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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Parcly Taxel, quid, PSPACEhard, iadvd, Watson
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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 $$\frac{x} 4 - 1< \lfloor \lfloor x/2 \rfloor/2 \rfloor \leq \frac{x}4 - \frac 1 2 \leq \frac x 4$$