4 ms·
FTA is massive overkill. For every number n, either n can be expressed as 2k for some k, or 2k+1 for some k, but not both (proof: by induction); in particular t
by Smaug123 2y ago
FTA is massive overkill. For every number n, either n can be expressed as 2k for some k, or 2k+1 for some k, but not both (proof: by induction); in particular the square root can too. If the square root is (2k+1), then the square is 4k^2 + 4k + 1 = 2(2k^2+2k) + 1, which is by definition odd, not even as we supposed.
- tines 2y agoTrue, but the FTA proof is just really intuitive for me and I like it.
- Hackbraten 2y agoThanks for the proof, it was fun to follow, and I agree that it's quite intuitive. I think that it would be helpful to mention why sqrt(y) must be an integer. (I know that it is, but it also feels a bit glossed over, given that all the other steps of the proof were explained so thoroughly.)
- thaumasiotes 2y agoThe FTA proof is the one that's obvious, though. If it's really easy to do something using a basic tool, why worry that the basic tool is complex to describe?