# [SOLVED] Ceiling Functions

• May 18th 2010, 10:32 AM
dwsmith
[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: $\displaystyle x\notin\mathbb{Z}$
$\displaystyle \left \lceil x+n \right \rceil=\left \lceil x \right \rceil+n\rightarrow \left \lceil k+x'+n \right \rceil=\left \lceil k+x' \right \rceil+n$
$\displaystyle k+1+n=k+1+n$
• May 18th 2010, 11:10 AM
chiph588@
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 $\displaystyle x+n\not\in\mathbb{Z} \implies \left \lceil x+n \right \rceil = \left \lfloor x+n \right \rfloor + 1 = x+n+\{x+n\}+1 = x+n+\{x\}+1 =$ $\displaystyle x+\{x\}+1+n = \left \lceil x \right \rceil +n$
• May 18th 2010, 11:15 AM
dwsmith
Quote:

Originally Posted by chiph588@
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?
• May 18th 2010, 11:19 AM
chiph588@
Quote:

Originally Posted by dwsmith
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)
• May 18th 2010, 11:23 AM
dwsmith
Quote:

Originally Posted by chiph588@
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.
• May 18th 2010, 11:41 AM
chiph588@
Quote:

Originally Posted by dwsmith
$\displaystyle \forall x\in\mathbb{R}$ 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?
• May 18th 2010, 11:42 AM
dwsmith
Quote:

Originally Posted by chiph588@
I got that far, but what is your reasoning after that?

The equality then holds since the LHS=RHS.
• May 18th 2010, 11:46 AM
chiph588@
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.
• May 18th 2010, 11:56 AM
dwsmith
Quote:

Originally Posted by chiph588@
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.