Real Number is Integer iff equals Ceiling

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Theorem

Let $x \in \R$.

Then:

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

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


Proof

Let $x = \left \lceil {x} \right \rceil$.

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


Now let $x \in \Z$.

We have:

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

As $x \in \inf \left({\left\{{m \in \Z: m \ge x}\right\}}\right)$, and there can be no lesset $n \in \Z$ such that $n \in \inf \left({\left\{{m \in \Z: m \ge x}\right\}}\right)$, it follows that:

$x = \left \lceil {x} \right \rceil$

$\blacksquare$