Base system with $b \in \mathbb N$ consists of $b$ digits $d\_0,d\_1,d\_2\dots d\_{b-1}$. A number $a$ is expressed by some weighted sum of (integer) powers of $b$, where the digits $d\_0,d\_1,d\_2\dots d\_{b-1}$ are the weights.

My question is, why do we, if $b$ is a natural number greater than $1$, select the weights or the digits to be whole numbers and why do we need $b$ of them? 

In the usual base $2$ system, with weights $0$ and $1$, number $15$ in base $10$, for example, would be expressed as $1111$.

However, if we say that the base two system has weights $0$ and $\frac{1}{3}$, which we name $a$, $15\_{10}$ would be $a0aa0a$, ie. $\frac{1}{3} 2^5+\frac{1}{3} 2^3+\frac{1}{3} 2^2+\frac{1}{3} 2^0$.

It is a tad longer than with the usual weights. Is that the only reason to use $0$ and $1$, instead of $0$ and $1/3$?

How about base $3$ with weights $0$ and $\frac{2}{5}$, denoted by $b$? For example $15\_{10}$ would be $bb0b.\overline {bbb}$ or $\frac{2}{5}3^3 + \frac{2}{5}3^2 + \sum\_{n=0}^{\infty} \frac{2}{5}3^{-n}$.

Can I choose the weights or base digits in any way I want and there would always be a way to write any number in any base using them? Or do some other complications arise?