Complement of a Set and Its Properties

The complement is always taken relative to U. Here A' = U − A = {2, 4, 6, 8}. Taking the complement of A' removes these even elements from U and returns the odd elements: (A')' = U − A' = {1, 3, 5, 7} = A. This is the double-complement law, (A')' = A. Therefore, option B is correct. Option A is only A', not the double complement.

The universal set consists of all integers from 0 through 10: U = {0,1,2,3,4,5,6,7,8,9,10}. The complement A' contains elements of U that are not in A. Since A contains all the even integers in this range, the remaining elements are the odd integers {1,3,5,7,9}. Thus A' = {1,3,5,7,9}, so option A is correct. Zero cannot appear in A' because it already belongs to A.

A set A and its complement A' divide the universal set U into two disjoint parts. Therefore, |A| + |A'| = |U|. Substituting the given values gives |A| + 6 = 15, so |A| = 15 − 6 = 9. Hence option B is correct. The answer cannot be 6 because that is the size of A', and it cannot be 15 because A is only one part of U. The complement relation also guarantees that the two cardinalities add to the universal-set cardinality.

The complement of A is defined relative to the universal set U: A'=U\A, or equivalently A'={x∈U : x∉A}. Since A is empty, none of the elements of U are removed. Consequently every element of U belongs to A', so A'=U={1,2,3,4,5}. The set {∅} is not the same as ∅, and 0 is not even an element of the given universal set.

The complement of \(A\) is formed by taking all elements of the universal set that are not in \(A\). Here \(A\) contains the perfect squares \(1,4,9\) from \(U\). Therefore, \(A'=U\setminus A=\{2,3,5,6,7,8\}\), which is precisely the set of non-perfect-square numbers in \(U\). Hence option B is correct; option A describes \(A\), not its complement.

The multiples of 4 in U are 4, 8, and 12, so A = {4,8,12}. The complement contains all members of U that are not in A. Removing these three elements from U gives A′ = {1,2,3,5,6,7,9,10,11}. Notice that the complement includes even numbers such as 2, 6, and 10 because they are not divisible by 4.

The complement A' contains all elements of the universal set U that are not in A. If A' is equal to the entire universal set U, then every element of U must be outside A. Therefore, A contains no elements and must be the empty set: A = ∅. If A were U, its complement would be empty instead. Hence, option A is correct.

The universal set U contains all 26 lowercase English letters. The set V contains the five vowels a, e, i, o, and u. Its complement V' therefore contains every lowercase letter that is not a vowel. Since V is a subset of U, |V'| = |U| − |V| = 26 − 5 = 21. Hence, option B is correct.

Within the universal set U, the numbers not divisible by 2 are the odd numbers, so A = {1,3,5,7,9}. The complement A′ contains every element of U that is not in A. Therefore, A′ consists of the even members of U, namely {2,4,6,8}; 10 is excluded because it is not in U.

The set A contains all prime numbers from 1 through 12: 2,3,5,7, and 11. The complement therefore contains every element of U that is not prime. Number 1 is neither prime nor composite, while 4,6,8,9,10, and 12 are composite. Hence A′ is {1,4,6,8,9,10,12}, described by option A.

A complement is always determined relative to the stated universal set. Here U is the set of natural numbers, and A contains the even natural numbers. Removing all even natural numbers from N leaves exactly the odd natural numbers, such as 1, 3, 5, 7, and so on. Therefore, A′ represents all odd natural numbers.

The complement A' contains exactly those elements of U that are not in A. Therefore A is obtained by removing A' from the universal set: A=U\A'. Removing 2, 4, and 6 from {1,2,3,4,5,6} leaves {1,3,5}. Hence A={1,3,5}, so option A is correct. The universal set is essential because complements are always relative to it.

The complement \(A'\) contains all elements of the universal set \(U\) that are not in \(A\). Therefore, its cardinality is found by subtracting the number of elements in \(A\) from the number in \(U\): \(n(A')=n(U)-n(A)=30-11=19\). Thus, option B is correct. The other options either give the size of \(A\), the size of \(U\), or an incorrect sum.

De Morgan’s law states that the union of the complements of two sets equals the complement of their intersection: A' ∪ B' = (A ∩ B)'. An element belongs to the left side if it is outside A or outside B, which means it cannot be simultaneously inside both A and B. Therefore, option A is correct.

First find the union using inclusion–exclusion: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 20 + 16 − 6 = 30. The elements in neither A nor B form the complement of A ∪ B in U. Therefore, their number is n(U) − n(A ∪ B) = 50 − 30 = 20. Hence, option B is correct; 30 is the union, not the neither region.

The complement of A ∩ B contains every element of the universal set except the elements common to A and B. Therefore, use the complement formula n((A ∩ B)') = n(U) − n(A ∩ B). With the given values, the result is 32 − 9 = 23. Hence, option B is correct. The intersection itself has 9 elements, while its complement has the remaining 23.

The complement A′ is determined relative to the universal set U. It contains all elements of U that are not in A. From U = {1, 2, 3, 4, 5}, remove 2 and 5, the elements of A. The remaining elements are 1, 3, and 4. Therefore, A′ = {1, 3, 4}. Without specifying U, the complement cannot be determined uniquely.

The complement \(A'\) contains all elements of the universal set \(U\) that are not in \(A\). Thus, for a finite universal set, \(n(A')=n(U)-n(A)\). Substituting the given values gives \(n(A')=38-16=22\). Therefore, option B is correct. The number 16 is the size of A itself, 38 is the size of the whole universal set, and 54 incorrectly adds the two quantities instead of finding the elements outside A.

The complement A' is defined relative to the universal set U. It contains every element of U that is not a member of A. Starting with U = {1, 2, 3, 4, 5, 6} and removing 1, 4, and 6 leaves 2, 3, and 5. Therefore A' = {2, 3, 5}. The answer would be empty only if A were equal to U.

First form the union by listing every element that occurs in A or B: A ∪ B = {a, b, c, d}. The complement of this union contains the elements of U that are absent from it. In U, those remaining elements are e and f. Hence (A ∪ B)' = {e, f}. This also agrees with De Morgan’s law: (A ∪ B)' = A' ∩ B'.

The universal set is divided into four regions: only A, the intersection A ∩ B, only B, and the region outside both sets. The first three regions contain 18, 12, and 20 elements, so n(A ∪ B) = 18 + 12 + 20 = 50. The outside region therefore contains 64 − 50 = 14 elements. Hence option A is correct.

A′ is the complement of A in the universal set U, so it contains precisely the elements of U that are not in A. Every element of U is therefore either in A or in A′. Their union covers the entire universal set, giving A∪A′=U. The empty set is their intersection, not their union. Thus option C is correct.

The complement A′ consists of elements in the universal set U that are not in A. Consequently, no element can belong to both A and A′ at the same time. Their common region in a Venn diagram is empty, so A∩A′=∅. Option A or A′ represents one whole region, while U represents the complete universal set, not the intersection. Therefore option C is correct.

First apply the inclusion-exclusion formula: n(A∪B)=n(A)+n(B)−n(A∩B)=45+40−15=70. The complement of A∪B contains all elements of U outside both sets. Therefore n((A∪B)′)=n(U)−n(A∪B)=100−70=30. Option C is the union size, not its complement, so option A is correct.

The governing concept is the cardinality of a complement in a finite universal set. For any subset X of U, n(X′) = n(U) − n(X), because the universal set is divided into X and the elements outside X. Taking X = A ∩ B gives n((A ∩ B)′) = 90 − 25 = 65. Option B is correct; 25 is the original intersection, 90 ignores the excluded part, and 115 exceeds the universal-set size.