Exercise 6.3.7 ($\alpha$ is rational iff there is some integer $k$ such that $\lfloor k! \alpha \rfloor = k! \alpha$)

Question

Prove that a number $\alpha$ is rational if and only if there exists a positive integer $k$ such that $\lfloor k \alpha \rfloor = k \alpha$. Prove that a number $\alpha$ is rational if and only if there exists a positive integer $k$ such that $\lfloor (k!) \alpha \rfloor = (k!) \alpha$.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

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

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

\begin{proof} 
\begin{enumerate}
\item[(a)]
let $\alpha$ be a real number.
\begin{itemize}
\item If $\alpha$ is rational, then $\alpha = p/q$, for some integers $p,q$ such that $q>0$. Then $q \alpha = p$, thus
$$\lfloor q \alpha floor = \lfloor pfloor = p = q \alpha.$$
Therefore there exists a positive integer $k$ ($k=q$) such that $\lfloor k \alpha floor = k \alpha$.
\item Conversely, suppose that there exists a positive integer $k$ such that $\lfloor k \alpha floor = k \alpha$. Then 
$$\alpha = \frac{\lfloor k \alpha floor}{k} \in \Q.$$
\end{itemize} 
\item[(b)]
\begin{itemize}
\item If $\alpha$ is rational, then $\alpha = p/q$, for some integers $p,q$ such that $q>0$. Then $q! \alpha = p (q-1)!$, thus
$$\lfloor q! \alpha floor = \lfloor p (q-1)! floor = p(q-1)! = q !\alpha.$$
Therefore there exists a positive integer $k$ ($k=q$) such that $\lfloor k! \alpha floor = k! \alpha$.
\item Conversely, suppose that there exists a positive integer $k$ such that $\lfloor k! \alpha floor = k! \alpha$. Then 
$$\alpha = \frac{\lfloor k !\alpha floor}{k!} \in \Q.$$
\end{itemize}

\end{enumerate}
\end{proof}

Exercise 6.3.7 ($\alpha$ is rational iff there is some integer $k$ such that $\lfloor k! \alpha \rfloor = k! \alpha$)

Answers

Comments

Exercise 6.3.7 ($\alpha$ is rational iff there is some integer $k$ such that $\lfloor k! \alpha \rfloor = k! \alpha$)

Answers

Comments

Add answer