Real Number is Ceiling minus Difference

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

Let $\ceiling x$ be the ceiling of $x$. Let $n$ be a integer.


 * $(1): \quad$ There exists $t \in \hointr 0 1$ such that $x = n - t$
 * $(2): \quad n = \ceiling x$

1 implies 2
Let $x = n - t$, where $t \in \hointr 0 1$.

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

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

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

2 implies 1
Now let $n = \ceiling x$.

Let $t = \ceiling x - x$.

Then $x = n - t$.

From Ceiling minus Real Number, $t = \ceiling x - x \in \hointr 0 1$

Also see

 * Real Number is Floor plus Difference
 * Definition:Fractional Part