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Topic:
[mathlearn] Algebra as a spatial motion game of logic, like chess or checkers
Replies:
4
Last Post:
Nov 27, 2003 10:39 AM




[mathlearn] Algebra as a spatial motion game of logic, like chess or checkers
Posted:
Nov 25, 2003 5:27 AM


Note: Rex asked for ideas on teaching algebra such as in
3x  2 = 5 2(x + 6) = 8(4  x) + 3 3(2x + 1)/5 = 9
So this message is intended for teachers of secondary school math and their students, especially the ones who struggle.
What follows is a way of thinking that many already know about, and some might think of as "blasphemy against the algebra god." I've actually gotten some responses from some other teachers along the lines of censorship: "Oh how horrible! This should not be allowed!" But in spite of the horrified fundamentalists, I've noticed that sharing this information actually helps many struggling algebra students.
I tell students that they can treat algebra as a spatial motion game played according to logical rules, like chess or checkers. This game is really just taking advantage of the fact that there are "optical illusions" of motion created by performing the same operation on both sides of an equation. Many people, minors and adults that were previously struggling with algebra, have told me that when viewed this way, algebra actually becomes "sort of fun and easy," really like playing a game. And, as in playing such visual games, those who get good at the game can see several moves ahead. When I mention this to students who see this for the first time, they quickly agree. The examples later below show this, I think, especially the more involved example involving only variables.
I'm always quite intrigued that many, many people, minors and adults, have never on their own noticed these "optical illusions" and how they could be taken advantage of before I pointed them out to them.
(One of the most notable responses came from a geometry student who, after taking two years of algebra and chemistry told me that he wished he saw this before chemistry, in that it would have really helped him deal with some rather complicated expressions where he had to solve for a variable in some unusual spot.)
Before I go further, there is a theoretical justification for looking at algebra this way. The theorem is this, without proof, in two parts:
Theorem (Transformation Law).
A) For all nonzero reals a and b, ax = xa = b has real number solution x = (1/a)b = b(1/a)
B) For all reals c and d, c+y = y+c = d has real number solution y = (c)+d = d+(c)
Note that we can interpret the Transformation Law verbally as "Take an element that is related by addition or multiplication to the rest of the expression on one side of the equation and move it to the other side, transforming it into its respective additive or multiplicative inverse, and rerelate it to that expression on that other side respectively by addition or multiplication."
NOTE: So what I'm addressing with the Transformation Law is just that part of algebra that is thought of as performing the same operation on both sides of an equation or, alternatively, moving and transforming an element from one side of the equation to the other.
There are two ways of viewing an equation:
1) We can always view an equation in the form of a sum set equal to a sum:" Two or more on the left side, two or more on the right side. Using the "[]" symbol to represent each addend as a place, it looks like . . . [] + [] = [] + [] . . . In each of these places, we will always have at least one element, which can of course be the additive identity 0. In each of these places, we can view each of these elements as factors, even if these elements are sums. So I will refer to the elements in this context as factors. Note also that any addend can be a fraction.
For 1), the spatial motion rule is. We can always move an addend from one place on one side of the equation another place on the other side of the equation, reversing is sign as we move it across the equals sign. This change of sign is a transformation of an element to its additive inverse as it moves to the other side of the equation. The goal is to end up with, on one side of the equation, the target factor or variable to be solved for in the numerator and 1 in the denominator, and on the other side of the equation, all the other factors.
We can move more than one addend at a time. This is key I think as to why many students find this method easier and more efficient.
2) We can always view an equation in the form of a ratio set equal to a ratio, thus having four "places:" Upper left (numerator), lower left (denominator), upper right (numerator), lower right (denominator). Using the "[]" symbol, it looks like [] / [] = [] / []. In each of these places, we will always have at least one element, which can of course be the multiplicative identity 1. We can view each of these elements as factors, even if these elements are sums. So I will refer to the elements in this context as factors.
For 2), the spatial motion rule is simple: We can always move any one or more factors diagonally but only diagonally, leaving 1, the multiplicative identity, behind where the moved factor or factors were. This change of level is a transformation of an element to its multiplicative inverse as it moves to the other side of the equation. The goal is to end up with, on one side of the equation, the target factor or variable to be solved for in the numerator and 1 in the denominator, and on the other side of the equation, all the other factors. The 1 in the denominator under the solved for variable can then be dropped.
We can move more than one factor at a time. This is key I think as to why many students find this method easier and more efficient.
(Notice that this is not crossmultiplying. Crossmultiplying is actually just a subset of this technique.)
Note: In the context of 2), I show them what the notation [] / [] can also be changed to the notation of a product, this being (1/[])*[] or []*(1/[]). They must develop a fluency in going from one notation to the other. I have had students come to me who were not taught these two ways. This I think is a weakness that has to be addressed.
EXAMPLES:
In the context of 2), note that a/100 = b/c can be a setup for many percentage problems. Instantiate two variables, then move factors according to the rule in one step, and we're done solving for the unknown.
Suppose c were the unknown and we instantiated as 37/100 = 69/c. Then in one step with the motion rule we have c/1 = (69*100)/37, and we are done solving for the unknown. Obviously, we still have to arithmetically simplify, but in terms of the algebra, we are done. I'm amazed at how many students never see this technique until shown it. Instead of one easy motion step, they are still stuck in thinking of three arithmetically involved steps: Multiply both sides by c, divide both sides by 37, and multiply both sides by 100. Should they be able to see how this way could be done as well? Yes. Understanding requires at least seeing it.
Note that these examples below show how important the distributive property is to prepare equations for performing the same operation on both sides or for the Transformation Law. I'll not simplify here, and leave at x/1. And also note that some steps that are written out here don't have to be written out. They are here just to illustrate.
i) 3x  2 = 5 => 3x = 5 + 2 1 transformation move => 3x/1 = 7/1 => x/1 = 7/3 1 transformation move
ii) 2(x + 6) = 8(4  x) + 3 => 2x + 12 = 32  8x + 3 => 8x + 2x = 35  12 2 transformation moves => 10x = 23 I think it good to explain this as an application of the distributive property => 10x/1 = 23/1 => x/1 = 23/10 1 transformation move
3(2x + 1)/5 = 9 => (6x + 3)/5 = 9/1 => (6x + 3)/1 = 9*5/1 1 transformation move => 6x + 3 = 45 => 6x = 45  3 1 transformation move => 6x/1 = 42/1 => x/1 = 42/6 1 transformation move
MORE INVOLVED EXAMPLE:
Before I show students how they can solve for i in just two transformation steps, they don't believe that it can be done. They are amazed when they see it, and start trying this technique on other more involved equations.
ab/cd = ef/(g(h+i)) => (h+i)/1 = (cdef)/(abg) 5 transformation moves => h+i = (cdef)/(abg) => i = (cdef)/(abg)  h 1 transformation move
I think that this shows the students that the solving for the variable part of performing the same operation on both sides via performing the transformation moves can be easier and more fun than the arithmetic and simplification parts.
Hope this helps.
Paul
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