Number greater than Integer iff Ceiling greater than Integer

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Theorem

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

Let $\ceiling x$ be the ceiling of $x$.

Let $n \in \Z$ be an integer.


Then:

$\ceiling x > n \iff x > n$


Proof

Necessary Condition

Let $x > n$.

By Number is between Ceiling and One Less:

$\ceiling x \ge x$

Hence:

$\ceiling x > n$

$\Box$


Sufficient Condition

Let $\ceiling x > n$.


We have that:

$\forall m, n \in \Z: m < n \iff m \le n - 1$

and so:

$(1): \quad \ceiling x - 1 \ge n$


Then:

\(\ds x\) \(>\) \(\ds \ceiling x - 1\) Number is between Ceiling and One Less
\(\ds \) \(\ge\) \(\ds n\) from $(1)$

$\Box$


Hence the result:

$\ceiling x > n \iff x > n$

$\blacksquare$


Sources