# Natural numbers

by tauon
Tags: natural, numbers
 P: 84 what is the convention you adhere to when it comes to natural numbers? for example there is a long standing debate about 0... should we define $$\mathbb N = \{0,1,2,...\}$$ or instead $$\mathbb N = \{1,2,3,...\}$$ and more about this, considering Peano's Axioms than we could choose $$\mathbb N =\{-7,-6,-5,...,0,1,2,3,....\}$$ -define the successor function $$\phi$$ as $$\phi (n) = n + 1$$ and "verifying" the axioms for this set is quite easy. so is there any real point(!) with the natural numbers? :D
 PF Patron Sci Advisor P: 1,767 There is, I think, no consistent convention. I prefer including 0 in N and calling {1,2,...} the "whole numbers" since you can't have "a whole nothing". Specifically, addressing your point about Peano's construction you start with 0 and the successor function and define the naturals as identified with the power of that function applied to 0. $$n \equiv \phi^n(0) = \phi(\phi(\phi(...\phi(0)...)))$$ (with n instances of phi) Since we include our starting point in our set we would allow that zeroth power. Invoking an inverse successor and axiom that the system is closed under it as well we get all the integers similarly defined notationally via symbolic powers. Your point about starting at -7 is just a relabeling of the starting point but breaking the power identification above. This identification really relates cardinal numbers (how many instances of phi) with the ordinals (how far past our starting point) and we see the distinction is in whether we are counting positions or counting steps. I think the main issue is whether we are using the natural numbers as cardinal or ordinal numbers. I would suggest the following convention and notation: The "natural cardinals" are $$\mathbb{N}=\{0,1,2, \ldots\}$$ since these are all possible cardinalities of finite sets. The "natural ordinals" are $$\mathcal{N}=\{1^{st},2^{nd},3^{rd},\ldots\}$$ since we don't want to bastardize the semantics by speaking of a zeroth element. As to their "point" yes, the main point is their use as a standard index set for countably infinite sets. As such either is fine but some formulas may appear simpler with either case as the index set to say avoid (n-1)'s or (n+1)'s in the expressions. Typically if you're going to invoke them by name or symbol you should footnote a clarification of your preferred convention. BTW: A typical construction of the cardinal numbers is to define the empty set as 0 and then n = {0,1,2,...(n-1)} defined recursively. The more universally accepted definition of the cardinal numbers is as the equivalence classes of sets of the same cardinality (using existence of a bijection as an equivalence relation). This can be troublesome because it invokes a "set of all sets" out of which these equivalence classes are pulled and that gets too close to Russell's paradox. At the very least you must first construct all sets before you can construct the cardinals. The nice thing about the naturals (including 0) is it is also the first transfinite cardinal in the above direct construction. The problem however is that this construction can't get past aleph0 since it is fundamentally ordinal in process. Hence the "equivalence class" definition better generalizes past the finite. The troubles are dealt with as I said by carefully constructing all your sets first then constructing their cardinality classes.
 P: 56 I don't think there's really any debate about it. It really doesn't matter whether or not you include 0, as long as you're consistent. As for Peano's Axioms, one of the axioms is to assume 0 (an arbitrary constant symbol) is a natural number. If you want to reaximoize (yes, i invented a word) to assume -7 is a natural number, and then use the successor function, that's fine.
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Emeritus
P: 16,094

## Natural numbers

 Quote by tauon and more about this, considering Peano's Axioms than we could choose $$\mathbb N =\{-7,-6,-5,...,0,1,2,3,....\}$$ -define the successor function $$\phi$$ as $$\phi (n) = n + 1$$ and "verifying" the axioms for this set is quite easy.
Even the axiom that says there is an element that isn't a successor? And what about the induction axiom?
 P: 403 It's true that many, apparently distinct, structures satisfy the Second-Order (where the induction axiom says "For any set X...") Peano Axioms; all three that you mention do (I would advise against the third one, because the negative integers are construted from the naturals). The point is that all these structures are isomorphic; therefore mathematically indistinguishable, so you must look for reasons in the definition of $$\mathbb N$$ from more basic entities (sets), or in their use in constructing other mathematical entities. So, regarding the inclusion of the zero or not, I prefer to include it, because: (1) If you construct the naturals from set theory, then you start from the empty set, which is taken as 0, and obtain the others by applying successive applications of the successor function. (2) Also from the Set Theory point of view, the naturals are a very special type of sets: the finite ordinals, and the first of these is the empty set (cardinals are a particular type of ordinals, not the reverse) so, it makes sense to start from zero to have a neat theory for the ordinals sets (that continues beyond $$\mathbb N$$ into the transfinite). (3) When you construct ($$\mathbb Z$$) from $$\mathbb N$$, it's easier (and neater) if you already have the 0. Why the two distinct "definitions"? The reason is historical: people who work in Analysis usually leave the 0 out, because they don't really mind how things below $$\mathbb R$$ are defined and they don't regard 0 as a "natural" quantity (this is how Poincaré, one of Peano's greatest detractors, viewed things). On the other hand, people more close to foundational issues definitely prefer the one with the 0, by the above reasons (and a few more).
P: 84
I'm sorry for the very very late reply (I've been busy).

 Quote by Hurkyl Even the axiom that says there is an element that isn't a successor?
of course. $$Ran \phi=\{-6,-5,...,0,1,2,...\}$$ i.e. $$\forall n\in\mathbb N, \phi(n)\not = -7.$$ qed
 Quote by Hurkyl And what about the induction axiom?
you only need to prove that the set is well-ordered , then the induction axiom follows from that.
P: 84
 Quote by JSuarez So, regarding the inclusion of the zero or not, I prefer to include it, because: (1) If you construct the naturals from set theory, then you start from the empty set, which is taken as 0, and obtain the others by applying successive applications of the successor function.
however ultimately irrelevant it may be (since we can always "shut up and compute"), I am inclined to refuse to call the set constructed starting with the empty set as a set of natural numbers (although it's all fine with the peano axioms)... "no elements" isn't really a number of elements, definitely not what we (or should I say "I" instead) intuitively think about when we ponder about natural numbers.

I like to say it's a set of ordinals (as you also pointed out), $$\omega$$

but again, perhaps this is ultimately pointless convention on my part (the inverse may also be true).

 Quote by JSuarez (3) When you construct ($$\mathbb Z$$) from $$\mathbb N$$, it's easier (and neater) if you already have the 0.
hmmm,
consider this construction

on $$\mathbb N \times \mathbb N$$ we define the equivalence relation $$\equiv$$ as:

$$(m,n)\equiv(m',n') \Leftrightarrow m+n'=n+m'$$

$$\mathbb Z$$ will be $$\frac{\mathbb N\times\mathbb N}{\equiv}$$ with addition and multiplication defined as:

$$\hat{(m,n)}+\hat{(m',n')}=\hat{(m+m',n+n')}$$
$$\hat{m,n)}*\hat{(m',n')}=\hat{(mm'+nn',mn'+nm')}$$

with $$\hat{(1,1)}\equiv 0$$ and $$\hat{(2,1)}\equiv 1$$

I think this is a pretty straightforward and neat construction without 0 as a natural number (appearing only for the integers)...
 P: 84 anyway, thanks everyone for the input... I guess I'm just circling around conventions.
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Emeritus
P: 16,094
 Quote by tauon of course. $$Ran \phi=\{-6,-5,...,0,1,2,...\}$$ i.e. $$\forall n\in\mathbb N, \phi(n)\not = -7.$$ qed
Oh whoops -- I misread and thought your set was left-infinite as well.
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The best way (and the actual definition of $$\omega$$) is the set of all finite ordinals, and $$\emptyset$$ is a finite ordinal. Now, given $$n \in \omega$$, every initial segment of it (that is, any $$m<n$$) will also be a finite ordinal, so it must belong to $$\omega$$, and $$0$$ is one of those initial segments.