Devlin's Right Angle, Part V

I think Mr. Devlin, in his follow-up article on the subject, provides the best transition from my previous post to this one (emphasis mine):

Over a century ago, mathematicians finally learned to sidestep that unanswerable "What is it?" question by adopting the axiomatic approach, where you simply specify the properties of numbers and the arithmetical operations, and concentrate on manipulating them according to those rules. . . .

At the turn of the twentieth century, an Italian mathematician called Peano did formulate what are often called the Peano axioms, but their purpose is to show how the positive whole numbers can be defined from first-order logic; they are not a descriptive axiom system that tells you how to work in the system.

Now, what Joe and Myrtle refer to when they call upon Peano and Edmund Landau to defend multiplication as repeated addition is probably almost exactly like what I will present below. Yet, I do not see the necessity of the concept of repeated addition in this definition—yes, it's still basically just a definition. And, after reading my previous post over again, I'm not convinced that Joe and I disagree all that much--at least mathematically--as he puts the word repeated in parentheses and describes repeated addition to represent multiplication as "informal."

So, let's start with a definition, taken from Mikusinski, but which I will translate for the uninitiated:

Denote by 1 the least element of N. If x is an element of N, then the least element of the set of all natural numbers greater than x is denoted by x + 1.

Translation: Let's just say that 1 is the least number of all the natural numbers. If some number (let's call it x) is a natural number, then we'll use the expression x + 1 to mean the natural number that comes right after x. So, essentially, the natural numbers start with 1 and go up by 1's forever, and there is no natural number between x and x + 1.

All right. Now on to the big show, the Induction Principle Theorem, for which I won't be showing a proof because we skipped over a lot to get here:

Let X be a subset of N such that

(a) 1 is an element of X;

(b) If x is an element of X, then x + 1 is an element of X.

Then X = N.

Translation: We have essentially defined the natural number system as starting with 1 and going up by 1's forever. So, imagine you could scoop up a bunch of natural numbers in a jar—you don't know how many. (We use N to refer to all the natural numbers and X to refer to what you scooped up.) What the Induction Principle Theorem says is that if you can prove that

(a) The number 1 is in your jar, and that

(b) No matter what natural number you name, both it and the natural number that comes right after it are in the jar,

Then your jar contains all the natural numbers.

I know that, in some sense, this all seems ridiculous—hey, and in some sense, it is. But I think the example shown here is a good example of how the Induction Principle can be used to check whether or not a statement applies to all natural numbers. In brief, you must check that the statement is true for the least element (1), then, given some natural number m, you must check that the statement works out for m + 1. Presto. It works for all natural numbers—certainly better than guessing.

When you're dealing with operations, like addition and multiplication, it's both different and the same. Different because you can't just throw an operation, pure and unadulterated, at the Induction Principle and ask, "Hey, does it work?" And the reason you can't do that is not because it's a stupid question—it's not. The reason you can't do it is because there is no such thing as an operation, pure and unadulterated. You have to define it first. Same because you still have to deal with the least element (1) and some x + 1 when you're dealing with the natural numbers.

So, let's make the following assumptions about the product of a and b (Note: These are assumptions. And they are not assumptions about "multiplying;" they are assumptions about the result of multiplying. And they are assumptions one would have to make, given the definition of the natural numbers.):

(1) a × 1 = a
(2) a(b + 1) = (ab) + a

Now, here's the freaking crazy part. We assume that (1) and (2) above are true. Then we simply imagine that we already have a jar that contains all the natural numbers, scooped up by the operation of a × b, with b being a natural number. All we have to do in order to show that what we have scooped up is indeed all the natural numbers is to find the number 1 and to find some b + 1.

According to our assumption, then, and (1), the number 1 is in our jar. That is, since we assume that a × b scoops up all natural numbers, and we assume that a × 1 = a, then 1 (substituted for b) is a natural number and must be in our jar.

Update: I'm deleting the part of the post that went on to describe how we find the b + 1, because it sucked.

Oy. More to come.

Part I | Part II | Part III | Part IV | Part V | Part VI | Finale