Let $x \in \Bbb R$, $n \in \Bbb N$, show that $n \lfloor x \rfloor \leq \lfloor nx \rfloor \leq n \lfloor x \rfloor + (n-1)$

Question

Problem: Let $x \in \Bbb R$, $n \in \Bbb N$, show that $n \lfloor x \rfloor \leq \lfloor nx \rfloor \leq n \lfloor x \rfloor + (n-1)$

I have one part of the inequality, namely that

since $ \lfloor x \rfloor \leq x$, then $n\lfloor x \rfloor \leq nx$, but $ n\lfloor x \rfloor \in \Bbb Z$, so $n \lfloor x \rfloor \leq \lfloor nx \rfloor$

The right side of the inequality I am not so sure about. Insights appreciated.

Answer 1

If $k \leq x <k+1$ then $nk \leq x <nk+n$. So $nx$ lies between $j$ and $j+1$ for some $j \in\{nk+0,nk+1,...,nk+n-1\}$. Can you now see the right hand inequality?