Is $(\mathbb{Q},<)$ elementary equivalent to $(\mathbb{R},<)$?

Question

I'm looking at those models over the language $\{<\}$. For the case of $L=\{+\}$ there's already an answer here.

Answer 1

Yes. The theory of dense linear orders without endpoints is complete, and both $(\mathbb{Q}, <)$ and $(\mathbb{R}, <)$ are models of this theory.