Complement of a Set and Its Properties

The universal set U contains the natural numbers 1 through 20, so |U| = 20. The positive factors of 20 are A = {1,2,4,5,10,20}, giving |A| = 6. Every element of U is either in A or in its complement, so |Aᶜ| = |U| − |A| = 20 − 6 = 14. Therefore option C is correct; six is the size of A, not Aᶜ.

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.

Relative to U, the complement of A is Aᶜ = {3,4,5}, because these are the elements of U that are not in A. The sets Aᶜ and A are disjoint, so removing A from Aᶜ removes nothing. Consequently, Aᶜ \ A remains {3,4,5}. Thus the result is option A. This also illustrates that subtracting a disjoint set leaves the original set unchanged.

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.

First find the difference A \ B. The common elements 4, 5, and 6 are removed from A, leaving A \ B = {1, 2, 3}. The complement is taken relative to U, so it contains every element of U except 1, 2, and 3. Hence (A \ B)ᶜ = {4, 5, 6, 7, 8, 9, 10}, which is option B. Option A is the difference itself, not its complement.

First take the complement of A in U: Aᶜ = {5, 6, 7, 8}. Its union with B = {3, 4, 5, 6} is {3, 4, 5, 6, 7, 8}. The complement of this union within U consists of the elements left out, namely {1, 2}. Thus (Aᶜ ∪ B)ᶜ = {1, 2}, so option A is correct. Option C is the union before complementation.

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.

The perfect squares among the integers from 1 through 15 are 1, 4, and 9, so n(A) = 3. The universal set U has 15 elements. Since Aᶜ contains all elements of U that are not in A, use n(Aᶜ) = n(U) − n(A) = 15 − 3 = 12. Therefore option B is correct. The value 3 is not listed, while 15 ignores the excluded squares.

The perfect squares in U = {1, 2, 3, …, 16} are 1, 4, 9, and 16, so A contains 4 elements. The universal set contains 16 elements altogether. Therefore the complement contains the remaining elements: n(Aᶜ) = n(U) − n(A) = 16 − 4 = 12. Hence option C is correct. Option A is n(A), and option D is the size of U, not its complement.

The set A contains 3 and every real number greater than 3, so A = [3, ∞). The complement is taken with respect to the universal set R, meaning we select all real numbers not belonging to A. These are precisely the real numbers less than 3. The endpoint 3 is excluded because 3 belongs to A. Therefore Aᶜ = {x ∈ R : x < 3}, so option B is correct.

The complement of A in U is Aᶜ = {7,8,9,10,11,12}. To find Aᶜ ∩ B, retain only the elements common to this complement and B. The elements of B are 2, 4, 6, 8, 10, and 12; among them, 8, 10, and 12 belong to Aᶜ. Hence Aᶜ ∩ B = {8,10,12}, making option A correct.

First form the union of all three sets. A contributes 1, 2, 3, 4; B adds 5 and 6; and C adds 7 and 8. Thus A ∪ B ∪ C = {1,2,3,4,5,6,7,8}. Since the universal set also contains 9, the only element outside this union is 9. Therefore (A ∪ B ∪ C)ᶜ = {9}, so option A is correct.

A number divisible by both 2 and 3 must be divisible by their least common multiple, 6. Between 1 and 30, the multiples of 6 are 6, 12, 18, 24, and 30, so n(A) = 5. The universal set has 30 elements, and Aᶜ contains all remaining elements. Therefore n(Aᶜ) = 30 − 5 = 25, so option B is correct.

The prime numbers in U are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29, so |A| = 10. The multiples of 3 from 1 to 30 are 3, 6, 9, 12, 15, 18, 21, 24, 27 and 30, so |B| = 10. Their only common element is 3, hence |A ∩ B| = 1. By inclusion-exclusion, |A ∪ B| = 10 + 10 − 1 = 19. Therefore, |(A ∪ B)'| = |U| − |A ∪ B| = 30 − 19 = 11. Thus, option A is correct.