Real Number is Floor plus Difference

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

Let $\floor x$ be the floor of $x$.

Then:
 * There exists an integer $n \in \Z$ such that for some $t \in \hointr 0 1$:
 * $x = n + t$


 * $n = \floor x$
 * $n = \floor x$

where:
 * $\hointr 0 1$ is the real interval half open on the right from $0$ to $1$
 * $\floor x$ is the floor of $x$

Sufficient Condition
Let there exist $n \in \Z$ such that $x = n + t$, where $t \in \hointr 0 1$.

We have that $1 - t > 0$.

Thus:
 * $0 \le x - n < 1$

Thus:
 * $n \le x < n + 1$

That is, $n$ is the floor of $x$.

Necessary Condition
Let $n = \floor x$.

Let $t = x - \floor x$.

Then $x = n + t$.

From Real Number minus Floor:
 * $t = x - \floor x \in \hointr 0 1$

and so:
 * $x = n + t: t \in \hointr 0 1$

Also see

 * Real Number is Ceiling minus Difference
 * Definition:Fractional Part