Date: Jan 2, 2013 2:43 PM Author: Paul A. Tanner III Subject: Re: A Point on Understanding On Tue, Jan 1, 2013 at 8:18 PM, Joe Niederberger <niederberger@comcast.net> wrote:

> PT III says:

>>To provide the reader with some context so that he or she can see Keith Devlin meant, to see that he is not confused at all, here is part of what Keith Devlin actually wrote:

>>"What Exactly is Multiplication?"

>>http://www.maa.org/devlin/devlin_01_11.html

>

> This is pretty funny and I can't pass it up. Following the above ref. to one of Keithe's columns, we find:

> Keith D. says:

> - --------------------------------------------------------

> The MIRA fallacy becomes very apparent when you consider my second example, where I take an elastic band of length 7.5 inches and stretch it by a factor of 3.8. The final length of the band is 28.5 inches. But what are the units? What goes after the number 3.8 in the calculation

>

> [3.8 - - -] x [7.5 INCHES] = 28.5 INCHES ?

>

> The answer is nothing. It has no units. In this case, the 3.8 is a dimensionless scaling factor.

> - ---------------------------------------------------------

>

> He is writing far too fast here to consider his words. If no units are involved ("it has no units") then stretched inches of rubber band and un-stretched must be exactly the same to any commercial buyer. Same with puffed rice and uncooked rice. An inch is an inch. A cup is a cup. If I sell you 1000 yards stretched inches of rubber band you have no recourse if you expected 1000 yards UN-stretched. The very "nature of multiplication" declares you have no case!

>

> Very funny indeed.

>

> Cheers,

> Joe N

?

If there is something funny here, then Devlin did not write it.

Devlin is simply saying that in some instances we can view one of the factors as a scaling factor that has no unit.

There really is nothing wrong with this since a quotient expression with units whose only units are such that we have the same unit in both the numerator and denominator is equal to an expression with no unit. By the substitution property of equality, we can use a parameter with no unit. See this as an example:

"About EVERYDAY MATHEMATICS: A Parent Resource Manual"

[Since the address is too long to give here without widening this page too much, go to the Google search engine and enter the above, with quotation marks, and click on the pdf link, which is one of the top hits.]

Quote on page 20:

"In multiplication scaling situations, a quantity is multiplied by the

ratio, which is called a scalar or scaling factor. The diagrams show that the scaling factor has no units. The unit of the product is the same as the unit of the other factor."

If the product is to have dimension 1 and not dimension 2, then we cannot use the same single unit for both factors, since multiplying the two units gives a squared unit, a product of dimension 2. This gives a motivation for why in some instances we can view one of the factors as a scaling factor that has no unit.

Back to what I wrote:

I reiterate everything I said in

"Re: A Point on Understanding"

http://mathforum.org/kb/message.jspa?messageID=7948301

and most especially this:

"Fact: For every ringoid that is under the non-dual distributive

property, because of the asymmetry of the non-dual distributive

property, the asymmetries in question exist and we can derive the

asymmetries in question using only algebraic properties.

To see these derivations, see my post

"Re: A Point on Understanding"

http://mathforum.org/kb/message.jspa?messageID=7945385

in which not only to I quote what Devlin actually said, I link to my post

"Re: Bringing the Discussion to Order"

http://mathforum.org/kb/message.jspa?messageID=7043327

in which I derive the behavior of multiplication as repeated addition

for every ringoid under the barest minimum of algebraic properties,

and adding more and more properties until we reach a field, that

barest minimum starting with the ringoid having only a multiplicative

identity, where, for all b in a specific subset of the set of all sums

in the ringoid and for all a, I derive in that minimal context the

equality

ab = a + a,

multiplication behaving as repeated addition.

In fact, even just the distributive property itself, which is the only

algebraic property that is part of the definition of a ringoid, is a

form of one operation as another operation.

For every ringoid that is under the dual distributive property and

that contains (ever merely) an additive identity, we can derive

a + b = aa,

addition behaving as repeated multiplication, using the same method I

used in that post above but with things appropriately reversed.

To sum up: For any ringoid (including the set of real numbers), there

is an asymmetry in the behavior of the two operations if and only if

the defining properties of the ringoid are such that they define some

asymmetry in question or are such that we can derive some asymmetry in

question. (The defining property most important in this regard is the

defining property of the distributive property: If it is an

asymmetrical non-dual distributive property, then we have the

asymmetries in question, and if it the symmetrical dual distributive

property, then we do not have the asymmetries in question unless there

is some additional defining properties of the given ringoid that

defines some asymmetry in question or from which can derive some

asymmetry in question.)"

And I reiterate everything I said in

"Re: A Point on Understanding"

http://mathforum.org/kb/message.jspa?messageID=7948471

and most especially this:

"I showed in my post above including the links to the proofs in question, such facts as there being a set of axioms including some definition of some operation from which we can derive some particular ringoid does not negate the fact that in all ringoids, one operation is not more fundamental than the other one - the fact that in all ringoids, the existence or nonexistence of the asymmetries in question between the operations (including when one operation can or cannot be derived from the other operation) can be shown to be entirely determined by the existence or nonexistence of asymmetries in the algebra."

Message was edited by: Paul A. Tanner III