Why there are real numbers with infinite digits, but no such natural numbers (or another reason why real numbers are uncountable) This question is me trying to understand (again) why there can be no one-to-one correspondence between the sets of natural and real numbers.
The source of confusion is this: if we abstract completely from the properties and operations on numbers and just think of them as strings of digits, it appears to be possible to list all the real numbers between $0$ and $1$ and all positive integers having the same digits in the same (or reverse) order, for example:
$$0.1 \to 1$$
$$0.01 \to 10$$
$$0.0101 \to 1010$$
$$0.110101 \to 101011$$
I use binary expansion for simplicity.
Since irrational numbers have infinite expansions, we would have to consider some kind of 'infinite natural numbers', like:
$$101110001\dots$$
or
$$\dots101110001$$
I think the main (and only) reason why such numbers don't exist is this requirement:


*

*We need the sets of real and natural numbers to be ordered.

*We need to be able to do arithmetics on real and natural numbers.


Obviously, if we consider any of two types of 'infinite natural numbers' and the usual absolute value, we find that it's impossible to order them, or do arithmetics with them.
The only way is to change the norm (to $p-$adic norm for example).
While with the real numbers it's still possible to order them, and do arithmetics with them when they have infinite amount of digits in their expansion (or at least in principle).

Is my reasoning correct? Are the two requirements (ordering and arithmetics) enough to explain why natural numbers with infinite amount of digits do not exist, while real numbers do?

This confusion is not restricted to me alone, see this paper for example
 A: The "signature property" of the natural numbers is the principle of induction for proofs (and its sibling, the principle of recursion for definitions).
Now how do we define the digit string that represents a natural number $n$ in the first place? The only sensible way is: by recursion: The number zero is represented by "$0$" and if $n$ is represented by a certain string of digits then its successor is represented as follows: If the representation ends in any digit other than $9$, we increase the last digit (according to the predefined sequence $0,1,2,3,4,5,6,7,8,9$); otherwise, if it ends in one or more $9$'s, but has more digits before that, replace this last block of $9$'s by $0$'s and increase the last digit before the block as described above; otherwise, if the representation of $n$ consists of $9$'s only, replace all $9$'s by $0$'s and precede the result by a $1$.
Now that we have given a definition of how to obtain a digit representation for each natural number, we can prove claims about these representations - by induction. One sees that the representation for $n+1$ has at most one digit more than that of $n$. And as the number zero uses only one digit (a finite amount), and a finite number plus one is still finite, we conclude by induction that the representation as defined above consists of finitely many digits.
