Rational and irrational numbers under base pi I am wondering, what would happen to the representation of a number like 2 in base pi? I know that things like $π^2$ would simply be 100, but what about numbers that are not of the form $π^n$ ? I think it ought to get tricky!
 A: See here for formal definition of representation in a non integral base. With such a definition $2$ in base $\pi$ is simply... $2$. This because $2<\pi$ hence it can be represented by itself.
In any base $\beta$ most numbers (all but a countable set) have an infinite representation. Only sums of powers of $\beta$ have a finite representation.
Here is an algorithm (written in python) to compute the representation in any base:
from math import *
import string

def repr(x,b,digits=30):
    """ represent @x in base @b """
    DIGITS = string.digits + string.ascii_uppercase
    n = int(floor(log(x) / log(b)))
    r = ''
    while n>0 or (len(r)<digits and x != 0.0):
        d = int(x/b**n)
        if n==-1:
            r += '.'
        r += DIGITS[d]
        x = x - d * b**n
        n -= 1
    if r=='':
        r= DIGITS[0]
    return r

for x,b in [(1973.0,10), (8.0,2), (10.0,pi)]:
    print x, 'in base', b, 'is written:', repr(x,b)

The output is:
1973.0 in base 10 is written: 1973
8.0 in base 2 is written: 100
10.0 in base 3.14159265359 is written: 100.01022122221121122001111210

A: It's not enough to just say "base $\pi$" - the concept isn't really defined until you specify what the digits are allowed to be. What seems to me to be the only reasonable option is to say that digits can be anything within $[0,\pi)$, but this then makes representing anything trivial, because for any $a>0$,
$$\Large \begin{align*}
a&=\pi^{\log_\pi(a)}=\pi^{\lfloor\log_\pi(a)\rfloor}\underbrace{\pi^{\log_\pi(a)-\lfloor\log_\pi(a)\rfloor}}_{\text{between 0 and }\pi}\\\\\\
&=\pi^{\log_\pi(a)-\lfloor\log_\pi(a)\rfloor}\underbrace{00\ldots0}_{\lfloor\log_\pi(a)-1\rfloor}\text{ in base }\pi
\end{align*}$$
So the representation of $2$ in base $\pi$ under this approach would just be $2$ (because $1<2<\pi$), while the representation of $4$ would be $\frac{4}{\pi}0$ (because $\pi<4<\pi^2$).
