Floor of Negative equals Negative of Ceiling

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

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

and so, by multiplying by -1:
 * $-x \ge -\left \lceil{x}\right \rceil > -x - 1$

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

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

Also see

 * Ceiling of Negative equals Negative of Floor