# Thread: Ceiling and Floor

1. ## Ceiling and Floor

$\displaystyle \displaystyle\left\lceil x\right\rceil=\left\lfloor x\right\rfloor +1$

Let $\displaystyle \displaystyle x=x'+d \ \ni \ x'\in\mathbb{Z} \ \mbox{and} \ 0<d<1$

$\displaystyle \displaystyle\left\lceil x'+d\right\rceil =x'+1$

How can I translate this into the RHS now?

2. Originally Posted by dwsmith
$\displaystyle \displaystyle\left\lceil x\right\rceil=\left\lfloor x\right\rfloor +1$
Of course that is not true if $\displaystyle x\in \mathbb{Z}$.

It is only true if $\displaystyle x\notin \mathbb{Z}$.
In which case it is well known that $\displaystyle \displaystyle\left\lceil x\right\rceil-\left\lfloor x\right\rfloor =1$

3. Originally Posted by dwsmith
Let $\displaystyle \displaystyle x=x'+d \ \ni \ x'\in\mathbb{Z} \ \mbox{and} \ 0<d<1$
This line should let you know x isn't an integer.

4. Originally Posted by dwsmith
This line should let you know x isn't an integer.
No should not!
That line contains the word 'let'. Which is not part of the statement.
You should correctly state what you mean to begin with.

5. I am absolutely content with you having others help if you are going to be so touchy and not aid in the question posed. You aren't obligated to help me. I am sure others need you.

Just to be clear by the term "you having others help" I by no means am imply for you to ask others. What I was trying to say in a nice manner was others can help me and I will be ok with it.

6. Originally Posted by dwsmith
$\displaystyle \displaystyle\left\lceil x\right\rceil=\left\lfloor x\right\rfloor +1$

Let $\displaystyle \displaystyle x=x'+d \ \ni \ x'\in\mathbb{Z} \ \mbox{and} \ 0<d<1$

$\displaystyle \displaystyle\left\lceil x'+d\right\rceil =x'+1$

How can I translate this into the RHS now?
I don't know what you're trying to ask.

What are you trying to translate into what?

7. I now have $\displaystyle \displaystyle \left\lceil x\right\rceil = \left\lceil x'+d\right\rceil=x'+1$

How can I make that the RHS now?

8. As in your other post, what is $\displaystyle \lfloor x\rfloor+1$? It is $\displaystyle \lfloor x\rfloor+1=\lfloor x'+d\rfloor +1=x'+1$.

9. Originally Posted by roninpro
As in your other post, what is $\displaystyle \lfloor x\rfloor+1$? It is $\displaystyle \lfloor x\rfloor+1=\lfloor x'+d\rfloor +1=x'+1$.
Huh?

Never mind figure it out.