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Thread: [SOLVED] Ceiling Functions

  1. #1
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    [SOLVED] Ceiling Functions

    $\displaystyle \left \lceil x+n \right \rceil=\left \lceil x \right \rceil+n$

    $\displaystyle x\in\mathbb{R}, n\in\mathbb{Z}$

    $\displaystyle x=k+x', 0\leq x'<1$

    Case 1: $\displaystyle x\in\mathbb{Z}$
    $\displaystyle \left \lceil x+n \right \rceil=\left \lceil x \right \rceil+n\rightarrow \left \lceil k+x'+n \right \rceil=\left \lceil k+x' \right \rceil+n$
    $\displaystyle k+n=k+n$

    Case 2: $\displaystyle x\notin\mathbb{Z}$
    $\displaystyle \left \lceil x+n \right \rceil=\left \lceil x \right \rceil+n\rightarrow \left \lceil k+x'+n \right \rceil=\left \lceil k+x' \right \rceil+n$
    $\displaystyle k+1+n=k+1+n$
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    If $\displaystyle x\in\mathbb{Z} $ then $\displaystyle x+n\in\mathbb{Z} \implies \left \lceil x+n \right \rceil = x+n = \left \lceil x \right \rceil + n $


    If $\displaystyle x\not\in\mathbb{Z} $ then $\displaystyle x+n\not\in\mathbb{Z} \implies \left \lceil x+n \right \rceil = \left \lfloor x+n \right \rfloor + 1 = x+n+\{x+n\}+1 = x+n+\{x\}+1 = $ $\displaystyle x+\{x\}+1+n = \left \lceil x \right \rceil +n $
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  3. #3
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    Quote Originally Posted by chiph588@ View Post
    If $\displaystyle x\in\mathbb{Z} $ then $\displaystyle x+n\in\mathbb{Z} \implies \left \lceil x+n \right \rceil = x+n = \left \lceil x \right \rceil + n $


    If $\displaystyle x\not\in\mathbb{Z} $ then $\displaystyle x+n\not\in\mathbb{Z} \implies \left \lceil x+n \right \rceil = \left \lfloor x+n \right \rfloor + 1 = x+n+\{x+n\}+1 = x+n+\{x\}+1 = $ $\displaystyle x+\{x\}+1+n = \left \lceil x \right \rceil +n $
    I was using some definition in my book so that is where I came up with x=k+x', so the question then becomes is what I have original correct? If not, why?
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by dwsmith View Post
    I was using some definition in my book so that is where I came up with x=k+x', so the question then becomes is what I have original correct? If not, why?
    To be honest, I was having trouble following your work clearly. Care to elaborate a bit?
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    Quote Originally Posted by chiph588@ View Post
    To be honest, I was having trouble following your work clearly. Care to elaborate a bit?
    $\displaystyle \forall x\in\mathbb{R}$ can be written as $\displaystyle x=k+x'$, where $\displaystyle k=\left \lfloor x \right \rfloor$ and $\displaystyle 0\leq x'<1$

    That is from the book.

    So then I did the substitution incorporating that definition.
    Last edited by dwsmith; May 18th 2010 at 11:30 AM. Reason: Too many periods.
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by dwsmith View Post
    $\displaystyle \forall x\in\mathbb{R}$ can be written as $\displaystyle x=k+x'$, where $\displaystyle k=\left \lfloor x \right \rfloor$ and $\displaystyle 0\leq x'<1$

    That is from the book.

    So then I did the substitution incorporating that definition.
    I got that far, but what is your reasoning after that?
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  7. #7
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    Quote Originally Posted by chiph588@ View Post
    I got that far, but what is your reasoning after that?
    The equality then holds since the LHS=RHS.
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    Ok I see it now, but for this to be a formal proof, you'll need more of an explanation of how you got to the last line of each case.
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  9. #9
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    Quote Originally Posted by chiph588@ View Post
    Ok I see it now, but for this to be a formal proof, you'll need more of an explanation of how you got to the last line of each case.
    Boo, I hate writing words.
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