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Math Help - Floor functions!

  1. #1
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    Floor functions!

    Given any real number x and positive integer n, prove that \left \lfloor\frac{\left \lfloor x\right \rfloor}{n}\right \rfloor = \left \lfloor\frac{x}{n}\right \rfloor

    Deduce that for any real number y and positive integers n and m, one has

    \left \lfloor\frac{\left \lfloor \frac{y}{m}\right \rfloor}{n}\right \rfloor = \left \lfloor\frac{\left \lfloor \frac{y}{n}\right \rfloor}{m}\right \rfloor

    I think the second part is very straightforward i.e. \left \lfloor\frac{\left \lfloor \frac{y}{m}\right \rfloor}{n}\right \rfloor = \left \lfloor\frac{\frac{y}{m}}{n}\right \rfloor = \left \lfloor\frac{\frac{y}{n}}{m}\right \rfloor = \left \lfloor\frac{\left \lfloor \frac{y}{n}\right \rfloor}{m}\right \rfloor


    I just can't prove the initial part.

    Any help appreciated!
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  2. #2
    Senior Member
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    Re: Floor functions!

    Let x=qn+r+a where 0\le r<n and 0\le a<1 and q and r are integers.

    I think this goes somewhere.
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  3. #3
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    Re: Floor functions!

    Cheers, think I get it.. So by putting x=qn+r+a as you suggest we have \left \lfloor x\right \rfloor = qn+r and as r<n,  \left \lfloor \frac{r}{n}\right \rfloor = 0 so \left \lfloor\frac{\left \lfloor x\right \rfloor}{n}\right \rfloor = q

    Now  \frac{x}{n} = q+ \frac{r+a}{n} as r\in Z and  r<n, and a<1 we have (r+a)<n so \left \lfloor q+ \frac{r+a}{n}\right \rfloor = q also.

    Therefore \left \lfloor\frac{\left \lfloor x\right \rfloor}{n}\right \rfloor = \left \lfloor\frac{x}{n}\right \rfloor
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