Integer equals Floor iff between Number and One Less

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

Let $\left \lfloor{x}\right \rfloor$ denote the floor of $x$.

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

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

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

From $n \le x$, we have by Number not less than Integer iff Floor not less than Integer:
 * $n \le \left \lfloor{x}\right \rfloor$

From $x - 1 < n$:
 * $x < n + 1$

Hence by Number less than Integer iff Floor less than Integer:
 * $\left \lfloor{x}\right \rfloor < n + 1$

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

and so:
 * $\left \lfloor{x}\right \rfloor \le n$

Thus as:
 * $n \le \left \lfloor{x}\right \rfloor$

and:
 * $\left \lfloor{x}\right \rfloor \le n$

it follows that:
 * $\left \lfloor{x}\right \rfloor = n$

Sufficient Condition
Let $\left \lfloor{x}\right \rfloor = n$.

Then:
 * $\left \lfloor{x}\right \rfloor \ge n$

By Number not less than Integer iff Floor not less than Integer:
 * $n \le x$

By definition of the floor of $x$:
 * $x < \left \lfloor{x}\right \rfloor + 1$

and so subtracting $1$ from both sides:
 * $x - 1 < \left \lfloor{x}\right \rfloor$

and so by hypothesis:
 * $x - 1 < n$

So:
 * $\left \lfloor{x}\right \rfloor = n \implies x - 1 < n \le x$

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