$\QQ$ is not a finitely generated group

Suppose for contradiction that the additive group \((\QQ,+)\) is generated by finitely many rationals \[ q_1,\ldots,q_r. \] Write each \[ q_i=\frac{a_i}{b_i} \] with \(a_i\in\ZZ\) and \(b_i\ge1\), and let \[ B=\operatorname{lcm}(b_1,\ldots,b_r). \] Then every integer linear combination of the \(q_i\) lies in \[ \frac1B\ZZ. \] Hence the subgroup they generate is contained in \(B^{-1}\ZZ\).

But \[ \frac{1}{2B}\in\QQ \] does not lie in \(B^{-1}\ZZ\), since \(1/(2B)=m/B\) would force \(m=1/2\notin\ZZ\). This contradicts generation of all of \(\QQ\). Therefore \((\QQ,+)\) is not finitely generated.