# Show that given any rational number $x$, there exists an integer $y$ such that $x^2y$ is an integer

I have a question:

Show that given any rational number $$x$$, there exists an integer $$y$$ such that $$x^2y$$ is an integer.

I am new to proofs, but I have the following so far:

Definitions:

$$x$$ is a Rational number

$$y$$ is an Integer

$$x^2y$$ is an Integer

Assume that $$x = \frac{m}{n}$$ where $$m$$ is an Integer and $$n$$ is an integer and $$n\neq 0$$

$$x^2 = \frac{m^2}{n^2}$$

$$x^2y = \frac{m^2}{n^2}y$$

I have no idea how to go beyond this to prove that $$x^2y$$ is an Integer in proof form. I think it is that $$\frac{m^2y}{n^2}$$ should not have a remainder, but how do I write it in proof form?

• lol just set y=n^2 Jan 29 '19 at 23:35

Write $$x=m/n$$ with integers $$m$$ and $$n,$$ and then choose $$y=n^2.$$
Then we have $$x^2 y = \frac{m^2}{n^2} \cdot n^2 = m^2,$$ which is an integer.