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:

$(1): \qquad \left \lfloor{x}\right \rfloor < n \iff x < n$

$(2): \qquad \left \lfloor{x}\right \rfloor \ge n \iff x \ge n$

$(3): \qquad \left \lfloor{x}\right \rfloor = n \iff x - 1 < n \le x$

$(4): \qquad \left \lfloor{x}\right \rfloor = n \iff n \le x < n + 1$

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left \lfloor{x}\right \rfloor\)	\(<\)	\(\displaystyle n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left \lfloor{x}\right \rfloor + 1\)	\(\le\)	\(\displaystyle n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(<\)	\(\displaystyle \left \lfloor{x}\right \rfloor + 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(<\)	\(\displaystyle n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Next suppose $x < n$.

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

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

$\blacksquare$

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

$\blacksquare$

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

$\blacksquare$

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

$\blacksquare$

Also see

• Range of Values of Ceiling Function

Sources

• Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (1968): $\S 1.2.4$: Exercise $3$