Exercise 4.1.15* (Prove $\left \lfloor \xi \right \rfloor + \left \lfloor \xi + \frac{1}{n} \right \rfloor + \cdots + \left \lfloor \xi + \frac{n-1}{n}\right \rfloor = \left \lfloor n \xi \right \rfloor$)

Question

If $n$ is any positive integer and $\xi$ any real number, prove that
$$\left \lfloor \xi \right \rfloor + \left \lfloor \xi  + \frac{1}{n} \right \rfloor + \cdots + \left \lfloor \xi  + \frac{n-1}{n}\right \rfloor = \left \lfloor n  \xi \right \rfloor.$$

Richard Ganaye · Accepted Answer

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(This is a generalization of Problem 6.)
\begin{proof} 
We write $\xi = m + 
u$, where $0 \leq 
u < 1$. Note that
\begin{align*}
[0, 1[ &= \bigcup_{0 \leq i < n} \left[ \frac{i}{n} , \frac{i+1}{n} ight[ &	ext{(disjoint union)}.
\end{align*}
So there is a unique $i \in [\![ 0, n [\![$ such that $
u \in  \left[ \frac{i}{n} , \frac{i+1}{n} ight[$. This integer $i$ satisfies
$$m +  \frac{i}{n} \leq \xi < m + \frac{i+1}{n},$$
thus 
$$nm + i \leq n \xi < nm + i + 1,$$
so
\begin{align}
\lfloor n \xi floor = nm + i.
\end{align}
Moreover, for every $j \in [\![ 0, n [\![$,
\begin{align}
m + \frac{i}{n} + \frac{j}{n} \leq \xi + \frac{j}{n} < m + \frac{i+1}{n} + \frac{j}{n}.
\end{align}

\begin{itemize}
\item If $0 \leq j < n - i$, then $j \leq n- i -1$, thus  $\frac{i+1}{n} + \frac{j}{n} < 1$, so by inequality (1)
\begin{align}
\xi + \frac{j}{n} < m+1.
\end{align}
Moreover $0 \leq  \frac{i}{n} + \frac{j}{n}$, so by inequality (1)
\begin{align}
m \leq \xi + \frac{j}{n} < m+1.
\end{align}
thus
\begin{align}
 \left \lfloor \xi  + \frac{j}{n}ight floor = m\qquad (0 \leq j < n-i).
\end{align}
\item If $n-i \leq j < n$, then $i+j\geq n$, thus $\frac{i}{n} + \frac{j}{n} \geq  1$, so by (1)
\begin{align}
&\xi + \frac{j}{n}  \geq m + \frac{i}{n} + \frac{j}{n} \geq m+1.
\end{align}
Moreover $j<n, i+1\leq n$, so
\begin{align}
\xi + \frac{j}{n} < m + \frac{i+1}{n} + \frac{j}{n} < m+ 2.
\end{align} 
Then the inequalities (6) and (7) give
$$m+1\leq \xi + \frac{j}{n} < m+ 2,$$
thus
\begin{align}
 \left \lfloor \xi  + \frac{j}{n}ight floor = m+1\qquad (n-i \leq j < n).
 \end{align}

\end{itemize}
Now, using (5) and (8),
\begin{align*}
\sum_{j = 0}^{n-1}\left \lfloor \xi  + \frac{j}{n}ight floor &=\sum_{j = 0}^{n-i-1}\left \lfloor \xi  + \frac{j}{n}ight floor + \sum_{j = n-i-1}^{n-1}\left \lfloor \xi  + \frac{j}{n}ight floor\
&= (n-i) m + i (m+1)\
&= nm + i\
&= \lfloor n \xi floor,
\end{align*}
by equation (1). This proves
$$\left \lfloor \xi ight floor + \left \lfloor \xi  + \frac{1}{n} ight floor + \cdots + \left \lfloor \xi  + \frac{n-1}{n}ight floor = \left \lfloor n  \xi ight floor.$$
\end{proof}

Exercise 4.1.15* (Prove $\left \lfloor \xi \right \rfloor + \left \lfloor \xi + \frac{1}{n} \right \rfloor + \cdots + \left \lfloor \xi + \frac{n-1}{n}\right \rfloor = \left \lfloor n \xi \right \rfloor$)

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Exercise 4.1.15* (Prove $\left \lfloor \xi \right \rfloor + \left \lfloor \xi + \frac{1}{n} \right \rfloor + \cdots + \left \lfloor \xi + \frac{n-1}{n}\right \rfloor = \left \lfloor n \xi \right \rfloor$)

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