Number not greater than Integer iff Ceiling not greater than Integer

Theorem

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

Let $\ceiling x$ be the ceiling of $x$.

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

Then:

$\ceiling x \le n \iff x \le n$

Proof

Necessary Condition

Let $\ceiling x \le n$.

By Number is between Ceiling and One Less:

$x \le \ceiling x$

Hence:

$x \le n$

$\Box$

Sufficient Condition

Let $x \le n$.

Aiming for a contradiction, suppose $\ceiling x > n$.

We have that:

$\forall m, n \in \Z: m < n \iff m \le n - 1$

Hence:

$\ceiling x - 1 \ge n$

and so by hypothesis:

$\ceiling x - 1 \ge x$

This contradicts the result Number is between Ceiling and One Less:

$\ceiling x - 1 < x$

Thus by Proof by Contradiction:

$\ceiling x \le n$

$\Box$

Hence the result:

$\ceiling x \le n \iff x \le n$

$\blacksquare$

Sources

• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $3 \ \text{(c)}$