Krdytkyu

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.

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 $y = frac{a}{b} cdot frac{z}{x}$.

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 $rmiff 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.

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.

A group theoretic proof: You know that if $G$ is a group and $Hneq G$ is one of its subgroups then $h in H$ and $y in Gsetminus H$ implies that $hy in Gsetminus H$. 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 $qy in Bbb{R}setminus Bbb{Q}$.

Let's see how we can modify your argument to make it perfect.

First of all, a minor picky point. You wrote
$$qy=frac{a}{b} qquadtext{where $a$ and $b$ are integers, with $b ne 0$}$$

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.

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.

$$ainmathbb{Q},binmathbb{R}setminusmathbb{Q},abinmathbb{Q}implies binmathbb{Q}impliestext{Contradiction}therefore abnotinmathbb{Q}.$$