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Thread: Floor function2

  1. #1
    Sea
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    Floor function2

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


    Show that:


    $\displaystyle
    \lfloor x\times y \rfloor \geq \lfloor x \rfloor \times \lfloor y \rfloor$
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  2. #2
    o_O
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    $\displaystyle x = -3, \ y = -4.5$

    $\displaystyle \lfloor xy \rfloor = \lfloor (-3)(-4.5) \rfloor = \lfloor 13.5 \rfloor = 13$
    $\displaystyle \lfloor x\rfloor \lfloor y\rfloor = \lfloor -3 \rfloor \lfloor -4.5 \rfloor = -3 \times -5 = 15$
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Sea View Post
    $\displaystyle \forall x,y \in \mathbb{R}$


    Show that:


    $\displaystyle
    \lfloor x\times y \rfloor \geq \lfloor x \rfloor \times \lfloor y \rfloor$
    As o_O pointed out this is not always true. So let us resrict our numbers. The case when either is zero is trivial, so fix $\displaystyle x,y\in\mathbb{R}^+$

    So as in past problems let $\displaystyle x=x_1+\varphi\quad x_1\in\mathbb{N}\wedge\varphi\in[0,1)$ and $\displaystyle y=y_1+\varphi'\quad y_1\in\mathbb{N}\wedge\varphi'\in[0,1)$

    So it is clear from the definition of the floor function that

    $\displaystyle \begin{aligned}\lfloor x\rfloor\cdot\lfloor y\rfloor&=\lfloor x_1+\varphi\rfloor\cdot\lfloor y_1+\varphi'\rfloor\\
    &=x_1\cdot y_1\end{aligned}$

    So we must just examine $\displaystyle \lfloor x\cdot y\rfloor$. Using the same tecnhique we get

    $\displaystyle \begin{aligned}\lfloor x\cdot y\rfloor&=\lfloor (x_1+\varphi)\cdot(y_1+\varphi')\rfloor\\
    &=\lfloor x_1\cdot y_1+x_1\cdot\varphi'+y_1\cdot\varphi+\varphi\cdot\ varphi'\rfloor\end{aligned}$

    Now since it is possible that $\displaystyle x_1\cdot \varphi',y_1\cdot\varphi>1$ it follows that

    $\displaystyle \lfloor x_1\cdot y_1+x_1\cdot\varphi'+y_1\cdot\varphi+\varphi\cdot\ varphi'\rfloor\geqslant\lfloor x_1\cdot y_1\rfloor$. And since the naturals are closed under multiplication it follows that $\displaystyle \lfloor x_1\cdot y_1\rfloor=x_1\cdot y_1=\lfloor x\rfloor\cdot\lfloor y\rfloor\leqslant \lfloor x_1\cdot y_1+x_1\cdot\varphi'+y_1\cdot\varphi+\varphi\cdot\ varphi'\rfloor=\lfloor x\cdot y\rfloor$. Get ridding of the useless terms yields the desired result.
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  4. #4
    Super Member PaulRS's Avatar
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    $\displaystyle
    x,y \in \mathbb{R}^ + \Rightarrow \left\lfloor {x \cdot y} \right\rfloor \geqslant \left\lfloor x \right\rfloor \cdot \left\lfloor y \right\rfloor
    $
    ________________________________________

    Consider that $\displaystyle
    a \geqslant b \Rightarrow \left\lfloor a \right\rfloor \geqslant \left\lfloor b \right\rfloor
    $ ( the floor function is non-decreasing)

    Since we are working with non-negative numbers: $\displaystyle
    \left. \begin{gathered}
    x \geqslant \left\lfloor x \right\rfloor \hfill \\
    y \geqslant \left\lfloor y \right\rfloor \hfill \\
    \end{gathered} \right] \Rightarrow x \cdot y \geqslant \left\lfloor x \right\rfloor \cdot \left\lfloor y \right\rfloor
    $ $\displaystyle
    \Rightarrow \left\lfloor {x \cdot y} \right\rfloor \geqslant \left\lfloor {\left\lfloor x \right\rfloor \cdot \left\lfloor y \right\rfloor } \right\rfloor = \left\lfloor x \right\rfloor \cdot \left\lfloor y \right\rfloor \square
    $
    Last edited by PaulRS; Dec 22nd 2008 at 05:40 PM.
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