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?
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.
Even numbers can be written as: 2n where n is any number. The two is a constant, the n is a variable.
2 * 1 = 1
2 * 2 = 4 2 * 3 = 6...
The division problem even/even can be written as 2n/2m = n/m where n and m are both variables. The result, n/m, does not tell us whether the result is even or odd. And the above examples confirm that the result can be either.
Is the result of odd/odd even or odd for all whole numbers?
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 the formula (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:
(2n-1)2m = _________________
Given the work done before, will this be even or odd?
What is the result of odd * odd using our new formula for odd numbers? Will this result be odd or even for all numbers?
Multiply out (2n - 1)(2m - 1).
When 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.
(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
The result matches only if every number in the first parenthesis is multiplied against every number in the second parenthesis.
This is breadfruit distribution: 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."
(Joe * James) + (Joe * Jacob) + (John * James) + (John * Jacob)
(a + b)(c + d) = ac + ad + bc + bd
Thus (2n - 1)(2m - 1) = 4nm - 2n - 2m + 1.
4nm is even (2 (2nm)), 2n and 2m are even, and an even minus an even is even. Hence 4nm - 2n - 2m 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.
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 ©1996 and ©1999 College of Micronesia - FSM. For further information on this project, contact email@example.com Designed and run originally on Micron Millenia P5 - 133 MHz with 32 MB RAM, Windows 95 OS.