Real Number is Ceiling minus Difference

Theorem
Let $$x \in \mathbb{R}$$.

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

Proof

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

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

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


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

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