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 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, $t = \ceiling x - x \in \hointr 0 1$

$\blacksquare$


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