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Thread: Floor functions!

  1. #1
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    Floor functions!

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

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

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

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


    I just can't prove the initial part.

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

    Let $\displaystyle x=qn+r+a$ where $\displaystyle 0\le r<n$ and $\displaystyle 0\le a<1$ 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 $\displaystyle x=qn+r+a$ as you suggest we have $\displaystyle \left \lfloor x\right \rfloor = qn+r$ and as $\displaystyle r<n, \left \lfloor \frac{r}{n}\right \rfloor = 0$ so $\displaystyle \left \lfloor\frac{\left \lfloor x\right \rfloor}{n}\right \rfloor = q$

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

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