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Math Help - Floor function2

  1. #1
    Sea
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    Floor function2

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


    Show that:


    <br />
\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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    x = -3, \ y = -4.5

    \lfloor xy \rfloor =  \lfloor (-3)(-4.5) \rfloor = \lfloor 13.5 \rfloor = 13
    \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
    \forall x,y \in \mathbb{R}


    Show that:


    <br />
\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 x,y\in\mathbb{R}^+

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

    \begin{aligned}\lfloor x\rfloor\cdot\lfloor y\rfloor&=\lfloor x_1+\varphi\rfloor\cdot\lfloor y_1+\varphi'\rfloor\\<br />
&=x_1\cdot y_1\end{aligned}

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

    \begin{aligned}\lfloor x\cdot y\rfloor&=\lfloor (x_1+\varphi)\cdot(y_1+\varphi')\rfloor\\<br />
&=\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 x_1\cdot \varphi',y_1\cdot\varphi>1 it follows that

    \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 \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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    <br />
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 <br />
    ________________________________________

    Consider that <br />
a \geqslant b  \Rightarrow \left\lfloor a \right\rfloor  \geqslant \left\lfloor b \right\rfloor<br />
( the floor function is non-decreasing)

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