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.
On
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] "
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.)