Devlin's Right Angle, Part VI
I fell far short of my goal of making myself completely clear at the end of that post, so I want to see if I can rescue the clarity of the presentation somewhat—for myself, if no one else.
Anyway, here it is, again:
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.It is very important to remember that we're not proving anything here, so to speak. As I mentioned in my previous post, we can use the Induction Principle to find out whether a statement like 1 + 2 + 3 + . . . n = n(n + 1) / 2 works for all natural numbers. We check that it works for 1, then we check whether or not it works for some b + 1. If the statement passes those tests, then we know that it works for all 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.
With an operation like multiplication, the same idea applies. However, with multiplication (or addition), we first have to have some guess about how that operation works on 1 and some b + 1. You can see this in (1) and (2) above. The first statement, (1), imagines how the operation applies to the number 1, and (2) imagines how the operation applies to some b + 1. But after we send those statements through the tests of the Induction Principle, the only relevant conclusion that will come out is whether or not they describe an operation that applies to all natural numbers. That's it.
So how do we take those hair-brained statements, imagining some operation called multiplication in the natural numbers, and show that they do indeed work for all natural numbers?
Here's how: First, we just imagine that we're right on the big idea here and that the operation of multiplication (a × b, where both a and b are natural numbers) does indeed apply to all the natural numbers—or, as I said before, the operation of multiplication "scoops up" all natural numbers. Second, we assume that the statements we made in (1) and (2) are true. And these statements tell us how multiplication applies to both the number 1 and some b + 1. Third, we need to show that, given these assumptions, our "multiplication jar," as it were, could contain the number 1 and some b + 1, which would prove that our jar contained all of the natural numbers.
(1) says that a × 1 = a. And since we've assumed that that is true and that a × b scoops up all natural numbers, which we assume are in our jar, then 1 must be a natural number (because we can just substitute b for 1), and, therefore, must be—or, rather, could be—in our jar.
(2) says that a(b + 1) = (ab) + a. Since we've already assumed that a × b scoops up all the natural numbers, obviously (b + 1) can be substituted for b, which means that b + 1 is a natural number and could be in the jar. The only problem is that one must show that a(b + 1) is "uniquely defined," which means that you must show, using axioms of addition already settled upon, that (ab) + a is uniquely defined, which it is, because of the uniqueness of addition. We do, indeed, need to rely upon addition when we define multiplication for the natural numbers.
However, there is no reason to believe that repeated addition is criterial. It should be noted that nowhere--nowhere--in the "definition" above was it necessary to make use of repeated addition to define multiplication. Nowhere.
The Thorn
From what I've seen, what most people get stuck by is the statement in (2) above--a(b + 1) = (ab) + a--what is known as the distributivity of multiplication over addition. So, let's say a = 4, and b = 5. We know that the product of 4 and 5 is 20, right? What the statement above says—if you read it cinemathematically--is that if you want to multiply 4, or a, by the next natural number after 5 (b + 1, or 6), you can multiply 4 and 5 and then just add another 4 [(ab) + a]. In other words, to find 4 × (5 + 1), first find 4 × 5, then just add 4.Now, of course, this "works" for all natural numbers. And, if you think about it, you could just continue this process repeatedly, using the second part of the definition for multiplication I showed above. In other words, you could find 4 × (5 + 2) by, again, first finding 4 × 5, and then adding 4 and then another 4. The product is 28, bravo!
Hey, you know what we could do? We could basically extend this definition—that is, we can say that "all anyone means when they say" a(b + 1) = (ab) + a is just (a + a + a . . . )btimes! That way, we can say that multiplication is just, by definition, repeated addition, right?
NO. NOPE. NYET. NOPEY-NOPE-NOPE.
Of course, the idea of repeated addition follows very directly and immediately—as a way to calculate multiplication problems—from a definition of multiplication on the natural numbers, but IT. IS. NOT. THE. DEFINITION. OF. MULTIPLICATION.
As Nunes and Bryant (1996!) argued (I'll get the source up soon):
Although there is a conceptual discontinuity between multiplication and addition, there is a procedural connection between these operations. Because multiplication is distributive with respect to addition, repeated addition can be used as a procedure to solve multiplication sums.
Here's another good quote from the same source (again, I'll put it up soon; keep in mind that this is from '01):
The English National Numeracy Strategy (DfEE, 1999, p. 14) suggests that pupils should be taught to understand multiplication as repeated addition. In contrast, the Japanese Association of Mathematical Instruction proposes that "repeated addition is a way to calculate multiplication, not a meaning of it" (Yamonoshita & Matsushita, 1996, p. 291).
Reference:Park, J. & Nunes, T. (2001) "The development of the concept of multiplication." Cognitive Development 16 (2001) 763-773
Part I | Part II | Part III | Part IV | Part V | Part VI
Labels: education, mathematics, research, textbooks





