Ceiling Function is Integer

Theorem
Let $f: \R \to \R$ be the mapping defined as:
 * $\forall x \in \R: f \left({x}\right) = \left\lceil{x}\right\rceil$

where $\left\lceil{x}\right\rceil$ denotes the ceiling of $x$.

Then $f \left({x}\right)$ is an integer.

Proof
By definition of ceiling function:


 * $\forall x \in \R: \left\lceil{x}\right\rceil = \inf \left({\left\{ {m \in \Z: m \ge x}\right\} }\right)$

That is, if $f \left({x}\right)$ is the ceiling of $x$, then:
 * $f \left({x}\right) \in \Z$

where $\Z$ denotes the set of integers.

Also see

 * Floor Function is Integer