Prove all 4 axioms of "less than" are necessary (for real numbers)

Question

One way to define an ordered field is as a field $F$ with a relation $<$ that satisfies:

1. For all $x,y \in F$, exactly one of $x<y$, $x=y$, $y<x$ holds.
2. For all $x,y,z \in F$, if $x<y$ and $y<z$, then $x<z$.
3. For all $x,y,z \in F$, if $x<y$, then $x+z < y+z$.
4. For all $x,y,z \in F$, if $x<y$ and $0<z$, then $x \cdot z < y \cdot z$.

When $F$ is the real numbers $\mathbb R$, the usual "less than" is the only relation that satisfies all four axioms. I expect that all four axioms are necessary to know this: if we drop one of them, then other relations may "sneak in". There are easy constructions for two of the axioms:

• If we drop axiom 1, then we can declare that $x<y$ for all $x,y \in \mathbb R$: this satisfies the other three axioms.
• If we drop axiom 4, then we can declare that $x<y$ when $x$ is greater than $y$ in the usual sense: this satisfies the other three axioms.

Are there easy constructions showing that the other two axioms are necessary?

(Again, I want to stick with the real numbers. There are interesting examples over finite fields; for example, when $F = \mathbb F_3$, let $x<y$ exactly when $y=x+1$, and axioms 1,3,4 hold (but not 2). I am not interested in these right now.)

Answer 1

Axioms 1 and 3 are equivalent to saying that you have a set $P$ of "positive" numbers such that $\mathbb{R}$ is partitioned into $P,\{0\},$ and $-P$, and then we say $x<y$ iff $y-x\in P$. Adding axiom 4 is then eqivalent to saying $P$ is closed under multiplication. So to satisfy axioms 1, 3, and 4, you just need a set $P\subset\mathbb{R}\setminus\{0\}$ closed under multiplication such that for all nonzero $x$, exactly one of $x$ and $-x$ is in $P$. The multiplicative group $\mathbb{R}\setminus\{0\}$ splits as a direct sum of $\{1,-1\}$ and $\mathbb{R}_+$, and $P$ just needs to be some other direct sum complement to $\{1,-1\}$. (Note that $P$ is automatically closed under multiplicative inverses: if $x\in P$ and $1/x\not\in P$ then $-1/x\in P$ so $x\cdot(-1/x)=-1\in P$ so $(-1)^2=1\in P$ which is a contradiction.) In particular, if we take a basis for $\mathbb{R}_+$ as a vector space over $\mathbb{Q}$, we could replace one basis element with its negative to get a different complement, and thus a relation that satisfies axioms 1, 3, and 4 but which has a different set of "positive" numbers.

(This is quite arguably not an "easy" construction. However, there is not really any easier option: we are asking for a group-homomorphism $\mathbb{R}_+\to\mathbb{R}\setminus\{0\}$, and any such homomorphism that is at all "nice" (e.g., measurable) is automatically continuous and so could only be the standard choice if it is to split the projection map. In particular, for instance, it is consistent with ZF that no such counterexample exists so axioms 1, 3, and 4 suffice.)

For axioms 1, 2, and 4, just take any total order on $\mathbb{R}$ for which $0$ is the greatest element. Then axiom 4 holds vacuously.