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

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.

If $$k \leq x then $$nk \leq x . 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?