Theorem 1   The set of rational numbers $\Q$ do not form a free $\Z$ -module.

Proof. First note that any two elements in $\Q$ are $\Z$ -linearly dependent. If $x=\frac{p_1}{q_1}$ and $y=\frac{p_2}{q_2}$ , then $q_1p_2x-q_2p_1y=0$ . Since basis elements must be linearly independent, this shows that any basis must consist of only one element, say $\frac{p}{q}$ , with $p$ and $q$ relatively prime, and without loss of generality, $q>0$ . The $\Z$ -span of $\{\frac{p}{q}\}$ is the set of rational numbers of the form $\frac{np}{q}$ . I claim that $\frac{1}{q+1}$ is not in the set. If it were, then we would have $\frac{np}{q}=\frac{1}{q+1}$ for some $n$ , but this implies that $np=\frac{q}{q+1}$ which has no solutions for $n,p\in\Z$ ,$q\in\Z^+$ , giving us a contradiction.