Real Number is Ceiling minus Difference

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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.


The following are equivalent:

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


Proof

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 $x$.

$\Box$

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, $t = \left \lceil {x}\right \rceil - x \in \left[{0 \,.\,.\, 1}\right)$

$\blacksquare$


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