How to prove that $|(kx\mod1)-(ky\mod1)|=k|x-y|$ under the following conditions:

Question

Suppose that $k\in\mathbb{N}_{\geq2}$ and $x,y\in[0,1)$. Also assume that $|x-y|\leq1/(2k)$ and $|kx\mod1-ky\mod1|\leq1/2$. How do I prove that $$|(kx\mod1)-(ky\mod1)|=k|x-y|?$$

Answer 1

Without loss of generality assume $y > x$. We have $$kx \text{ mod } 1 - ky \text{ mod } 1 = kx -ky \text{ mod 1 } = k(x-y) \text{ mod } 1,$$ and since $|x-y| \leq 1/(2k)$ we obtain that $|k(x-y)| \leq 1/2 < 1$. Thus, since $|k(x-y)| = k(x-y)$ by assumption, we get your desired conclusion.