Can every countable number be represented in all numeral systems?

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All numeral systems, including binary and ternary, can represent every countable number through a specific mathematical formulation. The representation of any number N can be expressed as a sum of powers of the base S, where the coefficients n_i are constrained within the range of the base. A proof by induction can be employed to show that if a number N has a numeral expansion, then N+1 will also have one. The key to this proof lies in demonstrating that the addition process, including carrying, concludes after a finite number of steps. Thus, every countable number can indeed be represented in any numeral system.
Wminus
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Hi. This might be a stupid question (I'm studying engineering :p), but how do you prove that all numeral systems (binary, ternary etc.) can represent every countable number?

I guess you will need to prove that any number ##N## can be written as ##N= S^0 n_0 + S^1 n_1 + S^2 n_2 + ...## where ##S## is the base of the numeral system, and ##n_i \in [0,max\{S\}]## with ##i \in \mathbb{N}##.

EDIT: fixed an error in my equation
 
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Is the question unclear in some way? btw n_i should be element of [0,S], not [0, max{S}]. Duno why I wrote max S, I guess I'm just exhausted due to the exams.
 
Wminus said:
Hi. This might be a stupid question (I'm studying engineering :p), but how do you prove that all numeral systems (binary, ternary etc.) can represent every countable number?

I guess you will need to prove that any number ##N## can be written as ##N= S^0 n_0 + S^1 n_1 + S^2 n_2 + ...## where ##S## is the base of the numeral system, and ##n_i \in [0,max\{S\}]## with ##i \in \mathbb{N}##.

EDIT: fixed an error in my equation
You can do a proof by induction--show that if some N has an expansion, then N+1 also has an expansion. Demonstrating that the process of "carrying" in addition terminates after a finite number of steps is sufficient.
 
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