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.

  • $\begingroup$ What do you mean by represent? Represent as a terminating or repeating decimal? $\endgroup$ – siegehalver Jul 27 '17 at 23:04
  • $\begingroup$ Able to be represented as a fraction, so I guess that means as a repeating decimal. $\endgroup$ – Andrew Eisenberg Jul 27 '17 at 23:06
  • $\begingroup$ Doesn't answer your question but this may be of interest: en.wikipedia.org/wiki/Golden_ratio_base $\endgroup$ – TomGrubb Jul 27 '17 at 23:06
  • $\begingroup$ Thanks. I will look. $\endgroup$ – Andrew Eisenberg Jul 27 '17 at 23:07
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    $\begingroup$ I believe any number that is not algebraic will not be able to represent any rational number. $\endgroup$ – user1952500 Jul 27 '17 at 23:14

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.)

  • $\begingroup$ Thank you. This makes sense, even without the formal proof. $\endgroup$ – Andrew Eisenberg Jul 28 '17 at 14:13

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