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Math Help - [SOLVED] Ceiling Functions

  1. #1
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    [SOLVED] Ceiling Functions

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

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

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

    Case 1: x\in\mathbb{Z}
    \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
    k+n=k+n

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


    If  x\not\in\mathbb{Z} then  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 =  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  x\in\mathbb{Z} then  x+n\in\mathbb{Z} \implies \left \lceil x+n \right \rceil = x+n = \left \lceil x \right \rceil + n


    If  x\not\in\mathbb{Z} then  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 =  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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  5. #5
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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?
    \forall x\in\mathbb{R} can be written as x=k+x', where k=\left \lfloor x \right \rfloor and 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 12:30 PM. 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
    \forall x\in\mathbb{R} can be written as x=k+x', where k=\left \lfloor x \right \rfloor and 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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