Solution
= Solution
Express that the order is a <dense order>:
$$
\boxed{\psi=\forall x\,\forall y\,\bigl(x<y\to\exists z\,(x<z\land z<y)\bigr).}
$$
In either real <interval>, $(x+y)/2$ lies strictly between any two distinct ordered elements and remains in the domain. In the <integers>, $x=0,y=1$ have no intervening <integer>, so this <first-order sentence> is false. The endpoints of a closed <interval> do not prevent a <dense order>.