How do I prove or disprove the following statements:$$(a) n  \mathbb N, k  \mathbb N, x  \mathbb R , \lfloor nx \rfloor  n \lfloor x \rfloor  k$$ and $$ (b)\exists k \in \mathbb N, \forall n \in \mathbb N, \forall x \in \mathbb R, \lfloor nx \rfloor  n \lfloor x \rfloor  k $$I've also been given the following three properties to use:$ (i) \forall x \in \mathbb R, \exists \epsilon \in R, 0  \epsilon < 1  x = \lfloor x \rfloor + \epsilon \\(ii) \forall x \in \mathbb Z, \forall y \in \mathbb R, \lfloor x + y \rfloor = x + \lfloor y \rfloor \\(iii) \forall x \in \mathbb Z, \lfloor x \rfloor = x $I know that we must negate the statement$(a)$ first in order to pave the way for a disproof(if it false) by proving: $$ \exists n \in \mathbb N, \forall k \in \mathbb N, \exists x \in \mathbb R, \lfloor nx \rfloor - n \lfloor x \rfloor > k $$Other than that I have no idea which property to use and in which order. The help would be really appreciated.
The floor function
proof techniques
You have been given that$$0\le\epsilon<1$$multiplying by $-1$\begin{equation}\therefore  0\ge-\epsilon>-1\tag{1}\end{equation}adding $x$$$\therefore x\ge x-\epsilon> x-1$$\begin{equation} \therefore x\ge \lfloor x \rfloor >x-1\tag{2}\end{equation}We can use this inequality $(2)$  to show that\begin{equation}\therefore nx\ge n\lfloor x\rfloor > nx - n\tag{3}\end{equation}adding $nx$ to inequality $(1)$ we get$$nx\ge nx-\epsilon > nx -1$$\begin{equation}nx\ge \lfloor nx \rfloor > nx -1 \tag{4}\end{equation}Using $(3)$ and $(4)$ we can write$$nx-1 - nx<\lfloor nx\rfloor - n \lfloor x\rfloor<nx - (nx - n)$$$$\therefore -1<\lfloor nx\rfloor - n \lfloor x\rfloor<n $$