- #36
NATURE.M
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So it turns out when I was writing the proof, I stumbled upon a little issue. Basically, in order to be able to use the IH on rfloor(√n) we require 3 ≤floor(√n) < n which is just not the case. Like take n = 3, then floor(√3) < 3. Even if you try to lower bound floor(√n) by 0 or 1 it doesn't work since then the log term in the predicate becomes undefined. In other words, P(n) just isn't true for n = 0, 1. This is rather unfortunate. I'm not really sure how to get around this ?haruspex said:A pleasure.