= Solution
Suppose a finite set of <rational numbers> $q_1,\ldots,q_m$ generated $\mathbb Q$ as a <module> over $\mathbb Z$. A positive common denominator $D$ makes every $Dq_i$ an <integer>. For every <integer> linear combination, therefore,
$$
D\sum_i n_iq_i\in\mathbb Z.
$$
The entire generated <submodule> is contained in $(1/D)\mathbb Z$. But $1/(2D)$ is a <rational number> outside that set: $1/(2D)=k/D$ would require the noninteger $k=1/2$. This contradicts generation of all of $\mathbb Q$. Consequently
$$
\boxed{\mathbb Q\text{ is not finitely generated as a }\mathbb Z\text{-module}.}
$$
Each finite set of generators imposes a common denominator; the full <rational number> system has no such bound.
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