Complement of a Set and Its Properties

Take each complement relative to U. Since A contains 1, 2, 3, and 4, A′ = {5, 6}. Since B contains 3, 4, and 5, B′ = {1, 2, 6}. The common element in A′ and B′ is therefore only 6. Thus A′ ∩ B′ = {6}, so option A is correct. This also agrees with De Morgan’s law: A′ ∩ B′ = (A ∪ B)′.

Relative to U, A′ = {2, 4, 6, 7}, because these elements are not in A. Similarly, B′ = {1, 4, 5, 7}. Taking their union gives {1, 2, 4, 5, 6, 7}. Therefore A′ ∪ B′ = {1, 2, 4, 5, 6, 7}, so option A is correct. De Morgan’s law provides the same result: A′ ∪ B′ = (A ∩ B)′.

The complement of A is defined relative to the universal set U. It consists of every element that belongs to U but does not belong to A, written as A′ = U \ A or A′ = {x : x ∈ U and x ∉ A}. The condition x ∈ U is essential; without it, the set could include elements outside the stated universe. Therefore option A is the precise definition.

By definition, A′ contains the elements of U that are not in A. Consequently, no element can belong to A and A′ at the same time. Their intersection is therefore empty: A ∩ A′ = ∅. The complementary identity for the union is A ∪ A′ = U, not the empty set. Also, A′ need not be a subset of A, so option A is the only statement that is always true.

Taking complements reverses the direction of set inclusion. Since every element of A is also in B, any element that is outside B must certainly be outside A. Thus every element of B′ belongs to A′, which proves B′ ⊆ A′. This is called the order-reversing property of complements. The statement does not require A = U or B = ∅, so option A is correct.

The complement A′ is defined as the set of elements of U that are not in A. Therefore every element of A′ already belongs to U, which directly gives A′ ⊆ U. The other statements are not always true: A′ may not be contained in A, U may contain elements of A, and A′ equals A only in special cases. Hence option A is correct.

A set and its complement together contain every element of the universal set, with no element omitted. This is the fundamental identity A ∪ A′ = U. Here U is explicitly given as {1, 2, 3, 4, 5, 6}, and A′ = {1, 6}; the remaining elements form A, namely {2, 3, 4, 5}. Their union is therefore the complete universal set U. Hence option A is correct.

A′ contains precisely the elements of U that are not in A. Therefore A and A′ are disjoint: an element cannot simultaneously belong to A and to its complement. This gives the fundamental identity A ∩ A′ = ∅, regardless of the particular elements in U or A′. In this question A′ = {2, 5}, so 2 and 5 are outside A, and no common element exists. Hence option A is correct.

The complement P′ contains every element of the universal set U that is not in P. Here, U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, while P = {2, 3, 5, 7}. Removing the elements of P from U leaves {1, 4, 6, 8, 9, 10}. Notice that 1 is included because 1 is not a prime number.

Since the universal set is U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, the complement S′ contains every element of U that is not in S. Removing 1, 4, and 9 from U leaves S′ = {2, 3, 5, 6, 7, 8}. Therefore, option A is correct. The complement always depends on the specified universal set.

The complement C′ is formed by selecting the elements that belong to U but do not belong to C. The universal set has four colors: red, blue, green, and yellow. Since C contains red and green, the colors left outside C but still inside U are blue and yellow. Thus C′ = {blue, yellow}, making option A correct.

The universal set U contains the natural numbers 1 through 15, so n(U) = 15. Set M contains exactly three distinct elements: 5, 10, and 15, so n(M) = 3. Because M is a subset of U, the number of elements in its complement is n(M′) = n(U) − n(M) = 15 − 3 = 12. Hence option A is correct.

For any set A contained in a universal set U, the complement A′ contains all elements of U outside A. Here, A′ = {3, 4, 5, 6}. Taking the union of A′ and A combines every element in either set: {3, 4, 5, 6} ∪ {1, 2, 7, 8} = U. Therefore, A′ ∪ A = U, which is option A.

With respect to U, the complement of A is A′ = {1, 2, 7, 8}. The set A = {3, 4, 5, 6} and its complement have no common elements. Therefore, their intersection is empty: A′ ∩ A = ∅. This is a standard complement property, A ∩ A′ = ∅, so option A is correct.

The relevant rule is that the complement of a set contains precisely the elements of U that are not in that set. Combining A and B gives A ∪ B = {1, 2, 3, 4, 5, 6, 7}; repeated elements 4 and 5 are written only once. Comparing this union with U = {1, ..., 9}, the missing elements are 8 and 9. Thus (A ∪ B)′ = {8, 9}. Option B is A without B, and option D is the common part.

The intersection A ∩ B consists of elements common to both sets. Comparing A = {1, 3, 5, 7, 9} with B = {2, 3, 5, 8}, the common elements are 3 and 5. Thus A ∩ B = {3, 5}. Removing these from U = {1, 2, 3, 4, 5, 6, 7, 8, 9} gives (A ∩ B)′ = {1, 2, 4, 6, 7, 8, 9}.

The governing definitions are A′ = U − A and X − Y = the elements in X that are not in Y. First, A′ = {1, 3, 5, 6}, because these are the elements of U outside A = {2, 4}. A set and its complement are disjoint, so none of the elements of A′ is removed when A is subtracted. Therefore A′ − A = A′ = {1, 3, 5, 6}. Option B is A itself, not its complement.

Relative to U, the complement of A is A′ = {2, 3, 4}. Since A and A′ are disjoint, none of the elements removed from A belongs to A′. Therefore, A − A′ leaves A unchanged, so A − A′ = A = {1, 5, 6}. Do not confuse this difference with A ∩ A′, which would be empty.

The complement A′ is formed relative to the universal set U. The elements of U that are multiples of 2 are A = {2,4,6,8,10}. Removing these elements from U leaves the odd numbers {1,3,5,7,9}. Therefore, option A is correct. The complement includes every element of U that does not belong to A.

The universal set is U = {1, 2, 3, ..., 20}. Since A contains the elements greater than 12, A = {13, 14, 15, 16, 17, 18, 19, 20}, which has 8 elements. The complement A′ contains all elements of U that are not in A, namely {1, 2, ..., 12}. Therefore, n(A′) = 20 − 8 = 12. The number 12 is not included in A because the condition is x > 12, not x ≥ 12.