How to prove that $0 \leq\lfloor nx \rfloor - n\lfloor x \rfloor \leq n-1$?

Question

$x \in \Bbb R, n \in \mathbb N_{\ne 0}$ , prove the following inequality :

$0 \leq\lfloor nx \rfloor - n\lfloor x \rfloor \leq n-1$

Here's my current try

$\lfloor x \rfloor\leq x < \lfloor x \rfloor +1$

$n\lfloor x \rfloor\leq nx < n\lfloor x \rfloor +n$

$\lfloor n\lfloor x \rfloor \rfloor\leq \lfloor nx \rfloor < \lfloor n\lfloor x \rfloor +n\rfloor$

$n\lfloor x \rfloor\leq \lfloor nx \rfloor < n\lfloor x \rfloor +n$

$0\leq \lfloor nx \rfloor - n\lfloor x \rfloor < n$

And this is where I'm stuck, I don't know how to progress further to get to $n - 1$

Answer 1

Since $⌊nx⌋−n⌊x⌋$ is an integer and is strictly less than $n$ (as you showed), it can be at most $n−1$.

Answer 2

Easier to assign variables, to make everything easier to decipher.

I am supposed to disect your work, but it simply gets confusing.

Let $F = \lfloor x\rfloor$ and let $r = x - F \implies 0 \leq r < 1.$

Then $nx = nF + nr,$ and $0 \leq nr < n \implies 0 \leq \lfloor nr\rfloor \leq (n-1)$.

Then $\lfloor nx\rfloor = nF + \lfloor nr\rfloor \implies$

$0 \leq \lfloor nx\rfloor - nF \leq (n-1)$.

---

Edit
The statement that
$\lfloor nx\rfloor = nF + \lfloor nr\rfloor$
Is based on the idea that if
$A = B + C,$ where $B$ is an integer, then $\lfloor A\rfloor = B + \lfloor C\rfloor.$