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
This is why it also works for √3, but doesn’t work for √5 or 1.5
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)