Integer equals Floor iff Number between Integer and One More

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

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

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

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

and so by hypothesis:
 * $\left \lfloor {x} \right \rfloor < n + 1$

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

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

So we have:
 * $n \le \left \lfloor{x}\right \rfloor$

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

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

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

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

and so by hypothesis:
 * $n \le x$

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

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

Thus:
 * $\left \lfloor{x}\right \rfloor = n \implies n \le x < n + 1$.

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