Complement of a Set and Its Properties

The complement A' relative to U consists of elements that are in U but not in A. Remove 10, 20, and 30 from U = {5, 10, 15, 20, 25, 30}. The elements left are 5, 15, and 25, so A' = {5, 15, 25}. Option B is A itself, while options C and D either retain or omit elements incorrectly.

The complement of A with respect to U is A' = U − A = {2, 3, 4, 5}. The set A contains {1, 6}, so A and A' have no common elements. Therefore, their intersection is empty: A' ∩ A = ∅. This is also the standard complement identity A ∩ A' = ∅. Option B is A' itself, not its intersection with A, so option C is correct.

First find the union: A ∪ B = {1, 2, 3, 4, 5}. The complement is taken relative to the universal set U, so remove every element of this union from U. The remaining elements are {6, 7, 8}. Hence (A ∪ B)' = {6, 7, 8}. Option A gives the union itself, while option C incorrectly includes 1 and 2; therefore option B is correct.

The digits in the decimal system form the universal set U = {0,1,2,3,4,5,6,7,8,9}. The complement A′ consists of every element of U that is not an element of A. Removing 2, 4, 6, and 8 from U leaves {0,1,3,5,7,9}. Therefore option A is correct. The zero must be included because 0 is also a digit.

The complement A′ contains the elements of the universal set U that are not in A. From U = {1,2,3,4,5,6}, remove the elements 1, 3, and 5. This gives A′ = {2,4,6}, which contains three elements. Equivalently, because A is a subset of U, n(A′) = n(U) − n(A) = 6 − 3 = 3. Therefore option B is correct.

The complement A′ is defined relative to the universal set U. It contains all elements of U that are not elements of A. Removing 2, 4, 6, and 8 from U gives A′ = {1, 3, 5, 7}. Therefore, 5 belongs to A′. The numbers 2, 4, and 8 are already in A, so none of them can be in its complement.

The complement of a set contains the elements of the universal set that are not in that set. When the set itself is the universal set U, every element under consideration is already in U. There is therefore no element of U outside U. Consequently, U′ = U \ U = ∅. Option A is U itself, not its complement, while the singleton options omit several elements without justification.

The complement A′ contains the elements of U that are not in A. Since A is a subset of U, we can use n(A′) = n(U) − n(A). Here n(U) = 6 and n(A) = 2, so n(A′) = 6 − 2 = 4. Directly, A′ = {3, 4, 5, 6}, which confirms the result. Therefore option C is correct; option D counts all elements of U without removing A.

The complement A′ contains every element of the universal set U that does not belong to A. Removing 3, 5, and 7 from U gives A′ = {1, 2, 4, 6}. The smallest element of this set is 1, so option A is correct. Numbers 3 and 5 are not in the complement because they already belong to A, while 2 is larger than 1.

The universal set U contains seven days, while A contains three specified days. The complement A′ consists of the days in U that are not in A: Tuesday, Thursday, Saturday, and Sunday. Therefore, |A′| = |U| − |A| = 7 − 3 = 4. Thus option B is correct; three is the size of A, not its complement.

The complement A′ contains exactly those elements of the universal set U that are not elements of A. Therefore, A′ = U − A. Starting with U = {a, b, c, d, e}, remove b and d because they belong to A. The elements left are a, c, and e, so A′ = {a, c, e}. Option B is A itself, option C is the entire universal set, and option D would occur only if A were equal to U.

The complement A′ consists of the elements of U that are absent from A. From U = {1, 2, 3, 4, 5, 6, 7, 8}, remove 1, 3, 5, and 7. This gives A′ = {2, 4, 6, 8}, which contains four elements. Equivalently, because U has 8 elements and A has 4 elements, n(A′) = n(U) − n(A) = 8 − 4 = 4. Therefore, option C is correct.

The double-complement law states that taking the complement twice returns the original set: (A′)′ = A. The first complement contains all elements of U that are not in A. Taking its complement removes those elements and retains exactly the elements originally in A. This result depends on using the same universal set U for both complements. Hence option A is correct, while A′, U, and ∅ are not generally equal to (A′)′.

A′ contains the elements of U that are not in A. If A′ equals the whole universal set U, then every element of U is outside A. Consequently, A contains no elements and must be the empty set: A = ∅. This also follows from the complement identity ∅′ = U. Option A would imply A′ = ∅, not U; option C is not a value for A, and option D is incorrect because the condition determines A uniquely.

For a finite set A contained in U, the elements of A and A′ together make up U and do not overlap. Therefore, n(U) = n(A) + n(A′), so n(A′) = n(U) − n(A). Substituting the given values gives n(A′) = 35 − 12 = 23. Thus, 23 elements belong to the complement. Option A is the size of A, option C is the size of U, and option D incorrectly adds the two quantities.

The complement A' consists of all elements of the universal set U that are not in A. When A is a subset of U, the cardinalities satisfy n(A') = n(U) − n(A). Substituting the given values gives n(A') = 60 − 28 = 32. Thus x = 32, so option B is correct. The result is also sensible because A and A' together contain all 60 elements of U.

First find the union of A and B by listing every element that appears in either set: A ∪ B = {1,2,4,7,8}. The complement of this union is taken with respect to the universal set U, so remove these elements from U. The remaining elements are U − (A ∪ B) = {3,5,6,9}. Therefore, option A is correct. Option B is the union itself, option C contains only the common element, and option D incorrectly leaves out 9.

The governing concept is complement relative to a specified universal set: A′ = U − A. Start with U = {1, 2, 3, 4, 5, 6, 7, 8} and remove every element of A, namely 2, 3, 5, and 7. The elements left are 1, 4, 6, and 8, so A′ = {1, 4, 6, 8}. Option B repeats A rather than finding its complement, and the other options omit valid remaining elements.

The complement \(A'\) contains every element of the universal set \(U\) that is not an element of \(A\). Starting with \(U=\{a,b,c,d,e,f\}\), remove \(a,c,f\), because those elements belong to \(A\). The remaining elements are \(b,d,e\). Hence \(A'=\{b,d,e\}\), making option A correct.

The complement \(A'\) contains all elements of the universal set \(U\) that are not present in \(A\). The universal set has 10 elements, while \(A\) has 3 elements: 1, 4, and 9. Therefore, \(n(A')=n(U)-n(A)=10-3=7\). In fact, \(A'=\{2,3,5,6,7,8,10\}\), which confirms that it contains seven elements. Hence, option C is correct.

A set and its complement are disjoint and together contain every element of the universal set. Therefore, their cardinalities satisfy \(n(A)+n(A')=n(U)\). Substituting the given values gives \(n(A)+19=80\), so \(n(A)=80-19=61\). Hence option C is correct. The answer cannot exceed 80 because \(A\subseteq U\).

The complement of A relative to U is defined as A′ = U \ A, meaning all elements of U that are not in A. Since A is empty, none of the elements of U are removed. Therefore A′ = U = {2, 4, 6}. Option A is A itself, option C contains an element not in U, and option D is a power set rather than a complement.

The complement of a set is taken with respect to the universal set \(U\). Thus, \(A'=U\setminus A=\{1,3,5,7,8\}\), because these elements are in \(U\) but not in \(A\). Similarly, \(B'=U\setminus B=\{1,2,3,5,7\}\). The intersection contains only elements common to both complements: \(A'\cap B'=\{1,3,5,7\}\). Therefore, option A is correct. This also agrees with De Morgan’s law: \(A'\cap B'=(A\cup B)'\).

The complement A' contains the elements of the universal set U that are not in A. Therefore, A consists of all elements of U that are not in A'. Subtracting {4, 10} from {2, 4, 6, 8, 10, 12} leaves {2, 6, 8, 12}. Thus option A is correct. The result also satisfies A ∩ A' = ∅ and A ∪ A' = U.

The governing definition is A′ = U − A: the complement contains precisely those elements of U that are absent from A. Removing 2, 4, 6, and 8 from U = {1, 2, 3, 4, 5, 6, 7, 8, 9} leaves the odd elements {1, 3, 5, 7, 9}. Hence option A is correct. Option B is A itself, option C is the whole universal set, and option D would apply only if A equalled U.