1. ## [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: $x\notin\mathbb{Z}$
$\left \lceil x+n \right \rceil=\left \lceil x \right \rceil+n\rightarrow \left \lceil k+x'+n \right \rceil=\left \lceil k+x' \right \rceil+n$
$k+1+n=k+1+n$

2. 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 $x+n\not\in\mathbb{Z} \implies \left \lceil x+n \right \rceil = \left \lfloor x+n \right \rfloor + 1 = x+n+\{x+n\}+1 = x+n+\{x\}+1 =$ $x+\{x\}+1+n = \left \lceil x \right \rceil +n$

3. Originally Posted by chiph588@
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?

4. 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?

5. Originally Posted by chiph588@
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.

6. Originally Posted by dwsmith
$\forall x\in\mathbb{R}$ 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?

7. Originally Posted by chiph588@
I got that far, but what is your reasoning after that?
The equality then holds since the LHS=RHS.

8. 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.

9. 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.