Proof of Elementary Order Arithmetic in an Ordered Field

Throughout we use the axioms of a field for algebraic identities in $F$, the fact that $\le$ is a total order, hence reflexive, antisymmetric, transitive and comparing any two elements, and conditions 1 and 2 in the definition of an ordered field.

Claim 1. Suppose $a<b$. Condition 1 gives $a+c\le b+c$. If $a+c=b+c$, then adding $-c$ to both sides gives $a=b$, contradicting $a\ne b$; hence $a+c<b+c$. Conversely, suppose $a+c<b+c$. The implication just proved, applied to $a+c<b+c$ with $-c$ in place of $c$, gives $(a+c)+(-c)<(b+c)+(-c)$, that is $a<b$.

Claim 2. Suppose $a\le b$ and $b<c$. Transitivity gives $a\le c$. If $a=c$, then $a\le b$ reads $c\le b$, and together with $b\le c$ antisymmetry gives $b=c$, contradicting $b\ne c$. Hence $a\ne c$ and $a<c$. Now suppose $a<b$ and $b\le c$. Transitivity gives $a\le c$. If $a=c$, then $b\le c$ reads $b\le a$, and together with $a\le b$ antisymmetry gives $a=b$, contradicting $a\ne b$. Hence $a<c$.

Claim 3. Suppose $a<b$ and $c\le d$. Claim 1 gives $a+c<b+c$. Condition 1 applied to $c\le d$ with $b$ added gives $c+b\le d+b$, that is $b+c\le b+d$. Claim 2 gives $a+c<b+d$.

Claim 4. Suppose $a\le b$. Condition 1 with $(-a)+(-b)$ added gives $a+((-a)+(-b))\le b+((-a)+(-b))$, that is $-b\le -a$. Applying this implication to $-b\le -a$ gives $-(-a)\le -(-b)$, that is $a\le b$; so the two inequalities are equivalent. For the strict form, note that $a=b$ holds if and only if $-a=-b$, since the additive inverse is its own inverse operation; combining this with the equivalence just proved gives that $a<b$ holds if and only if $-b<-a$.

Claim 5. Suppose $0<a$ and $0<b$. Condition 2 gives $0\le ab$. If $ab=0$, then multiplying by $a^{-1}$, which exists because $a\ne 0$, gives $b=a^{-1}(ab)=a^{-1}0=0$, contradicting $b\ne 0$. Hence $0<ab$.

Claim 6. In a field $1\ne 0$. Since $\le$ compares any two elements, either $0\le 1$ or $1\le 0$. Suppose $1\le 0$. Claim 4 gives $0\le -1$, so condition 2 gives $0\le (-1)(-1)$, and $(-1)(-1)=1$ in a field, so $0\le 1$. With $1\le 0$ antisymmetry gives $1=0$, a contradiction. Hence $0\le 1$, and since $1\ne 0$ we get $0<1$.

Claim 7. Suppose $0<a$. Then $a\ne 0$, so $a^{-1}$ exists, and $a^{-1}\ne 0$ because $aa^{-1}=1\ne 0$. Suppose $a^{-1}\le 0$. Claim 4 gives $0\le -a^{-1}$, so condition 2 gives $0\le a(-a^{-1})$, and $a(-a^{-1})=-(aa^{-1})=-1$ in a field, so $0\le -1$. Claim 4 then gives $1\le 0$, contradicting claim 6. Since $\le$ compares any two elements, $0\le a^{-1}$, and with $a^{-1}\ne 0$ this gives $0<a^{-1}$.

Claim 8. By claim 6, $0<1$. Claim 1, adding $1$, gives $1<1+1=2$, and claim 2 gives $0<2$. Hence $2\ne 0$, so $2^{-1}$ exists, and $0<2^{-1}$ by claim 7. Let $0<\varepsilon$. Claim 5 gives $0<\varepsilon\cdot 2^{-1}$. In a field $2^{-1}+2^{-1}=(1+1)\cdot 2^{-1}=2\cdot 2^{-1}=1$, hence

Finally, adding $\varepsilon\cdot 2^{-1}$ to both sides of $0<\varepsilon\cdot 2^{-1}$ gives, by claim 1,

Claim 9. Since $\le$ compares any two elements, $a\le b$ or $b\le a$. In the first case take $m=a$: then $m\le a$ by reflexivity and $m\le b$. In the second take $m=b$: then $m\le b$ by reflexivity and $m\le a$.

Claim 10. Suppose $a<b$ and $0<c$. Claim 1, adding $-a$, gives $a+(-a)<b+(-a)$, that is $0<b-a$. Claim 5 gives $0<c(b-a)$, and $c(b-a)=cb-ca$ in a field. Claim 1, adding $ca$, gives $0+ca<(cb-ca)+ca$, that is $ca<cb$.