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
Ffrom 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