In integer-base positional numeral systems, the notation of a number in base $n$ uses $n$ numerals. Base 2 uses the symbols 0 and 1, base 10 uses 0123456789, base 16 uses base 10 + ABCDEF. Although the choice of symbols for the numerals is arbitrary, the number of numerals (unique glyphs) is identical to the base. That still holds when the base is negative or even complex. However, non-integer bases exist. I understand how the definition of base $n$ can extend to the base of any real base $b$ (I suppose we need $|b|>1$ for the positional property to hold?), what determines the number of numerals used in a number system with a non-integer base? Is it $\lfloor b \rfloor$ or can we decide any number of numerals?
For example, imagine I decide I want to use base τ. How many numerals do I use?
==== number of numerals that can be digits ====
In the article sited the $k$th digit can but any integer from $0$ to $\frac x{\beta^k}$ which as $x < \beta^{k+1}$ would be any digit greater than equal to $0$ and less than $\beta$. If $\beta$ is not an integer that is from $0$ to $\lfloor \beta \rfloor < \beta$.
Or $\lfloor \beta \rfloor + 1= \lceil \beta \rceil$.
[Note; if $\beta$ is an integer then $\lfloor b \rfloor \not < \beta$. But $\beta = \lceil \beta \rceil$.]
==== number of digits ====
An $n + 1$ digit number $N$ will be such:
$\beta^n \le N < \beta^{n+1}$
so $n \le \log_{\beta} N < n+ 1$
So $n = \lfloor \log_{\beta} N \rfloor$
so then number of digits is $n+1 = \lfloor \log_{\beta} N \rfloor + 1$.
It doesn't matter if $\beta$ is an integer or not. It only matters that $\beta > 1$.