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