Is there a base (besides base 0) that cannot represent all rational numbers? My son is learning about different bases for numbers and he is wondering if there is any base for a number system that cannot represent all rational numbers?
I'm thinking specifically of irrational bases, like base $\pi$ or negative bases, like $-2$.
I am not a mathematician and so do not know how to prove this one way or the other.
 A: Any representation of a number $N$ in base $B$ of the form $N = A_{n-1}A_{n-2}...A_1A_0$ can be represented as a polynomial:
$$N = A_{n-1} \cdot B^{n-1} + A_{n-2} \cdot B^{n-2} + \dots + A_1 \cdot x + A_0$$
That is $x=B$ is a root of the polynomial $$f(x) = A_{n-1} \cdot x^{n-1} + A_{n-2} \cdot x^{n-2} + \dots + A_1 \cdot x + A_0 - N = 0$$
A transcendental number is a number that cannot be the root of such an equation. Hence any transcendental number will fit your requirements.
Examples: $\pi$, $e$, $e^\pi$, $2^\sqrt{2}$ etc.
(Unfortunately proving these examples require a decent level of Mathematical knowledge, so the proof may be beyond the current scope of your son.)
A: Base one can only represent integers, and some rational numbers are not integers. So, yes, there is indeed a base (besides base O) that cannot represent all rational numbers, and it is base one.
What I call 'base one', Wikipedia calls 'unary' and has this to say:
"The unary numeral system is the simplest numeral system to represent natural numbers:[1] to represent a number N, a symbol representing 1 is repeated N times.[2]
In the unary system, the number 0 (zero) is represented by the empty string, that is, the absence of a symbol. Numbers 1, 2, 3, 4, 5, 6, ... are represented in unary as 1, 11, 111, 1111, 11111, 111111, ...[3]
In the positional notation framework, the unary is the bijective base-1 numeral system. However, because the value of a digit does not depend on its position, one can argue that unary is not a positional system.[citation needed] "
