How do you order numbers?

Answer 1

This is a great question!

First off, we have to define what is meant by ordering numbers. It turns out that there is no possible way to prove that numbers even can be ordered, we simply must assume that they can be. Mathematically speaking, this means that the ordering of numbers must be an axiom, or an unprovable statement which is considered to be the truth.

This is the most famous axiom in modern mathematics, and also the most controversial. It is known by several different names: the axiom of choice, Zorn's lemma, Zermelo's theorem, and the well-ordering theorem, to name a few. It is also the ninth and final axiom of ZFC (see link below), which is the axiomatic set theory that we base our entire system of mathematics on.

The axiom of choice basically says that a set is defined as being well-ordered by a strict total order, if every non-empty subset of the set has a least element under the ordering. OK, there's our definition of order, but how do you actually construct a strict total order? Well, I'm going to show you how below, but it involves a little set theory. If you're unfamiliar with basic set theory, follow the corresponding link below.

First we need to make a partial ordering. Fortunately, that's already been done for the numbers that we use via the relation, "less than or equal to," symbolized as ≤. Here's how the partial ordering is made:

Let x, y, z Є N, where N is the set of natural numbers (0, 1, 2, 3, 4, ...). If N is partially ordered under the relation ≤, then the following three rules must hold.

1) x ≤ x for any x Є N

2) If x ≤ y and y ≤ x then x = y for any x, y Є N

3) If x ≤ y and y ≤ z then x ≤ z for any x, y, z Є N

Now, in order to turn this partial order into a strict total order, only one more thing is required.

If N is to be considered a strict total ordering under the relation ≤, then either x ≤ y or y ≤ x for all x, y Є N.

So, finally, is N well-ordered? Before I answer that, I'm going to give a quick example of a well-ordered set.

The set {1, 2, 3} is well-ordered under the relation ≤. Why? Well, let's look at every possible subset of {1, 2, 3}: {Ø}, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, and {1, 2, 3}. For {1, 2, 3} to be well-ordered, each one of these sets must have a "least" element under the relation ≤. Well the first four sets listed are trivial since they either have 0 or 1 element in them. The next three listed all have a least element, specifically 1, 1, and 2 (1 ≤ 2 but 2 is not ≤ 1, thus 1 has a lower ordering than 2, for example). Finally, the set {1, 2, 3} has 1 as a least element. So, every subset of {1, 2, 3} has a least element, therefore, by the definition written above, {1, 2, 3} is a well-ordered set under the relation ≤.

Well, it should be obvious that N is a well-ordered set since N is basically the set {1, 2, 3} extended out to infinity (with 0 included). I'm obviously not going to list every subset of N in order to show that they all have a least element, but if you'd like to try and find one that doesn't, be my guest.

Lastly, I'll toss you a (hopefully) curiosity-inducing bone. It turns out that the ordering of N can be extended further, into the realm of group theory in fact. Besides what is written above, N is also the only naturally ordered semigroup in mathematical existence, within isomorphism of course.