= Solution
Let $x,y\in\mathbb Q$. Choose a positive <integer> $D$ clearing their denominators and write $x=m/D$, $y=n/D$ with $m,n\in\mathbb Z$. If $x$ and $y$ are not both zero, then
$$
nx-my=0
$$
is a relation with <integer> coefficients not both zero. If both vanish, use $1\cdot x+0\cdot y=0$. Thus every pair is <linearly dependent> over $\mathbb Z$.
A <submodule> $M\subseteq\mathbb Q$ is a <torsion-free module>: if $kz=0$ for a nonzero <integer> $k$ and a <rational number> $z$, then $z=0$. If $M$ is also a <finitely generated module>, part (a), applied to the <Euclidean domain> $\mathbb Z$, makes it a <free module>. Its <rank of a module>[rank] cannot exceed one, since two different members of a <basis of a module> would be linearly independent, contradicting the relation just proved. Since $M\ne0$, its rank cannot be zero. Therefore
$$
\boxed{\operatorname{rank}_{\mathbb Z}M=1,\qquad M\cong\mathbb Z.}
$$
The denominator-and-<greatest common divisor> construction in part (c) gives an explicit version of this result for a <finitely generated subgroup of the rational additive group>.
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