Past exam of the algebra 2 course of the University of Oxford / 2015 / 2 / d / Solution

Suppose a finite set of rational numbers $q_1,\dots ,q_m$ generated $\mathbb{Q}$ as a module over $\mathbb{Z}$. A positive common denominator $D$ makes every $D{q}_i$ an integer. For every integer linear combination, therefore,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 $\mathbb{Q}$. ConsequentlyEach finite set of generators imposes a common denominator; the full rational number system has no such bound.