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 $$\sup$$.