)=x$. Hence $ab \in P$, i.e., $ab > 0$. For the "in particular" clause: the identity $(-a)(-b) = ab$ was derived above purely by field algebra together with part (c)'s computation (it holds for all $a, b \in F$, not just negative ones, since it's a pure algebraic identity — we only needed the sign hypothesis to conclude positivity, not to derive the identity itself). Setting $b=1$: $(-a)(-1) = a\cdot 1 = a$; combined with the general identity $(-a)(-1) = -(-a) \cdot$ [algebra] one recovers $(-1)a = -a$ directly from field axioms as noted above. $\blacksquare$ (O5)

Proof of (O6). Let $a \in F$. By trichotomy exactly one of $a \in P$, $a = 0$, $-a \in P$ holds.

• If $a \in P$ (i.e. $a > 0$): by (O5)(a) with $b = a$, $a \cdot a = a^2 \in P$, so $a^2 > 0$.
• If $a = 0$: then $a^2 = 0 \cdot 0 = 0$ by field algebra.
• If $-a \in P$ (i.e. $a < 0$): by (O5)(c) with $b=a$, $a \cdot a = a^2 \in P$, so $a^2 > 0$.

In every case $a^2 \ge 0$, and the only case giving $a^2 = 0$ is $a = 0$; conversely if $a=0$ then $a^2=0$. This proves the equivalence. For the consequence $1 > 0$: since $F$ is a field, $1 \neq 0$ (field axiom: $0 \neq 1$). Apply the above with $a = 1$: since $1 \ne 0$, $1^2 = 1 \cdot 1 = 1 \in P$ by the dichotomy just shown (as $1^2 > 0$ whenever $1 \ne 0$). Hence $1 > 0$. $\blacksquare$ (O6)

Proof of (O7). Suppose $a < b$, so $b - a \in P$. First claim: if $c > 0$, then $ac < bc$. Since $c \in P$ and $b - a \in P$, closure under multiplication gives $(b-a)c \in P$. By the distributive law, $(b-a)c = bc - ac$. Hence $bc - ac \in P$, i.e. $ac < bc$. Second claim: if $c < 0$, then $ac > bc$. By (O4), $-c \in P$. Since $b - a \in P$, closure under multiplication gives $(b-a)(-c) \in P$. By field algebra, $(b-a)(-c) = -(b-a)c = -(bc - ac) = ac - bc$. Hence $ac - bc \in P$, i.e., $bc < ac$, i.e., $ac > bc$. $\blacksquare$ (O7)

Proof of (O8). Suppose $0 < a < b$. First, we show $a^{-1}, b^{-1}$ exist and are positive. Since $a \neq 0$ (as $a \in P$ and $0 \notin P$ by trichotomy applied to $0$: if $0\in P$ then also $-0=0\in P$, contradicting exactly-one-of-three unless we note $0$ satisfies the middle clause, so $0 \notin P$), $a$ has a multiplicative inverse $a^{-1}$ in the field $F$. We claim $a^{-1} > 0$. By trichotomy applied to $a^{-1}$, either $a^{-1} \in P$, or $a^{-1}=0$, or $-a^{-1} \in P$.

• $a^{-1} = 0$ is impossible: it would give $a \cdot a^{-1} = a \cdot 0 = 0$, contradicting $a\cdot a^{-1}=1\ne 0$.
• If $-a^{-1} \in P$: since also $a \in P$, closure under multiplication gives $a(-a^{-1}) \in P$. But $a(-a^{-1}) = -(a\cdot a^{-1}) = -1$. So $-1 \in P$. But also $1 \in P$ by (O6), so by closure under addition $1 + (-1) = 0 \in P$. This contradicts trichotomy at $0$ (we showed $0 \notin P$ above). Contradiction, so this case is impossible.

Hence by elimination $a^{-1} \in P$, i.e. $a^{-1} > 0$. The identical argument (verbatim, replacing $a$ by $b$) gives $b^{-1} > 0$. Now from $a < b$, multiply both sides by $c := a^{-1}b^{-1}$. Since $a^{-1} > 0$ and $b^{-1} > 0$, closure under multiplication gives $c = a^{-1}b^{-1} \in P$, i.e., $c > 0$. By (O7) (first claim) applied with this positive $c$,

Simplify each side by field algebra: $a\cdot a^{-1}b^{-1} = (a a^{-1})b^{-1} = 1\cdot b^{-1} = b^{-1}$, and $b \cdot a^{-1}b^{-1} = a^{-1}(b b^{-1}) = a^{-1}\cdot 1 = a^{-1}$. So the inequality reads

i.e. $\frac{1}{b} < \frac{1}{a}$. Combined with $b^{-1}>0$ shown above, we get $0 < \frac1b < \frac1a$. $\blacksquare$ (O8)