So to my first question: is there a proof that the null set is divisible into subsets? — alan1000

b is divisible by a if and only if there exists an integer k such that b = ak

Is there an integer k that satisfies the equation 0 = 2k? Yes; 0. So 0 is divisible by 2.

Second question: let it be given that zero is divisible. It is evident that division by two will yield a remainder of zero. But division by three also yields a remainder of zero. Is a remainder of zero therefore a sufficient criterion of even parity?

The definition of an even number is a number evenly divisible by 2. Whether or not this number is also evenly divisible by 3 is irrelevant. 6 is an even number and is evenly divisible by 3.

Third question: let it be given that zero is an even number, and the null set, consequently, is a member of the set of even numbers. But the null set is also a member of the set of odd numbers. What is the correct resolution of this paradox?

0 isn't an odd number.

Final (optional) question for bonus marks: why does The Philosophy Forum spellchecker represent "Peano" as a spelling error?

The forum doesn't have a spellchecker. It's your browser that's doing the spellchecking.

Thanks for your reply, Michael.

I accept its validity, but how does one express this in set theory? Within the context of this discussion, at least, it implies that 0/2 yields the set {0,0}, in the same way that 4/2 would yield {2,0}. This seems non-intuitive (because the null set can contain only one member) and violates the rule against duplication of elements. This latter rule suggests that {0,0} must collapse to {0}, meaning that the act of division has failed - there are not two identifiable subsets in the null set.

0 isn't the null set {}, it's the cardinality of the null set, the number of elements of that set.

(And given that, you can define 1 as the cardinality of the set that contains only the null set as a member, so {{}}; 2 would be the cardinality of the set {{{}},{}}, and so on.)

The null set has no members, so it is identical to all its subsets.

Within the context of this discussion, at least, it implies that 0/2 yields the set {0,0}, — alan1000

In the standard set-theoretic construction of the rational numbers, each number is represented by an equivalence class of ordered pairs (p,q) where the numerator p is any integer and the denominator q is a positive integer. The equivalence relation is that

(p1,q1)=(p2,q2) iff p1 q2 = p2 q1.

Hence zero is represented by the equivalence class

{ (0,q) : q is a positive integer}

Hence 0/2, which is zero, is represented by that equivalence class.

The null set (empty set) does not represent any element of the set-theoretic representation of the rational numbers. It does however come into the representation of the integers, in which the empty set represents zero.

By the way, the standard way to represent an ordered pair (p,q) in set theory is as {p,{p,q}}. This distinguishes p as the only one out of p and q that is both an element of the mother set and an element of a proper subset of the mother set. The ordered pair (p,p) is represented by {p,{p}} which the assiduous set theory student will recognise to be different from {p}.

Is there a specific operation in mathematics for which the even parity of zero is a prerequisite — alan1000

Yes. Alternating series of the form:

sum (k=0 to infinity) a(k) x^k (-1)^k

are common. The Taylor series for e^(-x) is an example. The terms of even (odd) parity are positive (negative). Hence the term with index k=0 is positive, having even parity.

Thanks for that, chaps, you have certainly cleared a few points for me there and given me something to think about. I didn't know about the spellchecker either!

0 isn't the null set {}, it's the cardinality of the null set, the number of elements of that set. — Srap Tasmaner

As I understand it, on the standard set-theoretic construction of the integers, it's both. The cardinality of a set is n just in case the set is equivalent to n, i.e. in the equivalence class of n consisting of all those sets that can be put into one-to-one correspondence with it. So the cardinality of the empty set is 0 because the empty set is 0, and it's trivially in a singleton equivalence class with itself.

Yeah, I almost went back to change what I said there about cardinality. Then I decided that what I had written was probably already too much.

One other question, if I can test your patience a bit further, The Peano axioms (at least as we have them today) tell us that the series of cardinal numbers is generated from 0 and and every number in the series inherits all of the properties of 0 inductively. But this is clearly not the case with even parity. One might reply that it's the property of "parity" itself which is inductive, but there's something uncomfortable about the idea of an inductive property which can, as it were, change its nature from one number to another. Any comments?

Since parity is not mentioned in the axioms, and is therefore not an essential property of a well-formed number, my own instinct is to argue that it's no more than a useful descriptive feature, an artefact of certain kinds of arithmetical operation.

The Peano axioms (at least as we have them today) tell us that the series of cardinal numbers is generated from 0 and and every number in the series inherits all of the properties of 0 inductively. — alan1000

I think you might be misreading the axiom of induction. It doesn't say that every natural number has all the properties of 0. It says

IF 0 has the property
AND IF n having the property implies that n's successor has the property
THEN every natural number has the property.

Is this what you're talking about? It's how induction works in mathematics. Show that you can start, and then show how you can always continue from one to the next. If you can do both of those, you get to claim you can do it for everything, which amounts to claiming it's done. Does that make sense?

Bear with me, everybody, if I'm always about six replies behind; I'm trying to keep up...

I understand the points made about the cardinality of the null set, because of course the null set does indeed contain one member - itself - and I should have kept that more clearly in mind. But my understanding (probably incorrect!) is that the null set is treatable as having cardinality 1 only in the special context of establishing a purely logical definition of the series of natural numbers.

Outside of that context, once we begin to talk about bundles of objects, relationships, and all of that exciting stuff, the cardinality of the null set assumes a value of 0 to make its properties consistent with those which we expect from the series of natural numbers as a whole. But that's something I'm not at all clear about: when the null set should or should not be counted among the cardinality of a set. For example, {x,y,z} has cardinality 3. The null set is not counted. But its power set has a cardinality of 8 and the null set IS counted.

Moving on, is it meaningful to say that we can divide into 2 equal subsets a set which contains only 1 member? And if it is, in the case of the null set, how to resolve the paradox that division by 2 yields 3 subsets of equal cardinality (if we are to stipulate a remainder of 0)? Is it not a condition of even parity that there should be only 2 subsets of the same cardinality?

Zero may well be an even number. But I sense something specious or at least simplistic about an argument such as "0 is even because 0 divided by 2 is 0, with a remainder of 0". It takes too many short cuts. It smells too much of selective attention. The traditional tests of even parity, applied to 0, turn up outcomes with anomalous features which do not occur in the case of any other natural number, and which require rationalisation. The anomalies are too glaring to be self-evidently consistent with the inductive nature of the series of natural numbers, and surely suggest a need for modification or re-definition of the tests.


In parenthesis, may I say, I like this forum. I have tried other forums before with, frankly, mixed results... I thank everybody for the courteous, intelligent and relevant replies posted.

I am not sure I can agree without further clarification, Srap.

The key word is "implies". What does it signify exactly? Does it mean "logically entails", or something less?

If we take it to mean "logically entails", and n=0, then it follows that all of the posterity of 0 has all of the properties as 0, insofar as these are essential to the definition of "number".

There are some properties of numbers which cannot be hereditary in this way, for example, the property of being <1, which applies only to 0 with no implication that it could apply to any of its successors. But such a property cannot be a part of the definition of "number", obviously.

Sorry Srap, I posted that last message accidentally while in draft.

I was working from Russell's articulation of the axiom, which says (Introduction To Mathematical Philosophy, Routledge 2000 paperback edition, p6):

"Any property which belongs to 0, and also to the successor of every number which has the property, belongs to all numbers".

I take this to mean that if P is a property of n, and also of n+1, then P is a property of ALL of the numbers, provided that n=0 is a value of n.

I did miss one important qualification. It is not enough that the property should attach to 0 alone; it must also attach to 1. If that condition is met, it attaches to all numbers, by definition.

The stipulation that the property must belong to 0, as well as to the successor of every [other] number which has the property, I assume is designed to eliminate trivial properties, such as ">10", which clearly could not apply to all of the numbers.

the null set does indeed contain one member - itself - — alan1000

No no no. You have to be clear about the distinction between "being a member of ..." and "being a subset of ..." The null set has no members. It is a set, though.

when the null set should or should not be counted among the cardinality of a set. For example, {x,y,z} has cardinality 3. The null set is not counted. But its power set has a cardinality of 8 and the null set IS counted. — alan1000

Right, because the null set is not a member of the set {x,y,z} but it is a subset. Cardinality is the number of members; the cardinality of the power set is the number of subsets.

I'm going to leave the rest of your questions here for you to work through once you're clearer about sets and membership. Keep in mind that a set might be a member of another set: the power set has sets as members. They are not subsets of the power set (except for the null set--think that through). Subsets of the power set will be sets of sets.

On induction, you're getting there. In practice, it would work like this: you show that some property applies to 0; then you show that if you assume it applies to n, it can be shown to apply to n+1. To make that second deduction, the one that shows you can continue, you want to use an unknown, arbitrary n, because you don't want to inadvertently rely on any peculiarity of the number you chose. (0 and 1 are pretty special, so you'll stay away.)

Sorry Srap, I posted that last message accidentally while in draft. — alan1000

Btw, you can't delete posts, but you can edit them after you've posted. You could replace the whole thing with "[deleted]" or something.

Thank you for your comments, Srap. Much for me to think about there. Your clarification of the distinction between "subset" and "member of a set" is particularly valuable.

For the moment, I hold to these conclusions:

(1) Neither "even parity" nor "odd parity" fits with the axiom of induction, in the sense that neither can be a property both of n and n+1.

(2) Nor can "parity" (as a general principle) be an essential property of "number", because to say that a number has the property of being either "e" (even) or "not-e" (odd) is trivial and non-informative. (But I need to think more about that; I sense that this argument is probably fallacious).

(3) Consequently the answer to the question whether 0 has even parity, or odd, or no parity, or universal parity, is not self-evident and must be argued.

(4) The usual arguments for the even parity of 0 are facile, self-serving, and question-begging. There are certain mathematical contexts where it is convenient to assume that 0 has even parity, but it does not follow that 0 MUST have even parity. If it does, it must be proved from set theory, or from the axioms of arithmetic, or better still both.

The usual arguments for the even parity of 0 are facile, self-serving, and question-begging. There are certain mathematical contexts where it is convenient to assume that 0 has even parity, but it does not follow that 0 MUST have even parity. If it does, it must be proved from set theory, or from the axioms of arithmetic, or better still both. — alan1000

I don't see why they are facile. Given a language L = {+, *, 0, 1} and the normal axioms for the natural numbers, we can define am unary predicate E(x) as (I'll write in prose in order to avoid using LaTeX):

E(x) iff there is a y such that 2*y=x.

It's an easy theorem of number theory that, for every x, 0x=0. So, as previously remarked, 2*0=0, whence (by existential generalization), there is a y such that 2*y=0, so, by definition, E(0).

You seem to be confused by the idea of division. Division (or any other mathematical operation, for the matter) should not be understood too literally, i.e. as a way of breaking down an entity into constituent parts, in such a way that every even number could be somehow broken down into two halves. Rather, as has already been pointed out, one should instead define a relation "x divides y" as holding between two numbers iff there is a (natural) number n such that n*x = y. If you want to define this inside ZFC, you will need first to define what "*" means; this is easily done (for the finite cardinals) as "|A|*|B| = |AxB|", where "AxB" is the cartesian product of the two sets (i.e. the set {<x,y> : x belongs to A and y belongs to B}). Our definition of division can now be established for the finite cardinals, which gives the result that 0 is even (given the definitions). Notice that, given this definition, since the cartesian product of any set with the empty set is also empty, our previous theorem (for every natural number n, n*0 = 0) is now just a special case of a more general theorem, according to which, for every cardinal k, k*0=0.

Finally, notice that it's not entirely uncontroversial that the natural number 0 just is the empty set. We can definitely say that the empty set models the natural number 0 inside ZFC, but it's not entirely clear if it is 0. Some would prefer to identify 0 with a sui generis abstract object (say, by employing Fregean abstraction---cf. neo-Fregean approaches to arithmetic) or with an isomorphism type or a role in a structure, or (...).

Perhaps what I have to say is childish but kindly answer my question.

From the very basic math books I've read the invention of zero is celebrated as a great acheivement in the field. I have a vague idea of what that means. For instance the trivial? solution to 2 - 2 is zero. Also I think the concept of negative numbers would probably not exist without zero to connect the positive integers and negative integers.

However, we can't divide by zero. Neither is there an accepted solution to 0^0. My understanding is division and exponents are very fundamental operations in math and so, I reckon this difficulty with zero with respect to fundamental math operations gets transmitted, like a disease, into all branches of math where zero is a number.

My question is:

Isn't zero more of a problem than a solution?

I don't see the problem. Let d(x, y) be the result of dividing x by y, i.e. d(x, y) is the unique z such that y*z=x.. Then it's true that d(x, y) is undefined when y=0, as for any x not equal to 0, there is no z such that 0*z=x. But so what? How is this a problem?

for any x not equal to 0, there is no z such that 0*z=x. But so what? How is this a problem? — Nagase

I thought not having a solution to a mathematical problem is, well, a problem itself. For instance, before zero became a number 2 - 2 had no solution. Zero was invented and now 2 - 2 = 0. Fine. However, 4 ÷ 0 has no solution. So, doesn't this take the punch out of zero's use. It solved some problems but created new ones.

Also, zero is nothing. And, mathematically, there's no solution to 4 ÷ 0. Put differently, the solution to 4 ÷ 0 is nothing. But nothing in mathematics is, well, zero. So, I shouldn't be completely off the mark in saying 4 ÷ 0 = 0.

I thought not having a solution to a mathematical problem is, well, a problem itself. For instance, before zero became a number 2 - 2 had no solution. Zero was invented and now 2 - 2 = 0. Fine. However, 4 ÷ 0 has no solution. So, doesn't this take the punch out of zero's use. It solved some problems but created new ones. — TheMadFool

But (using my previous notation) d(4, 0) is not a "problem", it is a functional term, so it requires no "solution". Suppose I define a function f on the natural numbers as such: f(x) = 1 if x is an even number, otherwise it is undefined. There's no "problem" here about f(1) which would require some kind of "solution".

Also, zero is nothing. And, mathematically, there's no solution to 4 ÷ 0. Put differently, the solution to 4 ÷ 0 is nothing. But nothing in mathematics is, well, zero. So, I shouldn't be completely off the mark in saying 4 ÷ 0 = 0. — TheMadFool

But 0 is not <nothing>, it is something, namely a number, and moreover the identity element in the additive group of integers, just as the identity function f(x)=x is the identity element in the symmetric group of (say) a given set S. Just as the identity function f(x)=x over a given set S is a distinct mathematical object, so is 0.

But 0 is not <nothing>, it is something — Nagase

Then why don't I feel ecstatic about someone gifting me $0?

Then why don't I feel ecstatic about someone gifting me $0? — TheMadFool

Suppose I gave you $1. Does that mean I also thereby gave you the number 1?

Suppose I gave you $1. Does that mean I also thereby gave you the number 1? — Nagase

But you gave me something.