Let $G$ be a torsion-free abelian group of having $n$ number of maximally rationally independent elements $r_{1}, r_{2}, ..., r_{n}$ and assume that $G$ is not finitely generated. Is this correct to say $G$ is a $\mathbb Q$-vector space having basis $\{r_{1}, r_{2}, ..., r_{n}\}$?
If this is correct, then I am looking to write for every $g\in G,$ there exist integers $m$ and $m_{1}, m_{2},..., m_{n}$ such that $mg = m_{1}r_{1} + m_{2}r_{2} + \cdots + m_{n}r_{n}.$ Thank you.
If $G=\mathbb Z[\frac12]$, then $G$ is torsion-free and $\{1\}$ is a maximal rationally independent set. Furthermore, $G$ is not finitely generated as you already know from this answer.
It's not hard to prove that $G$ is not a $\mathbb Q$-vector space: since $(1,x)\mapsto x$ for all $x\in G$ then we should have $(q,x)\mapsto qx$ for all $q\in\mathbb Q$ and $x\in G$, and this is impossible.
Instead, you can say that for every $g\in G$ there exist integers $m$ and $m_{1},..., m_{n}$ such that $mg = m_{1}r_{1} + \cdots + m_{n}r_{n}$. This follows easily once you notice that $g,r_1,\dots,r_n$ are not rationally independent (why?).
