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The set of natural numbers is infinite and countable. Ok. But think of an object with infinite digits (141258173412873....). Is it a natural number?


Edit: What i found confusing was the fact that, since $\mathbb{N}$ is an infinite set, an object with infinite digits should be also a number and should belong to $\mathbb{N}$. I know this is a naive view. But now things are clearer to me, thanks to your answers! If i had to explain to a person not (too) educated in mathematic what $\mathbb{N}$ (the set of natural numbers) is, i would start with this:

consider the following algorithm (procedure) to construct $\mathbb{N}$={1,2,3,4....}:

  1. num = 1
  2. $\mathbb{N}$ is the empty starting set of numbers
  3. put num in $\mathbb{N}$
  4. num = num + 1
  5. repeat from 3

Now, does $\mathbb{N}$ has objects with an infinite number of digits? No. The procedure goes on forever, but everytime we add a number to $\mathbb{N}$ (step 3), the number we are adding has a finite number of digits.

This view is only slightly different from other answers given to my original question, but i think it is simple enough to explain why a procedure that goes forever and build objects with an increasing number of digits does not produce a set with objects with an infinite number of digits.

  • 4
    By definition, a natural number has a finite number of digits, so no.2011-08-17
  • 3
    No, all natural numbers are defined to be finite. That doesn't mean your thing doesn't make sense, but it's not a natural number.2011-08-17
  • 4
    No. The natural numbers are $1$, $2$, $3$, and so on, and that's all. Your expression, if we interpret $\dots$ to mean that the digits go on forever, does not represent a natural number.2011-08-17
  • 0
    Do you allow leading zeros?2011-08-17
  • 1
    As others have said, no, it is not a natural number. You may be interested in the [p-adics](http://en.wikipedia.org/wiki/P-adic_number) though. You can have infinitely many digits to the left of the decimal point (although, only finitely many to the right). Also, all digits are in $0$ through $p-1$, instead of $0$ through $9$. Addition and multiplication behave pretty much the same as usual, just extended infinitely: $\ldots 44444 + 1 = 0$, in the $5$-adics.2013-09-03

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