Floor equals Ceiling iff Integer

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 = \begin{cases}

\left \lceil {x}\right \rceil & : x \in \Z \\ \left \lceil {x}\right \rceil - 1 & : x \notin \Z \\ \end{cases}$

or equivalently:
 * $\left \lceil {x}\right \rceil = \begin{cases}

\left \lfloor {x}\right \rfloor & : x \in \Z \\ \left \lfloor {x}\right \rfloor + 1 & : x \notin \Z \\ \end{cases}$

where $\Z$ is the set of integers.

Proof
From Integer Equals Floor And Ceiling, we have that:
 * $x \in \Z \implies x = \left \lfloor {x}\right \rfloor$;
 * $x \in \Z \implies x = \left \lceil {x}\right \rceil$.

So $x \in \Z \implies \left \lfloor {x}\right \rfloor = \left \lceil {x}\right \rceil$.

Now let $x \notin \Z$.

From the definition of the floor function:


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

From the definition of the ceiling function:


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

Thus $\left \lfloor {x} \right \rfloor < x < \left \lceil {x} \right \rceil$.

Hence the result, from the definition of $\inf$ and $\sup$.