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

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 Paul A. Tanner III Posts: 5,920 Registered: 12/6/04
[math-learn] 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 re-relate 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 cross-multiplying. Cross-multiplying 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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Date Subject Author
11/25/03 Paul A. Tanner III
11/26/03 Gerald Harnett
11/27/03 Paul A. Tanner III
11/27/03 Rex Boggs
11/27/03 Paul A. Tanner III