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 Integer equals Floor iff between Number and One Less we have:
 * $x - 1 < \left \lfloor{x}\right \rfloor \le x$

and so, by multiplying both sides by -1:
 * $-x + 1 > -\left \lfloor{x}\right \rfloor \ge -x$

From Integer equals Ceiling iff between Number and One More we have:
 * $\left \lceil{x}\right \rceil = n \iff x \le n < x + 1$

Hence:
 * $-x \le -\left \lfloor{x}\right \rfloor < -x + 1 \implies \left \lceil{-x}\right \rceil = -\left \lfloor{x}\right \rfloor$

Also see

 * Floor of Negative equals Negative of Ceiling