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

and so:
 * $$-x + 1 > -\left \lfloor{x}\right \rfloor \ge x$$

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

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

Also see

 * Ceiling of Negative equals Negative of Floor