Complement of a Set and Its Properties

A set and its complement have two defining properties: their union is the universal set, and their intersection is empty. The equations A ∪ A′ = U and A ∩ A′ = ∅ state exactly these properties. Thus A′ contains every element of U that is not in A, so A′ is the complement of A. Therefore, option C is correct.

The complement operation is reversible because the complement of a complement is the original set. Starting with \(A'=B'\), take complements on both sides: \((A')'=(B')'\). Thus \(A=B\). The other statements do not necessarily follow from equality of the complements, so option C is correct.

The universal set contains the integers from 0 through 15. Among the options, 2, 3, and 13 are prime, so they belong to A. The number 9 is composite because \(9=3\times3\), and it is in U but not in A. Therefore, 9 belongs to the complement \(A'=U\setminus A\), making C correct.

The complement of a set consists of all elements of the universal set that are not in that set. Here A contains the natural numbers up to 100 that are not divisible by 2. Therefore, A' contains precisely those elements of U that are divisible by 2, namely the even natural numbers from 1 to 100.

The relevant governing concept is the double-complement law. For any subset A of a universal set U, A' contains exactly the elements of U outside A. Taking the complement again removes those outside elements and restores the original members of A, so (A')' = A. Although A is the set of even numbers from 2 through 18, the identity does not require listing them. Therefore option A is correct; A' is only the first complement, while U and ∅ are not generally the result.

A′ consists of all elements of the universal set U that do not belong to A. Assuming A ⊆ U, the elements of U are divided into two disjoint parts: A and A′. Hence, |U| = |A| + |A′|, so n = r + |A′|. Rearranging gives |A′| = n − r. Thus option B is correct.

For every element of the universal set U, either it belongs to A or it does not belong to A. The elements not in A form A′, so combining A and A′ gives the whole universal set: A ∪ A′ = U. No element can simultaneously belong to A and its complement, so A ∩ A′ = ∅. Therefore, option A is correct.

A′ contains the elements of U that are not in A. Therefore, A is obtained by removing 1, 5, 7, and 11 from U. The remaining elements are 2, 3, 4, 6, 8, 9, 10, and 12. Hence A = U − A′ = {2, 3, 4, 6, 8, 9, 10, 12}, so option A is correct. The answer depends on the stated universal set.

The difference A − B consists of elements that belong to A but do not belong to B; equivalently, A − B = A ∩ B'. If this set is empty, there is no element of A outside B. Therefore every element of A must be in B, which is exactly the statement A ⊆ B. The other conclusions do not necessarily follow.

One of De Morgan’s laws states that the complement of an intersection equals the union of the complements: (A ∩ B)' = A' ∪ B'. The other related law is (A ∪ B)' = A' ∩ B'. Thus, complementation reverses the operation between intersection and union. Option B fails to change the operation, option C uses the wrong operation, and option D contradicts the double-complement law.

A set and its complement are disjoint by definition, so A ∩ A′ is always the empty set. Also, A ∪ A′ = U and (A′)′ = A are standard complement identities. Since the universal set U is non-empty, the empty set cannot equal U. Therefore, A ∩ A′ = U is always false.

The universal set U has 18 elements. The given complement A′ contains 7 elements. Since A and A′ partition U into disjoint parts, |A| + |A′| = |U|. Therefore, |A| = 18 − 7 = 11. The actual elements of A are the members of U that are not listed in A′, but only the cardinality is required.

The difference A − B contains the elements that are in A but not in B. Since A = {1, 2, 3, 4, 5, 6} and the common elements with B are {4, 5, 6}, we get A − B = {1, 2, 3}. The complement is taken with respect to U = {1, 2, …, 12}, so every element of U except 1, 2, and 3 remains. Thus (A − B)′ = {4, 5, 6, 7, 8, 9, 10, 11, 12}.

The complement \(B'\), relative to \(U\), contains the elements not listed in \(B\); here these are the odd numbers from 1 to 15. From \(A=\{1,4,9,16\}\), the elements that are not in \(B\) are 1 and 9. Hence \(A\cap B'=\{1,9\}\), so option A is correct.

The complement is defined by \(A'=U\setminus A\), so it contains the elements of \(U\) that are outside \(A\). If this complement is empty, no element of \(U\) lies outside \(A\). Since \(A\) is a subset of \(U\), it must contain every element of \(U\), and therefore \(A=U\).

De Morgan's law states that the complement of a union equals the intersection of the complements: (A ∪ B)' = A' ∩ B'. This identity is valid whether or not A and B overlap. The given condition A ∩ B = ∅ only tells us that A and B are disjoint; it does not change the De Morgan identity. Therefore, option A is the unique correct answer.

The universal set U contains the integers from 1 through 21. Its odd elements are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, and 21, so there are 11 odd numbers. Among these, 3, 9, 15, and 21 are divisible by 3 and therefore belong to A, not A′. Removing them leaves 11 − 4 = 7 odd elements in A′. Hence, option B is correct.

A set and its complement are disjoint, so A ∩ A' = ∅ for every set A. Since the universal set U is explicitly non-empty, ∅ cannot equal U. The other statements are standard complement identities: A ∪ A' = U, (A')' = A, and the complement of U relative to U is ∅. Therefore, statement C cannot always be true.

The universal set U contains all integers from 1 through 24, so |U| = 24. The given complement A′ contains 8 elements. Since A and A′ are disjoint and together make U, |A| + |A′| = |U|. Therefore, |A| = 24 − 8 = 16. Hence, option C is correct. This uses the complement cardinality property for a finite universal set.

A consists of the numbers in U that are not divisible by 12. Therefore, its complement A′ consists exactly of the numbers in U that are divisible by 12. The positive multiples of 12 not exceeding 144 are 12, 24, 36, …, 144. Their number is 144 ÷ 12 = 12, because the kth multiple is 12k and 12k ≤ 144 gives k ≤ 12. Hence |A′| = 12, so option C is correct.

The complement of a set is taken with respect to the universal set. Every element of the universal set belongs to \(U\), so there is no element left outside it; consequently, the complement of the universal set is empty: \(U'=\varnothing\). Thus the given statement identifies \(U\) as the universal set. The other options are not generally implied.

The complement of A ∪ B contains the elements of U that are divisible by neither 3 nor 5. The largest element of U is 45, but it is divisible by both 3 and 5. The next number, 44, is not divisible by 3 and is not divisible by 5. Therefore 44 belongs to the complement and is its greatest element. Option A is correct.

If the union of A and B is empty, neither set can contain any element; otherwise that element would belong to their union. Thus A = ∅ and B = ∅. Their complements relative to the universal set U are both U. Consequently A' ∩ B' = U ∩ U = U. This also follows directly from De Morgan’s law: A' ∩ B' = (A ∪ B)' = ∅' = U. Therefore option A is correct.

By De Morgan’s law, A' ∩ B' = (A ∪ B)'. The given condition therefore says that the complement of A ∪ B is the entire universal set U. The only set whose complement is U is the empty set, because ∅' = U. Taking complements on both sides also gives A ∪ B = ∅. Thus option A is the only correct conclusion.

The prime numbers in \(U\) are \(2,3,5,7,11,13,17,19\). The complement contains every element of the stated universal set that is not prime. Thus it contains 1 and all composite numbers from 4 to 20: \(A'=\{1,4,6,8,9,10,12,14,15,16,18,20\}\). Remember that 1 is neither prime nor composite, but it is still in the complement.