# Ceiling is between Number and One More

## Theorem

$x \le \ceiling x < x + 1$

where $\ceiling x$ is the ceiling of $x$.

## Proof

$\ceiling x - 1 < x \le \ceiling x$

Thus by adding $1$:

$x + 1 > \paren {\ceiling x - 1} + 1 = \ceiling x$

So:

$x \le \ceiling x$

and:

$\ceiling x < x + 1$

as required.

$\blacksquare$