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 $k \leq x <k+1$ 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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.