Complement of a Set and Its Properties

The multiples of 5 in U are 5, 10, and 15, so A = {5, 10, 15}. The complement is found by taking U − A. Removing these three multiples from the numbers 1 through 15 leaves {1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14}. Therefore option A is correct; option B lists A itself, not its complement.

The complement Aᶜ contains the elements of U that are not in A. Set A contains all five listed vowels, while U also contains the letters b, c, and d. Removing the vowels from U leaves exactly {b, c, d}. Therefore option A is correct. This example also shows why the universal set must always be specified before finding a complement.

First form the union by collecting every element appearing in A or B: A ∪ B = {1, 2, 3, 7, 8}. The complement of this union contains the elements of U that are absent from the union. Removing 1, 2, 3, 7, and 8 from U leaves {4, 5, 6}. Hence (A ∪ B)ᶜ = {4, 5, 6}, so option A is correct.

The intersection A ∩ B contains elements common to both sets. The only common element is 6, so A ∩ B = {6}. To find its complement relative to U, remove 6 from U. The result is {1, 2, 3, 4, 5, 7, 8, 9}. Therefore option A is correct, while option B gives the intersection before taking its complement.

A and its complement Aᶜ are disjoint, and together they make the universal set U. Therefore their cardinalities satisfy n(A) + n(Aᶜ) = n(U). Substituting the given values gives n(A) + 31 = 72, so n(A) = 72 − 31 = 41. Thus option A is correct. The other values represent either a sum or one of the given cardinalities.

The complement Aᶜ contains all elements of the universal set U that are not in A. Since A = (3, 8], the number 3 is excluded from A and therefore belongs to Aᶜ. The number 8 is included in A, so it must be excluded from the complement. Thus, within U = [0, 12], the complement is [0, 3] ∪ (8, 12].

First list the integers in the universal set: U = {-3,-2,-1,0,1,2,3,4}. The complement Aᶜ consists of elements of U that are absent from A. Removing -2, 0, 2, and 4 leaves {-3,-1,1,3}. Therefore option A is correct. The universal set is essential because a complement is always defined relative to it.

Students who do not like mathematics form the complement Aᶜ of A. When A is a subset of the universal set U, the cardinality rule is n(Aᶜ) = n(U) - n(A). Substituting the given values gives n(Aᶜ) = 55 - 37 = 18. Thus 18 students do not like mathematics; 37 is the size of A and 55 is the total class size.

By definition, the complement B = Aᶜ contains every element of the universal set U that is not in A. Therefore A and B have no common element, so A ∩ B = ∅. Together, A and its complement contain every element of U, so A ∪ B = U. These two identities characterize a complement and make option A the only valid statement.

By De Morgan’s law, Aᶜ ∩ Bᶜ = (A ∪ B)ᶜ. The union A ∪ B is {2, 3, 4, 6, 8, 9, 10, 12}. Removing these elements from the universal set U leaves {1, 5, 7, 11}. Therefore, Aᶜ ∩ Bᶜ = {1, 5, 7, 11}. Notice that elements such as 6 and 12 are excluded because they belong to both A and B.

Apply De Morgan’s law: Aᶜ ∪ Bᶜ = (A ∩ B)ᶜ. The common elements of A and B are A ∩ B = {4,5}. The complement of {4,5} relative to U contains every other element of U, namely {1,2,3,6,7,8,9,10}. Hence option A is correct. This also shows why the intersection, not the union, is needed first.

The universal set contains the 20 natural numbers from 1 through 20. The primes in this range are 2, 3, 5, 7, 11, 13, 17, and 19, so n(A) = 8. Since the complement contains all non-prime members of U, n(Aᶜ) = n(U) - n(A) = 20 - 8 = 12. Number 1 is not prime and is correctly included in the complement.

The condition x² ≤ 4 is equivalent to -2 ≤ x ≤ 2. Because x must be an integer, A = {-2, -1, 0, 1, 2}. The universal set contains every integer from -4 through 4: {-4, -3, -2, -1, 0, 1, 2, 3, 4}. The elements of U that are absent from A are -4, -3, 3, and 4. Hence Aᶜ = {-4, -3, 3, 4}.

The complement consists of the points in U that do not belong to A. In A = (1, 4], the endpoint 1 is excluded, so 1 belongs to the complement. The endpoint 4 is included, so it does not belong to the complement; the interval resumes immediately after 4. Therefore, relative to U = [-2, 6], Aᶜ = [-2, 1] ∪ (4, 6].

The complement is formed relative to U = (-5,5]. In A = [-1,3), the endpoint -1 is included, so it must be excluded from Aᶜ. The endpoint 3 is excluded from A, so it must be included in Aᶜ. Thus the part before A is (-5,-1), and the part after A is [3,5]. Therefore Aᶜ = (-5,-1) ∪ [3,5].

Since A is a subset of B, every element of A is also in B. Taking complements with respect to the same universal set reverses the direction of inclusion: if X ⊆ Y, then Yᶜ ⊆ Xᶜ. Therefore, A ⊆ B implies Bᶜ ⊆ Aᶜ. Equality is not guaranteed unless A and B are equal, and the other options contradict complement laws.

For complements taken with respect to the same universal set, inclusion reverses when complements are taken. The statement Aᶜ ⊆ Bᶜ can be viewed as Yᶜ ⊆ Xᶜ, which implies X ⊆ Y in reverse notation; hence B ⊆ A. The original inclusion A ⊆ B is not logically forced, and neither equality nor disjointness follows from the given condition.

The complement \(A^c\) consists of exactly those elements of the universal set \(U\) that are not elements of \(A\). Therefore, no element can belong to both \(A\) and \(A^c\), so their intersection is empty: \(A\cap A^c=\varnothing\). The related identity is \(A\cup A^c=U\), not \(A\). Thus, option A is the only universally valid statement.

First apply the inclusion–exclusion formula to the complements: \(n(A^c\cup B^c)=n(A^c)+n(B^c)-n(A^c\cap B^c)=40+55-20=75\). By De Morgan’s law, \(A^c\cup B^c=(A\cap B)^c\). Hence \(n((A\cap B)^c)=75\), and therefore \(n(A\cap B)=n(U)-75=100-75=25\).

With respect to \(U\), \(A^c=\{5,6,7,8,9\}\) and \(B^c=\{1,2,3,7,8,9\}\). The difference \(A^c-B^c\) retains elements in \(A^c\) that are not in \(B^c\), giving \(\{5,6\}\). Equivalently, \(X-Y=X\cap Y^c\), so \(A^c-B^c=A^c\cap B\). Thus option A is correct.