MATH Section 2.2

Suppose that the universal set is U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Express the given set with a bit string where the ith bit in the string is 1 if i is in the set and 0 otherwise. {3, 7, 8} Multiple Choice 00 1000 1100 10 1001 0001 01 1100 1110 10 1010 1010 Explanation The given set contains the elements 3, 7, and 8. The universal set has 10 elements, so the bit string has 10 bits. The bits at the 3rd, 7th, and 8th positions will be 1 and the remaining will be 0. Solution The bit string representation of the set {3, 7, 8} is 00 1000 1100.

00 1000 1100

Consider the sets A = {a, b, c, d, e}, B = {b, c, d, g, p, t, v}, C = {c, e, i, o, u, x, y, z}, and D = {d, e, h, i, n, o, t, u, x, y}. Each of the given sets can be represented using a bit string of length 26, where ith bit represents the ith letter of the alphabet. Match each set on the left to the corresponding bit string on the right. (You must provide an answer before moving to the next part.) 1. D ={d, e, h, i, n, o, t, u, x, y} 2. C ={c, e, i, o, u, x, y, z} 3. A ={a, b, c, d, e} 4. B ={b, c, d, g, p, t, v} Explanation We represent the sets by bit strings of length 26, where the first position represents the letter a of the alphabet, second position represents the letter b, and so on. While converting a set to a bit string, the corresponding position of each element in the set represents 1 and the remaining will be 0. Solution The sets A, B, C, and D can be converted to bit strings as follows: A ={a, b, c, d, e} → 11 1110 0000 0000 0000 0000 0000 B ={b, c, d, g, p, t, v} → 01 1100 1000 0000 0100 0101 0000 C ={c, e, i, o, u, x, y, z} → 00 1010 0010 0000 1000 0010 0111 D ={d, e, h, i, n, o, t, u, x, y} → 00 0110 0110 0001 1000 0110 0110

1. 00 0110 0110 0001 1000 0110 0110 2. 00 1010 0010 0000 1000 0010 0111 3. 11 1110 0000 0000 0000 0000 0000 4. 01 1100 1000 0000 0100 0101 0000

Match the given expressions to the conclusion that can be drawn from this statement. 1. A− B= B− A 2. A− B= A 3. A∩ B= A 4. A∪ B= A 5. A∩ B= B∩ A Explanation A∩ B= AA⁢∩ B⁢= A : A⊆ BA⁢⊆ B A∩ B= AA⁢∩ B⁢= A : A⊆ BA⁢⊆ B A− B= AA⁢− B⁢= A : A ∩ B= ∅A⁢ ∩ B⁢= ∅ A∩ B= B∩ AA⁢∩ B⁢= B⁢∩ A : This is always true A− B= B− AA⁢− B⁢= B⁢− A : A= BA⁢= B Solution A∩ B= AA⁢∩ B⁢= A : A⊆ BA⁢⊆ B A∩ B= AA⁢∩ B⁢= A : A⊆ BA⁢⊆ B A− B= AA⁢− B⁢= A : A ∩ B= ∅A⁢ ∩ B⁢= ∅ A∩ B= B∩ AA⁢∩ B⁢= B⁢∩ A : This is always true A− B= B− AA⁢− B⁢= B⁢− A : A= BA⁢= B

1. A= B 2. A ∩ B= ∅ 3. A⊆ B 4. B⊆ A 5. This is always true.

Prove De Morgan's law by showing that A∪ B¯¯¯¯¯¯¯¯¯A⁢∪ B¯ = A¯¯¯A¯ ∩ B¯¯¯∩ B¯ if A and B are sets. Identify the the unknowns X, Y, Z, P, Q, and R in the given membership table. ABA ∪∪ BA ∪ B¯¯¯¯¯¯¯¯¯¯A⁢ ∪ B¯A¯¯¯A¯B¯¯¯B¯A¯¯¯∩B¯¯¯A¯∩B¯111X000101001Q0110Z00000Y1PR Multiple Choice X = 1, Y = 0, Z = 1, P = 1, Q = 0, R = 1 X = 0, Y = 1, Z = 1, P = 1, Q = 1, R = 0 X = 0, Y = 1, Z = 0, P = 0, Q = 1, R = 1 X = 0, Y = 1, Z = 1, P = 1, Q = 0, R = 1 Explanation Negation of 0 is 1. Negation of 1 is 0. Solution X = 0, Y = 1, Z = 1, P = 1, Q = 0, R = 1

X = 0, Y = 1, Z = 1, P = 1, Q = 0, R = 1

Identify the correct steps involved in proving that if A, B, and C are sets, then |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|. (Check all that apply.) Count the elements in each of the sets and add. Addition of the elements gives us all the elements in the union, but we have overcounted. Addition of the elements gives us all the elements in the intersection, but we have overcounted. Each element in A ∩ B, A ∩ C, and B ∩ C has been counted twice. Each element in A ∩ B, A ∩ C, and B ∩ C has been counted thrice. Subtract the cardinalities of the intersections to make up for the overcount. The elements of A ∩ B ∩ C have now been counted three times and subtracted three times. The elements of A ∩ B ∩ C have now been counted two times and subtracted three times. We adjust by adding back the cardinality of A ∩ B ∩ C. Explanation To count the elements of A ∪ B ∪ C, we proceed as follows. First, we count the elements in each of the sets and add. This certainly gives us all the elements in the union, but we have overcounted. Each element in A ∩ B, A ∩ C, and B ∩ C has been counted twice. Therefore, we subtract the cardinalities of these intersections to make up for the overcount. Finally, we have compensated a bit too much, since the elements of A ∩ B ∩ C have now been counted three times and subtracted three times. We adjust by adding back the cardinality of A ∩ B ∩ C.

check the following: Count the elements in each of the sets and add. Addition of the elements gives us all the elements in the union, but we have overcounted. Each element in A ∩ B, A ∩ C, and B ∩ C has been counted twice. Subtract the cardinalities of the intersections to make up for the overcount. The elements of A ∩ B ∩ C have now been counted three times and subtracted three times. We adjust by adding back the cardinality of A ∩ B ∩ C.

Click and drag the given steps to their corresponding step number to prove the given statement. A ∩ B ∩ C ⊆ A ∩ B. Explanation Suppose x ∈ A ∩ B ∩ C. Then x is in all three of these sets. In particular, it is in both A and B. Thus, x is in A ∩ B.

correct blocks and order (going down): 1. Suppose x ∈ A ∩ B ∩ C. 2. Then x is in all three of these sets. 3. In particular, it is in both A and B. 4. Thus, x is in A ∩ B.

Click and drag the steps on the left to their corresponding step number on the right to prove the given statement. (A ∩ B) ⊆ A Explanation If x is in A ∩ B, x is in A and x is in B by definition of intersection. Thus x is in A. Hence proved.

correct blocks and order (going down): step 1: If x is in A ∩ B, x is in A and x is in B by definition of intersection. step 2: Thus x is in A.

Click and drag the given steps to their corresponding step number to prove the given statement. (A - B) - C ⊆ A - C. Explanation Suppose that x ∈ (A - B) - C. Then x is in A - B but not in C. Since x ∈ A - B, x ∈ A. x ∈ A and x ∉ C. This shows that x ∈ A - C.

correct blocks and order (going down): step 1: Suppose that x ∈ (A - B) - C. step 2: Then x is in A - B but not in C. step 3: Since x ∈ A - B, x ∈ A. step 4: x ∈ A and x ∉ C. This shows that x ∈ A - C.

Click and drag the steps on the left to their corresponding step number on the right to prove the given statement. A - B ⊆ A Explanation Suppose x is in A - B. By definition, x is in A and x is not in B. By simplification, x is in A. Since the assumption that an element exists in A ∩ (B - A) leads to a contradiction, A ∩ (B - A) must be empty. Hence proved.

correct blocks and order (going down): step 1: Suppose x is in A - B. step 2: By definition, x is in A and x is not in B. step 3: By simplification, x is in A.

Click and drag the steps on the left to their corresponding step number on the right to prove the given statement. A ⊆ (A ∪ B) Explanation Suppose x ∈ A. A ∪ B consists of elements in A or B. Thus, x ∈ A implies x ∈ A ∪ B.

correct blocks and order (going down): step 1: Suppose x ∈ A. step 2: A ∪ B consists of elements in A or B. step 3: Thus, x ∈ A implies x ∈ A ∪ B.

Suppose that A is the set of sophomores at your school, B is the set of students in discrete mathematics at your school, and the universal set U is the set of all students at your school. Match the sets given in the left to their symbolic expression in the right. 1. The set of students at your school who either are sophomores or are taking discrete mathematics 2. The set of sophomores at your school who are not taking discrete mathematics 3. The set of students at your school who either are not sophomores or are not taking discrete mathematics 4. The set of sophomores taking discrete mathematics in your school Explanation The set of sophomores taking discrete mathematics in your school : A ∩ B The set of sophomores at your school who are not taking discrete mathematics : A - B The set of students at your school who either are sophomores or are taking discrete mathematics : A ∪ B The set of students at your school who either are not sophomores or are not taking discrete mathematics : A− ∪ B−A−⁢ ∪ B− Solution The set of sophomores taking discrete mathematics in your school : A ∩ B The set of sophomores at your school who are not taking discrete mathematics : A - B The set of students at your school who either are sophomores or are taking discrete mathematics : A∪BA∪B The set of students at your school who either are not sophomores or are not taking discrete mathematics : A− ∪ B−A−⁢ ∪ B−

1. A∪B 2. A - B 3. A−∪B− 4. A ∩ B

Let A = {1, 2, 3, 4, 5} and B = {0, 3, 6}. Match the given sets on the left to the sets on the right. 1. A ∩ B 2. B − A 3. A ∪ B 4. A − B Explanation Use the properties of a set. A ∪ B = {0, 1, 2, 3, 4, 5, 6} A ∩ B = {3} A − B = {1, 2, 4, 5} B − A = {0, 6} Solution A ∪ B = {0, 1, 2, 3, 4, 5, 6} A ∩ B = {3} A − B = {1, 2, 4, 5} B − A = {0, 6}

1. {3} 2. {0, 6} 3. {0, 1, 2, 3, 4, 5, 6} 4. {1, 2, 4, 5}

Let A = {a, b, c, d, e} and B = {a, b, c, d, e, f, g, h}. Match the given sets in the left to the sets in the right. 1. A ∩ B 2. A − B 3. A ∪ B 4. B − A Explanation Let A and B be sets. The union of the sets A and B, denoted by A ∪ B, is the set that contains those elements that are either in A or in B, or in both. The intersection of the sets A and B, denoted by A ∩ B, is the set containing those elements in both A and B. A ∪ B = {a, b, c, d, e, f, g, h} A ∩ B = {a, b, c, d, e} A − B = Ø B − A = {f, g, h} Solution A ∪ B = {a, b, c, d, e, f, g, h} A ∩ B = {a, b, c, d, e} A − B = Ø B − A = {f, g, h}

1. {a, b, c, d, e} 2. Ø 3. {a, b, c, d, e, f, g, h} 4. {f, g, h}

Suppose that the universal set is U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Express the given set with a bit string where the ith bit in the string is 1 if i is in the set and 0 otherwise. {1, 3, 6, 10} Multiple Choice 00 1110 0000 10 1010 1010 01 1100 1110 10 1001 0001 Explanation The given set contains the elements 1, 3, 6, and 10. The universal set has 10 elements, so the bit string has 10 bits. The bits at the 1st, 3rd, 6th, and 10th positions will be 1 and the remaining will be 0. Solution The bit string representation of the set {1, 3, 6, 10} is 10 1001 0001.

10 1001 0001