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