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
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $3 \ \text{(d)}$