My question is really as to whether I can consider the following results as proof of one another, since $(2a)$&$(2b)$ cannot be true unless $(0)$ is true, and vice versa.
So if for below I prove $(0)$ using the elementary means in Real Analysis, can I also then consider this to be proof for $(2a)$&$(2b)$?
I am looking to prove with the standard epsilon approach that:
$$x \in \mathbb R \Rightarrow\lim _{n\rightarrow \infty }\Bigl({\frac { \lfloor nx \rfloor }{n} }\Bigr)=x $$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(0)}$$
Which has become necessary in the following context:
It's well known that in the p-adic coefficient of a natural number: $$d_{n,p,j}= \Bigl\lfloor{n\,{p}^{\,j- {\lfloor \ln_p(n)\rfloor} -1}} \Bigr\rfloor-p\Bigl\lfloor{n\,{p}^{\,j- {\lfloor \ln_p(n)\rfloor} -2}} \Bigr\rfloor$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(1)}$$ as seen in the p-adic expansion of that number: $$n=\sum_{j=1}^{\lfloor \ln_p(n)\rfloor+1}d_{n,p,j}\,p^{\,\lfloor \ln_p(n)\rfloor+1-j}\quad\forall n \in \mathbb N$$
can likewise be used to separate any positive real number into it's integer and fractional parts:
$$\lfloor x\rfloor=\sum_{j=1}^{\lfloor \ln_p(x)\rfloor+1}d_{x,p,j}\,p^{\,\lfloor \ln_p(x)\rfloor+1-j}\quad\forall x \in \mathbb R^{+}$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(2a)}$$ $${\{x}\}=\sum_{j=\lfloor \ln_p(x)\rfloor+2}^{\infty}d_{x,p,j}\,p^{\,\lfloor \ln_p(x)\rfloor+1-j}\quad\forall x \in \mathbb R^{+}$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(2b)}$$
And because the corresponding finite sum for the above always computes a value much like the one I am trying to prove: $$\sum_{j=1}^{n}d_{x,p,j}\,p^{\,\lfloor \ln_p(x)\rfloor+1-j}=\frac{\lfloor Nx\rfloor}{N}$$ Then if it is true that taking the limit $n \rightarrow \infty$ results in convergence to $x$ in the manner described above, this means that proving $(0)$ to be true also concurrently proves $(2a)$ & $(2b)$ to be true.