Could you please confirm if this proof is correct?
Theorem: If $q \neq 0$ is rational and $y$ is irrational, then $qy$ is irrational.
Proof: Proof by contradiction, we assume that $qy$ is rational. Therefore $qy=\frac{a}{b}$ for integers $a$, $b \neq 0$. Since $q$ is rational, we have $\frac{x}{z}y=\frac{a}{b}$ for integers $x \neq 0$, $z \neq 0$. Therefore, $xy = a$, and $y=\frac{a}{x}$. Since both $a$ and $x$ are integers, $y$ is rational, leading to a contradiction.
Best Answer
As I mention here frequently, this ubiquitous property is simply an instance of complementary view of the subgroup property, i.e.
THEOREM $\ $ A nonempty subset $\rm\:S\:$ of abelian group $\rm\:G\:$ comprises a subgroup $\rm\iff\ S\ + \ \bar S\ =\ \bar S\ $ where $\rm\: \bar S\:$ is the complement of $\rm\:S\:$ in $\rm\:G$
Instances of this are ubiquitous in concrete number systems, e.g.