# if $\left\lfloor x/2\right\rfloor$ is odd, show that $\left\lfloor \left\lfloor x/2\right\rfloor /2\right\rfloor = \left\lfloor x/4 \right\rfloor$ [closed]

Let $x$ be real numbers, if $\left\lfloor x/2\right\rfloor$ is odd, show that $\left\lfloor \left\lfloor x/2\right\rfloor /2\right \rfloor = \left\lfloor x/4 \right\rfloor$

## closed as off-topic by Parcly Taxel, quid♦, PSPACEhard, iadvd, WatsonSep 8 '16 at 8:49

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• It's true for all $x$, isn't it? – TonyK Sep 7 '16 at 13:26

If $\left\lfloor \frac{x}{2}\right\rfloor$ is odd, then there exist $m$ such that $2m+1\leq \frac{x}{2}<2m+2$ and so $\left\lfloor \frac{x}{2}\right\rfloor=2m+1$. Now, $\left\lfloor \frac{ \left\lfloor x/2 \right\rfloor }{ 2 }\right\rfloor=\left\lfloor m+\frac{1}{2}\right\rfloor=m$. On the other hand
$$m+\frac{1}{2} \leq \frac{x}{4} < m+1,$$
Then $\lfloor \frac{x}{4} \rfloor = m$, which is indeed equal to $\left\lfloor \frac{ \left\lfloor x/2\right\rfloor}{2}\right\rfloor$.
Start from the basic property of floor : $\lfloor x \rfloor$ is the only integer such that $x-1<\lfloor x \rfloor\leq x$.
So $x/2 - 1 < \lfloor x/2 \rfloor \leq x/2$. If $\lfloor x/2 \rfloor$ is odd, then $\lfloor \lfloor x/2 \rfloor/2 \rfloor= \lfloor x/2 \rfloor/2 -1/2$. Since we have $$\frac{x/2 - 1} 2 - \frac 1 2< \lfloor x/2 \rfloor/2 -1/2 \leq \frac{x/2}2 - \frac 1 2$$ we can derive $$\frac{x} 4 - 1< \lfloor \lfloor x/2 \rfloor/2 \rfloor \leq \frac{x}4 - \frac 1 2 \leq \frac x 4$$