Let $y$ be irrational. Consider $15y$:

If $15y$ were rational, then $15y = \frac{p}{q}$ for integers $p, q$, which would give $y = \frac{p}{15q}$, making $y$ rational. This contradicts the given.

Therefore $15y$ must be irrational.

Note: $15y$ could be positive or negative (depending on $y$), so we cannot determine its sign. The only statement that must be true is that $15y$ is irrational.