Open Query 21

course Mth151

I emailed you about taking the Chapter 1 test but I was wondering if you ever got it? Thanks 4/6 around 10:30

Email is OK but forms are more reliable. You should always use the form; email is OK as a backup.

I don't show any emails from you since January 6.

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: It is a closed operation because they are all from the original set of numbers

1 1 3 5 7

3 3 1 7 5

5 5 7 1 3

7 7 5 3 1

confidence rating #$&* 2

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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):OK

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Self-critique rating #$&*2

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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=5, b=4, c=6

5+(4*6)= 29 (5+4) * (5+6)=99

confidence rating #$&* 2

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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): Did a have to equal 2 and b equal 5 and c equal 7 for this problem?

No. a = 2, b = 5 and c = 7 is only one possible example. Your solution is another example, and is every bit as good.

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Self-critique rating #$&*2

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

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Your solution: For (A U B) ^ (A U C) we would first shade all of A and B, for A U (B ^ C) we would shade all of A in addition to the part of B that overlaps C, then all of A and C, and our set would be described by the overlap between the two shadings. We would have all of A, and the overlap between B and C.

confidence rating #$&* 2

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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). **

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