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.
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You can directly divide by $q$ assuming the fact that $q \neq 0$.
Suppose $qy$ is rational then, you have $qy = \frac{m}{n}$ for some $n \neq 0$. This says that $y = \frac{m}{nq}$ which says that $\text{y is rational}$ contradiction.
On
I don't think it correct. It seems like a good idea to indicate both x as an integer, and z as a non-zero integer. Then you also want to "solve for" y, which as Eric points out, you didn't quite do.
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A group theoretic proof: You know that if $G$ is a group and $H\neq G$ is one of its subgroups then $h \in H$ and $y \in G\setminus H$ implies that $hy \in G\setminus H$. Proof: suppose $hy \in H$. You know that $h^{-1} \in H$, and therefore $y=h^{-1}(hy) \in H$. Contradiction.
In our case, we have the group $(\Bbb{R}^*,\cdot)$ and its proper subgroup $(\Bbb{Q}^*,\cdot)$. By the arguments above $q \in \Bbb{Q}^*$ and $y \in \Bbb{R}\setminus \Bbb{Q}$ implies $qy \in \Bbb{R}\setminus \Bbb{Q}$.
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Let's see how we can modify your argument to make it perfect.
First of all, a minor picky point. You wrote $$qy=\frac{a}{b} \qquad\text{where $a$ and $b$ are integers, with $b \ne 0$}$$
So far, fine. Then come your $x$ and $z$. For completeness, you should have said "Let $x$, $z$ be integers such that $q=\frac{x}{z}$. Note that neither $x$ nor $z$ is $0$." Basically, you did not say what connection $x/z$ had with $q$, though admittedly any reasonable person would know what you meant. By the way, I probably would have chosen the letters $c$ and $d$ instead of $x$ and $z$.
Now for the non-picky point. You reached $$\frac{x}{z}y=\frac{a}{b}$$ From that you should have concluded directly that $$y=\frac{za}{xb}$$ which ends things, since $za$ and $xb$ are integers.
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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.
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If you want to play a bit with logical arguments, I offer you an alternative proof based on the contrapositive of the statement you are trying to show. Instead of proving $P\to Q$, I'll prove $\lnot Q\to \lnot P$.
We want to show that
$$(\forall x)(\forall y)[x\in \mathbb{Q}^* \land y\not \in\mathbb{Q} \to xy\not\in\mathbb{Q}],\quad(*)$$
since any number can either be rational or irrational; $\mathbb{Q}^*=\mathbb{Q}-\{0\}$. But
\begin{align*} x\in \mathbb{Q}^* \land y\not \in\mathbb{Q} \to xy\not\in\mathbb{Q} \equiv \lnot\left(xy\not\in\mathbb{Q}\right)\to \lnot\left(x\in \mathbb{Q}^* \land y\not\in\mathbb{Q}\right)\\ \equiv xy\in\mathbb{Q}\to \left(x\not\in \mathbb{Q}^* \lor y\in\mathbb{Q}\right),\quad\quad(**) \end{align*}
The original problems translate to proving (*). For, let $z = xy$ and assume $x\in \mathbb{Q}^+$. Then
$$ z\frac{1}{x}= (xy)\frac{1}{x} = y\in \mathbb{Q}. $$
We used the fact that $x\neq 0$ has an inverse and that the product of two rationals is rational. This proves that (**) is true, so its equivalent version (*) must also be true. Q.E.D.
It's wrong. You wrote $\frac{x}{z}y = \frac{a}{b}$. That is correct. Then you said "Therefore $xy = a$. That is wrong.
You need to solve $\frac{x}{z}y = \frac{a}{b}$ for $y$. You get $y = \frac{a}{b} \cdot \frac{z}{x}$.