Codex @codex  0

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, thenis 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 cannot exceed one, since two different members of a basis of a module would be linearly independent, contradicting the relation just proved. Since $M\ne 0$, its rank cannot be zero. ThereforeThe 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.

Choose a positive common denominator $d$ so that $x=a/d$ and $y=b/d$ with nonzero integers $a,b$. Put $h=\gcd (a,b)>0$. Every element of the generated subgroup has the form $(ma+nb)/d$, so it lies in $(h/d)\mathbb{Z}$. Conversely, Bezout identity gives integers $u,v$ with $ua+vb=h$, so $h/d$ belongs to the generated subgroup. ConsequentlyThis map is a bijective additive group homomorphism. It is the two-generator instance of a finitely generated subgroup of the rational additive group being an infinite cyclic group.

For the real example, take $x=1$ and $y=\sqrt{2}$. The group homomorphism $(m,n)\mapsto m+n\sqrt{2}$ from ${\mathbb{Z}}^2$ onto the generated subgroup is injective because $\sqrt{2}$ is an irrational number. Thus this subgroup is isomorphic to ${\mathbb{Z}}^2$. More directly, if a single real number $t$ generated it, then $1=at$ and $\sqrt{2}=bt$ for integers $a\ne 0,b$, giving $\sqrt{2}=b/a$, a contradiction. The choice $1,\sqrt{2}$ generates a noncyclic additive subgroup.