Real Number is Ceiling minus Difference
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Theorem
Let $x \in \R$ be a real number.
Let $\ceiling x$ be the ceiling of $x$.
Let $n$ be a integer.
The following statements are equivalent:
- $(1): \quad$ There exists $t \in \hointr 0 1$ such that $x = n - t$
- $(2): \quad n = \ceiling x$
Proof
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 $x$.
$\Box$
2 implies 1
Now let $n = \ceiling x$.
Let $t = \ceiling x - x$.
Then $x = n - t$.
From Ceiling minus Real Number, $t = \ceiling x - x \in \hointr 0 1$
$\blacksquare$