The goal was to get students to think about numbers and math in a way that they probably hadn't thought about before. Each exercise had the students first tackle the problem and propose answers as well as proofs.
The first day of class I actually began with the following:
When an even and an even combine, they form another even.
When an even and an odd combine, they form an odd.
When an odd and an odd combine, they form an even.
Even's are preferentially formed in two out of three instances! Some examples of this exploitation:
4 + 6 =10 even
5 + 2 =7 odd
but 7+ 11 = 18 even
Does this exploitation continue into subtraction?
Does this exploitation continue into multiplication?
For homework the first day of class, the following assignment was made?
In the following examples only consider problems where the answer is a whole number. If the answer can never be a whole number then write that as the answer.
Note that 24/8 = 3 12/6 = 2 Even over even can be even or odd.
The next day the concept that even numbers can be represented by 2n was given. This will require explaining just what is an even number and how the expression 2n operates.
Even/even can be written as 2n/2m = n/m. The result, n/m, does not tell us whether the result is even or odd.
Is the result of odd/odd even or odd for all whole numbers? The students should have tackled this first.
27/9 = 3
Any division problem of the form a/b = c can be rewritten as a = b*c. Thus, for the above problem, 27 = 9*3. Using this we can write:
odd/odd = ? or odd = odd*?
Odd numbers can be represented by (2n-1). Check it out:
2(0)-1 = -1
2(1)-1 = 1
2(2)-1 = 3
2(3)-1 = 5
2(4)-1 = 7
The equation odd = odd*? demands that the result of odd*? be odd. Consider first whether odd*even is odd:
Have the students work the following out. Circulate in the class to see what errors including distributional errors are being made. Bear in mind that the students have NOT been specifically taught distribution. This was the first activity in the term. These students are not a tabula rasa. They have had ten to twelve years of mathematics. The hypothesis is that they are very confused. The reversed order of this distribution is sure to throw many of our students.
(2n-1)2m = 4mn - 2.
Will 4mn be even or odd? (This can be based on the earlier work of even*even and even*odd above). Will subtracting two result in a change?
? cannot be even as it would generate an even result. Can the unknown be odd?
Have the students multiply out (2n-1)(2m-1). This may seem completely out of place in a prealgebra course, these students are theoretically prealgebraic. This is not actually true. They have done high school algebra, but their knowledge is riddled with gaps, misconceptions, and confusion. I have found that most can handle (x+3)(x+4), yet (2n-1)(2m-1) will likely confuse them. If, as I walk around, I see that (2n-1)(2m-1) is not being accomplished, then I would turn to the board and have them do (x+2)(x+3). I have found that the bulk of the students can handle that. I then ask how they did it. Then we apply that extant knowledge to the binomial at hand. It is important that you have let the students first try (2n – 1)(2m – 1) without any instruction. This creates a student stake in the outcome of the below: the student pays attention to see if they were right or wrong.
At this point I usually introduce the class breadfruit distribution below.
Binomial times BinomialWhen multiplying a binomial by a binomial the distribution pattern is more complex. Each element in the first binomial must be multiplied against each element in the second binomial.
Consider:
(2 + 3)(4 + 5)
This is equal to:
(2 + 3)(4 + 5) = (5)(9) = 45
To get the same answer by distributing, the 2 must be multiplied against both the 4 and 5 AND the 3 must also be multiplied against the 4 and 5. The resulting multiplications are then added up.
(2 + 3)(4 + 5)
(2*4) + (2*5) + (3*4) + (3*5)
8 + 10 + 12 + 15
18 + 27
45
The result matches only if every number in the first parenthesis is multiplied against every number in the second parenthesis.
This is usually introduced as a FOIL distribution, however FOIL is not extendable to trinomial. Think of this as a breadfruit distribution instead: Joe, John, James, and Jacob are brothers. Suppose the brothers Joe and John each have breadfruit. When they distribute the breadfruit both Joe and John must give breadfruit to James and John:
(Joe and John) (James and Jacob)
(Joe gives to James) and (Joe to Jacob) and (John to James) and (John to Jacob)
Call it "From each, To each."
This becomes in symbols:
(Joe * James) + (Joe * Jacob) + (John * James) + (John * Jacob)
In symbols:
(a + b)(c + d) = ac + ad + bc + bd
The reason for introducing this as breadfruit distribution is that FOIL does not work for (a + b + c)(d + e + f), while breadfruit distribution always works.
Thus (2n – 1)(2m – 1) = 4nm – 2n – 2n + 1.
4nm is even (2 (2nm)), 2n and 2m are even, and an even minus an even is even. Hence 4nm – 2n – 2n is even. Add one to any even number and the result is an odd number. Thus odd times odd must be odd for all numbers. This satisfies our original demand, therefore odd/odd is always odd and will never be even for whole number solutions.
Break the students up into groups. Remind them of some of the rules available to them, 2n is even, 2n – 1 is odd, and a/b = c is the same as a = b × c. Have the students work on odd/even and even/odd.
Assign completion of the above as homework. Warn them that fractional answers are not even nor odd and hence must be thrown out as examples.
The next day look over what the students have produced and proceed from there.
Developed by Dana Lee Ling with the support and funding of a U.S. Department of Education Title III grant and the support of the College of Micronesia - FSM. Notebook material 1999 College of Micronesia - FSM. For further information on this project, contact dleeling@comfsm.fm Designed and run on Micron Millenia P5 - 133 MHz with 32 MB RAM, Windows 95 OS.