Inherently discrete concepts

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

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Would you mind removing that **AWESOME** nonsense? Seriously distracting. –  t.b. Oct 23 '11 at 18:11
@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
@GM2001: I completely agree. It's easier to blame the crowd. –  Asaf Karagila Oct 23 '11 at 18:39
@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
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

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...)

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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

Turing machine is a discrete model.

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I don't see how this could work for this question, since fuzzy theorists have talked about "fuzzy Turing machines". –  Doug Spoonwood Oct 26 '11 at 20:23
Given that nothing seems to stop fuzzy theorists from introducing "fuzzy X", where X is anything, maybe the answer to this question is that there are no inherently discrete concepts. –  leslie townes Oct 26 '11 at 21:41
@leslietownes You are right, or no, maybe. –  Hagen von Eitzen May 22 at 21:17
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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 8:46

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.

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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 4:27
@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 8:57
@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
@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
@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 2:02
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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 $\Bbb{R}$.

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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 9:15
@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
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

Frankly, I'm really unsure what you mean by "discrete only". I mean, in principle any set can be given a topology, so obviously discrete in the sense of noncontinuous only functions isn't possible.

Also, any set can be partially ordered with a properly chosen relation by Axiom of Choice or even stronger, well ordered, by the well ordering theorum. So this property of the integers (the positive integers only for the case of well ordering) be extended arbitrarily to any set within any model of set theory that accepts the Axiom of Choice. (The complex numbers under the usual orderings are not partially ordered, but in principle, we could impose an ordering.)

Combining these 2 concepts gives us the concept of a net, which allows us to generalize the concept of a sequence to arbitrarily large sets with domains that are partially ordered.

So to be honest,I don't see how we can have "non-discrete" sets of the kind you describe.

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-1 : This does not seem at all relevant for the question. –  Adam Smith Oct 26 '11 at 2:13
@Mathemagician1234: Which concept in your answer is one that a) is usually only defined for the natural numbers or integers and b) has resisted attempts to extend the definition to a larger domain? (Also, please correct your spelling of theorem.) –  Zev Chonoles Oct 26 '11 at 3:56
@Mathemagician1234 : You are wrong about discreteness. It has many meanings, but none of those meanings have anything to do with continuity. For subsets $S$ of a topological space $X$, you say that $S$ is discrete if the induced topology on $S$ is the discrete topology. This is certainly true for the integers (inside $\mathbb{R}$ or $\mathbb{C}$). –  Adam Smith Oct 26 '11 at 18:07
The current version of this question does not say anything about discreteness. The title really shouldn't. This answer should at least get updated in light of that fact. –  Doug Spoonwood Oct 26 '11 at 20:34
@Mathemagician1234 : If you think it is such a bad question, then why bother trying to answer it? And please stop accusing me of downvoting you multiple times. I have only one account on math.se, and thus can only vote once per post. –  Adam Smith Oct 27 '11 at 2:04