Range of Values of Floor Function

Theorem
Let $x \in \R$ be a real number and let $\left \lfloor{x}\right \rfloor$ be the floor of $x$.

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

Then the following results apply:
 * $(1): \quad \left \lfloor{x}\right \rfloor < n \iff x < n$
 * $(2): \quad \left \lfloor{x}\right \rfloor \ge n \iff x \ge n$
 * $(3): \quad \left \lfloor{x}\right \rfloor = n \iff x - 1 < n \le x$
 * $(4): \quad \left \lfloor{x}\right \rfloor = n \iff n \le x < n + 1$

Proof
We are going to use throughout the fact that $\forall m, n \in \Z: m < n \iff m \le n - 1$.

Proof of Result 1
Let $\left \lfloor{x}\right \rfloor < n$.

Then:

Next suppose $x < n$.

Then as $\left \lfloor {x} \right \rfloor \le x$ it follows that $\left \lfloor {x} \right \rfloor < n$.

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

Proof of Result 2
Let $\left \lfloor{x}\right \rfloor \ge n$.

Then as $x \ge \left \lfloor{x}\right \rfloor$ it follows that $x \ge n$.

Now let $x \ge n$.

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

Then $\left \lfloor{x}\right \rfloor + 1 \le n$ and so $\left \lfloor{x}\right \rfloor + 1 \le x$, which is a contradiction of $\left \lfloor{x}\right \rfloor + 1 > x$.

Thus by Proof by Contradiction, $\left \lfloor{x}\right \rfloor \ge n$.

So $\left \lfloor{x}\right \rfloor \ge n \iff x \ge n$.

Proof of Result 3
Suppose $\left \lfloor{x}\right \rfloor = n$.

Then $\left \lfloor{x}\right \rfloor \ge n$ and so by result 2, $n \le x$.

Also, we have that $x - 1 < \left \lfloor{x}\right \rfloor = n$ and so $x - 1 < n$.

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

Now suppose $x - 1 < n \le x$.

From $n \le x$, we have by result 2 that $n \le \left \lfloor{x}\right \rfloor$.

From $x - 1 < n$ we have that $x < n + 1$.

Hence by result 1 we have $\left \lfloor{x}\right \rfloor < n + 1$ 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 $\left \lfloor{x}\right \rfloor = n$.

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

So $\left \lfloor{x}\right \rfloor = n \iff x - 1 < n \le x$.

Proof of Result 4
Suppose $\left \lfloor{x}\right \rfloor = n$.

We have already shown that $n \le x$ (from result 2).

We also have that $\left \lfloor{x}\right \rfloor + 1 = n + 1$.

But from above, we have $x < \left \lfloor {x} \right \rfloor + 1$, and so $x < n + 1$.

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

Now suppose $n \le x < n + 1$.

We have already shown that $n \le x \implies n \le \left \lfloor{x}\right \rfloor$ by result 2.

In result 3 we saw that $x < n + 1 \implies \left \lfloor{x}\right \rfloor \le n$.

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

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

Also see

 * Range of Values of Ceiling Function