Complement of a Set and Its Properties

The complement is found by removing the elements of \(B\) from the universal set: \(B^c=U\setminus B=\{b,c,d\}\). This set has three elements, so option B is correct. The answer is not 5, because 5 is the cardinality of the entire universal set, not the number of elements outside \(B\).

The empty set contains no elements, so there is no element of \(U\) to remove when forming its complement. By definition, \(A^c=U\setminus A\). With \(A=\varnothing\), this becomes \(\varnothing^c=U\setminus\varnothing=U\). Thus, the entire universal set is the complement of the empty set.

The complement of \(A\) contains the elements of the universal set that are not in \(A\). If \(A=U\), every element of the universal set is already in \(A\), so no element remains outside it. Therefore, \(A^c=U\setminus U=\varnothing\). Notice that \(\{U\}\) would be a singleton containing the set \(U\), not the empty set.

First form the union by listing each element that appears in either set: \(A\cup B=\{1,2,3,5,7,8\}\). To find its complement, select the elements of \(U\) that are absent from this union. The only such elements are 4 and 6. Therefore, \((A\cup B)^c=\{4,6\}\), making option A correct. Option C is the intersection \(A\cap B\), not the complement.

The complement Aᶜ contains all elements of the universal set U that are not members of A. Since A = {p,r,t}, removing these elements from U leaves Aᶜ = {q,s}. Among the given choices, q belongs to this complement, whereas p, r, and t are already elements of A. Hence option C is the only correct answer.

By definition, the complement Aᶜ consists of the elements of the universal set U that are not in A. Therefore, the statement x ∈ Aᶜ means both x ∈ U and x ∉ A. Option A contradicts the definition, option C places x outside the universal set, and option D would require x to belong to both A and its complement, which is impossible.

The multiples of 3 that lie in U are 3, 6, 9, and 12, so A = {3,6,9,12}. The complement contains every element of U that is not a multiple of 3. Removing these four multiples from U gives Aᶜ = {1,2,4,5,7,8,10,11}. Thus option B is correct; options A and D list only multiples of 3.

The prime numbers in U are 2, 3, 5, and 7, because each has exactly two positive divisors. Thus A = {2,3,5,7}. The complement contains the remaining elements of U, namely {1,4,6,8,9}. Notice that 1 is neither prime nor composite, so it belongs to the complement. Therefore option B is correct.

The complement is defined by Aᶜ = U − A. If Aᶜ is empty, there is no element of U outside A. Therefore A must contain every element of U, which means A = U. Option A would make the complement equal to U, not empty. Also, A ∩ U = A, so option D would be empty only if A itself were empty. Hence option B is the unique correct answer.

Since Aᶜ = U − A, the complement can equal the entire universal set only when A contains no elements. Thus A must be the empty set, ∅. Equivalently, the complement of ∅ is U. Option A would give an empty complement, while A ∪ U is always U, so option D cannot be true for a nonempty universal set. Therefore option B is correct.

The universal set has n(U) = 8 elements, while the complement has n(Aᶜ) = 3 elements. Since A and Aᶜ partition U, n(A) + n(Aᶜ) = n(U). Therefore n(A) = 8 − 3 = 5. Equivalently, A consists of the remaining elements {2,3,5,6,8}, which also confirms that its cardinality is 5. Hence option B is correct.

The difference U \ A consists of the elements that belong to U but do not belong to A. From U = {1,2,3,4,5,6}, remove 2, 3, and 5. The remaining set is {1,4,6}. By the definition of a complement relative to U, this remaining set is Aᶜ. Therefore U \ A = Aᶜ, so option B is correct. The actual elements are {1,4,6}.

The complement Aᶜ contains all elements of U that are not in A, so Aᶜ = {2,3,5,6}. Taking the union with A gives {2,3,5,6} ∪ {1,4,7} = U. Thus Aᶜ ∪ A contains every element of the universal set, and its cardinality is |U| = 7. This is the standard identity A ∪ Aᶜ = U, so option C is correct.

Relative to U, the complement of A is Aᶜ = {1,3,5,7}. No element can belong to both A and its complement: A contains {2,4,6}, while Aᶜ contains {1,3,5,7}. Therefore their intersection is Aᶜ ∩ A = ∅. The empty set has cardinality zero, so the number of elements is 0 and option A is correct.

The complement of A is formed by selecting from U every element that is not in A. Since U contains 1 through 8 and A contains 1, 2, 5, and 8, the elements left outside A are {3,4,6,7}. This set is exactly C. Hence C = Aᶜ relative to U, making option A correct. The other options either equate C with A, all of U, or the empty set.

The complement Aᶜ is taken relative to U, so it contains the elements of U that are absent from A. Set A contains all the odd numbers from 1 through 10: 1, 3, 5, 7, and 9. The remaining elements of U are the even numbers {2,4,6,8,10}. Since B = Aᶜ, B equals this set, making option A correct.

First list the integers in the stated universal set: U = {−3,−2,−1,0,1,2,3}. The complement Aᶜ contains exactly those members of U that are not in A. Since A contains −3, −1, 1, and 3, the remaining elements are −2, 0, and 2. Therefore Aᶜ = {−2,0,2}, so option A is correct.

The complement Aᶜ relative to U consists of all elements in U that are not in A. Removing {1,2,3,4} from U = {1,2,3,4,5,6,7,8} leaves {5,6,7,8}, which is exactly B. Hence B = Aᶜ. Also, A and B are disjoint, so their intersection is empty; this shows why option D is false. Therefore option A is correct.

The governing ideas are complement and set difference with respect to the universal set U. Since U={1,2,3,4,5} and A={1,2}, the complement is Aᶜ=U\A={3,4,5}. The difference A\Aᶜ means the elements that belong to A but do not belong to Aᶜ. A set and its complement are disjoint, so neither 1 nor 2 occurs in Aᶜ. Consequently, removing Aᶜ from A removes nothing, and A\Aᶜ remains {1,2}. Thus option A is correct. Option B is Aᶜ itself. Option C would describe A∩Aᶜ, which is empty, not the difference in the stated order. Option D is the entire universal set and incorrectly includes 3, 4, and 5, which are not elements of A.

In a finite universal set, A and its complement Aᶜ are disjoint and together make U. Therefore their cardinalities satisfy n(A) + n(Aᶜ) = n(U). Substituting the given values gives n(A) + 12 = 30, so n(A) = 30 − 12 = 18. Thus option B is correct. The values 12 and 30 are the complement and universal-set sizes, not n(A), while 42 incorrectly adds the quantities.

The complement Aᶜ contains all elements of U that are not in A. Here, A contains the even numbers 2, 4, 6, and 8, while 9 belongs to U but not to A. Thus 9 ∈ Aᶜ. Statement B is false because 2 is in A, statement C is false because 4 is in A, and statement D incorrectly gives A instead of its complement.

The sets A = {a, b, c} and D = {d, e, f} have no common element, so A ∩ D = ∅. Their union is {a, b, c, d, e, f} = U. A set that is disjoint from A and, together with A, covers U is precisely the complement of A. Therefore D = Aᶜ, making option A correct. D is neither equal to A nor empty, and it is clearly a subset of U.

Taking ℕ here as the positive natural numbers, U = {1, 2, 3, …, 12}. Numbers not divisible by 2 are the odd numbers, so A = {1, 3, 5, 7, 9, 11}. The complement contains the remaining elements of U, namely the even numbers {2, 4, 6, 8, 10, 12}. Therefore option B is correct; option A lists A itself, not its complement.

The complement Aᶜ contains all elements of U that are absent from A. Removing 2, 3, 5, and 7 from U leaves {1, 4, 6, 8, 9, 10}, which is exactly B. Therefore, B = Aᶜ. The sets are disjoint and their union is U, but they are not equal and their intersection is not U.

Because x is an integer and x² ≤ 9, we have −3 ≤ x ≤ 3. Therefore, A = {−3, −2, −1, 0, 1, 2, 3}. The universal set is U = {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}. The complement A' consists of every element of U that is not in A. Removing the seven elements of A leaves {−5, −4, 4, 5}. Hence, option A is correct. The complement must always be taken with respect to the stated universal set U.