How to show that the expression can be made arbitrarily close to 0?

56 Views Asked by At

Let $x\in\mathbb{R}$ and $\lfloor x \rfloor = \max\{z\in\mathbb{Z} : m\leq x\}.$

Let $z\in\mathbb{Z}.$

How to show that the expression $-\sqrt{2}z+\lfloor \sqrt{2}z \rfloor$ can be made arbitrarily close to $0$ (due to the choice of $z$)?