[Math] A “number” with an infinite number of digits is a natural number

elementary-number-theory

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.

Best Answer

The principle of mathematical induction says that if $S$ is a subset of the natural numbers such that:

  • $1\in S$; and
  • If $n\in S$, then $n+1\in S$;

then $S=\mathbb{N}$; that is, $S$ is the set of all natural numbers.

Let $S=\{k\in\mathbb{N}\mid k\text{ has only finitely many digits when written in base 10 with no leading 0s}\}$.

Clearly, $1\in S$. If $n\in S$, then $n$ can be written with a finite number of digits, say $k$. Then $n+1$ can be written with either $k$ digits as well, or, in the worse case scenario (when $n = \underbrace{9\cdots99}_{k\text{ digits}}$) with $k+1$ digits; either way, if $n\in S$ then $n+1\in S$.

By induction, we conclude that $S=\mathbb{N}$. That is, every natural number has only finitely many digits when written in base 10 with no leading 0s.

So an infinite string of digits (omitting the "silly" possibility of infinitely many leading 0s), whatever it may be, is not a natural number.

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