Exercise 4.1.6 (Prove that $\left \lfloor x \right \rfloor + \left \lfloor x + \frac{1}{2}\right \rfloor = \lfloor 2x \rfloor $)

Question

For any real number $x$ prove that $\left \lfloor x \right \rfloor  + \left \lfloor x + \frac{1}{2}\right \rfloor = \lfloor 2x \rfloor $.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

\begin{proof} We write $x = n + 
u$, where $n \in \Z$ and $0 \leq 
u < 1$, so that $ \lfloor x floor = n$.

\begin{itemize}

\item If $0 \leq 
u < 1/2$, then
$$\begin{array}{llll}
n & \leq x + \frac{1}{2} &= n + 
u + \frac{1}{2} & < n+1,\
2n &\leq 2x &= 2n + 2
u &< 2n+1.
\end{array}
$$
Therefore
$$ \left \lfloor x + \frac{1}{2}  ight floor = n,\qquad  \lfloor 2x floor = 2n,$$
thus
$$\left \lfloor x ight floor  + \left \lfloor x + \frac{1}{2}ight floor = 2n = \lfloor 2x floor.$$

\item If $1/2 \leq 
u < 1$, then
$$\begin{array}{llll}
n + 1 & \leq x + \frac{1}{2} &= n + 
u + \frac{1}{2} & < n+2,\
2n+1 &\leq 2x &= 2n + 2
u &< 2n+2.
\end{array}
$$
Therefore
$$\left \lfloor x + \frac{1}{2}  ight floor= n+1, \qquad  \lfloor 2x  floor = 2n+1,$$
thus
$$\left \lfloor x ight floor  + \left \lfloor x + \frac{1}{2}ight floor = 2n + 1 = \lfloor 2x floor.$$

\end{itemize}
\end{proof}

Exercise 4.1.6 (Prove that $\left \lfloor x \right \rfloor + \left \lfloor x + \frac{1}{2}\right \rfloor = \lfloor 2x \rfloor $)

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Exercise 4.1.6 (Prove that $\left \lfloor x \right \rfloor + \left \lfloor x + \frac{1}{2}\right \rfloor = \lfloor 2x \rfloor $)

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