# 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.

• By definition, a natural number has a finite number of digits, so no.
– Joe
Aug 17, 2011 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. Aug 17, 2011 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. Aug 17, 2011 at 16:16
• Do you allow leading zeros? Aug 17, 2011 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. Sep 3, 2013 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.

• The most common definition of "finite" is based on the definition of natural number, though Jun 26, 2020 at 18:45

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.

• 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. Aug 18, 2011 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). Oct 4, 2011 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! Sep 21, 2015 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? Oct 28, 2015 at 4:06
• Countability is defined by using the definition of natural number Jun 26, 2020 at 18:04

Let me first comment on your definition of $\Bbb N$. There are many ways to define $\Bbb N$, and there's one quite close to your intuitive understanding of what $\Bbb N$ is. The axioms are called the Peano axioms, and I'll state them here:

• $1$ is a natural number ($1\in\Bbb N$)

Then we define an $S(n)$ function (known as the successor function, but know that although intuitively, this should be $n+1$, the axioms say nothing about what $+$ means, and only uses $S$ to define the natural numbers)

• For all natural $n$, we know $S(n)$ must be a natual number ($n\in\Bbb N\implies S(n)\in\Bbb N$)
• For all natural $n$ and $m$, $n=m$ if and only if $S(n)=S(m)$ (that is to say, $S$ is an injection).
• For every natural $n$, $S(n)=1$ is false (that is, no natural number has $1$ as successor).

Now the important thing to note at first, is that the axioms say nothing about $2$, $3$, $1000$, or any other "numbers" (strings of digits); it merely shows us the structure of the natural numbers. To make it easier for ourselves, people've thought up symbols for each of these elements to be able to express them, but we could just as well give them other symbols. We could even give one symbol to each natural number, while maintaining the structure (for example, we can call $23$ the symbol $\circ$ and $24$ the symbol $\square$ and then $S(\circ)=\square$). As long as we follow the rules that are the axioms, we are still working with the same set, $\Bbb N$, but simply with a different representation of it. Now if we were to give each number its own symbol, it makes total sense that each number has a finite amount of "digits" - the word "digit" doesn't even mean anything here, since each number is one symbol anyway. See how we can have infinitely many natural numbers, while none has an infinitely long representation?

Furthermore, when you write $141258173412873\cdots$, you're not writing down an infinite number, since it must end somewhere (at the decimal dot), and since it starts, it cannot be infinite. But then you might ask, but what about $\cdots 141258173412873$? Is that a natural number? The question I would ask you then is: which one do you mean? If you were to write numbers this way, you cannot have a unique representation of each one. Also (going back to the axioms), how many times would you have to do $S$ on $1$ to get to that number? The axioms only state things about doing $S$ a finite amount of times, and your number would then require to do $S$ exactly $\cdots 141258173412872$ times, but this is not finite!

I think what you were stuck on before (and hopefully aren't anymore, after my and many other's attempts to explain it) is to wrap your head around the concept of infinity. How can you have infinitely many, but all finite numbers? It takes time and a lot of thought to get familiar with it, but I'm sure you have the motivation and the mind to do this. My main point is, don't think about natural numbers (or numbers from any set) as strings of digits, but rather as mathematical objects defined by some rules (axioms). The strings of digits are merely a common representation of them.

• "The axioms only state things about doing S a finite amount of times". Please correct me if I am wrong but in your axioms here and the axioms in Wikipedia I cannot find that $S$ can only be done a finite amount of time. Jun 26, 2020 at 18:09

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.

• 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. Sep 21, 2015 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.