# 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

$\blacksquare$