Prove that at an arbitrarily chosen x the sequence $\frac{\lfloor(nx)\rfloor}{n}$ is bounded.

Question

I solved it using a programming code. The upper boundary is $x$, and the lower one is $\lfloor(nx)\rfloor$. I want to know if my answer is right and if there is a better way to solve it, because all I did was putting so many inputs and observing the answer.

Answer 1

Hint:

For $n \in \mathbb{N}, x \in \mathbb{R}$ we have

$\lfloor nx \rfloor \geq n \lfloor x \rfloor$

and of course

$\lfloor x \rfloor \leq x$

and thus

$ \frac{ \lfloor nx \rfloor }{ n } \leq \frac{nx}{n} = x$

$ \frac{ \lfloor nx \rfloor }{ n } \geq \frac{n \lfloor x \rfloor}{n} = \lfloor x \rfloor$