Range of Values of Ceiling Function

Theorem
Let $x \in \R$ be a real number and let $\left \lceil{x}\right \rceil$ be the ceiling of $x$.

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

Then the following results apply:


 * $(1): \qquad \left \lceil{x}\right \rceil > n \iff x > n$
 * $(2): \qquad \left \lceil{x}\right \rceil \le n \iff x \le n$
 * $(3): \qquad \left \lceil{x}\right \rceil = n \iff x \le n < x + 1$
 * $(4): \qquad \left \lceil{x}\right \rceil = n \iff n - 1 \le x \le n$

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 $\left \lceil{x}\right \rceil > n$.

Then:

Next suppose $x > n$.

Then as $\left \lceil {x} \right \rceil \ge x$ it follows that $\left \lceil {x} \right \rceil > n$.

So $\left \lceil{x}\right \rceil > n \iff x > n$.

Proof of Result 2
Let $\left \lceil{x}\right \rceil \le n$.

Then as $x \le \left \lceil{x}\right \rceil$ it follows that $x \le n$.

Now let $x \le n$.

Suppose $\left \lceil{x}\right \rceil > n$.

Then $\left \lceil{x}\right \rceil - 1 \ge n$ and so $\left \lceil{x}\right \rceil - 1 \ge x$, which is a contradiction of $\left \lceil{x}\right \rceil - 1 < x$.

Thus by proof by contradiction, $\left \lceil{x}\right \rceil \le n$.

So $\left \lceil{x}\right \rceil \le n \iff x \le n$.

Proof of Result 3
Suppose $\left \lceil{x}\right \rceil = n$.

Then $\left \lceil{x}\right \rceil \le n$ and so by result 2, $x \le n$.

Also, we have that $x + 1 > \left \lceil{x}\right \rceil = n$ and so $x + 1 > n$.

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

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

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

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

Hence by result 1 we have $\left \lceil{x}\right \rceil > n - 1$ and so $\left \lceil{x}\right \rceil \ge n$.

Thus as $n \le \left \lceil{x}\right \rceil$ and $\left \lceil{x}\right \rceil \le n$ it follows that $\left \lceil{x}\right \rceil = n$.

Thus $n \iff x \le n < x + 1 \implies \left \lceil{x}\right \rceil = n$.

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

Proof of Result 4
Suppose $\left \lceil{x}\right \rceil = n$.

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

We also have that $\left \lceil{x}\right \rceil - 1 = n - 1$.

But from above, we have $x > \left \lceil {x} \right \rceil - 1$, and so $x > n - 1$.

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

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

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

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

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

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