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Are there any concepts which are naturally defined only for the integers and so far has resisted any attempts at extension to other fields such as rationals or reals?

Does not meet criteria:
Cardinalities of sets
n! / Gamma function
Differentiation / Fractional differentation

Community
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GM2001
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    ...I would say there is always an implicit challenge to analytically continue any function that is supposedly only defined for integer arguments to complex values. – J. M. ain't a mathematician Oct 23 '11 at 13:18
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    Would you mind removing that **AWESOME** nonsense? Seriously distracting. – t.b. Oct 23 '11 at 18:11
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    @GM2001: 40 views in 5 hours is not "exponentially". In any case I have removed it. (Everyone: Please see my comment below before voting to close.) – Zev Chonoles Oct 23 '11 at 18:14
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    @GM2001: View count is of course important for the exposure of the question, but more important is to write a good question with a reasonable title. Otherwise it's just spam... Even if every person on the planet sees your question, what good is it if no one is willing to answer? – Asaf Karagila Oct 23 '11 at 18:16
  • The notion of being divisible goes away when you jump from the integers to rationals because you jump to a field and it makes the idea of divisibility trivial. – tomcuchta Oct 23 '11 at 18:30
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    Essentially this question is the complement of this MO question. I'm not so sure it should be closed, though I will make it CW. – Zev Chonoles Oct 23 '11 at 18:32
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    @GM2001: I completely agree. It's easier to blame the crowd. – Asaf Karagila Oct 23 '11 at 18:39
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    "View count" seems like a decent answer. – Adam Saltz Oct 23 '11 at 19:11
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    By the way, GM2001, it is a common belief that closed questions get more views (as people want to see what is it that was closed) so you should be happy about that, I guess. – Asaf Karagila Oct 23 '11 at 20:08
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    @GM2001: Exactly how do you think that throwing a snit and insulting people will advance your cause? – Arturo Magidin Oct 23 '11 at 20:14
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    @IAmBrianDawkins But what if I have only read half of the question? – Phira Oct 24 '11 at 20:30
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    I have to say: The comments of GM2001 and the title thing are immmature and out of place, but I don't think that the question as it is now in its sixth version should be closed, so I am voting to reopen. – Phira Oct 24 '11 at 20:32
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    I have reopened the question. I agree with Phira, I think people were reacting to GM2001's behavior, instead of the question content. I encourage anyone who votes to close again to explain their reasoning in the comments. – Zev Chonoles Oct 25 '11 at 16:21
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    I don't see how there can be a reasonable answer to this question. Anyone who hedges that a particular discrete concept cannot be extended to the continuous world is asking for trouble, thanks to Clarke's First Law: When a distinguished but elderly scientist states that something is possible, he is almost certainly right. When he states that something is impossible, he is very probably wrong. – Srivatsan Oct 25 '11 at 16:32
  • Too add on Srivatsan's comment, I also don't see how any tag other than [soft-question] would be fitting here. It is certainly not philosophy, nor set theory or number theory. I also fail to see the connection to discrete mathematics other than the title. If anything, it is the opposite - the question requests for things which are no longer discrete. – Asaf Karagila Oct 25 '11 at 21:36
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    @Asaf What? No. It asks for things which are only discrete. – GM2001 Oct 25 '11 at 22:06
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    @GM2001: The primes and numbers of the form $\frac{1}{p}$ for a prime $p$ will generate the rational numbers using multiplication. – J126 Oct 25 '11 at 22:43
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    I want to say that the exterior derivative operation defined on differential forms will meet the criteria... could "half of an exterior derivative" be defined? What on earth could it be? What would be, say, the 1/2 exterior derivative of "dx dy" or of "f'(x) dx"? – tomcuchta Oct 25 '11 at 22:48
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    How about prime numbers, is there a similar atomic set for the real numbers for which we can generate all reals through some operation analogous to multiplication? – GM2001 Oct 26 '11 at 08:09
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    @Sriva And yet you state that a reasonable answer to this question is impossible. You should learn from your own quote. – GM2001 Oct 26 '11 at 16:00
  • @GM2001 That's a fair and nice point actually (thanks for pointing it out!), something I have been wondering about myself. In fact, to quote my own comment from the chat room a few hours back, "Perhaps someone will actually write a good answer to the question, making the comment bite the dust. It will be like Clarke's First Law coming back to bite it's own tail.". – Srivatsan Oct 26 '11 at 16:08

5 Answers5

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I've not yet seen, that the "rank of a matrix" has been interpolated to fractional ranks (but I'm surely not very profound with literature...)
(side remark: also your question focuses on whether it has "resisted to attempts" to interpolate ... such attempts may or may not exist, but to know this needed even a bigger radius of insight into literature and non-literature manuscripts...)

Gottfried Helms
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    For arbitrary matrices I basically agree. But for matrices representing projections, the rank can be computed from the matrix trace and vice versa--- and there are generalizations of the matrix trace that take values in $[0, \infty)$ (e.g. "traces" on type II von Neumann algebras). And it's not uncommon to think of such things as "continuous" analogues of (or extensions of) the "discrete" dimension function. – leslie townes Oct 26 '11 at 21:39
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Turing machine is a discrete model.

Nobody
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The Arithmetic derivative.

Mark
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    After the classical "derivative" was generalized to fractional derivatives - is there any obvious argument (or even a worked out proof), that the according generalization is not applicable to "numberderivatives"? (As I understood the question it asks for examples, where the interpolation is (or is very likely) inherently not possible due to its characteristic as containing a discrete concept.) – Gottfried Helms Oct 26 '11 at 13:59
  • @$\text{}$Mark, @Gottfried: FYI this is a special instance of a derivation. – anon Oct 29 '11 at 08:46
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I don't know if this works, because someone may actually have done something along these lines. I'm by no means an expert on mathematics in general. That said, some of the concepts of number theory may well work. For example, what would the concept of a prime number, composite number, "versatile" number (highly composite number), etc. mean in terms of real numbers (or rational numbers for that matter)? 5 has only 2 positive integer factors (other than 1), but it has an infinity of factors in just the rational numbers. So, if such an extension of the concept of prime number or composite number exists, I don't know how it works at least, and it would take some explaining.

The Fundamental Theorem(s) of Arithmetic don't seem extendable to the reals, since we don't seem to have the notion of a prime number in the reals, in the sense that some numbers exist in the real numbers which have exactly two factors.

I'll add that I know of at least two concepts which can get defined for the integers, but simply can't get extended to the rationals or reals (and never will legitimately).

For any given number n, there exists a least number o, such that o>n in the integers, where ">" indicates the usual ordering relation of "greater than". There does not exist any such number in the rationals or reals. One might say that in the integers, every integer has a distinct-least-upper-bound or distinct-supremum.

For any given number m, there exists a greatest number l, such that m>l in the integers. There does not exist any such number in the rationals or reals. One might say that in the integers, every integer has a distinct-greatest-lower-bound or distinct-infimum.

In other words, given a "direction" which either takes towards larger numbers, or smaller numbers, for any number x, there exists a "next" number "y" and a previous number "v". This does not hold true for the reals or rationals.

Doug Spoonwood
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  • What about Dedekind domains and their component prime ideals in commutative rings? They seem to possess precisely the factorization property of prime numbers in the integers described by the Fundamental Theorem of Arithmetic,as in every integral domain,its nonzero proper ideals factor into products of prime ideals. Since the real numbers are a field, they are trivially a Dedekind domain, aren't they? So this seems to extend an abstract version of the Fundamental Theorem of Arithmetic to any field! – Mathemagician1234 Oct 26 '11 at 04:27
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    @Mathemagician1234: I'm not sure if your comment on $\mathbb{R}$ being a Dedekind domain is accurate, but I know that the idealic structure of a ring is not the same as the original ring itself (unique factorization may hold in the former and not the latter for example), so you're not really addressing $\mathbb{R}$ but a 'higher-elevation' algebraic structure. – anon Oct 26 '11 at 08:57
  • @anon: While indeed you are correct the question only used $\mathbb R$ as an example. Saying that the primes are not extended to $\mathbb R$ does not mean they are not extended to rings larger than the integers, which they do, as we know today. – Asaf Karagila Oct 26 '11 at 12:33
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    @Mathemagician1234 : I don't think you understand Dedekind domains. It is true that fields are Dedekind domains, but for trivial reasons -- their only proper ideals are the zero ideal. Thus unique factorization of ideals for them is entirely trivial and without content. The interesting Dedekind domains are not fields, but rings of integers in algebraic number fields. – Adam Smith Oct 26 '11 at 18:05
  • @Adam I knew that,but I'm not that familiar with those subrings of algebraic number fields and their properties. So I stuck with a trivial case rather then to try and discuss something I really don't know much about.I thought it was sufficient to make the point-namely,that the fundamental theorem of arithmetic CAN-in a sense-be generalized to other rings. – Mathemagician1234 Oct 26 '11 at 21:02
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    @Mathemagician1234 : But it doesn't generalize to $\mathbb{R}$ or to fields except in a trivial sense. It's like someone asking you for an interesting topological space, and you mentioning the one-point space (or even worse, the empty set). – Adam Smith Oct 27 '11 at 02:02
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The principle of mathematical induction.

Reduction modulo $n$, as a map of rings. It exists only as a map of additive groups for $\Bbb{Q}$ and $\Bbb{R}$.

zyx
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    Induction can be generalized both to partial orders, transfinite ordinals and even to the real numbers. – Asaf Karagila Oct 29 '11 at 09:15
  • @Asaf: Q and R are not ordinals. There is no generalization of induction to the real numbers that give a method comparable to induction on integers, or can really be considered a true generalization. There is solution of differential equations, but this corresponds to induction for a very limited set of propositions. There are topological arguments that an open property true at one point is true on R, which is not quite induction. AC-based well ordering arguments are not usable as "induction". – zyx Oct 29 '11 at 18:15
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    I did not mean any well-ordering arguments by my previous comments. This question on MathOverflow and this question on this site is what I meant. – Asaf Karagila Oct 29 '11 at 18:28
  • @Asaf, the Real Induction principle there (e.g. in Pete Clark's notes) is the same as the open set argument and the examples use it in the same way. I don't see how the axioms for poset induction at MO allow a double induction. At the end of the day you get "topological" induction with the same set of examples and essentially the same proofs as existed without induction. – zyx Oct 29 '11 at 19:14