Real Number is Floor plus Difference

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

Then:
 * $x = n + t: n \in \Z, t \in \left[{0 \,.\,.\, 1}\right) \iff n = \left \lfloor {x}\right \rfloor$

where $\left \lfloor {x}\right \rfloor$ is the floor of $x$.

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

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

Thus:
 * $n = \sup \left({\left\{{m \in \Z: m \le x}\right\}}\right) = \left \lfloor {x}\right \rfloor$

Now let $n = \left \lfloor {x}\right \rfloor$.

From Real Number Minus Floor, $x - \left \lfloor {x}\right \rfloor \in \left[{0 \,.\,.\, 1}\right)$.

Here we have $\left \lfloor {x}\right \rfloor = n$.

Thus $x - \left \lfloor {x}\right \rfloor \in \left[{0 .. 1}\right) \implies x - n = t$, where $t \in \left[{0 \,.\,.\, 1}\right)$.

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