Complement of a Set and Its Properties

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.

A complement relative to U is normally introduced for a set A that is contained in U. Here A contains d, but d is not an element of U; instead, U contains e, which is not in A. Thus A is not a subset of U, so the stated complement setup is invalid under the usual school-level definition. The other options are false: U is not a subset of A, A is not equal to U, and A is not empty.

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.

To count students in A ∪ B, use the inclusion–exclusion formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Thus, n(A ∪ B) = 18 + 22 − 7 = 33. The universal set has 50 students, so those outside the union number 50 − 33 = 17. The intersection is subtracted because students belonging to both sets were counted twice.

The complement A′ consists of all elements of the universal set U that are not in A. If A′ = ∅, there is no element of U outside A. Since A is already a subset of U, it must contain every element of U. Therefore, A = U. The condition does not imply that either set is empty; it means A covers the whole universal set.

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.

Taking N here as the positive natural numbers, U = {1,2,3,4,5,6,7,8,9,10,11,12}. The multiples of 3 in U are A = {3,6,9,12}. The complement A' contains every element of U that is not a multiple of 3, namely {1,2,4,5,7,8,10,11}. Therefore option B is correct; it has 12 − 4 = 8 elements.

Complements must be taken relative to the universal set U. Thus A' = U − A = {4,5,6}, while B' = U − B = {1,2,6}. The elements common to both complements are found by intersection: {4,5,6} ∩ {1,2,6} = {6}. Therefore, A' ∩ B' = {6}. This also agrees with De Morgan's law: A' ∩ B' = (A ∪ B)'.

First find the intersection of A and B. The only element common to A = {a,b} and B = {b,c} is b, so A ∩ B = {b}. The complement is taken relative to U = {a,b,c,d}; therefore, (A ∩ B)' consists of every element of U except b. Hence, (A ∩ B)' = {a,c,d}, which is option B.

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.

Taking ℕ as the positive natural numbers, U = {1,2,...,20}, so n(U) = 20. The positive divisors of 20 are 1, 2, 4, 5, 10, and 20, giving n(A) = 6. The complement A′ contains the members of U that are not divisors of 20. Hence n(A′) = 20 − 6 = 14, so option C is correct.

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.

Assuming N denotes the positive natural numbers, U = {1,2,3,...,15}, so n(U) = 15. The multiples of 5 that lie in U are A = {5,10,15}, giving n(A) = 3. Since A′ contains the elements of U outside A, n(A′) = n(U) − n(A) = 15 − 3 = 12. Thus option D is correct.