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 $x/2$, then $\lfloor x/2\rfloor$, then $\lfloor x/2\rfloor/2$, and finally $\lfloor\lfloor x/2\rfloor/2\rfloor$.

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 $4n+x$ and $4n+2+x$ where $n\in\mathbb{Z}$ and $0\le x<2$. Then, obviously, $0\le x/2<1$ and $0\le x/4<1/2$.

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

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 $4n+x$ and $4n+2+x$ where $n\in\mathbb{Z}$ and $0\le x<2$. Then, obviously, $0\le x/2<1$ and $0\le x/4<1/2$.
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 $\pi$ divided by 4? It may make sense to say that the ratio of $\pi$ and 4 is 0 since $\pi$ < 4 and the remainder is $\pi$. As I said, remainders are rarely considered for real numbers.

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