4 ms·
For clarity: the proof hinges on (√2 - 1) < 1 (so that k’ < k) and √2 being a square root so that you can multiply k’ by it and indeed get 2 k - k √2. This is
by jacobn 4y ago
For clarity: the proof hinges on (√2 - 1) < 1 (so that k’ < k) and √2 being a square root so that you can multiply k’ by it and indeed get 2 k - k √2.
This is why it also works for √3, but doesn’t work for √5 or 1.5
- deleted 4y ago[deleted]
- perihelions 4y agoRight. It can also generalize for any non-integral square root of an integer √n. Let a be the integer ⌊√n⌋, such that a < √n < a+1 -- strict inequalities, because √n is non-integral. So 0 < √n - a < 1 and the same proof works with k' = √n k - a k (instead of √2 k - 1 k) = k(√n - a) (so 0 < k' < k) √n k' = n k - a k √n (is integral; k' ∈ K)