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 $z$)?
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 $z$)?
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