Real Number is Ceiling minus Difference

Theorem
Let $x \in \R$ be any real number.

Then:
 * $x = n - t: n \in \Z, t \in \left[{0 \,.\,.\, 1}\right) \iff n = \left \lceil {x}\right \rceil$

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

Proof
Let $x = n - t$, where $t \in \left[{0 \,.\,.\, 1}\right)$.

Now $1 - t > 0$, so $n - 1 < x$.

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

Now let $n = \left \lceil {x}\right \rceil$.

From Ceiling minus Real Number:
 * $\left \lceil {x}\right \rceil - x \in \left[{0 \,.\,.\, 1}\right)$

Here we have $\left \lceil {x}\right \rceil = n$.

Thus:
 * $\left \lceil {x}\right \rceil - x \in \left[{0 \,.\,.\, 1}\right) \implies n - x = t$

where $t \in \left[{0 \,.\,.\, 1}\right)$.

So:
 * $x = n - t$

where $t \in \left[{0 \,.\,.\, 1}\right)$.