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Clearly we have equality for k=1. But k cannot be equal to 1 as per your assumption that n<m. So you need to still prove a base case for the induction step.
by kleiba 4y ago
Clearly we have equality for k=1.
But k cannot be equal to 1 as per your assumption that n<m. So you need to still prove a base case for the induction step.
- diffeomorphism 4y agoNah, k=1 is the trivial base case and is allowed. The "assumption" is no such thing, but just notational convenience. One can of course tidy that up (e.g. n\leq m and trivial case is obvious), but for a random HN post it seems not worth the effort.
- kleiba 4y agoWell, I guess you're right. I mean, we're looking for two disctinct numbers as the final solution, but that doesn't mean we can't use k=1 in the proof.