Originally Posted by

**Archie** I honestly don't understand what your concern is. If I demonstrate that $\displaystyle \mathcal{P}(1)$ is true and that $\displaystyle \mathcal{P}(\tfrac1n) \implies \mathcal{P}(\tfrac1{n+1})$ and that $\displaystyle \mathcal{P}(\tfrac m a) \implies \mathcal{P}(\tfrac{m+1}a)$ for all natural numbers $\displaystyle a,m,n$ it follows that $\displaystyle \mathcal{P}(\frac m n)$ is true for all positive rationals $\displaystyle \frac m n$. If some condition exists that makes one of those proofs fail, that is a failure of the particular proof not of the concept.

It's no different to demonstrating that $\displaystyle \mathcal{Q}(1,1)$ is true and that $\displaystyle \mathcal{Q}(1,n) \implies \mathcal{Q}(1,n+1)$ and that $\displaystyle \mathcal{Q}(m,a) \implies \mathcal{Q}(m+1,a)$. If that doesn't prove that $\displaystyle \mathcal{Q}(m,n)$ is true for all natural numbers $\displaystyle m,n$, then induction on the natural numbers is broken.