= Solution
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>. Consequently
$$
\boxed{N=(h/d)\mathbb Z\cong\mathbb Z,\qquad k\longmapsto kh/d.}
$$
This 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 $y=\sqrt2$. The <group homomorphism> $(m,n)\mapsto m+n\sqrt2$ from $\mathbb Z^2$ onto the generated <subgroup> is injective because $\sqrt2$ 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 $\sqrt2=bt$ for <integers> $a\ne0,b$, giving $\sqrt2=b/a$, a contradiction. \b[The choice $1,\sqrt2$ generates a noncyclic additive <subgroup>.]
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