Real Number is Floor plus Difference

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

Let $\left \lfloor {x}\right \rfloor$ be the floor of $x$. Let $n$ be a integer.


 * 1) There exists $t \in \left[{0 \,.\,.\, 1}\right)$ such that $x = n + t$
 * 2) $n = \left \lfloor {x}\right \rfloor$

1 implies 2
Let $x = n + t$, where $t \in \left[{0 \,.\,.\, 1}\right)$.

Because $0\leq t < 1$, we have:
 * $0 \leq x-n < 1$

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

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

2 implies 1
Now let $n = \left \lfloor {x}\right \rfloor$.

Let $t = x - \left \lfloor {x}\right \rfloor$.

Then $x = n + t$.

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

Also see

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