Complement of a Set and Its Properties

The positive multiples of 8 from 1 through 80 are 8, 16, 24, 32, 40, 48, 56, 64, 72, and 80, so A has 10 elements. The complement contains all remaining elements of U. Therefore n(Aᶜ) = n(U) − n(A) = 80 − 10 = 70. Option B counts A itself, not its complement.

If every element of \(A\) is also in \(B\), then any element outside \(B\) must certainly be outside \(A\). Thus, the elements of \(B^c\) are a subset of the elements of \(A^c\), giving \(B^c\subseteq A^c\). Complementation reverses the direction of a subset relation; this is called order reversal.

The governing cardinality rule for a finite universal set is n(Aᶜ) = n(U) − n(A), because U is partitioned into A and its complement. Substituting the given values gives n(Aᶜ) = 40 − 18 = 22. Thus 22 elements are outside A but still inside U, so option B is correct. Option A repeats n(A), option C repeats the size of U, and option D incorrectly adds the two quantities.

Treat the complete class as the universal set, with 60 students in total. The cricket players form a subset containing 35 students, so the non-players form its complement. Assuming each student is counted once, the complement has 60 − 35 = 25 students. Therefore option A is correct. Option B is the number who play cricket, option C is the total class size, and option D results from an incorrect addition.

Using the usual school convention ℕ = {1,2,3,...}, the universal set is U = {1,2,3,4,5,6,7,8,9,10}. The even elements form A = {2,4,6,8,10}. The complement relative to U consists of the remaining elements, which are precisely the odd numbers {1,3,5,7,9}. Hence option B is correct; option A is A itself.

The universal set is the set of all integers. A contains the positive integers, namely integers greater than zero. The complement must therefore contain every integer that is not positive: zero and all negative integers. This set is described by x ≤ 0, so option B is correct.

The interval A = (2,5) contains real numbers strictly greater than 2 and strictly less than 5; its endpoints are excluded. Therefore, in the universal set ℝ, the complement contains every real number at most 2 and every real number at least 5. In interval notation this is (−∞,2] ∪ [5,∞). Option A wrongly excludes the endpoints, while option D is A itself.

The closed interval A = [−1,4] includes both endpoints −1 and 4 as well as every real number between them. Consequently, its complement in ℝ must exclude both endpoints and contain only numbers less than −1 or greater than 4. Therefore Aᶜ = (−∞,−1) ∪ (4,∞), so option A is correct.

A set and its complement partition the universal set: they contain no common elements and together contain every element of U. Since Aᶜ = {2,5,8}, obtain A by removing these three elements from U = {1,2,3,4,5,6,7,8,9}. The remaining elements are {1,3,4,6,7,9}. Thus option B is correct. Option A is the given complement, not A itself.

The complement of A is defined relative to a specified universal set U. It consists of exactly those elements that belong to U but do not belong to A. Therefore, in set-builder notation, Aᶜ = {x : x ∈ U and x ∉ A}. Option B states both required conditions and is correct.

The complement Aᶜ contains exactly those elements of the universal set U that are not in A. Therefore, an element cannot belong to both A and Aᶜ at the same time. The two sets are disjoint, so their intersection contains no element: A ∩ Aᶜ = ∅. Hence option D is correct; A, Aᶜ, and U represent other set expressions, not this intersection.

For every element of the universal set U, exactly one of two possibilities holds: it belongs to A, or it does not belong to A and therefore belongs to Aᶜ. Thus A together with Aᶜ covers every element of U. By the complement law, A ∪ Aᶜ = U. Option D is correct; the empty set would describe A ∩ Aᶜ, not their union.

The complement Bᶜ is taken with respect to U, so Bᶜ = U − B = {1,2,3,8,9,10}. Now intersect this set with A = {1,2,3,4,5}. The common elements are only 1, 2, and 3. Therefore, A ∩ Bᶜ = {1,2,3}, which is option A. The elements 4 and 5 are excluded because they belong to B.

With respect to U, Aᶜ = {4,5,6} and Bᶜ = {1,2,6}. Taking their union gives every element appearing in either complement: {1,2,4,5,6}. This also follows from De Morgan’s law, Aᶜ ∪ Bᶜ = (A ∩ B)ᶜ, because A ∩ B = {3} and its complement in U is {1,2,4,5,6}. Hence option A is correct.

Calculate each complement within U: Aᶜ = {3,5,6} and Bᶜ = {1,3,5}. Their common elements are 3 and 5, so Aᶜ ∩ Bᶜ = {3,5}. De Morgan’s law gives the same result because Aᶜ ∩ Bᶜ = (A ∪ B)ᶜ, A ∪ B = {1,2,4,6}, and its complement is {3,5}. Therefore option A is correct.

The complement Aᶜ contains every element of the universal set U that is not present in A. Removing 1, 2, 3, and 4 from U gives Aᶜ = {5,6,7,8,9,10}. Since Aᶜ is a subset of B, every one of these six elements must occur in B. B may contain additional elements, but it cannot omit any of them; therefore option B is the required minimum set.

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.

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.

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 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.