Results 1 to 6 of 6

Thread: Floor function

  1. #1
    Sea
    Sea is offline
    Junior Member Sea's Avatar
    Joined
    Dec 2008
    From
    Turkey
    Posts
    54

    Floor function

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

    Show that:

    $\displaystyle \lfloor x+y \rfloor \geq \lfloor x \rfloor
    + \lfloor y \rfloor$



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

    where $\displaystyle \lfloor x \rfloor$ is the floor function, the greatest
    integer less than $\displaystyle x$.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,410
    Thanks
    1
    Let: $\displaystyle x = \lfloor x \rfloor + a $ and $\displaystyle y = \lfloor y \rfloor + b$ for some $\displaystyle 0 \leq a,b < 1$.

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

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

    Take the floor function of both sides and with the fact, the conclusion follows.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1
    Quote Originally Posted by Sea View Post
    $\displaystyle \forall x,y \in \mathbb{R}$
    Show that: $\displaystyle \lfloor x+y \rfloor \geq \lfloor x \rfloor
    + \lfloor y \rfloor$.
    The rigor required by the instructor/text may well vary.
    But this is the general idea.
    $\displaystyle \left\lfloor x \right\rfloor \leqslant x\;\& \,\left\lfloor y \right\rfloor \leqslant y\, \Rightarrow \,\left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor \leqslant x + y$.
    Because, $\displaystyle \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor $ is an integer and $\displaystyle \left\lfloor {x + y} \right\rfloor $ is the greatest integer which does not exceed $\displaystyle x + y$ it follows that $\displaystyle \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor \leqslant \left\lfloor {x + y} \right\rfloor $.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Sea
    Sea is offline
    Junior Member Sea's Avatar
    Joined
    Dec 2008
    From
    Turkey
    Posts
    54
    Thanks a million.......

    I understand....

    And... Can you help me?

    __________________________________________________

    Show that:


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



    __________________________________________________ _


    Show that:


    $\displaystyle \lfloor
    \frac{\lfloor x \rfloor }{n} \rfloor = \lfloor
    \frac{x}{n} \rfloor +$ where $\displaystyle x \in \mathbb{R}, \ n \in \mathbb{Z^+}$
    Last edited by Sea; Dec 21st 2008 at 04:42 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Sea View Post
    Show that:


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

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

    Show that

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

    Now for the second part notice that $\displaystyle \left\lfloor\frac{x}{n}\right\rfloor=\left\lfloor\ frac{x_1}{n}+\frac{\varphi}{n}\right\rfloor=\left\ lfloor\frac{x_1}{n}\right\rfloor+\left\lfloor\frac {\varphi}{n}\right\rfloor$. Now conclude the answer by noticing that since $\displaystyle n\in\mathbb{Z}^+\implies\frac{\varphi}{n}\in[0,1)$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Sea
    Sea is offline
    Junior Member Sea's Avatar
    Joined
    Dec 2008
    From
    Turkey
    Posts
    54
    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......
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Floor Function
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: Nov 14th 2010, 11:44 AM
  2. an identity of floor function
    Posted in the Number Theory Forum
    Replies: 9
    Last Post: Aug 8th 2010, 08:36 AM
  3. Help with floor function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 3rd 2010, 11:51 AM
  4. Replies: 1
    Last Post: Dec 3rd 2009, 08:45 AM
  5. Floor function
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Jun 23rd 2009, 10:03 PM

Search Tags


/mathhelpforum @mathhelpforum