Proving the negation of a conditional using proof by contradiction

Question

CONTEXT: Question made up by uni maths lecturer

Prove the following statement using a proof by contradiction:

• For all nonzero rational numbers $x$, if $y$ is irrational then $\frac{x}{y}-3$ is irrational.

I have found the negation of the statement to be:

• There exists a nonzero rational number $x$ such that $y$ is irrational and $\frac{x}{y}-3$ is rational.

I'm a bit stuck on the proof.

So far I have written $\frac{x}{y}-3=\frac{a}{b}$ such that $a$ and $b$ are integers ($b$ is nonzero) which is just using the definition of a rational.

So far (but I'm not sure this will get me anywhere) I have written $\frac{x}{y}-3$ as $\frac{x-3y}{y}$
which I thought I might be able to show is not equal to $\frac{a}{b}$, which would mean it's irrational and would lead to a contradiction but I'm not sure how to do this.

If anyone has a better idea on how to set up the proof by contradiction, I'm all ears.

UPDATE:

I manipulated $\frac{x}{y}-3=\frac{a}{b}$ to get $s=\frac{bx}{a+3b}$ which is rational, contradicting the negation we took to be true.

Thanks everyone for your help!

Answer 1

Hint:

From the equation If $\frac{x}{y}-3=\frac{a}{b}$, try to calculate what $y$ is equal to, and prove that $y$ is rational. This will contradict with the assumption that $y$ is irrational.