Contrapositive and Contradiction · Challenge problem

A slick proof that $\sqrt2$ is irrational: if $\sqrt2 = \dfrac pq$ with $q$ the SMALLEST possible positive denominator, then $\sqrt2 = \dfrac{2q - p}{p - q}$, with a smaller denominator. Apply the map $\dfrac pq \mapsto \dfrac{2q - p}{p - q}$ to the approximation $\dfrac{99}{70}$. What fraction results?