Integer Equals Floor And Ceiling

Theorem

Let $x \in \R$.

Then:

• $x = \left \lfloor {x} \right \rfloor \iff x \in \Z$
• $x = \left \lceil {x} \right \rceil \iff x \in \Z$

where $\left \lfloor {x} \right \rfloor$ is the floor of $x$ and $\left \lceil {x} \right \rceil$ is the ceiling of $x$.

Proof

• Let $x = \left \lfloor {x} \right \rfloor$.

As $\left \lfloor {x} \right \rfloor \in \Z$, then so must $x$ be.

• Now let $x \in \Z$.

We have:

$\left \lfloor {x} \right \rfloor = \sup \left({\left\{{m \in \Z: m \le x}\right\}}\right)$

As $x \in \sup \left({\left\{{m \in \Z: m \le x}\right\}}\right)$, and there can be no greater $n \in \Z$ such that $n \in \sup \left({\left\{{m \in \Z: m \le x}\right\}}\right)$, $x = \left \lfloor {x} \right \rfloor$.

• The result for $\left \lceil {x} \right \rceil$ follows similar lines.

$\blacksquare$

Sources

• Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (1968): $\S 1.2.4$