Proving basic floor function inequality: $-1 \lt \lfloor 2x \rfloor - 2 \lfloor x \rfloor \lt 2$

Question

As a direct consequence of the definition of $\lfloor x \rfloor $ I know that $$2x-1 \lt \lfloor 2x \rfloor \le 2x$$ and $$2x-2 \lt 2\lfloor x\rfloor \le 2x$$ How can I use these to show that $-1 \lt \lfloor 2x \rfloor - 2 \lfloor x \rfloor \lt 2$? I'm a bit rusty with inequalities. Any help is appreciated.

Answer 1

$$-2x\le-2\lfloor x\rfloor<2-2x$$ For the left hand side of the wanted inequality, using the left hand side of the first and the third inequality: $$\begin{align*} \lfloor2x\rfloor-2\lfloor x\rfloor &> (2x-1)-2\lfloor x\rfloor \\ &\ge (2x-1)-2x \\ &= -1 \end{align*}$$

And similar for the second wanted inequality.

Answer 2

Let $n = \lfloor x\rfloor$ and $f=x-n = x-\lfloor x\rfloor$, i.e. $x=n+f$. Then $$\begin{align*} \lfloor 2x\rfloor-2\lfloor x\rfloor&= \lfloor 2n+2f\rfloor - 2\lfloor n+f\rfloor\\ &=2n+\lfloor 2f\rfloor - 2n\\ &= \lfloor 2f\rfloor \end{align*}$$ Since $0\le f<1$, $0\le 2f<2$ and then $\lfloor 2f\rfloor$ can only be $0$ or $1$.

Hence we have a stronger result: $$0\le\lfloor 2x\rfloor-2\lfloor x\rfloor\le1$$