# Proving that $\lfloor \lceil x \rceil \rfloor = \lceil x \rceil$ for all real numbers $x$.

Let $\mathsf{LHS} = \lfloor \lceil x \rceil \rfloor$ and $\mathsf{RHS} = \lceil x \rceil$.

Let us call $n = \lceil x \rceil$.

Case 1: $n$ is even, i.e., there exists a $k \in \mathbb{Z}$ such that $n = 2 k$.

I’m not sure how to go on or if I’m setting this up right.

• What is ⎤? Floor? Ceil? – Joao Nov 14 '14 at 6:59
• You have to show that, if $n=\lceil x\rceil$, then $\lfloor n\rfloor=n$? If $n$ is the floor of some number then $n$ is its own ceiling? If $n$ is in the range of the floor function, then $n$ is a fixed point of the ceiling function? It would help a lot if I knew (a) what is the range of the floor function, and (b) what are the fixed points of the ceiling function. Beats me. You have any ideas? – bof Mar 1 '15 at 8:30
• This question seems a bit simple and should probably be considered off topic or simple. Ceiling returns an integer. Floor only changes a number when it is not an integer. This is really a question of data types. Ceiling changes a number from R -> N. And then floor only does something when changing R -> N. It's a data type converter function. If it's the right data type then casting is literally redundant. The question is nice but I think it's one that CRUDE would say shouldn't be here, sadly. – The Great Duck Mar 4 '16 at 5:08

$\lceil x\rceil$ is an integer, and if $n$ is an integer then $\lfloor n\rfloor=n$.
First notice that $\left \lfloor{n}\right \rfloor = n$ for all integers $n$. And $\lceil x \rceil$ is an integer