Real Number is Integer iff equals Ceiling
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Theorem
Let $x \in \R$.
Then:
- $x = \ceiling x \iff x \in \Z$
where $\ceiling x$ is the ceiling of $x$.
Proof
Let $x = \ceiling x$.
As $\ceiling x \in \Z$, then so must $x$ be.
Now let $x \in \Z$.
We have:
- $\ceiling x = \inf \set {m \in \Z: m \ge x}$
As $x \in \inf \set {m \in \Z: m \ge x}$, and there can be no lesser $n \in \Z$ such that $n \in \inf \set {m \in \Z: m \ge x}$, it follows that:
- $x = \ceiling x$
$\blacksquare$