prove $\lceil{x}\rceil=-\lfloor-x\rfloor$ i am trying to prove that $\lceil{x}\rceil=-\lfloor-x\rfloor$, but having difficulties to prove. the definitions are: 
$\lceil{x}\rceil:=m-1<x\leq m$ and $\lfloor{x}\rfloor:=n\leq x<n+1$.  
how can i prove the statement above, if needed, by using the definitions? i dont know how to change the inequalities when i negate the $x$ back and forth. i am thankful for any hand for help
 A: If $m < x \leq m+1$, where $m \in \mathbb{Z}$, then $\lceil x \rceil = m+1$. Now $-m-1 \leq -x <-m$.
Hence, $\lfloor -x \rfloor = -m-1$.
Hence, $-\lfloor -x \rfloor = m+1 = \lceil x \rceil$
A: The definitions are a bit confusing to me; perhaps you mean to say that $\lceil x \rceil = m$ when $m - 1 < x \le m$, and $\lfloor x \rfloor = n$ when $n \le x < n + 1$.
The key is that $a < (\le) b$ if and only if $-a > (\ge) -b$. That is, when you multiply each part by $-1$ you have to change the direction of the inequalities as well.
Let $n = \lfloor -x \rfloor$. Then $n \le -x < n + 1$, and using the sign-switching property above, $-n - 1 < x \le -n$. Can you see now how $- \lfloor -x \rfloor = -n = \lceil x \rceil$?
A: Both functions $\lceil x \rceil, -\lfloor -x \rfloor$ satisfy $f(x+1)=f(x)+1$, so it is enough to check what happens when $0\le x <1$ (because any such $f$ satisfies $f(x)=f(\{x\})+\lfloor x \rfloor$, i.e. their values are determined by their behavior on $[0,1)$). 
On $[0,1)$ both functions equal 1, hence they coincide everywhere.
A: This is not a substitute for the perfectly good answers already given, but if you’re visually oriented, a picture can help make this very clear. In the figure below $x$ is a positive real number, and $n=\lfloor x\rfloor$. Multiplying by $-1$ pivots the whole picture $180^\circ$ degrees about $x=0$, something that even I, with virtually no visual imagination, can easily visualize.
Start with the relationship between $n$ and $x$: $n$ is the nearest integer to the left of $x$, so after the rotation $-n$ is the nearest integer to the right of $-x$. In other words, $-n=\lceil-x\rceil$. Rotate the relationship between $-x$ and $\lceil-x\rceil=-n$ about $0$ again, and of course you get the original relationship between $x$ and $\lfloor x\rfloor=n$ back, but you’re also multiplying by $-1$ so you also get the relationship $-\lceil-x\rceil=-(-n)=n$. In other words, $-\lceil-x\rceil=\lfloor x\rfloor$.

Of course the same reasoning works, both visually and mathematically, when $x<0$.
