Elementary Order Arithmetic in an Ordered Field

Let $F$ together with $\le$ be an ordered field, with additive identity $0$ and multiplicative identity $1$, and with the addition and multiplication of its underlying field; its order $\le$ is in particular a total order. For $a,b\in F$ write $a<b$ to mean that $a\le b$ and $a\ne b$, write $-a$ for the additive inverse of $a$, write $a-b$ for $a+(-b)$, and write $a^{-1}$ for the multiplicative inverse of $a$ when $a\ne 0$. Set $2=1+1$.

Then the following hold for all $a,b,c,d\in F$.

1. (Strict compatibility with addition) $a<b$ if and only if $a+c<b+c$.

2. (Mixed transitivity) If $a\le b$ and $b<c$, then $a<c$; and if $a<b$ and $b\le c$, then $a<c$.

3. (Addition of inequalities) If $a<b$ and $c\le d$, then $a+c<b+d$.

4. (Sign reversal) $a\le b$ if and only if $-b\le -a$; and $a<b$ if and only if $-b<-a$.

5. (Products of positive elements) If $0<a$ and $0<b$, then $0<ab$.

6. (Positivity of the unit) $0<1$.

7. (Inverses of positive elements) If $0<a$, then $a^{-1}$ exists and $0<a^{-1}$.

8. (Halving) $0<2$, so $2^{-1}$ exists; and if $0<\varepsilon$, then $0<\varepsilon\cdot 2^{-1}$, $\varepsilon\cdot 2^{-1}<\varepsilon$, and $\varepsilon\cdot 2^{-1}+\varepsilon\cdot 2^{-1}=\varepsilon$.

9. (Least of two elements) There is $m\in F$ such that $m\le a$, $m\le b$, and either $m=a$ or $m=b$.

10. (Strict compatibility with multiplication) If $a<b$ and $0<c$, then $ca<cb$.