Example of a torsion-free abelian group of rank zero

In Derek Robinsons A course in the theory of groups, on page 98 the rank of an abelian group is defined:

Let $$G$$ be an abelian group and let $$S$$ be a nonempty subset of $$G$$. Then $$S$$ is called lineary independent [...] if $$0 \notin S$$ and, given distinct elements $$s_1, \ldots, s_r$$ of $$S$$ and integers $$m_1, \ldots, m_r$$, the relation $$m_1 s_1 + \ldots + m_r s_r = 0$$ implies that $$m_i s_i = 0$$.

Note that this is slightly different than the usual definition in the sense that he just asserts the implication $$m_i s_i = 0$$, and not that the coefficients $$m_i$$ should vanish. Then he goes on:

If $$p$$ is a prime and $$G$$ an abelian group, the $$p$$-rank of $$G$$ $$r_p(G)$$ is defined as the cardinality of a maximal independent subset of elements of $$p$$-power order. Similarly the $$0$$-rank or torsion-free rank $$r_0(G)$$ is the cardinality of a maximal independent subset of elements of infinite order. Also important is the Prüfer rank, often just called the rank of $$G$$ $$r(G) = r_0(G) + \max_{p} r_p(G).$$

Then he proves that these notions are well-defined. Later on in a section on torsion-free groups of rank $$1$$, he introduces the notion of the type of an abelian group, and proves that two torsion-free abelian groups of rank $$\le 1$$ are isomorphic if and only if the y have the same type.

If we assume $$G$$ is torsion-free, I think the $$p$$-rank vanishes for every $$p$$ as there is no element of order $$p$$ by definition. Hence $$r(G) = r_0(G)$$. But in this case, how should a torsion-free group of rank zero look like? Let $$G$$ be any abelian torsion-free subgroup and $$g \in G$$, then $$g$$ itself is lineary indepedent, hence $$r(G) \ge 1$$. Or do I miss anything? Could anybody give me an example of an torsion-free abelian group of rank $$0$$? I am afraid I have misunderstood something because I cannot come up with one...

First, note that any 1-element set other than $$\{0\}$$ is linearly independent: the condition is "$$ms = 0$$ implies $$ms = 0$$" which is clearly true! (I think you've realised this, but it's worth pointing out explicitly).
You are right in saying that if $$G$$ is torsion-free then $$r(G) = r_0(G)$$, and if we then take $$g \neq 0$$ then $$\{g\}$$ is indeed linearly independent and so $$r(G) \geq 1$$ - and so if we know that $$r(G) = 0$$ then no such $$g$$ exists, and $$G = \{0\}$$.
It may help to realise that the rank of an abelian group is analogous to the dimension of a vector space - and a $$0$$-dimensional vector space is also just $$\{0\}$$.