What is worthy of investigation?

We describe in dialogue form a possible way of discovering and investigating 10-adic numbers starting from the naive question about a “largest natural number”. Among the topics we pursue are possibilities of extensions to transfinite 10-adic numbers, 10-adic representations of rational numbers, zero divisors, square roots and 10-adic roots of higher degree of natural numbers, and applications of 10-adic number representation in computer arithmetic. The participants of the dialogue are idealized embodiments of different philosophical attitudes towards mathematics. The article aims at illustrating how these attitudes interact, in both jarring and stimulating ways, and how they impact mathematical development. Moreover, the article demonstrates how different attitudes translate into mathematical heuristics. In our opinion this is also relevant for teaching mathematics as it raises consciousness for reflections on how mathematical investigations work.

how different philosophical attitudes interact, how they get in each other's way, but how they can also stimulate each other. Moreover, we want to exemplify how they impact the course of mathematical investigations on unknown territory. The idea for this came about from a conversation the second author had on a mathematical enrichment camp in Kyoto with a senior high-school student, Yuki Kitamura, who was investigating 10-adic roots of natural numbers (within a school project) without prior knowledge about the theoretical background and without using common terminology. The discussion between the author and the student evolved from mutual misunderstanding in the beginning to an enriching experience for both sides in the end. 1 The different philosophical attitudes, which actually came to light within only two people, are embodied here by five characters, each of whom idealizes a particular stance. Note that these (and other) attitudes could also interact within a single person's mind. Moreover, note that as the characters serve to represent stances and no actual individuals, we take it in stride that some of them, in particular Nana, mathematically evolve in fast motion during the discussion. That goes along with the fact that the whole investigation will be presented in an idealized way, and can therefore seem to be developing implausibly quickly in some places.
We believe this article to be worthy of reading for mathematics instructors, learners of mathematics, and anybody who is interested in the philosophy of mathematics. By reflecting about the participants of the dialogue, mathematics instructors can boost their awareness of the variety of stances towards mathematical thinking. They are invited to think about how these can be beneficial for mathematical development and for the learning of mathematics. Moreover, having different stances in mind, mathematical instructors could try to trigger certain of them purposeful within their classes. Learners of mathematics can get acquainted to different attitudes towards mathematics, and should ask themselves which of them they have already discovered in their own mind, and which of them were sleeping thus far. In future mathematical investigations, they could try to consciously take different perspectives from time to time. By doing this, they enable themselves to switch between different work modes in order to make progress. To point out this "heuristical aspect" of the philosophy of mathematics is another aim of this article. For those who are interested in the philosophy of mathematics in general, and its heuristical aspects in particular, the text can serve as an "appetizer" with several suggestions for further reading. We now introduce the characters.
Nana represents a naive attitude. She does not question whether the expressions she writes are meaningful. She happily constructs and operates with concepts without worrying whether her objects are well-defined or the conditions for her operations are satisfied. For example, she has the opinion that every sequence of digits is a number. Although this might seem a bit careless, it turns out to be the initial spark and a driving force of the whole investigation. In the history of mathematics, the "naive" approach of putting difficulties with precision, formalization or even downright contradictions aside and working with an intuitive understanding has frequently turned out to be quite fruitful. Examples include the initial uses of imaginary numbers in algebra in the 16th century (see, e.g., [16,26] p. 99), the use of infinitesimals in the dawn of analysis ( [4]), the nonchalance of Euler and many of his contemporaries in attributing values to divergent series ( [3]) or "naive" set theory as introduced in many mathematics beginner's lectures. Constance is of a conservative persuasion. 2 She is attached to the traditional scope of subject matter and relates expressions to it without considering alternative interpretations or deviations. Consequently, she argues that infinite numbers are no numbers, taking it for granted that such numbers must be natural. Reila adheres to critical realism. 3 She insists that symbols must mean -refer tosomething and that this "something" does not simply coincide with the rules for the use of symbols or the structural properties of the symbol combinations thus obtained, but she is open to the possibility that they mean something other than originally intended or that one must first seek a meaning to a given system of symbols. 4 Reila, however, would not consider a purely formal system sufficient in itself. Fortune, being a formalist 5 , lets the symbols stand for themselves and has no need for an object reference. Once the rules of manipulation are established, she is satisfied -one must be able to deal with the expressions, then for her the meaning consists in the rules for their use. 6 Although imagining a referent for a mathematical concept may be inspiring or heuristically helpful, such interpretations or intuitions are notoriously unreliable and should therefore not form a part of mathematics proper in her view. 7 Preston approaches mathematics from a practitioner's perspective. She understands mathematical theories as tools that should be suitable to be beneficially applied in other fields, inside or outside mathematics. Although rarely formulated 2 Although one can hardly say that they form a philosophical school, conservative stances have always played a role in the history of mathematics. For instance, one may think of the famous Kronecker quote, according to which the natural numbers were created by god, while everything else is man-made, or the resistance with which Cantor was met when he introduced infinite ordinals. 3 The corresponding school of thought would be mathematical "platonism". See, e.g, [2]. 4 In this sense, her position resembles that of Gödel, who was, on the one hand, an adherent of mathematical realism, (see, e.g., [15,28]) while at the same time emphasizing the possibility to interpret formal expressions in several different ways; for example, the famous Gödel sentence that is used in the incompleteness theorem can at the same time be understood as a rather involved statement about natural numbers and as a statement about provability in a certain formal system. 5 For more on the formalistic viewpoint in the philosophy of mathematics see, e.g., [23]. 6 See, e.g., Wittgenstein, [30], p. 7: "Words and chess pieces are analogous; knowing how to use a word is like knowing how to move a chess piece. (...) The meaning of a word is to be defined by the words for its use, not by the feeling that attaches to the words." 7 For a formalist approach to mathematics see, e.g., Hilbert [11], p. 18 (translation in [6], p. 1122) "Because I take this standpoint, the objects [Gegenstände] of number theory are for me (...) the signs themselves (...). (...) The solid philosophical attitude that I think is required for the grounding of pure mathematicsas well as for all scientific thought, understanding, and communication -is this: In the beginning was the sign. (...) These number-signs [Zahlzeichen], which are numbers and which completely make up the numbers, are themselves the object of our consideration, but otherwise they have no meaning [Bedeutung] of any sort." and defended as a systematic position, this utilitarian attitude towards mathematics, which sees its value in its applications, is of course widespread.
Note that the participants should not be read as representatives of schools of the philosophy of mathematics, but rather as representing philosophical attitudes as they enter the working practice of mathematicians. Therefore, they are caricatures of thought tendencies of which most mathematicians probably have several. For some modern, differentiated philosophical perspectives on mathematical practice see, e.g., [14].
The dialogue is inspired both by Imre Lakatos' "Proofs and Refutations" [13] and by Detlef Spalt's "Vom Mythos der Mathematischen Vernunft" ("On the Myth of Mathematical Reason", [25]), which also contains a mathematical investigation conducted by proponents of various directions in the philosophy of mathematics. The decision to present our article in dialogue form is strongly influenced by these sources of inspiration. By reflecting this decision beyond the obvious reasons 8 we came across the interesting paper [9], where possible purposes for which Lakatos may have employed the dialogue form are discussed. The paper sees the main purpose in the fact that the dialogue form "accurately reflects the role of criticism in the growth of knowledge" ( [9], p. 9), which the author calls "congruence of content and form". Moreover, as a side effect it is mentioned that the dialogue form is apt to veil the author's position, which in this case "ironically helps to deflect criticism of Lakatos"s own ideas' ( [9], p. 10). We think that both "congruence of content and form" and something similar to "veiling the authors" position' (in the sense of not taking position for one of the stances, i.e., in the sense of a pluralistic metaposition 9 ) apply also to this article.

The question and a first try
Four mathematics students -Constance, Fortune, Reila, and Preston -are sitting at the cafeteria together with a friend and his daughter Nana, a schoolchild.

The question about the largest natural number is posed and defended
against the standard "C1"-argument.
Friend: You folks are smart at math, aren't you? Reila: Sure, why? Friend: The other day Nana asked me a mathematical question and I wasn't sure how to respond. I wonder what your answer would be. Fortune: Shoot, Nana! 8 As obvious reasons to write such an article in dialogue form we count: different stances become more palpable through impersonating them, it is easier to illustrate opposed notions through a dispute between human beings about opposed opinions, the text becomes more entertaining and evokes emotional reactions by the readers more easily. 9 See, e.g., [31] for a pluralistic position on the foundation of mathematics.

Nana:
What's the largest number? Constance: (smiles knowingly) See, Nana, there is no such number. Nana: Why's that? There must be one! Constance: Okay, then tell me the largest number you know! Nana: (ponders for a while) Thousand millions! Constance: Well, I would call that a billion, and it's not the largest number, because it's less then a billion and one. Nana: Then I say a thousand billions! Constance: (smiles complacently) That's a trillion, which is less than a trillion and one. Nana: Then I say a trillion trillions! Constance: Well, whatever that number's name is, I say a trillion trillions and one.
Look Indeed, I see that I was right. The result is 0, so you don't get a larger number by adding 1. My number is the largest number. Constance: I don't agree. There is only one number that gives 0 when added to 1.
And that number is 1. So your number would be 1, and that's not so large. Or your number is no number at all, and that would be my preferred conclusion.

Going transfinite:
A new intuitive concept is proposed, analyzed, and dismissed. Thereby, methods, criteria, and concepts arise that will foster the subsequent development.

A new naive idea is confronted with naive criticism.
Nana: (Stares unhappily at the sheet of paper, shakes her head.) I think we did it wrong! Reila: What do you mean? Nana: The :::999 C 1 D 0 thing. It's wrong. Constance: (sighs) It's not, Nana. You applied the laws of arithmetic, and four mathematicians approved your calculation. 10 This idea can also be found on the website http://www.gtodoroff.de/mathe.htm of Georg Todoroff, a well known "alternative mathematician", who then goes on to conclude that the reciprocal of 999... must be the smallest positive real number and that countable unions of countable sets are uncountable. It would be an interesting instance of the demarcation problem (see e.g. https://plato.stanford.edu/entries/pseudoscience/) to consider the difference between courageous naivety and getting lost in pseudo-science.

Reila:
Nana, what do you think would be the "right" answer? Nana: We have 9 C 1 D 10, and 99 C 1 D 100, and 999 C 1 D 1000 ... Constance: Oh no. I think I know what's coming next... Nana: ...and it goes on like that, the 9s become 0s and there is a new 1... Constance: Please don't say what I think you want to say... Nana: So :::999 C 1 must be 1....000 -a 1 followed by infinitely many 0s. Constance: Look, Nana. That's not sensible. You cannot have a 1 followed by infinitely many 0s. Nana: Why not? It makes a lot of sense. It was really strange that ...999 should be such a small number, but now I understand that it is actually as huge as it should be. Constance: Because there is no position infinity in a number representation. There is a first, a second, a third, a hundreds, a billions. But these are all finite.
None of them has infinitely much space to its right. But no place to write your 1! Nana: Well, now there is one. I just made it. Constance: (waves hands in desperation) Nana, that's absurd! I mean, where would you even write your 1? There would be no place in the universe left for it once all of your 0s are there! Nana: I could make the zeroes very small. Reila: Even very small zeroes would not help much when you want them to be all of the same size. But if, for example, the ith zero was of size 2 i ...

2.3.2
Examples are used to get a grasp of the new area by determining arithmetic laws. Along the way, in striving for precision, notation is developed that will outlast the attempt.
Fortune: Before Constance suffers a heart attack, let me try to get this a bit clearer: I will write s for infinitely many repetitions of a digit string s to the left. Now Nana proposes 9 C 1 D 1 0. I wonder if that works out. Nana, what is 9 C 1? Nana: That's easy: Which is what we should expect; after all, it is just 9 1 C 1, i.e., 10 1, and multiplication by 10 shifts everything by one place to the left and appends a 0 on the right. Constance: Would you please tell me which digit is "shifted by one place to the left" to yield your new leading digit with its infinitely many successors? Nana: Oh, come on, you know what I mean. You are just being stubborn. Constance: No, in fact, I don't.
Fortune: I believe Constance. The point is that she doesn't even want to know, due to her tendency to believe that everything that is worth knowing is already known. But she still has a point: A simple "shifting" does not account for the effect that Nana wants to achieve. Constance: Nor will anything else. Fortune, I respect your good will to extract meaning from even the most absurd proposals, but this is really crossing the border. There is just no reason to add this 1, nor a place for it. Nana: But it has to be there! There are all of these carries, and we cannot simply ignore them. And if we can say that there are infinitely many digits, why can't we say that there is an infinite position? Fortune: Constance, I respect your critical attitude insofar it triggers progress by forcing us to be precise, but I reject it when it is used to discourage research just because things seem unusual. Maybe Nana has something in her mind that makes sense; then it is worth investigating. It is of course always possible that things turn out to be inconsistent -in fact, your beloved traditional theories fare no better in this respect -, but that is the result of an investigation and not something we can see before we even try. Reila: Right. We have to get the picture of what Nana is either seeing or believes to be seeing; then we need to find out whether it is actually a picture of something or a mere illusion. Fortune: Not at all. I distrust your so-called "seeing"; it turns out to yield mere illusions too often to be more than a source of inspiration at best. The right way to proceed is to describe as precise as possible how Nana deals with her new objects, and then to consider whether the manipulation rules thus extracted are consistent. Reila: I disagree with you on the status of seeing things. Sure, there are misperceptions, but that does not prevent perception from being the way that things occur to us. On the other hand, your focus on formalisms may simply produce pedantic descriptions of nothing. But your method has helped as a vehicle of seeing things before, and I am willing to learn. So please go on. Fortune: Right; so back to business. Nana, that is an interesting point; so you propose to focus on the carry. Let me see. What is 89 C 01? Nana: Do you really need to ask? Clearly, that is 90. Fortune: No leading digit here? After all, we get a carry 1 infinitely often. Nana: No, because they always go away afterwards. Fortune: I see. OK, in the addition of two such numbers, no carries greater than 1 can occur. So you are saying: When the carry is eventually constantly 1 when carrying out the addition from right to left, there is a leading 1, and if it is 0 unboundedly often, there isn't? Nana: (frowns) I guess. Fortune: I see.

Reila:
No, Fortune -you do not "see" anything, if we believe in your story. That is a typical example of the weakness of your method. Now you have an addition rule, but still nothing to add with it. However, I think I can help now: What Nana seems to be considering are simply digit strings whose length are transfinite ordinals. Are you folks familiar with those?
Heads are shaken from right to left and back, hands are waved in typical "more or less"-gestures. Reila: Well, just read them up, then. It's fascinating stuff. 11 Roughly, an ordinal is just the set of its predecessors. So the first ordinal, which has no predecessors, is ¿, which we call 0, then comes f¿g, also known as "1", followed by f¿; f¿gg, which is 2, and so on. Whenever we have such an initial segment of the ordinals, we regard it as a new set to get the next ordinal; thus, in particular, we obtain the first transfinite ordinal ! D f0; 1; 2; 3; :::g; the numbers we talked about before are digit strings of length Ä !. But the ordinals go on after that: After ! comes ! [ f!g D f0; 1; 2; ::: Nana simply proposes to consider digit strings of length ! C 1. Fortune: I see. That is indeed nice. But is it enough? What about, say 10 1 0 or 10000 1 0? Reila: Oh right. That should probably be 10 0 and 10000 0. So we should look at digit strings of length < ! 2 to ensure closure under addition. So we have a structure .T; C/, where T is the set of functions from ordinals to f0; 1; :::; 9g and C is defined like the before discussed addition on the !-part and like ordinary finite addition on the transfinite part, with the extra point that, if the carry in the finite part eventually becomes constantly 1, we add another 1 to the transfinite part. Fortune: Right, that seems to be what Nana wants.
Nana looks totally perplexed.

The acceptability of an extension of a definition is discussed controversially between a conservative and a conventionalist point of view. The preservation of basic properties arises as a criterion (and fails twice).
Constance: But mathematics is still not Santa Claus who gives you what you want. All you have is a set and a function. You call that weird function an "addition", but does it have any properties of addition? Is it commutative, associative? What about subtraction? Is there, for all a, b, a c such that a C c D b or b C c D a? Probably, you want to define subtraction in a similar way to addition by "generalizing" the usual algorithm to your 11 A good place to do so would be [5], Chap. 2. transfinite monstrosities. But will this have the required property that a C .b a/ D b? Fortune: Neither is mathematics a folklore festival. Terms and symbols have the meaning we give them by definition. The new addition is what we defined it to be, nothing else. You, Constance, may be used to the word "addition" as refering to an operation having certain properties, but your conditioning is not my concern. Reila: I disagree. Given that this operation arose as and is intended to be an extension of addition, it is very natural to ask what of the properties of addition hold true in the new sense. So these are good questions. But they are not very hard to answer. Constance: OK, I'll grant you commutativity, which is indeed obvious. After all, both transfinite and finite addition are commutative, and the carries also do not depend on the order of the summands. Reila: Right; so let us consider associativity. Say we have three of these trans- where a 0 , a (possibly empty) digit string is the transfinite initial part of a and a 1 is the digit string of the last ! (or less) digits of a; and likewise for b and c. Now, like in Constance's argument, associativity works for a 0 , b 0 and c 0 and likewise for a 1 , b 1 and c 1 ; so the last ! many digits of a C.b Cc/ and .a C b/ C c will both agree with .
; so all that remains to be considered is the contribution of the carry. Constance: Excellent idea. I propose to consider the carry for the case a D 69, b D 92 and c D 26. Then .a C b/ is calculated by ::: 6 9 6 9 6 9 6 9 C ::: 9 1 2 1 9 1 2 1 9 1 2 1 9 1 2 1 ::: So .a C b/ C c gives us 1 ::: 6 2 6 2 6 2 6 2 C ::: 2 6 2 6 2 6 2 6 1 ::: And this is 1 87. Nana, why don't you compute a C .b C c/? Nana: Why would I? It is clear that the result will be 1 87, as you just demonstrated. Reila: No, Nana -we cannot just take associativity for granted, we are checking it. Let me do it: First, we obtain b C c: ::: 9 2 9 2 9 2 C ::: 2 6 1 2 6 1 2 6 ::: K And next, a C .b C c/ yields: ::: 6 9 6 9 6 9 C ::: Well, look: If we add 9 C 1, there is always a 1, but it is not the same 1: The first 1 vanishes in the second position, the second in the third and so on. So, they all vanish, but still, the leftmost one remains... Constance: ...stubbornly stuck in your head. Nana: So, I think we have, for example, 08 C 02 D 1 10, not 10, as I first thought. And then, for Constance's example, we get 1 87 for a C .b C c/, so all is well again. Fortune: I think what Nana is proposing here is a new definition of addition, in which the roles of 0 and 1 are switched with respect to the carry, compared with our first attempt: If the carry is 1 unboundedly often, we add 1 to the transfinite part, and only when it is constantly 0 we have a "transfinite carry" of 0. Constance: And again, this is a non-starter. Just let a D 65, b D 56, c D 05.
Then, by your new rules, we have .a C b/ C c D 1 726 and a + (b + c) = 2 726.

Constituting a new area
The new theory is probed from various angles. Internal consistency is identified as a criterion and used as an argument in favor of the new area. After a short moment of silence...

Nana:
Then let's do it the way Constance suggested. Constance: What did I suggest? Nana: That 9 C 1 D 0.
Constance: (eyes bulging) I didn't suggest that! I said that to convince you that your infinite 9-line is nonsensical. Nana: If I'm not mistaken, we then don't have the . because it led to a non-associative "addition". In mathematics, when a theory is extended, we have to do it in a way that ensures that the old rules still apply. And that's obviously not given in this case. So I don't see the point. Reila: I agree with Fortune, and indeed I must correct you Constance. There are loads of examples where a theory is extended and the old rules have to be revised. For instance, in the naturals a product is never smaller than any of its factors. This rule neither holds for integers nor rationals. Does this imply that we shouldn't extend the additive semigroup .N; C/ to .Z; C/ and so on? Constance: But where's the difference? A moment ago you rejected Fortune's proposal because it failed to preserve a crucial law for addition. And now you're happy to admit a strong violation of another such property. Reila: Associativity is an "internal" property of addition: We do not need any concepts besides addition itself to express it. Thus, there was no hope to obtain a satisfying theory. In contrast, the ordering seems to be external to addition. Indeed we know examples, such as the complex numbers, where addition and everything else works just fine, although no ordering is possible. So the difference is that you can separate addition from ordering but you can't separate addition from addition. We should rather care for internal consistency. Our new use of symbols evokes some expectations, or anticipations, and we should investigate whether they are fulfilled or not. Fortune: Thanks for your support, Reila. Let's go ahead. Anyway, I still agree with Constance that there should be a reasonable meaning for a theory to be worth studying. But there still could be a sensible interpretation for our new "infinite numbers"; maybe we just haven't found it yet. And of course, that meaning must be different from the usual meaning in the decimal system... Preston: And by meaning you mean some sort of application? Reila: Not necessarily... Fortune: We will never know if we keep discussing and don't start analyzing. Constance: Then go ahead, for god's sake! Reila: Before we go ahead, I'd like to mention another intriguing consistency, namely that 9 D 1 fits perfectly with the geometric series. We have :::9999 D 9 :::111 D 9 1 X kD0 10 k D 9 1 1 10 D 1; you see? Constance: And again, it's simply wrong, because you ignore the divergence of the series. Reila: But shouldn't you be happy because a rule from an old theory generalizes to our new theory? Constance: I still wouldn't call it a new theory, but must admit that your observation is at least not a bad omen.

Further trust in the new area is gained by observing that known domains can be interpreted in it and a precise semantics is established.
Reila: Good girl. Anyway, we found out that 9 seems to be another representation of 1. I wonder if there are other non-natural numbers that can be represented by our new "numbers". Nana: Of course! While you folks were discussing esoteric stuff I found out that all negative integers have such a representation. For instance, 17 equals 983, because 983 C 17 D 0. So, any negative integer, say x, is represented by 9 followed by something. And that something is 100 x if x is a two-digit number, 1000 x, if x a three-digit number, and so on. Reila: Well done, Nana. What do we want to consider next? Fractions? Fortune: That shouldn't be too hard to find out. Let's take one third for a start.
We simply have to look for a number with infinitely many digits to the left which gives 1 when multiplied by 3. The last digit, say a 1 , has to satisfy 3a 1 Á 1 .mod10/. The only possible choice is a 1 D 7, and as 3 7 D 21 this gives a carry of 2. Therefore, 3a 2 C 2 Á 0 .mod10/, and we obtain a 2 D 6. Again, we have the carry 2, and obtain a 3 D 6, and so on. Thus, 1=3 D 67. Reila: Not bad, thereby we also have 1=3 D 3, which is certainly intriguing. Let us determine 1=7. We have 7a 1 Á 1 .mod10/, so a 1 D 3. Next we have 7a 2 C 2 Á 0 .mod10/, so a 2 D 4 and the next carry is 3. Thus, 7a 3 C 3 Á 0 .mod10/, which implies a 3 D 1.
Reila is scribbling and murmuring for some time...

Reila:
That's what I thought: We have 1=7 D 2857143 and therefore, 1=7 D 142857. Preston: And that's just the same block of repeating digits as in the ordinary decimal representation of 1=7, namely 0.142857. Reila: There is a simple reason for that phenomenon. If you multiply 0.285714 by 7, you get 0,999.... and this equals 1, which explains the ordinary decimal expansion of 1=7. But that means also that you get ...999 if you multiply 285714 by 7. And as we believe in 9 D 1, we obtain that 1=7 has the same block of digits repeatedly to the left as the decimal representation of 1=7 has to the right. Fortune: Well done, Reila. And this easily generalizes, as it does not hinge on the numerator to be 1. I claim that a 1 :::a k D 0; a 1 :::a k . Reila: Can you prove it? Fortune: Sure. We have 0; a 1 :::a k D a 1 :::a k 10 k 1 , where a 1 :::a k is meant to be the number with the digits a 1 ; :::a k in decimal representation and not a product or so. Reila: Of course, go ahead. Fortune: Therefore, 0; a 1 :::a k D a 1 :::a k 1 10 k . Thus, to prove my claim, I simply have to show that a 1 :::a k .1 10 k / D a 1 :::a k . But this is obvious since a 1 :::a k 10 k D a 1 :::a k 0:::0 "ƒ‚… k .

Reila:
Well deduced. Fortune: Thanks. I'd like to summarize what we have achieved so far. Formally we consider the set L D f.a n / n2N ja n 2 f0; 1; :::; 9g for all ng. 14 For simplicity, we denote a 2 L without parentheses and from right to left as a D :::a 4 a 3 a 2 a 1 . We define addition and multiplication as in the naturals, where carries can go on "infinitely" to the left. The naturals can be identified as a subset of L, by noting that for a finite digit string a k :::a 1 we can write 0a k :::a 1 ; in particular we have 0 D 0 2 L. Each negative integer can be represented by an element of L. Thus, .L; C; / is a commutative unitary ring. Preston: It's so easy to make you happy. But an algebraic structure isn't worth anything in its own right if it has no use. Fortune: And that's simply not true. Haven't we just mentioned non-euclidean geometries? How many times in history has science benefited greatly from theories that have been developed and studied previously by mathematicians in their own right? Reila: Calm down, folks. We haven't wasted too much effort on our new theory until now, have we? What about inverse elements of multiplication? Fortune: We have already seen that there is an inverse element of 7, but I think we get problems in general. Take 2 for instance. There can't be an element a D :::a 2 a 2 a 1 2 L with 2a D 1 because this would imply 2a 1 Á 1 .mod10/, which cannot be. Reila: We could allow finitely many digits right of the decimal point, then 1=2 D :::000.5 would have its ordinary representation. Constance: Could we please stop constantly changing the prerequisites we agree on! Fortune: Constance is right, let's stick to the set L for the time being. Preston: I think among the rationals only pure recurring decimals can be represented by elements of L. Nana: Why's that? Preston: It's the same argument as for 1=2. Considering a rational decimal that is not pure repeating means considering a reduced fraction m=n with gcd.n; 10/ > 1. Now, let's assume there was an element a D :::a 3 a 2 a 1 2 L with m=n D a. If n was a multiple of 2 or 5, we'd have m D n m=n D n :::a 3 a 2 a 1 . Thus, the last digit of m would be a multiple of 2 or 5, which contradicts that m=n is reduced. Fortune: But we could still get a field if we allowed finitely many digits right of the decimal point... Constance: grunts Reila: We keep that in mind, Fortune. Right now, I have another question. We have seen different "new" factorizations of 1, namely 3 :::6667 D 1 or 7 142857 D 1. I wonder if we can also find "new" factorizations of other numbers...

Another peculiarity is discovered. The discussion whether it is a flaw or a property and whether to continue the investigation leads our friends to explicating criteria for assessing research directions
Constance: Oh, right. How about, for example, 0? Nana: You are just being silly. You can't multiply two non-zero numbers and get zero! K Reila: Let's see, Nana. Indeed, that is a good question: Can we factor 0 in some non-trivial way? In other words, are there zero divisors in L? Preston: Ok, let's start by assuming we have a D :::a 3 a 2 a 1 ; b D :: You can't just assume that! That is the weird question we are trying to answer! Preston: You're right, Nana. But the point of this assumption is not to take an answer to this question for granted, but to get the investigation going: My plan is to derive as much information as I can on these fictitious a and b; maybe we get enough information to actually find an example. And once we have that, we can forget the assumption, because the question is answered. Fortune: Of course, it could also happen that we just run into contradictions, and then we have seen that such a and b do not exist, as you suspect, Nana. Preston: Right. Anyway, let's start. We obtain immediately that a 1 b 1 Á 0 .mod10/. We want a and b to be different from 0, so let us ensure this by stipulating a 1 ¤ 0 and b 1 ¤ 0. Then we must have a 1 D 2 and b 1 D 5. Or vice versa, but let's continue with this choice of a 1 ; b 1 . Following the ordinary multiplication algorithm this gives a carry of 1, so we can conclude that a 1 b 2 C a 2 b 1 C 1 Á 0 .mod10/ in the next step. In other words, 5a 2 C 2b 1 Á 9 .mod10/.

Reila:
This congruence has several solutions, any odd digit as a 2 combined with b 2 2 f2,7g will do the job. Preston: So let's make a choice, say a 2 D 1; b 2 D 2, and see if we can find a 3 and b 3 . Constance: Hmm..., let's see if we can simplify this. Until now, we did two steps following the multiplication algorithm to find zero divisors. More general -and regardless of the algorithm -in step n we want to find two n-digit numbers a; b with ab Á 0 .mod10 n /. It's clearly sufficient that a Á 0 .mod2 n / and b Á 0 .mod5 n /.

Reila:
I see what you are getting at, Constance. We could do this inductively. Assume that we have found a D a n :::a 1 and b D b n :::b 1 with a Á 0 .mod2 n / and b Á 0 .mod5 n /. How can we find a nC1 and b nC1 such that 10 n a nC1 C a Á 0 .mod2 nC1 / and 10 n b nC1 C b Á 0 .mod5 nC1 /? Preston: For a nC1 it's pretty easy. Write a D 2 n a 0 , so 10 n a nC1 C a D 2 n .5 n a nC1 C a 0 /. As we only have to ensure that the expression in parentheses is even, we can choose any even digit as a nC1 if a 0 is even, and any odd digit otherwise. Reila: Well done, Preston. That coincides perfectly with our earlier result and generalizes easily to b nC1 . Write b D 5 n b 0 , so 10 n b nC1 C b D 5 n .2 n b nC1 C b 0 / and we have to choose b nC1 such that the expression in parentheses is congruent to 0 modulo 5. As 3 is inverse to 2 modulo 5 this means that we have to choose b nC1 according to b nC1 Á 3 n . b 0 / .mod5/. And of course there are always two such digits.
Constance: Zero divisors galore! In particular this means that there is not even an ordering of your domain of calculation that is compatible with multiplication. Nana: Why not? Constance: Consider a product ab D 0 with a; b ¤ 0. Then a; b have to be positive or negative. If both were positive the product would have to be positive as well, but it is zero. Similar problems arise whatever "signs" a and b have. Reila: We started looking for the largest natural number and ended up discovering a domain where "larger" doesn't even make sense. 15 Isn't it fascinating? Constance: Fascinating? It only shows how pointless your funny "new numbers" are. Fortune: Constance, I find it really strange how you could think that something is pointless because it is new. The existence of zero divisors is just another property of our new numbers, not a reason to reject them. And concerning the ordering, I reply with a quote from Lord of the Rings: "We have not found what we sought, but what have we found?" 16 Constance: I understand that this is best defended by a quote from a fantasy story, just as in the case of your non-associative "addition". Fortunately, I have the voice of innocence on my side in both cases: Nana made it quite clear that she intended her "numbers" to have neither violations of non-associativity of addition nor zero divisors. I mean, just look at this mess! We started out searching the "largest natural number", we found a candidate, which, however, turned out to be not quite so large, but negative. We changed the rules numerous times to make even something as basic as associativity of addition work, and as soon as we introduce multiplication, the next disaster happens. What does it require for you to accept that there is just nothing there? Fortune: A contradiction. Which is not just something that contradicts your or Nana's expectations, but an internal contradiction. Things can be arbitrarily strange, but they cannot contradict themselves. Constance: Seriously, Fortune? That is all that is required to make you happy? I can tell all kinds of crazy stories and you will just listen as long as I don't contradict myself? Fortune: You should not underestimate the criterion of consistency. Staying consistent is quite hard, just think of Frege. 15 This was actually pointed out by one of the participants of an enrichment course in which the authors treated 10-adic numbers with school students. 16 [29], p. 191.
Constance: A crazy story remains a crazy story, even if it is hard to tell it. But I wonder: Why do you even insist on consistency? Given your focus on "manipulation rules", I would expect you to go the final mile and simply accept inconsistency as "just another property"? 17 Fortune: Don't be silly, Constance. From an inconsistent system, we can easily derive anything. They are just trivial and therefore provide no worthwhile subject of research. 18 Constance: Aha! So your ultimate criterion turns out to be hedonistic: You accept that an object exists if it is in your interest to do so. Fortune: Not quite. Math is the study of formal systems. You won't hear me talking about "existence" of mathematical objects in any sense that transcends derivability in the formal system we happen to be currently working in. Inconsistent systems are, indeed, just formal systems with a certain property. My "hedonism", as you call it, only comes in when I decide not to work with them. Certainly, you will not claim that my decision to work on one mathematical topic rather than another would be part of mathematics. You, on the other hand, seem to have a much stronger concept of existence, and the single "criterion", if you can call it that, I have seen so far from you is that you only accept that X exists if you learned about X more than 10 years ago. Constance: It is not the worst criterion that something has withstood the test of time, and certainly more than just subjective interest. Ultimately, I don't really care whether the reasons for rejecting nonsense are considered to be part of mathematics or not. Well-established theories prevail and flourish, while fashionable nonsense is being proposed, gathers a brief interest for the sake of curiosity and is then put aside. Just look at your non-associative addition: No one else but you was willing to go ahead with that. The topic was dropped, just as these new "numbers" with their zero divisors will be dropped and forgotten. Right, Reila? Reila: Not quite. It is indeed a bit much to accept a non-associative operation as a generalization of addition, but structures with zero divisors are quite sensible. Just think of residue rings modulo non-primes or matrix rings. The question is, does that lead us anywhere? Preston: The real question is, is it good for anything? I am quite willing to accept very strange and even inconsistent theories when they can be put to good use. If it can, it would be crazy not to use it. It will be used, we will get used to it, and with every year, it will sound less crazy. The expositions will become more and more down-to-earth, more and more people will know about it, and after a few decades, folks like Constance will scoff at any criticism of such a well-established theory. On the other hand, if it cannot be applied, it is a pointless discussion who has the better fantasy universe. Fortune: We have to leave it for another time whether or not "applicability" is a good criterion for the quality of scientific theories. As a criterion for or against investigating a certain topic, it is no great help: We can only apply what we can handle, and learning how to handle something is just what the investigation is about. I propose to continue research rather than wasting our time on meta-discussions. Does anyone have a question for us?

Continuing the investigation reveals that the new field offers the opportunity of original and challenging mathematical work. Although the resistance against the new area decreases the question for applications comes to the fore again.
Reila: I do. We have learned that some rationals are not in L and some are. I wonder if L contains any irrational numbers. What do you think? Constance: I think it contains nothing but thoroughly irrational numbers, except for the few naturals you kindly offered a place in your realm of madness. Reila: Any constructive comments? Nana: Well, you can't just have whatever you like. For instance, p 2 cannot be in L. If a 2 L ends with an even digit, its square will be divisible by 4, while if it ends with an odd digit, the square will be odd. So a 2 D 2 is not possible. Preston: That's right. What about p 5? Let's assume we have a D :::a 3 a 2 a 1 2 L with a 2 D 5. From a 2 1 Á 5 .mod10/ we can conclude a 1 D 5. Fortune: So we need a 2 . If we want .:::a 3 a 2 a 1 / 2 D :::005 we should have .10a 2 C a 1 / 2 Á 5 .mod100/. Hmm... that doesn't look too good. We have .10a 2 C a 1 / 2 D 100a 2 2 C 20a 1 a 2 C a 2 1 Á 25 .mod100/. That was bad luck, I guess. Let's check another square root, say p 11... Constance: Hang on! I still distrust this whole business, but just trying out random numbers will certainly not get us anywhere. We should conceptualize this a little bit. Let q 2 N be no perfect square. For p q to be in L it is apparently necessary that q is a quadratic residue modulo 10 n for every n, because in each step we have to find a n-digit natural number x with x 2 Á q .mod10 n /. Fortune: Okay, let's see. By the Chinese Remainder Theorem 19 , q is a quadratic residue modulo 10 n if and only if it is a quadratic residue modulo 2 n and modulo 5 n .
Reila: But isn't that just a reformulation of the problem? We still have to find solutions to these congruences for all natural numbers. Preston: You're wrong, Reila. There's a very handy result from number theory that does the job now, namely Hensel's Lemma 20 . It immediately implies that if there is a positive integer k such that f .x/ Á 0 .modp k / has a solution x that satisfies f 0 .x/ 6 Á 0 .modp/, then f .x/ Á 0 .modp n / has a solution for every positive integer n. Here, f should be a polynomial with integer coefficients and p prime. Reila: I see.
We apply this to f .x/ D x 2 q. So if for instance x 2 Á q .mod5/ has a solution x with 2x 6 Á 0 .mod5/ we can conclude that q is a quadratic residue modulo 5 n for every n. Constance: Not bad. Let us for the time being stipulate that q 6 Á 0 .mod5/. Then we obtain that q is a quadratic residue modulo n for all n if and only if q Á˙1 .mod5/.

Reila:
Fine, but what about the congruences modulo powers of 2? As we always have 2x Á 0 .mod2/, we can't use Hensel's Lemma for them. Preston: I remember another Corollary 21 that follows from Hensel's Lemma, a variant that is tailor-made for square roots of integers modulo powers of 2. It says for odd q that if x 2 Á q .mod8/ has a solution, then x 2 Á q .mod2 n / has a solution for every n.

Reila:
That's nice! Thus, if we also stipulate q 6 Á 0 .mod2/, by remembering that odd squares are always congruent to 1 modulo 8 we obtain that q is a quadratic residue modulo 2 n for all n if and only if q Á 1 .mod8/. Constance: OK, let me summarize: We've found out that a natural number q that is no perfect square and that satisfies gcd.q; 10/ D 1 has a square root in L if and only if q Á 1 .mod40/ or q Á 9 .mod40/. Fortune: I'm not sure about the if part. Reila: Why not? Fortune: Having a finite sequence of length k for every k does not mean to have one infinite sequence. Nana: I don't understand a word! Fortune: Ok, let me explain. We have a quite interesting issue here, and I'm sure you can understand the problem. We are interested in sequences with a certain property, and Reila says that we can find such sequences of arbitrary but finite length. But what we need is to find an infinite sequence with that property. Nana: Isn't that the same? Constance: For every k there is a strictly increasing sequence of length k of natural numbers that is bounded. Do you conclude from this fact that there is an infinite strictly increasing sequence of natural numbers that is bounded? Nana: Ah, I see the difference now! Reila: (stares into space, deep in thought) 20 For a good presentation of Hensel's Lemma and the mentioned corollary see [12], pp.101-116. 21 For this corollary see also [12], pp.101-116.

Prof. König comes around the corner and beckons. 22 Reila:
Now I got it! For every k there is a sequence s k of digits of length k with the desired property, namely that s 2 k Á q .mod10 k /. Each of these has a first digit and there are only finitely many first digits, so there is one digit d 1 that occurs infinitely often. There are infinitely many sequences of length k with the desired property having the first digit d 1 , and each of these has a second digit. So there is one second digit d 2 that occurs infinitely often. Continuing in this way we construct an infinite sequence of digits d i , which will have the desired property. Constance: Well done! And what about numbers q with gcd.q; 10/ ¤ 1? Preston: I used the time you spent on figuring out the last step to think about that. Essentially we already have the right condition, see?
Preston hands a sheet of paper to the others. After a short while of silent reading everybody nods in agreement. 23

Preston:
Well done! I wonder what we can do about roots of higher degree. Reila: Well, for third roots, things just get easier because we can simply apply Hensel's lemma. For simplicity, let's again stipulate that gcd.q; 10/ D 1. Now, for some such natural number q, we want to solve x 3 q D 0 modulo 2 and modulo 5. Say f .x/ D x 3 q. If these congruences are solvable, we then only need to check that a solution x satisfies f 0 .x/ ¤ 0, i.e., 3x 2 ¤ 0 modulo 2 and 5, but modulo 2 and 5, this is just saying that gcd.x; 10/ D 1, which is clear anyway. Preston: For example, we have 1 3 Á 3 (mod 2) and 2 3 Á 3 (mod 5), so we know that there must be a third root of 3.

Reila:
Actually, any odd number has a third root modulo 2, and, since every residue modulo 5 is a third power ... Constance: ... since gcd.3; .5// D gcd.3,4/ D 1 and by the Chinese remainder theorem... Reila: ... we obtain third roots for every natural number that has no common divisor with 10. And, by Constance's remark, the same will work for any odd exponent that is not divisible by 5. But what about, for example 4 and 5? Preston: I am sure they can be brought home with quite similar techniques. But now that I have suspended my interest in applications for quite a while due to Fortune's call for patience, I wonder: After quite a bit of investigation of these new numbers, are they actually good for anything? 22 König's Lemma states that an infinite, finitely branching tree must have an infinite branch. It is an ubiquitous tool in many areas of mathematics, though it is rarely made explicit; see [24]. In this case, our tree would consist of finite sequences of digits s such that s 2 Á q (mod 10 jsj ), where jsj denotes the length of s and we freely confuse digit sequences with the decimal numbers they represent. 23 A natural number q that is no perfect square has a square root in L if and only if it is of the form q D 2 2`52k q 0 , where gcd.q 0 ; 10/ D 1, k;`2 N 0 , and q 0 Á 1 .mod40/ or q 0 Á 9 .mod40/. This follows from the above seen arguments and the fact that natural numbers that have a square root in L cannot contain odd powers of 2 or 5 in their prime factorization. The latter statement is a generalization of the fact that that 2 cannot have a square root in L (see Nana's argument on the bottom of page 13) and can be derived similarly.
Constance: Good point, Preston. I let myself be persuaded to go along with the investigation up to here, but I still think that it was only worth the effort if this new form of number representation has some useful implications in usual fields of mathematics. Fortune: Although I don't agree with you, I'm not opposing our moving the investigation in that direction now. Preston, have you already something in mind?

Tooling the toys: Applications
It turns out that the new concepts are indeed applicable. A small defect triggers rethinking the approach in a very fruitful direction.

Preston:
The whole thing looks very similar to the method of complements used in computer science. Constance: Our number representations reminded me of that, too. The difference is that the method of complements comes with the restriction to a finite subset of the rationals. With our number representation we don't have this restriction anymore. We can represent all rationals. Fortune: That's not true for the set L we agreed on, whose members are supposed to have no digits after the decimal point. We've discussed that we cannot represent any reduced fraction whose denominator contains one of the prime factors 2 or 5. Preston: That's true, so to get our hands on, let's remove this constraint. I suggest to consider all numbers that can be represented with infinitely many digits, but only finitely many may occur after the decimal point. Fortune: Alright, but there's another problem. As these representations have infinite lengths, we cannot use them for computer arithmetic, unless we truncate them at some point. So we haven't gained anything, have we? Preston: I see the problem, but there's an easy solution. If we assume in addition that the digits to the left are repeating from some point on, each representation can be written down in finite length by using Reila's arrow notation, and still the set of numbers we can represent contains at least all rationals. Reila: To be more precise, it's exactly the rationals that we consider then. Fortune: Sure? Reila: Consider such a representation, say a 1 :::a k b 1 :::b l :c: This equals 10 k a 1 :::a k C b 1 :::b l C 0.c 1 :::c r : And as we already know that a 1 :::a k D 0.a 1 :::a k is rational, all three summands are rational. Everybody nods in agreement. Preston: Okay, so we have a new number representation to do computer arithmetic on the rationals. That's a very good suggestion, Fortune. But considering that we started looking for the largest natural number and ended up accelerating machine arithmetic, don't you think that we have deserved our afternoon? Fortune: You're right. But tomorrow we should start analyzing our new number representation with a prime base. 25 Reila: I'm in. Actually, I still have a few things bugging me: We investigated and even applied this stuff, but we still don't know what it is. For example how should we interpret the fact that intuitively, the sequence 9; 99; 999; ::: seems to converge to 1? I have an idea which I would expect to advance our investigation. 26 Nana: Maybe I will stop by again. You folks are smart but sometimes a tad narrow-minded, probably could use my help. What do you think, mom? Friend: Certainly. I think that today's discussion has been a valuable lesson. Did you notice that everyone had to put aside their interests at some point for the discussion to proceed? But everybody contributed something and got something out of it: Preston bugged us to look for applications and sure enough we found some. Fortune's open-mindedness for everything that can be described by rules prevented the discussion from petering out and led to precise calculation rules. Reila mediated the exchange between Fortune and Constance, whose "extreme" positions would be less fruitful on their own, and in the end she got her interpretations. And 24 For a good survey on the applications of p-adic numbers in various fields see [22], where, among many other things, it is mentioned that p-adic numbers are employed in Andrew Wiles' famous proof of Fermat's Last Theorem. For more information on how Wiles' proof uses p-adic numbers see [7]. For a more detailed discussion of the usefulness of the here considered 10-adic number representation with respect to computer science see [10]. It is argued there that this "novel system for representing the rational numbers [...] allows exact arithmetic, and approximate arithmetic under programmer control [..., and] is superior to existing coding methods because the arithmetic operations take particularly simple, consistent forms. These attributes make the new number representation attractive for use in computer hardware." [10], p. 1. 25 For a good introduction on p-adic numbers see [8]. 26 For the definition of the so called p-adic absolute value and applications thereof, see, e.g., [8].
even Constance, in her critical attitude, contributed important counterexamples and I'm sure there will be benefits for "classical mathematics" that she would appreciate 27 . Nana: Great. But I still don't know the largest number! Funding Open Access funding enabled and organized by Projekt DEAL.
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Conflict of interest
The authors declare that they have no conflict of interest.