If $0\leq x\leq1/2$, then why $\inf_{k\in\mathbb{Z}}|x+k|=x$?

Question

Suppose that $0\leq x\leq1/2$. Then how do I formally prove the (rather intuitive) identity $$\inf_{k\in\mathbb{Z}}|x+k|=x?$$ It is easy to see that $\inf_{k\in\mathbb{Z}}|x+k|\leq x$, even without the bounds on $x$. But I don't know how to prove the other inequality. Any help would be greatly appreciated! Thanks in advance!

Or more generally: If $|x|\leq1/2$, then why $\inf_{k\in\mathbb{Z}}|x+k|=|x|$?

Answer 1

Is this obvious? You know that $|x+k|=|x|$ when $k=0$. On the other hand, when $|x|\leq 1/2$, then for any $k\in\mathbb Z$ which $k\neq0$, we have $|x+k|\ge |k|-|x|\ge 1-1/2=1/2\ge |x|$. Thus for any $k$, we have $|x+k|\ge |x|$ and equality when $k=0$. Thus $\inf_{k\in\mathbb Z}|x+k|= |x|$.