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.


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!

  • $\begingroup$ Write it like this: $\frac{x}{y} = \frac{a}{b}-3$. The right side is then rational by definition... $\endgroup$ – Max Apr 8 at 10:32
  • 1
    $\begingroup$ Cross multiply $\frac{x-3y}{y}=\frac{a}{b}$ to get $y=\frac{bx}{a+3b}$ which contradicts the assumption of $y$ as irrational. $\endgroup$ – 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.


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