[SOLVED] Ceiling Functions

dwsmith

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Mar 2010
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\(\displaystyle \left \lceil x+n \right \rceil=\left \lceil x \right \rceil+n\)

\(\displaystyle x\in\mathbb{R}, n\in\mathbb{Z}\)

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

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: \(\displaystyle x\notin\mathbb{Z}\)
\(\displaystyle \left \lceil x+n \right \rceil=\left \lceil x \right \rceil+n\rightarrow \left \lceil k+x'+n \right \rceil=\left \lceil k+x' \right \rceil+n\)
\(\displaystyle k+1+n=k+1+n\)
 

chiph588@

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Sep 2008
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Champaign, Illinois
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 \(\displaystyle x+n\not\in\mathbb{Z} \implies \left \lceil x+n \right \rceil = \left \lfloor x+n \right \rfloor + 1 = x+n+\{x+n\}+1 = x+n+\{x\}+1 = \) \(\displaystyle x+\{x\}+1+n = \left \lceil x \right \rceil +n \)
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
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?
 

chiph588@

MHF Hall of Honor
Sep 2008
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Champaign, Illinois
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? (Itwasntme)
 

dwsmith

MHF Hall of Honor
Mar 2010
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Florida
To be honest, I was having trouble following your work clearly. Care to elaborate a bit? (Itwasntme)
\(\displaystyle \forall x\in\mathbb{R}\) 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.
 
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chiph588@

MHF Hall of Honor
Sep 2008
1,163
429
Champaign, Illinois
\(\displaystyle \forall x\in\mathbb{R}\) 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?
 

chiph588@

MHF Hall of Honor
Sep 2008
1,163
429
Champaign, Illinois
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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dwsmith

MHF Hall of Honor
Mar 2010
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582
Florida
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.