Real Number is Ceiling minus Difference

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

Let $\left \lceil {x}\right \rceil$ be the ceiling of $x$. Let $n$ be a integer.


 * 1) There exists $t \in \left[{0 \,.\,.\, 1}\right)$ such that $x = n - t$
 * 2) $n = \left \lceil {x}\right \rceil$

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

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

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

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

2 implies 1
Now let $n = \left \lceil {x}\right \rceil$.

Let $t = \left \lceil {x}\right \rceil - x$.

Then $x = n - t$.

From Ceiling minus Real Number, $t = \left \lceil {x}\right \rceil - x \in \left[{0 \,.\,.\, 1}\right)$

Also see

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