Are $(\Bbb R,+)$ and $(\Bbb Q, +)$ elementary equivalent? Let $\mathscr L$ be the first-order language having a binary function, denoted by $+$, as signature.

We consider the two $\mathscr L$-structures $(\Bbb R, +)$ and $(\Bbb Q, +)$, where $+$ denotes the usual addition. Are they elementary equivalent?

Clearly they are not isomorphic as $\mathscr L$-structures. I know a similar example (with a binary relation $≤$) : $(\Bbb R, ≤)$ and $(\Bbb Q, ≤)$ are elementary equivalent.
I believe that the answer to my main question is "yes", but I don't know how to prove it.
Any hint is welcome. Thanks!

Edit: Let $\mathscr L'$ be the first-order language having two binary functions as signature.
Then the two $\mathscr L'$-structures $(\Bbb R, +,\cdot)$ and $(\Bbb Q, +,\cdot)$ (where $+$ denotes the usual addition and $\cdot$ denotes the usual multiplication) are not elementary equivalent.
The formula
$\newcommand{\_}{\;\;}$
$$\exists t \_ [\exists x \_ (x+x=x \_\wedge\_ \exists y \_ y \cdot y = y \_\wedge\_ \neg(x=y)) \_\wedge\_
t \cdot t = y+y]$$
is satisfied in $(\Bbb R, +,\cdot)$ by picking $t=\sqrt 2,x=0,y=1$, but not in $(\Bbb Q, +,\cdot)$.
 A: As @Wore pointed out not quite correctly in a comment, both of these structures are models of the same well-known complete theory: the theory of torsion-free divisible Abelian groups. That theory has the following axioms:


*

*$+$ forms an Abelian group

*(axiom scheme where $n$ varies over positive integers) $\forall x : \exists y : \sum_{k=1}^n y = x $

*(axiom scheme where $n$ varies over positive integers) $\forall x: \sum_{k=1}^n x = 0 \rightarrow x = 0$

*$\exists x : x \neq 0$


(Note that I've added $0$ as a constant symbol to make the axioms easier to state correctly, but it doesn't matter. You can rewrite "$t = 0$" as "$t + t = t$" whenever it occurs, for any term $t$.)
All you need to do is show that the above axiomatizes a complete theory; it is clear that both $(\mathbb{R}, +)$ and $(\mathbb{Q},+)$ are models.
There are two main approaches to showing that the above axiomatizes a complete theory. The first approach is to show that two uncountable models of the same cardinality are isomorphic; Vaught's test then implies completeness. The second approach is to use your favorite technique to show that the theory has quantifier elimination; you can then observe that the theory is complete for quantifier-free sentences (because you can't say anything interesting without quantifiers in this theory), and therefore complete.
Hint for the Vaught's test approach: you can think of any model of this theory as a vector space over $\mathbb{Q}$, and vice versa. Two vector spaces are isomorphic if they have the same dimension (which is possibly an infinite cardinal), and the dimension of an uncountable vector space over $\mathbb{Q}$ is just its cardinality.
The quantifier elimination approach doesn't really need a further hint. Read about any standard technique for showing quantifier elimination, and then use it directly to show QE for this theory as an exercise.
