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 $$x$$.

Then the following results apply:


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

Proof
From the definition of the ceiling function:


 * $$\left \lceil {x} \right \rceil = \inf \left({\left\{{m \in \Z: m \ge x}\right\}}\right)$$

That is, $$\left \lceil {x} \right \rceil$$ is the smallest integer greater than or equal to $$x$$.

Thus we have that:


 * $$\left \lceil {x} \right \rceil$$ is an integer;
 * $$\left \lceil {x} \right \rceil \ge x$$;
 * $$\left \lceil {x} \right \rceil - 1 < x$$ and so $$\left \lceil {x} \right \rceil < x + 1$$.

Also, we have 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 $$\left \lceil{x}\right \rceil = n \iff n - 1 < x \le n$$.