Can the rational and real ordered fields define only two linear orders?

Question

This is two questions in one. In the structure $(\mathbb{R};+,-,*,0,1,\leq)$, can you only define two linear orders, namely the standard order and its converse? If not, can someone define another linear order? I am not asking that the order be compatible with the field operations. My second question is the same, but about the structure $(\mathbb{Q};+,-,*,0,1,\leq)$.

Answer 1

You can define, easily, the order $x<y\iff (x<y\land(0<x<1\leftrightarrow 0<y<1)\lor(0<y<1\land(x\leq 0\lor 1\leq x)$ which is the order you get by shifting the whole interval $(0,1)$ to the top.

In the case of the rationals, this is isomorphic to the original order, but in the case of the real numbers it is not, since it's not a Dedekind-complete order.

You can furthermore shift the interval $[0,1]$ to the top, then in both cases this is a non-isomorphic order since it has a maximal element. You can also take some finitely many combinations of clever extensions of that idea.

In the case of $\Bbb Q$, since the integers are definable, we can even take them out to get the order $\Bbb{Q\setminus Z+Z}$, and pretty much do anything we want all the way to, but not including, $\Bbb Q+\omega_1^{CK}$, since all the ordinals below it are definable over $\Bbb N$. And again, we can be clever about it and push some intervals on top, etc. etc.