Thread: Proving two floor problems equal each other

1. Proving two floor problems equal each other

Excuse the confusing text but I don't know how to input math texts on this forum (yet).

So my problem is:
Prove that for positive real number x,

floor(floor(x/2)/2)=floor(x/4).

I figured I have to prove cases when the remainder of x/4 is 0,1,2 or 3.
MY question is, how do I get started?

2. It helps to draw a graph: first of $\displaystyle x/2$, then $\displaystyle \lfloor x/2\rfloor$, then $\displaystyle \lfloor x/2\rfloor/2$, and finally $\displaystyle \lfloor\lfloor x/2\rfloor/2\rfloor$.

The remainder is not often used for the division of real numbers. I believe it is sufficient to consider two cases: when the function argument has the form $\displaystyle 4n+x$ and $\displaystyle 4n+2+x$ where $\displaystyle n\in\mathbb{Z}$ and $\displaystyle 0\le x<2$. Then, obviously, $\displaystyle 0\le x/2<1$ and $\displaystyle 0\le x/4<1/2$.

3. Originally Posted by emakarov
It helps to draw a graph: first of $\displaystyle x/2$, then $\displaystyle \lfloor x/2\rfloor$, then $\displaystyle \lfloor x/2\rfloor/2$, and finally $\displaystyle \lfloor\lfloor x/2\rfloor/2\rfloor$.

The remainder is not often used for the division of real numbers. I believe it is sufficient to consider two cases: when the function argument has the form $\displaystyle 4n+x$ and $\displaystyle 4n+2+x$ where $\displaystyle n\in\mathbb{Z}$ and $\displaystyle 0\le x<2$. Then, obviously, $\displaystyle 0\le x/2<1$ and $\displaystyle 0\le x/4<1/2$.
We never learned on how to draw graphs of floor or ceiling functions so I can't really prove it that way. I asked my teacher and he said I was on the right track with just proving cases where the remainder is either 0,1,2 or 3, but I still don't know how to start off. How can I create a function that always has a remainder? I'm really lost on this question

4. The remainder of a real number divided by 4 is a real number from 0 (included) to 4 (excluded). It is not just 0, 1, 2 or 3. For example, what is the remainder of $\displaystyle \pi$ divided by 4? It may make sense to say that the ratio of $\displaystyle \pi$ and 4 is 0 since $\displaystyle \pi$ < 4 and the remainder is $\displaystyle \pi$. As I said, remainders are rarely considered for real numbers.

To continue my suggestion, suppose that $\displaystyle x=4n+y$ where $\displaystyle n\in\mathbb{Z}$ and $\displaystyle 0\le y<2$. Then $\displaystyle x/2=2n+y/2$. Since $\displaystyle 0\le y/2<1$, $\displaystyle \lfloor x/2\rfloor=2n$. Therefore, $\displaystyle \lfloor\lfloor x/2\rfloor/2\rfloor=n$. Now you show that $\displaystyle \lfloor (4n+y)/4\rfloor=n$. Also, show that $\displaystyle \lfloor\lfloor x/2\rfloor/2\rfloor=\lfloor x/4\rfloor$ when $\displaystyle x=4n+2+y$ where $\displaystyle n\in\mathbb{Z}$ and $\displaystyle 0\le y<2$.