
Re: Simplifying Algebraic Expressions with Subtracted Expressions
Posted:
Nov 14, 2013 6:47 PM


On Thu, 14 Nov 2013 14:56:12 0700, Joe Niederberger <niederberger@comcast.net> wrote:
> I'm not arguing that your interpretations are not just fine  I just > wonder if that's what she meant. She seems to be saying there is no > manipulation of "the unknown quantity" (x) in an algebraic fashion, > either mentally or on paper. I'm skeptical.
I don't think "manipulation of 'the unknown quantity'" is what makes algebra.
In Ma's solution, there's no *formal* manipulation of a symbol that used consciously to represent an unknown number. It's the conscious formal, symbolic representation of an unknown, together with its subsequent manipulation, that turns the solution into an algebraic one.
Manipulating one or more unknowns, in and of itself, isn't algebra. Representing the unknowns symbolically, and then manipulating those symbols without having to think about what they represent is algebra.
Consider the problem: "A sundae with a cherry costs six dollars. The sundae alone costs five dollars more than the cherry. What does the cherry cost?"
I can say: The cost of the sundae alone is five dollars plus the cost of the cherry, so the six dollar cost of the sundae with a cherry is five dollars plus twice the cost of the cherry. Hence the cherry costs fifty cents.
That wasn't algebra. We might call it prealgebra, or even protoalgebra. But it ain't algebra, because there's no conscious use of symbols for unknowns, nor formal manipulation of such symbols. In fact, I'd guess it was the need to simplify tortuous reasoning like that above that ultimately led to algebra's invention.
Or I can say: Let x be the cost of the cherry and let y be the cost of the sundae. Then y  x = 5, while y + x = 6. Now, without thinking about the sundae or the cherry, I go on to say, so (y  x) + (y + x) = 11, or 2 y = 11. That means that y = 11/2, whence x = 1/2. So the cherry costs half a dollar.
That was algebra.
I think that the nonalgebraic solution shows deeper insight. Algebra is for problems where insight isn't likely to be deep enough.
(I can also do what most beginning algebra students do: I can say "There are two numbers in the problem, five and six. It's obvious that I want to subtract the smaller from the larger. The cherry costs a dollar." This solution points to a reading problemnot a mathematics problem.)
 Lou Talman Department of Mathematical * Computer Sciences Metropolitan State University of Denver <http://rowdy.msudenver.edu/~talmanl>

