Integer equals Ceiling iff Number between Integer and One Less

Theorem
Let $x \in \R$ be a real number.

Let $\left \lceil{x}\right \rceil$ be the ceiling of $x$.

Let $n \in \Z$ be an integer.

Then:
 * $\left \lceil{x}\right \rceil = n \iff n - 1 < x \le n$

Necessary Condition
Let $n - 1 < x \le n$.

From Number not greater than Integer iff Ceiling not greater than Integer:
 * $x \le n \implies \left \lceil{x}\right \rceil \le n$

From Number is between Ceiling and One Less:
 * $x \le \left \lceil{x}\right \rceil$

and so:
 * $n - 1 < \left \lceil{x}\right \rceil$

We have that:
 * $\forall m, n \in \Z: m - 1 < n \iff m \le n$

and so:
 * $n \le \left \lceil{x}\right \rceil$

So we have:
 * $\left \lceil{x}\right \rceil \le n$

and:
 * $n \le \left \lceil{x}\right \rceil$

Thus:
 * $n - 1 < x \le n \implies \left \lceil{x}\right \rceil = n$

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

From Number is between Ceiling and One Less:
 * $x \le \left \lceil{x}\right \rceil$

and so:
 * $x \le n$

Also from Number is between Ceiling and One Less:
 * $\left \lceil{x}\right \rceil - 1 < x$

and so:
 * $n - 1 < x$

Thus:
 * $\left \lceil{x}\right \rceil = n \implies n - 1 < x \le n$.

Hence the result:
 * $\left \lceil{x}\right \rceil = n \iff n - 1 < x \le n$