Asgt 21 Query

course MTH 151

11/10/09 7:07 AM

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

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

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Your solution:

I think that the identity property is satisfied here, and that the identity element is 1 because the first row and col are equal to the elements.

confidence rating:0

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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. **

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Self-critique (if necessary):

Oh ok I see this better now. I was right about the identity, but I didnt think about the diagonal which showed an inverse. Also the commutative and associative.

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Self-critique Rating:3

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

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Your solution:

a = -2

b = 1

c = 0

-2 + (1*0) = -2

(-2 + 1) * (-2 + 0) = -1 * -2 = 2

confidence rating:3

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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. **

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Self-critique (if necessary):

Ok

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Self-critique Rating:

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

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Your solution:

a U (b ^ c) would be all of A and the b C intersection shaded.

(a U b) ^ (a U c) would be the following:

The first set would be all of a and b

The second set would be all of a and c

Since both sets include all of a, then a remains shaded. Also, the intersection of b and c is shaded in both sets so it remains shaded also.

So yes, it is distributive,

confidence rating:3

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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). **

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

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