Additive group of reals saturated Let $(\mathbb{R}, +, 0)$ be the additive group of reals. Is this structure $\aleph_0$-saturated? 
I don't really see how to go about showing this. To show it is not saturated, it is enough to exhibit a type omitted in $(\mathbb{R}, +, 0)$. The interesting statements we can make about groups are usually to do with torsion or divisibility, but I can't see a way to find a type omitted in $(\mathbb{R}, +, 0)$ from that. 
 A: Note, as in Zarathustra's answer, that $(\mathbb{R},0,+)$ carries the structure of a $\mathbb{Q}$-vector space. 
Let $A$ be a finite subset of $\mathbb{R}$. What are the possible types over $A$? 
For every $\mathbb{Q}$-linear combination of elements of $A$, there is a type $p(x)$ which is completely determined by the formula asserting that $x$ is equal to that linear combination (this is not strictly a formula in the language $\{0,+\}$, but it is equivalent to one by clearing denominators and interpreting scalar multiplication by integers as repeated addition). All of these types are realized in $\mathbb{R}$, since $\mathbb{R}$ is closed under $\mathbb{Q}$-linear combinations..
I claim there is just one other type over $A$. It suffices to show that if $a$ and $b$ are any two elements of some elementary extension $R$, such that neither is in the $\mathbb{Q}$-linear span of $A$, then $a$ and $b$ have the same type over $A$. This follows by from the fact that there is an automorphism of $R$ fixing $A$ and moving $a$ to $b$, obtained by linear algebra: pick a basis for the span of $A$, then extend this basis to a basis for $R$ in two different ways, one containing $a$ and one containing $b$. 
This last type is also realized in $\mathbb{R}$, since the span of $A$ is countable, while $\mathbb{R}$ is uncountable. Any element of $\mathbb{R}$ which is not in the span of $A$ satisfies it. 
Note that the same argument works whenever $|A|<|\mathbb{R}|$, which shows that this structure is saturated, not just $\aleph_0$-saturated. 
A: Every (nontrivial) torsion-free divisible group has a structure of $\Bbb Q$-vector space. This shows that the theory $T$ of torsion-free divisible group is uncountably categorical (because a vector space is identified up to isomorphism by its dimension, and all the uncountable $\Bbb Q$-vector spaces have the same dimension). Suppose that $(\Bbb R,+,0)$ is not $\aleph_0$-saturated. Then some type $p$ would be omitted in $(\Bbb R,+,0)$, but by Lowenheim-Skolem one would be able to find a model of $T$ of size $|\Bbb R|$ that realizes $p$, and this model would not be isomorphic to $(\Bbb R,+,0)$*. This contradicts the fact that $T$ is $|\Bbb R|$-categorical.
*One has to use the ultrahomogeneity of $(\Bbb R,+,0)$ for this argument to work, as explained by Alex in the comments below. Ultrahomogeneity of $(\Bbb R,+,0)$ follows from linear algebra: given two families $\overline a$ and $\overline b$ that satisfy the same formulas, there exists an automorphism of $(\Bbb R,+,0)$ that takes $\overline a$ to $\overline b$.
