# Proving the negation of a conditional using proof by contradiction

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.

• Write it like this: $\frac{x}{y} = \frac{a}{b}-3$. The right side is then rational by definition... – Max Apr 8 at 10:32
• Cross multiply $\frac{x-3y}{y}=\frac{a}{b}$ to get $y=\frac{bx}{a+3b}$ which contradicts the assumption of $y$ as irrational. – NewBornMATH Apr 8 at 10:33
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.
If you solve for x you have $$(\frac{a}{b}+3)y$$. However this is a rational (nonzero) multiplied by an irrational, which is irrational. This contradicts that x is supposed to be a nonzero rational.