Zero is even? How about roulette?

Cecil's Columns/Staff Reports
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u and v are not necessarily equal. For instance, both 1 and -1 are units in the integers.

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RM Mentock I know that 1 and -1 are units. If you look at what John Stracke posted his lesson

I think you can see if u is 1 and v is -1 then pp is not 1[sup]1[/sup](-1)[sup]-1[/sup]p[sup]2[/sup]. That is in his factorazation of pp, u and v must be the same unit. I did not mean all units are equal.

Virtually yours,

DrMatrix

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There is an argument that 0 is not one of the “evens” for the same sort of reason that 1 is not one of the “odds”. While the set of integers, for example, appears to have alternating even and odd numbers, with 1 odd and 0 even; a person might define the set of plural natural numbers as split into even & odd halves, this way:
“Even” numbers can be divided into two equal parts. “Odd” numbers can be split into two parts iff those parts are unequal.

I read once (IIRC, and don’t know whether to trust the source) that some ancient thinker used these definitions for, well, even and odd, basically–1 and 0 got left out.

Thus TWO is the first even number, and numerologically feminine(?); and THREE is the first odd number, and numerologically masculine(?).

Yes, numbers can be interpreted in weird ways.

Some of them more frightening than others…

“In my nightmares I am chased by algorithms”–crewman Celes, ST:V

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foolsguinea
If you consider zero to be a number then why can’t zero be split into two parts with zero in each part? or split one into two parts first part with one and the second part with zero?

I know what you mean about frightening. Numerology and people who take it seriously scare me!

Virtually yours,

DrMatrix

26

For you algebraists out there: You all are forgetting what the definition of a prime really is. Sure, it used to be something about divisors of positive integers, but the full definition has nothing to do with the natural numbers. Here it goes:

Start with a ring. That’s a set R with two binary operations (+ and *). R is an Abelian group over +. * has several properties:

There is an element 1 such that x1 = 1x = 1 for all x in R.

  • is associative.

  • is distributive over +; that is, for all a, b, c in R:
    a*(b+c) = ab + ac
    (a+b)c = ac + b*c

Now, we define the term “ideal”. Let I be a non-empty subset of R. Let a, b be any members of I.

I is an ideal if:
xa and ax are members of I for all x in R.
a+b and a-b are members of I.

I is a proper ideal if it is a proper subset of R.

The ideals of the integers are usually written as iZ, where i is a non-negative integer, and Z denotes the set of integers. By definition, iZ = {x in Z | x = i*y for some y in Z}. (This is equivalent to the statemenet that Z is a “principle ideal domain”.) So, the ideal 0Z is given by the set {0}. The ideal 1Z = Z is the set of integers, {0, 2, -2, 4, -4, 6, …}. This should clear up the question about whether 0 or negative numbers can be even. In ring theory, they can!

Now on to primes. A prime ideal P is a proper ideal (i.e, not the ring itself!!!) with the property that if ab is in P, then either a or b is in P. Thus, 0Z = {0} is a prime ideal, since if a and b are integers, and ab = 0, then a=0 or b = 0 (or both). However, the ideal 4Z = {0, 4, -4, 8, -8, 12, …} is NOT a prime ideal, since -2*2 = -4, which is a member of 4Z, while neither 2 nor -2 are members of 4Z.

What, then, is a prime number? Well, that’s simple: in the integers, p is a prime number if pZ is a prime ideal. Note that this means that 0 is prime, since 0Z = {0} is a proper ideal of Z, and has the property that a*b is a member of 0Z only if a=0 or b=0. 1, however, is NOT prime, since 1Z = Z is not a proper ideal.

So, there you have it. The prime numbers are as follows: {0, 2, -2, 3, -3, 5, -5, 7, -7, 11, -11, …}. How 'bout them apples. Of course, conventionally (for general applications), we only consider the positive primes, since 0Z is a trivial case, and -aZ = aZ, which makes the negative primes redundant.

So, there is the number theory answer (as best I remember it - it’s been more than two years since I took a course in algebra – I looked this up on the Web at http://www.math.niu.edu/~beachy/abstract_algebra/index.html)..)

  • Joe.

p.s. I hope I didn’t make any mistakes. If I did, go check out any basic abstract algebra book, such as the one I mentioned, or the classic “Abstract Algebra” by Herstein.