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Math Help - Proving two floor problems equal each other

  1. #1
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    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?
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  2. #2
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    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.
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  3. #3
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    Quote Originally Posted by emakarov View Post
    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
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  4. #4
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    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.
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