Complement of a Set and Its Properties

The complement Aᶜ of A in the universal set U is the set of all elements of U that are not in A. The condition A ∩ B = ∅ says that A and B have no common elements, while A ∪ B = U says that together they contain every element of U. Thus B contains exactly those elements of U that are outside A. Therefore, B = Aᶜ, so option A is correct.

A set and its complement are disjoint, as shown by A ∩ Aᶜ = ∅, and their union is the entire universal set, as shown by A ∪ Aᶜ = U. For two disjoint finite sets, the cardinality of their union equals the sum of their cardinalities. Therefore, n(A ∪ Aᶜ) = n(A) + n(Aᶜ). Since A ∪ Aᶜ = U, the result is n(A) + n(Aᶜ) = n(U).

First, Aᶜ relative to U is {6, 7, 8}. The double-complement law states that taking the complement twice returns the original set, so (Aᶜ)ᶜ = A = {1, 2, 3, 4, 5}. Now intersect this set with {2, 4, 6, 8}. The common elements are only 2 and 4, because 6 and 8 are not in A. Therefore, the required set is {2, 4}, which is option A.

The complement Aᶜ contains all elements of U that are not in A. Therefore, Aᶜ = {1,3,4,6,7,9,10}, which has 7 elements. Since 5 is not in Aᶜ, taking the union with {5} adds one new element. Thus Aᶜ ∪ {5} has 7 + 1 = 8 elements, so option A is correct.

The complement of \(A\) is formed by taking all elements of the universal set that are not in \(A\). Therefore, \(A^c=U\setminus A=\{1,3,5,7,8,9\}\). The expression then asks us to remove \(\{1,3,5\}\) from this complement. The elements left are \(\{7,8,9\}\), so option A is correct. Option B stops after finding the complement and does not perform the set difference.

A number divisible by both 2 and 3 must be divisible by their least common multiple, 6. Within U = {1,2,...,12}, the numbers divisible by 6 are 6 and 12, so A = {6,12}. The complement contains every other element of U: {1,2,3,4,5,7,8,9,10,11}. Hence option A is correct.

Since \(C=A^c\), first find the complement of \(A\) in the universal set: \(C=\{2,4,6,8\}\). Taking the complement of \(C\) again gives all elements of \(U\) that are not in \(C\), namely \(\{1,3,5,7\}\). Thus \(C^c=(A^c)^c=A\), so option A is correct. This illustrates the double-complement identity: the complement of a complement is the original set.

A partition of U consists of non-overlapping subsets whose union is U. The equation A ∩ Aᶜ = ∅ shows that A and its complement are disjoint, while A ∪ Aᶜ = U shows that together they contain every element of U. Thus A and Aᶜ form a partition of U, making option A correct.

A set and its complement are always disjoint: A ∩ Aᶜ = ∅. If A = Aᶜ, then this would imply A ∩ A = ∅, so A would have to be empty. However, the complement of the empty set is U, and because U is nonempty, ∅ ≠ U. Therefore equality A = Aᶜ is impossible, so option A is correct.

The complement of A is defined relative to a specified universal set: Aᶜ = U \ A. Consequently, even if the elements of A remain unchanged, changing U can add or remove elements from the complement. For example, if U expands by adding an element not in A, that element enters Aᶜ. Hence Aᶜ may change, so option A is correct.

A complement depends on the universal set being used. Relative to \(U_2\), the complement of \(A\) is \(U_2\setminus A=\{3,4,5,6\}\). Relative to \(U_1\), it is \(U_1\setminus A=\{3,4\}\). Taking the first complement minus the second gives \(\{3,4,5,6\}\setminus\{3,4\}=\{5,6\}\). Hence option A is correct. The result shows why the universal set must always be specified.

First find the complement of \(A\) in \(U\): \(A^c=U\setminus A=\{5,6,7,8,9,10\}\). Intersecting this with \(B=\{3,4,5,6\}\) gives \(A^c\cap B=\{5,6\}\), because 5 and 6 are the only elements common to both sets. Finally, take the complement of \(\{5,6\}\) in \(U\): \(U\setminus\{5,6\}=\{1,2,3,4,7,8,9,10\}\). Therefore, option A is correct.

The complement of A with respect to U is Aᶜ={1,3,5,7,9}. Therefore, Aᶜ∪B={1,2,3,4,5,7,9}. The elements of U not in this union are 6 and 8, so (Aᶜ∪B)ᶜ={6,8}. The same result follows directly from De Morgan’s law: (Aᶜ∪B)ᶜ=A∩Bᶜ. Since Bᶜ={5,6,7,8,9}, the intersection with A is {6,8}.

Aᶜ∩Bᶜ contains numbers from 1 through 18 that are neither divisible by 3 nor even. The odd numbers in this range are 1,3,5,7,9,11,13,15,17. Removing the odd multiples of 3, namely 3,9, and 15, leaves {1,5,7,11,13,17}. Thus the set has 6 elements. Equivalently, De Morgan’s law gives Aᶜ∩Bᶜ=(A∪B)ᶜ.

Use De Morgan’s law: Aᶜ∪Bᶜ=(A∩B)ᶜ. The common elements of A and B are A∩B={2,4,6}, so the intersection has 3 elements. Since U has 12 elements, its complement contains 12−3=9 elements. Directly, the union of the two complements contains every element except 2, 4, and 6, confirming that the answer is 9.

With respect to U, Aᶜ={4,5,6,7,8}, because these are the elements not in A. Similarly, Bᶜ={1,2,3,6,7,8}. The common elements of these two complements are 6, 7, and 8. Hence Aᶜ∩Bᶜ={6,7,8}. De Morgan’s law provides the same result: Aᶜ∩Bᶜ=(A∪B)ᶜ, and A∪B={1,2,3,4,5}.

Let n(Aᶜ)=x. The given relation says n(A)=3x. Because A and Aᶜ partition the finite universal set, their cardinalities add to n(U): n(A)+n(Aᶜ)=64. Substituting gives 3x+x=64, so 4x=64 and x=16. Therefore n(Aᶜ)=16. The value 48 is n(A), not the requested complement cardinality.

Let n(A)=x. Then the condition n(Aᶜ)=2n(A) gives n(Aᶜ)=2x. A set and its complement partition U, so n(A)+n(Aᶜ)=n(U). Hence x+2x=54, or 3x=54. Dividing by 3 gives x=18. Therefore n(A)=18, while the complement has 36 elements. This also confirms that the two cardinalities add to 54.

The set \(A^c\cap B^c\) contains numbers in U that are divisible by neither 2 nor 3. Listing the integers from 1 to 18 and removing all even numbers and all multiples of 3 leaves \(\{1,5,7,11,13,17\}\). This set has six elements. Equivalently, De Morgan’s law gives \(A^c\cap B^c=(A\cup B)^c\), which describes numbers divisible by neither divisor.

First combine the sets: \(A\cup B=\{1,2,4,6,8,9,10,12,14,16\}\). The complement contains every element of U not appearing in this union. Checking the integers 1 through 16 leaves \(\{3,5,7,11,13,15\}\). Therefore option A is correct. Notice that repeated elements such as 4 and 16 are written only once in a union.

The set \(B\) consists of all odd numbers from 1 to 20. The odd prime numbers in this range are 3, 5, 7, 11, 13, 17, and 19. Removing these from B leaves the odd, non-prime numbers \(\{1,9,15\}\). Note that 1 is not prime, because a prime number must have exactly two distinct positive divisors.

The condition \(x^2<9\) means \(-3<x<3\). Because x is restricted to integers, \(A=\{-2,-1,0,1,2\}\). The universal set contains all integers from -5 through 5, so removing A leaves \(A^c=\{-5,-4,-3,3,4,5\}\). The endpoints -3 and 3 are included in the complement because their squares equal 9, not a value less than 9.

The complement is taken inside the universal interval \([-3,7]\). Set A contains every real number strictly between -1 and 4, but it excludes the endpoints -1 and 4. Therefore those endpoints belong to the complement. The portions outside A, while remaining inside U, are \([-3,-1]\) and \([4,7]\), so \(A^c=[-3,-1]\cup[4,7]\).

The complement must contain points of U that are not in A. Since A=[0,2] includes both 0 and 2, those endpoints must be excluded from the complement. The left part is therefore (-4,0), while the right part is (2,6]. The endpoint -4 is excluded because it is excluded from U, and 6 is included because U includes 6. Hence option A is correct.

Taking the complement reverses the direction of set inclusion. Since A ⊆ B, every element outside B is also outside A, so Bᶜ ⊆ Aᶜ. Similarly, B ⊆ C gives Cᶜ ⊆ Bᶜ. Combining these results gives Cᶜ ⊆ Bᶜ ⊆ Aᶜ. Therefore, option A is correct. Equality is not necessary, and the intersection of two complements cannot generally be the whole universal set.