Are there any normal interpretations of $(=, <)$ with countable domain that are elementary equivalent but not isomorphic to $\mathbb{Q}$? To $\mathbb{Q}^+ = \mathbb{Q} \cap [0; +\infty)$?

First, I know that $(=, <)$ on $\mathbb{Q}$ allows quantifier elimination, so, intuitively, the only expressible closed formulas are those that express the axioms of $=$ and $<$ and the properties of $\mathbb{Q}$ (namely, having no minimum and maximum elements and being dense). So any other elementary equivalent normal interpretation should have a linear order, no min/max elements and be dense. So, the claim basically boils down to the question: is there a set $M$ having these properties but not isomorphic to $\mathbb{Q}$?

I tried proving that there is none by contradiction by assuming that for every bijection $\alpha : \mathbb{Q} \rightarrow M$ there exist $x, y \in \mathbb{Q} : x < y \land \alpha(x) > \alpha(y)$, but I couldn't come to any and got a bit stuck.

I also tried proving that there is none by taking some bijection $\alpha$ and then trying to build a "fixed" bijection $\alpha'$ that preserves the order, but it gets messy too.

I haven't paid as much attention to $\mathbb{Q}^+$ yet, but I seem to be able to prove that it also allows quantifier elimination after constant $0$ is added to the signature, so, intuitively, after this it should be the same as with $\mathbb{Q}$.

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    $\begingroup$ Nope. A countable dense linear order with no first or last element is isomorphic to $\mathbb Q$; with first but no last element it's isomorphic to $\mathbb Q^+.$ A theorem of Cantor. $\endgroup$ – bof Jun 23 '18 at 2:46
  • $\begingroup$ Ah, right, I guess it can be proved somewhat rigorously by more or less constructing the required isomorphism by recursion on the subset of $\mathbb{Q}$, thanks! $\endgroup$ – 0xd34df00d Jun 23 '18 at 12:23

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