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Math Help - Floor function

  1. #1
    Sea
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    Floor function

    \forall x,y \in \mathbb{R}

    Show that:

    \lfloor x+y  \rfloor \geq \lfloor x \rfloor<br />
+ \lfloor y \rfloor



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

    where \lfloor x \rfloor is the floor function, the greatest
    integer less than x.
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  2. #2
    o_O
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    Let: x = \lfloor x \rfloor + a and y = \lfloor y \rfloor + b for some 0 \leq a,b < 1.

    Fact: \lfloor m + n \rfloor = \lfloor m \rfloor + n where m \in \mathbb{R}, \ n \in \mathbb{Z}

    Consider: x + y = \lfloor x \rfloor + \lfloor y \rfloor  + (a + b)

    Take the floor function of both sides and with the fact, the conclusion follows.
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  3. #3
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    Quote Originally Posted by Sea View Post
    \forall x,y \in \mathbb{R}
    Show that: \lfloor x+y  \rfloor \geq \lfloor x \rfloor<br />
+ \lfloor y \rfloor.
    The rigor required by the instructor/text may well vary.
    But this is the general idea.
    \left\lfloor x \right\rfloor  \leqslant x\;\& \,\left\lfloor y \right\rfloor  \leqslant y\, \Rightarrow \,\left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  \leqslant x + y.
    Because, \left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor is an integer and \left\lfloor {x + y} \right\rfloor is the greatest integer which does not exceed x + y it follows that \left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  \leqslant \left\lfloor {x + y} \right\rfloor .
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  4. #4
    Sea
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    Thanks a million.......

    I understand....

    And... Can you help me?

    __________________________________________________

    Show that:


    \lfloor m + n \rfloor = \lfloor m \rfloor  n where m \in \mathbb{R}, \ n \in \mathbb{Z}



    __________________________________________________ _


    Show that:


    \lfloor <br />
\frac{\lfloor  x \rfloor }{n} \rfloor = \lfloor <br />
\frac{x}{n} \rfloor + where x \in \mathbb{R}, \ n \in \mathbb{Z^+}
    Last edited by Sea; December 21st 2008 at 04:42 PM.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Sea View Post
    Show that:


    \lfloor m + n \rfloor = \lfloor m \rfloor n where m \in \mathbb{R}, \ n \in \mathbb{Z}
    I think this is supposed to be \lfloor m+n \rfloor=\lfloor m \rfloor +n. A quick counterexample to what you have written is \lfloor \sqrt{2}+3 \rfloor=4\ne\lfloor\sqrt{2}\rfloor\cdot 3=3.

    So let m=m_1+\varphi\quad m_1\in\mathbb{Z},\varphi\in[0,1). So \lfloor m+n\rfloor=\lfloor m_1+n+\varphi\rfloor, and since m_1,n\in\mathbb{Z}\wedge\varphi\in[0,1) we can conclude by the defintion of the floor function that \lfloor m_1+n+\varphi\rfloor=m_1+n. Now lets look at the left side. Defining m the same as above we can see that \lfloor m\rfloor+n=\lfloor m_1+\varphi\rfloor+n and once again because m_1\in\mathbb{Z}\wedge\varphi\in[0,1) we can conclude that \lfloor m\rfloor+n=\lfloor m_1+\varphi\rfloor+n=m_1+n. Finally \lfloor m+n\rfloor=m_1+n=\lfloor m\rfloor +n

    Show that

    \left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor=\left\lfloor\frac{x}{n}\  right\rfloor\quad x\in\mathbb{R}\wedge n\in\mathbb{Z}^+
    Similar to before define x=x_1+\varphi\quad x\in\mathbb{Z}\wedge\varphi\in[0,1). So once again by the definition of the floor function \lfloor x_1+\varphi\rfloor=x_1. So \left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor=\left\lfloor\frac{\lfloo  r x_1+\varphi\rfloor}{n}\right\rfloor=\left\lfloor\f  rac{x_1}{n}\right\rfloor

    Now for the second part notice that \left\lfloor\frac{x}{n}\right\rfloor=\left\lfloor\  frac{x_1}{n}+\frac{\varphi}{n}\right\rfloor=\left\  lfloor\frac{x_1}{n}\right\rfloor+\left\lfloor\frac  {\varphi}{n}\right\rfloor. Now conclude the answer by noticing that since n\in\mathbb{Z}^+\implies\frac{\varphi}{n}\in[0,1)
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  6. #6
    Sea
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    I write false....... But you write true......



    I'm sorry for this and I thank you so much... for answering my
    question after all...



    Thanks a million......
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