Complement of a Set and Its Properties

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}.

For a finite universal set, A and Aᶜ are disjoint and together contain every element of U. Therefore, n(A)+n(Aᶜ)=n(U). Here n(U)=10 and Aᶜ={1,4,9}, so n(Aᶜ)=3. Thus n(A)=10−3=7. In fact, A is {2,3,5,6,7,8,10}, which visibly contains seven elements. Option B counts the complement rather than A.

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.

By definition, an element belongs to Aᶜ exactly when it belongs to U but does not belong to A. An element in A∩Aᶜ would therefore have to satisfy both x∈A and x∉A at the same time. This is logically impossible, so no element can be common to A and Aᶜ. Consequently, A∩Aᶜ=∅ for every subset A of U, regardless of whether A is empty or equal to U.

The complement of A relative to U contains every element of U that is not in A. Removing 1, 2, and 3 from U gives Aᶜ={4,5,6,7,8}. This is also exactly U\A and the set described by {x∈U | x∉A}. Option A, {4,5,6,7}, omits 8, which belongs to U and is not in A; therefore it is not the complement.

De Morgan’s law states that the complement of a union is the intersection of the complements: (A∪B)ᶜ=Aᶜ∩Bᶜ. An element is outside A∪B precisely when it is outside A and outside B, so it must belong to both Aᶜ and Bᶜ. The other statements either keep the wrong operation or incorrectly place complements, so they are not valid identities in general.

Because \(B=A^c\), set \(B\) contains exactly those elements of the universal set that are not in \(A\). Here, \(B=\{2,4,6,8,10\}\). Combining \(A\) and \(B\) includes every element from 1 through 10, so \(A\cup B=U\). This is the complement law: a set and its complement together cover the whole universal set.

The complement of \(A\) in \(U\) is \(B=\{1,3,5,7,9\}\). The sets \(A\) and \(B\) have no common elements: every element of \(U\) belongs to exactly one of them. Therefore their intersection is empty, \(A\cap B=A\cap A^c=\varnothing\). This is the standard complement identity and not merely a result of the particular numbers used.

The union combines all elements of the two sets: \(A\cup B=\{1,2,3,4,5,6,7,8\}=U\). The complement of a set consists of elements of U that are outside that set. Since the union already equals all of U, there are no elements left outside it. Consequently, \((A\cup B)^c=U^c=\varnothing\), making option A correct.

For any set \(A\) defined inside a universal set \(U\), its complement \(A^c\) contains precisely the elements of U that are not in A. Thus no element can belong to both sets, so \(A\cap A^c=\varnothing\). At the same time, every element of U belongs to A or to its complement, so \(A\cup A^c=U\). Therefore option A states both correct properties.

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.