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It's an interesting exercise to find the right generalization of this proof to sqrt(n) for arbitrary numbers n that are not perfect squares, and for kth roots f
by wging 2y ago
It's an interesting exercise to find the right generalization of this proof to sqrt(n) for arbitrary numbers n that are not perfect squares, and for kth roots for m >= 2. I.e. prove that if kth_rt(n) is rational, then n is a perfect kth power (or equivalently, that if n is not a perfect kth power, then kth_rt(n) is irrational).
(I'm talking about adapting the ideas of this divisibility-based proof. abstractbill's post https://news.ycombinator.com/item?id=41314547 https://news.ycombinator.com/item?id=41314547 about Conway's method, https://www.youtube.com/watch?v=wNOtOPjaLZs https://www.youtube.com/watch?v=wNOtOPjaLZs, is a completely different (and very cool) way to do this that I hadn't seen before today.)
- jfengel 2y agoI remember being extremely struck by how general this proof was. In fact it made me suspicious that it would even apply to perfect squares. Running the proof on the square root of 4 took a little while to sink in.
- Smaug123 2y ago(Important thing to do while understanding any theorem! Test the boundaries, discover what happens when you relax every hypothesis.)