Ceiling is between Number and One More

Theorem

 * $x \le \left \lceil {x} \right \rceil < x + 1$

where $\left \lceil {x} \right \rceil$ is the ceiling of $x$.

Proof
From Number is between Ceiling and One Less:
 * $\left \lceil {x} \right \rceil - 1 < x \le \left \lceil {x} \right \rceil$

Thus by adding $1$:
 * $x + 1 > \left({\left \lceil {x} \right \rceil - 1}\right) + 1 = \left \lceil {x} \right \rceil$

So:
 * $x \le \left \lceil {x} \right \rceil$

and:
 * $\left \lceil {x} \right \rceil < x + 1$

as required.