Ceiling of Negative equals Negative of Floor

Theorem
Let $x \in \R$ be a real number.

Let $\left \lfloor {x}\right \rfloor$ be the floor of $x$, and $\left \lceil {x}\right \rceil$ be the ceiling of $x$.

Then:
 * $\left \lceil {-x}\right \rceil = - \left \lfloor {x}\right \rfloor$

Proof
From Range of Values of Ceiling Function we have:
 * $x \le \left \lceil{x}\right \rceil < x + 1$

and so:
 * $-x \ge -\left \lceil{x}\right \rceil < -x - 1$

From Range of Values of Floor Function we have:
 * $\left \lfloor{x}\right \rfloor = n \iff x - 1 < n \le x$

And so $-x - 1 < -\left \lceil{x}\right \rceil \le x \implies -\left \lceil{x}\right \rceil = \left \lfloor{-x}\right \rfloor$.

Also see

 * Floor of Negative equals Negative of Ceiling