# Is there any proof that all natural numbers have a finite number of digits? [closed]

Numbers like $\cdots 999$ and $\cdots 777$ are not natural numbers. Ok! But why that's?

$\cdots 999$ has infinite number of digits. A natural number must have a finite number of digits.

My question is:

From where come that law? There are some proof of this, or it is just a "accept by default" statement.

Hi @mike4ty4!

I already know about $p$-adic, and have a little fun with them. And thanks by your explanation.

The definition that you made, is very, very, very good.

Actually I am confused exactly with this def.

We have $10$ "symbols": $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ to represent a digit.

Then, with two digits we can form $10^2$ different numbers: from 0 to 99. And with $8$ digits we can form $10^8$ numbers: from $0$ to $99999999$.

What is making me crazy is that with this line of argumentation, I can say that: with $n$ digits I can form $10^n$ different numbers.

And since the number of digits in a natural number is finite, then i can form $10^{finite}$ different numbers. Well,

How can i say that there are infinitely many natural numbers, if I must believe that $10^{finite}$ is a finite quantity ?

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## closed as not a real question by Andres Caicedo, Gottfried Helms, Did, Will Jagy, Henry T. HortonDec 9 '12 at 0:04

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Every natural number $n$ has less than or equal to $n$ digits. As $n$ is finite, each natural number has a finite number of digits. –  01000100 Dec 8 '12 at 8:29
Induction. ${}$ –  Qiaochu Yuan Dec 8 '12 at 8:52
It's a matter of definition. Asking "Why can't there be a natural number with infinitely many digits?" is like asking "Why can't there be liquid snow?" — it's because, if it were liquid, it would be called rain, not snow. –  Ilmari Karonen Dec 8 '12 at 10:56
Your mistake is that 'finite' is an adjective, but you're treating it like a number. –  Hurkyl Dec 8 '12 at 11:05
You can form only so many numbers with any given amount of digits. But there is no limit to how many digits you can have. So there can still be infinitely many numbers. –  mike4ty4 Dec 8 '12 at 11:27

A decimal representation of a natural number $n$ is defined to be a sequence $d_{k-1}d_{k-2}\cdots d_1d_0$ such that

$$n = \sum_{i=0}^{k-1} d_i 10^i$$

and $0 \le d_i \le 9$. The $d_i$ are the digits of $n$. By this definition, there are no infinite decimal representations. What we need to show is that every natural number can be represented this way. We can do this with an induction argument. We note that $0$ can be represented this way with $k = 1$ and $d_0 = 0$. Suppose that a natural number $n$ can be represented this way. Then, we have

\begin{align}n + 1 &= \left(\sum_{i=0}^{k-1} d_i 10^i\right) + 1 \\ &= \left(\sum_{i=1}^{k-1} d_i 10^i\right) + \left(d_0 + 1\right)\end{align}

So, we can form a sequence $d'_{k-1}d'_{k-2}\cdots d'_1 d'_0$ with $d'_0 = d_0 + 1$, to represent $n + 1$. But $d'_0$ may be equal to $10$ this way. In that case, we subtract 10 from it and add 1 to $d'_1$. If that also equals 10, we subtract 10 from it and add 1 to $d'_2$. We keep doing this until we either have no more 10s or $d_{k-1} = 10$, in which case we subtract 10 and introduce a new position $d_k = 1$. None of these operations changes the value of the sum, since if we subtract 10 from $d_i$ and add 1 to $d_{i+1}$, we have added $(-10) 10^i + 10^{i+1} = -10^{i+1} + 10^{i+1} = 0$ to $n$. So we now have a representation of $n$ in the given form, as a decimal representation with a finite number of digits. There are no natural numbers that require an infinite expansion.

Now, what if we did try to make an infinite expansion anyway? Well, we could not assign a natural-number value to the sum unless all but finitely many of the digits were zero. This can be shown with another simple induction argument to prove $10^n > n$ for all naturals $n$, so the sum would have to exceed every natural number, which is not possible.

Bonus! There is, it turns out, a system of "numbers" in which "infinite expansions" of the type you suggest do have a meaning, and it's called the "p-adic numbers". You can see more about it here:

However, in this case, we use a prime, non-decimal base.

These are not the natural numbers, however. But I figured it was worth a mention, since they pertain to the kind of expansion you mention and you might find it interesting.

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Hello Mike ! Thanks by your answer. I tried comment here, but this field is to short. I put some detail of my headache in the body of my text. Can you explain me what i am doing (thinking) wrong ? –  fer Dec 8 '12 at 9:30
$p$-adic numbers have a possibly infinite expansion, true. But natural numbers, seen as $p$-adic numbers, still do have a finite $p$-adic expansion. –  Andrea Mori Dec 8 '12 at 18:47
@Andrea Mori: True. But my point was the two systems "$p$-adic numbers" and "natural numbers" were not the same thing, and the asker was asking about natural numbers. –  mike4ty4 Dec 8 '12 at 22:16

There are many answers to your question (already), because there are many ways to define or axiomatize the natural numbers. The first issue here is whether "natural number" is a primitive notion or is defined from other concepts. If it's a primitive notion, then one must next ask what are the appropriate axioms governing it. Among the answers already given are some taking the induction principle as axiomatic, and one taking the least number principle as axiomatic. Let me follow the other path, namely to define the natural numbers, as is the custom in set-theoretic foundations of mathematics. I'll assume that we've already chosen a set-theoretic representation of "zero" and "immediate successor". (It doesn't matter much what these are; a fairly standard choice is to represent 0 by the empty set and, if a number is represented by a set $x$, then its immediate successor is represented by $x\cup\{x\}$.) Then one defines "x is a natural number" to mean that $x$ is an element of every set $A$ that (1) has zero as an element and (2) contains the immediate successors of all its elements.

If one works from this definition, then, after defining what decimal expansions are and what finiteness means, one can prove that every natural number has a finite decimal expansion by observing that the set of those numbers that do have a finite decimal expansion (1) contains 0 and (2) contains the immediate successor of each of its members. So this set is an $A$ as in the definition of "natural number", and therefore, by that definition, all natural numbers are elements of it.

(Notice that the set-theoretic definition of "natural number" essentially builds the induction principle into the definition, so the proof in the preceding paragraph is "morally the same" as proofs using induction as an axiom.)

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I think with ...999 you mean a number $n$ in decimal represantation which satisfies

$$n = \lim_{k \to \infty} \left(\sum_{i=0}^k 9 \cdot 10^{i}\right)$$

Since the sum converges to infinity, $n$ would be an 'infinite number'. But all natural numbers are finite by construction.

HTH, Maikel.

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To answer your followed up question: With a finite representation you can represent countable infinite many numbers, since the countable union of finite sets is countable. I think thats where you are getting stuck. Sorry, i dont know how to comment your post directly... –  manado Dec 8 '12 at 9:41
Hello manado. That is the point. Where "all natural numbers are finite by construction" can be verified ? –  fer Dec 8 '12 at 12:02
Usually natural numbers are just defined to be THE finite sets, and so finite by definition or construction, which means: $1 = \{ \emptyset \}, 2 = \{ \emptyset, 1 \} = \{ \emptyset, \{ \emptyset \} \}, 3 = \{ \emptyset, 1, 2 \}$ and so on. Thus $n = \{ \emptyset, \ldots, n-2, n-1 \}$. A set $F$ is called finite, if it is bijective to a number $n$ as I defined here... –  manado Dec 8 '12 at 15:23
Btw: you can take in this definition a limes and get the set of all natural numbers. Maybe you want to look at ordinal numbers, if you are interested in infinite numbers. –  manado Dec 8 '12 at 15:28

Each positive natural number $n$ has at most $n$ digits, so if you know that $n$ must be finite, you can conclude that $n$ has finitely many digits.

The crux of the issue, then, seems to be the apparent confusion over how there can be infinitely many natural numbers, while at the same time each natural number is finite.

I'll try to explain this to you, but first, there are some things I absolutely must know:

(1) What definition of "finite" are you using?

(2) What definition of "natural number" are you using?

(3) Are you familiar with the Principle of Mathematical Induction or the Well-Ordering Principle?

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DISCLAIMER : This answer is only half serious.

I give two variations of the same idea:

1st proof : A non-empy set of the natural numbers $\Bbb N$ has a minimum. So assume that the set $$\cal N_\infty=\{\text{n\in\Bbb N requiring infinitely many digits}\}$$ is non empty. Let $n_\infty=\min\cal N_\infty$. Then $N=n_\infty-1$ can be written using finitely many digits. Let's write numbers in base $N$. Then $$N=10\qquad\text{and so}\qquad n_\infty=11.$$ Since we can tranform the expression in base $N$ of numbers into the expression in base $10$ and finite expression are tranformed into finite expressions, this is a contradiction. Therefore $$\cal N_\infty=\emptyset$$

2nd proof : by induction. Note that $1$ has a finite expression. So suppose $n$ has a finite expression. By the same trick used in the 1st proof $n+1$ has a finite expression. Thus, we can conclude that all numbers in $\Bbb N$ have finite decimal expression by Peano's axioms.

** POST SCRIPTUM ** I just noticed that some of the above is included already in some of the answer given previously. Sorry, no plagiarism intended, just had fun writing this.

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The basic fact is $|A|<|{\cal P}(A)|$, where ${\cal P}(A)$ denotes the power set of $A$. For general sets $A$ you need Cantor's diagonal argument to prove this, but for finite sets $A$ the observation $$\{2^0, 2^1,\ldots 2^n\}\subset\{1,2,3,\ldots,2^n\}$$ suffices. It follows that $n+1\leq 2^n$, or $n\leq 2^n-1$, for all $n\geq0$. As the $2^n$ binary sequences $${\bf b}:\quad [n]\to \{0,1\}, \quad k\mapsto b_k$$ encode all natural numbers $\nu$ from $0$ up to $2^n-1$ (inclusive) via $$\nu({\bf b}):=\sum_{k=1}^n b_k\ 2^{k-1}$$ we may conclude that any given $n\geq0$ has a binary expansion of the form $n=\sum_{k=1}^n b_k\ 2^{k-1}$ and so can be written as a binary number with at most $n$ digits.

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Try unbounded as opposed to infinite-- all natural numbers have finite number of digits but the number of digits in a natural number is unbounded.

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