Number not less than Integer iff Floor not less than Integer

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

Let $\floor x$ denote the floor of $x$.

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

Then:
 * $x \ge n \iff \floor x \ge n$

Necessary Condition
Let $\floor x \ge n$.

By definition of the floor of $x$:
 * $x \ge \floor x$

Hence:
 * $x \ge n$

Sufficient Condition
Let $x \ge n$.

$\floor x < n$.

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

Hence:
 * $\floor x + 1 \le n$

and so by hypothesis:
 * $\floor x + 1 \le x$

This contradicts the definition of the floor of $x$:
 * $\floor x + 1 > x$

Thus by Proof by Contradiction:
 * $\floor x \ge n$

Hence the result:
 * $\floor x \ge n \iff x \ge n$