Number not less than Integer iff Floor not less than Integer

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

Necessary Condition
Let $\left \lfloor{x}\right \rfloor \ge n$.

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

Hence:
 * $x \ge n$

Sufficient Condition
Let $x \ge n$.

$\left \lfloor{x}\right \rfloor < n$.

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

Hence:
 * $\left \lfloor{x}\right \rfloor + 1 \le n$

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

This contradicts the definition of the floor of $x$:
 * $\left \lfloor{x}\right \rfloor + 1 > x$

Thus by Proof by Contradiction:
 * $\left \lfloor{x}\right \rfloor \ge n$

Hence the result:
 * $\left \lfloor{x}\right \rfloor \ge n \iff x \ge n$