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Thread: Divisibility 11

  1. #1
    Sea
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    Divisibility 11

    Show that: $\displaystyle a, k\in \mathbf{Z},$


    $\displaystyle a=2k \Leftrightarrow a-2\| \frac{a}{2}\| = 0 $


    .............................................

    Symbol...
    Example:

    $\displaystyle \| 5,3\|=5 $

    $\displaystyle \| -2,2\|=-3 $
    Last edited by Sea; Dec 20th 2008 at 12:56 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Sea View Post
    Show that: $\displaystyle a, k\in \mathbf{Z},$


    $\displaystyle a=2k \Leftrightarrow a-2\| \frac{a}{2}\| = 0 $


    .............................................

    Symbol...
    Example:

    $\displaystyle \| 5,3\|=5 $

    $\displaystyle \| -2,2\|=-3 $
    In more familiar notation:

    Show that: $\displaystyle a, k\in \mathbb{Z},$


    $\displaystyle a=2k \Leftrightarrow a-2 \left\lfloor \frac{a}{2}\right\rfloor = 0 $

    where $\displaystyle \lfloor x \rfloor$ is the floor function, the greatest integer less than $\displaystyle x$.

    CB
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    In more familiar notation:

    Show that: $\displaystyle a, k\in \mathbb{Z},$


    $\displaystyle a=2k \Leftrightarrow a-2 \left\lfloor \frac{a}{2}\right\rfloor = 0 $

    where $\displaystyle \lfloor x \rfloor$ is the floor function, the greatest integer less than $\displaystyle x$.

    CB
    Suppose $\displaystyle a=2k$ for some $\displaystyle k \in \mathbb{Z}$ then:

    $\displaystyle
    a-2\left\lfloor \frac{a}{2}\right\rfloor=a-2 \lfloor k \rfloor=a-2k=0
    $

    Which proves

    $\displaystyle a=2k \Rightarrow a-2 \left\lfloor \frac{a}{2}\right\rfloor = 0 $

    To complete the proof assume that: $\displaystyle a-2 \left\lfloor \frac{a}{2}\right\rfloor = 0$ and show that this implies that there exists a $\displaystyle k \in \mathbb{Z}$ such that $\displaystyle a=2k$

    CB
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  4. #4
    Sea
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    I didn't know this words means in english...I'm sorry for this and I thank you so much... for answering my question after all...



    Thanks a million......
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