my question is the following: given the additive group of rational numbers, i.e. $Q = \langle {\mathbb Q},+,0\rangle$ and $T$ the theory of $Q$, how can I find (explicitly) a 2-type which is not realised in $Q$?
My guess is the following: $$\{ \neg(n\cdot v_1 = m\cdot v_2) \mid m,n\in{\mathbb N}^+\}$$ (where $n\cdot v_1$ and $m\cdot v_2$ are just shorthand for $v_1+\ldots +v_1$ and $v_2+\ldots+v_2$)
but this type is realised in $Q$ since i can take e.g. $v_1$=positive and $v_2$=negative; this would realise the type.
Any suggestions?
Hint: The type you've written says that $v_1$ and $v_2$ do not have a common natural multiple. Can you write a type that says that $v_1$ and $v_2$ do not have a common integer multiple? Use the formula that says $v_1=-v_2$.