# A “number” with an infinite number of digits is a natural number?

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.

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By definition, a natural number has a finite number of digits, so no. – Joe Aug 17 '11 at 16:14
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. – Billy Aug 17 '11 at 16:15
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. – André Nicolas Aug 17 '11 at 16:16
Do you allow leading zeros? – Raphael Aug 17 '11 at 21:53
As others have said, no, it is not a natural number. You may be interested in the p-adics 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. – Henry Swanson Sep 3 '13 at 22:50

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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Let us consider the set of all "objects with infinite digits" and call it $S$. You have said that $\mathbf{N}$ is countable. Is $S$? If you consider Cantor's diagonal argument, you will quickly see that $S$ is an uncountable set. Thus, at the very least, there are infinitely more elements of $S$ than there are natural numbers.

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So, does that prove these things aren't natural numbers, or does it prove Luca is wrong in asserting the set of natural numbers is countable? Don't worry, I know the answer to that one, but possibly this is what Luca is worried about. – Gerry Myerson Aug 18 '11 at 1:43
It only proves that there exist numbers in $S$ that are not in $N$, so $S \not\subseteq N$. Others have shown that $S \not\supseteq N$ and moreover that $S \cap N = \varnothing$. A set is countable (by definition) if it can be placed into one-to-one correspondence with $N$, so therefore $N$ is countable (trivially). – kbolino Oct 4 '11 at 16:13
why are you considering such a set S? What if I just want to consider a few objects with infinite digits, not all of the possible ones? Your argument does not prove that such objects will not be natural numbers! – Luisinho Sep 21 '15 at 22:13
@Luisinho True enough, but how do you even "consider a few objects with infinite digits"? Could you name or describe one such object, so that we might consider it concretely? – kbolino Oct 28 '15 at 4:06

The set $\mathbb{N}$ of natural numbers is usually defined using the successor function: for natural number $n$, $S(n) = n+1$. We then define the natural numbers as follows:

1. $0 \in \mathbb{N}$
2. For all $n \in \mathbb{N}$, $S(n) \in \mathbb{N}$.

So for your "number" $(141258173412873....)$ to be in $\mathbb{N}$, we would have to have a natural number $m$ so that $S(m) = (141258173412873....)$. Try to find this number and you see that it cannot exist, since "adding $1$" makes no sense when there is no $1$'s place to add the $1$ to.

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I think the right way to think about objects of infinite "length" would be putting the dots on the left! This being done, your argument doesn't work. – Luisinho Sep 21 '15 at 22:35

You can prove by induction that $N < 10^N$ is true for any natural number N. Then N has at most N digits, a finite number. Well, you have to believe that natural numbers are finite, but since they all are obtained by adding 1 to 1 a finite number of times, I think we agree at this point.

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Every natural number has an infinite number of digits. E.g. 1 = ...00000001.

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This is beside the point; the question is clearly about nonzero digits. – Qiaochu Yuan Aug 17 '11 at 22:48
It would have been better to use $12.99999\dots$. – André Nicolas Aug 18 '11 at 0:09