#$&* course Mth 151 1/29 4 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a**Z' = {a,g}, the set of all elements of the universal set not in Z. Y ^ Z' = {a}, since a is the only element common to both Y and Z'. So (Y ^ Z') U X = {a, c, e, g}, the set of all elements which lie in at least one of the sets (Y ^ Z') U X. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This one was muddled in my brain. Not difficult, but took some organizing to make sense of the information. Once I get used to the “not not’s” and the “not not not’s” in this segment I should not have a problem. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qGive the intersection of the two sets Y and Z ' YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y^Z’= {a} confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a**Z' = {a,g}, the set of all elements of the universal set not in Z. Y ^ Z' = {a}, since a is the only element common to both Y and Z'.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 2.3.32 (formerly 2.3.30). This was not assigned, but you answered a series of similar questions and should be able to give a reasonable answer to this one: Describe in words (A ^ B' ) U (B ^ A') YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A set that is the intersection of A and the compliment of B in union with a set that is the intersection of B and the compliment of A confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** a description, not using a lot of set-theoretic terms, of (A ^ B' ) U (B ^ A') would be, all the elements that are in A and not in B, or that are not in A and are in B Or you might want to say something like 'elements which are in A but not B OR which are in B but not A'. STUDENT SOLUTION WITH INSTRUCTOR COMMENT:everything that is in set A and not in set B or everything that is in set B and is not in set A. INSTRUCTOR COMMENT: I'd avoid the use of 'everything' unless the word is necessary to the description. Otherwise it's likely to be misleading. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Got it, just with the textbook terminology ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q2.3.53 (formerly 2.3.51) Is it always or not always true that n(A U B) = n(A)+n(B)? This was not among the assigned questions but having completed the assignment you should be able to answer this. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It is always true. Whoops I see that it is not ALWAYS true. It depends on the sets themselves to determine that answer. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** This conclusion is contradicted by many examples, including the one of the dark-haired and bright-eyed people in the q_a_. Basically n(A U B) isn't equal to n(A) + n(B) if there are some elements which are in both sets--i.e., in the intersection. } MORE DETAIL: The statement can be either true or false, depending on the sets A and B; it is not always true. The statement n(A U B) = n(A)+n(B) means that the number of elements in A U B is equal to the sum of the number of elements in A and the number of elements in B. The statement would be true for A = { c, f } and B = { a, g, h} because A U B would be { a, c, f, g, h} so n(A U B) = 5, and n(A) + n(B) = 2 + 3 = 5. The statement would not be true for A = { c, f, g } and B = { a, g, h} because A U B would be the same as before so n(AUB) = 5, while n(A) + n(B) = 3 + 3 = 6. The precise condition for which the statement is true is that A and B have nothing in common. In that case n(A U B) = n(A) + n(B). A more precise mathematical way to state this is to say that n(A U B) = n(A) + n(B) if and only if the intersection A ^ B of the two sets is empty. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): In my head I gave it a way that would be true, but I did not consider the possibility of a repeated number. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 2.3.60 X = {1,3,5}, Y = {1,2,3}. Find (X ^ Y)' and X' U Y'. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (X^Y)’= {2, 4, 5} X’ U Y’= {2, 4, 5} confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** X ^ Y = {1,3} so (X ^ Y) ' = {1,3}' = {2, 4, 5}. (X ' U Y ' ) = {2, 4} U {4, 5} = {2, 4, 5} The two resulting sets are equal so a reasonable conjecture would be that (X ^ Y)' = X' U Y'. ** STUDENT QUESTION: Where did the 4 come from? INSTRUCTOR RESPONSE: I believe this problem, as stated in the text, indicates that the universal set is {1, 2, 3, 4, 5}. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q2.3.72 A = {3,6,9,12}, B = {6,8}. What is A X B and what is n(A X B)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A x B is each element of A paired with each element of B A x B= {(3, 6), (3, 8), (6, 6), (6, 8), (9,6), (9, 8), (12, 6), (12, 8)} And n(A x B) is the number of elements times the other n(A x B)= (4 x 2)= 8 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** (A X B) = {(3,6),(3,8),(6,6),(6,8),(9,6),(9,8),(12,6), (12,8)} (B X A) = (6,3),(6,6),(6,9),(6,12),(8,3),(8,6),(8,9),(8,12)} How is n(A x B) related to n(A) and n(B)? n(S) stands for the number of elements in the set S, i.e., its cardinality. n(A x B) = n(A) * n(B) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Did I miss the answer to the second part of this question or did I get it wrong?
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Given Solution: `a** everything in A and everything in B would be shaded. The rest of the universal set (the region outside A and B but still in the rectangle) wouldn't be. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): By “entire diagram” I meant just the circle part, I did not shade in the rectangle. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 2.3.100 Shade (A' ^ B) ^ C YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The shaded area is the area that is common to B and C, but not A. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** you would have to shade every region that lies outside of A and also inside B and also inside C. This would be the single region in the overlap of B and C but not including any part of A. Another way to put it: the region common to B and C, but not including any of A ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was unsure if shading all that was not A would include shading the entirety of regions B and C as well. Obviously it was not. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 2.3.108. Describe the shading of the set (A ^ B)' U C. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Here the shaded region is that of C that does not intersect with either A nor B confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** All of C would be shaded because we have a union with C, which will include all of C. Every region outside A ^ B would also be shaded. A ^ B is the 'overlap' region where A and B meet, and only this 'overlap' would not be part of (A ^ B) '. The 'large' parts of A and B, as well as everything outside of A and B, would therefore be shaded. Combining this with the shading of C the only the part of the diagram not shaded would be that part of the 'overlap' of A and B which is not part of C. ** STUDENT QUESTION I think I understand because the ‘ was outside the ( ) then only the answer to A^B would be prime. And so my answer is wrong to the extent that the larger regions of A &B would also be shaded, but had it been (AUB)’ no part of either A or B would have been Shaded? INSTRUCTOR RESPONSE Exactly. Very good question, which you answered very well. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I failed to shade the other regions of A and B because, I thought “not A” and “not B” intersected also meant “not A” and “not B” at all. ------------------------------------------------ Self-critique Rating:
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Given Solution: `a** Student Answer and Instructor Response: (B'^C')^A Instructor Response: Good. Another alternative would be A - (B U C ), and others are mentioned below. COMMON ERROR: A ^ (B' U C') INSTRUCTOR COMMENT: This is close but A ^ (B' U C') would contain all of B ^ C, including a part that's not shaded. A ^ (B U C)' would be one correct answer. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q2.3.114 Largest area of A shaded (sets A,B,C). Write a description using A, B, C, subset, union, intersection symbols, ', - for the shaded region. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (C’ ^ B’) ^ A confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Student Answer and Instructor Response: (B'^C')^A Instructor Response: Good. Another alternative would be A - (B U C ), and others are mentioned below. COMMON ERROR: A ^ (B' U C') INSTRUCTOR COMMENT: This is close but A ^ (B' U C') would contain all of B ^ C, including a part that's not shaded. A ^ (B U C)' would be one correct answer. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!