Query 21

#$&*

course Mth 151

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

021. `query 21

*********************************************

Question: `q4.4.6 star operation [ [1, 3, 5, 7], [3, 1, 7, 5], [5, 7, 1, 3], [7, 5, 3, 1]]

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

1 3 5 7

1 1 3 5 7

3 3 1 7 5

5 5 7 1 3

7 7 5 3 1

The operation would end up being closed because all the numbers in the results table are in the original set of numbers as well.

The operation has an identity because when 1 is paired with any other number, it doesn’t change.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a** Using * to represent the operation the table is

* 1 3 5 7

1 1 3 5 7

3 3 1 7 5

5 5 7 1 3

7 7 5 3 1

the operation is closed, since all the results of the operation are from the original set {1,3,5,7}

the operation has an identity, which is 1, because when combined with any number 1 doesn't change that number. We can see this in the table because the row corresponding to 1 just repeats the numbers 1,3,5,7, as does the column beneath 1.

The operation is commutative--order doesn't matter because the table is symmetric about the main diagonal..

the operation has the inverse property because every number can be combined with another number to get the identity 1:

1 * 1 = 1 so 1 is its own inverse;

3 * 3 = 1 so 3 is its own inverse;

5 * 5 = 1 so 5 is its own inverse;

7 * 7 = 1 so 7 is its own inverse.

This property can be seen from the table because the identity 1 appears exactly once in every row.

the operation appears associative, which means that any a, b, c we have (a * b ) * c = a * ( b * c). We would have to check this for every possible combination of a, b, c but, for example, we have (1 *3) *5=3*5=7 and 1*(3*5)=1*7=7, so at least for a = 1, b = 3 and c = 5 the associative property seems to hold. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

self-critique rating #$&*: OK

*********************************************

Question: `q4.4.24 a, b, c values that show that a + (b * c) not equal to (a+b) * (a+c).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

They are not equal to each other due to the parenthesis. For example,

a=1 b=2 c=3

1 + (2 x 3) = 7

(1 + 2) x (1 + 3) = 12

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a** For example if a = 2, b = 5 and c = 7 we have

a + (b + c) = 2 + (5 + 7) = 2 + 12 = 14 but

(a+b) * (a+c) = (2+5) + (2+7) = 7 + 12 = 19. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

Self-critique Rating: OK

*********************************************

Question: `q4.4.33 venn diagrams to show that union distributes over intersection

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A U (B^C) = Shade in A, the overlapping of B and C.

(A U B) ^ (A U C) = Shade in A, B, C, and where A and B meet and where A and C meet.

The set is A U (B^C).

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a** For A U (B ^ C) we would shade all of A in addition to the part of B that overlaps C, while for (A U B) ^ (A U C) we would first shade all of A and B, then all of A and C, and our set would be described by the overlap between these two shadings. We would thus have all of A, plus the overlap between B and C. Thus the result would be the same as for A U (B ^ C). **

"

&#This looks good. Let me know if you have any questions. &#

#$&*