Ceiling of x+m over n/Proof 1
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Theorem
Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
- $\ceiling {\dfrac {x + m} n} = \ceiling {\dfrac {\ceiling x + m} n}$
where $\ceiling x$ denotes the ceiling of $x$.
Proof
| \(\ds \ceiling {\dfrac {x + m} n}\) | \(=\) | \(\ds -\floor {-\dfrac {x + m} n}\) | Floor of Negative equals Negative of Ceiling | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\floor {\dfrac {-x - m} n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\floor {\dfrac {\floor {-x} - m} n}\) | Floor of $\dfrac {x + m} n$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\floor {\dfrac {-\ceiling x - m} n}\) | Floor of Negative equals Negative of Ceiling | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\floor {-\dfrac {\ceiling x + m} n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \ceiling {\dfrac {\ceiling x + m} n}\) | Floor of Negative equals Negative of Ceiling |
$\blacksquare$
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 $35$ (Answers to Exercises)