Floor of Half of n+m plus Floor of Half of n-m+1

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Theorem

Let $n, m \in \Z$ be integers.

$\floor {\dfrac {n + m} 2} + \floor {\dfrac {n - m + 1} 2} = n$

where $\floor x$ denotes the floor of $x$.


Proof

Either $n + m$ or $n - m + 1$ is even.

Thus:

$\dfrac {n + m} 2 \bmod 1 + \dfrac {n - m + 1} 2 \bmod 1 = \dfrac 1 2 < 1$

and so:

\(\ds \floor {\dfrac {n + m} 2} + \floor {\dfrac {n - m + 1} 2}\) \(=\) \(\ds \floor {\dfrac {n + m} 2 + \dfrac {n - m + 1} 2}\) Sum of Floors not greater than Floor of Sum
\(\ds \) \(=\) \(\ds \floor {\dfrac {n + m + n - m + 1} 2}\)
\(\ds \) \(=\) \(\ds \floor {n + \dfrac 1 2}\)
\(\ds \) \(=\) \(\ds n\)

$\blacksquare$


Sources

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