Real Number is Floor plus Difference

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

Let $$\left \lfloor {x}\right \rfloor$$ be the floor function of $$x$$.

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

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

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