Complement of a Set and Its Properties

A set and its complement must satisfy two fundamental conditions. First, they are disjoint, so they have no common element and A∩A^c=∅. Second, together they contain every element of the universal set, so A∪A^c=U. These two identities provide a complete and reliable check. The other options contradict standard complement laws or state unrelated claims.

The complement A^c consists of all elements that belong to the universal set U but do not belong to A. Starting with U={2,3,4,5,6}, remove the elements 3 and 6 because they are in A. The remaining elements are 2, 4, and 5, so A^c={2,4,5}. Option B is A itself, option C incorrectly retains elements of A, and option D is not empty.

The complement A^c contains exactly those elements of U that are absent from A. The universal set lists p, q, r, s, and t, while A contains p, s, and t. Removing these three elements from U leaves q and r. Therefore, A^c={q,r}, making option A correct. Option B repeats A, option C incorrectly includes p, and option D leaves out q.

Membership in A^c means that x belongs to the universal set U but does not belong to A. This follows directly from the definition A^c=U−A. Therefore both statements in option A must be true simultaneously. Option B places x in A, which contradicts complement membership; option C contradicts x being in U; and option D violates the disjointness of A and A^c.

The complement is defined as A^c=U−A, the set of elements in U that are not in A. Here A and U contain exactly the same four elements: 5, 10, 15, and 20. Since every element of U has already been included in A, no element remains outside A. Consequently, A^c is the empty set, so option A is correct.

The complement A^c contains all elements of U that are not in A. Since A is the empty set, it contains no elements to remove from U. Therefore every element of U remains in the complement, giving A^c=U={7,8,9}. This is the standard identity ∅^c=U. Options C and D contain only part of U, while option B incorrectly repeats the empty set.

For a finite universal set, A and A^c together contain every element of U and have no common elements. Hence n(A)+n(A^c)=n(U), so n(A^c)=n(U)−n(A). Substituting the values gives n(A^c)=36−14=22. Thus option A is correct. Option C is the size of A, option D is the size of U, and option B is an incorrect sum rather than the required difference.

The complement Aᶜ contains the elements of the universal set U that are not in A. Therefore, A is obtained by removing the elements of Aᶜ from U. Removing 1 and 7 from {1, 2, 3, 4, 5, 6, 7} leaves A = {2, 3, 4, 5, 6}. Hence option A is correct. This also illustrates the identity (Aᶜ)ᶜ = A.

The complement Aᶜ is defined with respect to the stated universal set U. It consists of every element in U that does not belong to A. Since A contains the odd members 1, 3, 5, and 7, the remaining elements of U are 2, 4, 6, and 8. Therefore Aᶜ = {2, 4, 6, 8}, so option A is correct.

By definition, Aᶜ = U − A, so we list the elements of U and exclude every element belonging to A. Removing 1, 4, and 9 from U = {1, 2, 3, 4, 5, 6, 7, 8, 9} leaves {2, 3, 5, 6, 7, 8}. Option B is A itself, not its complement, while the other options either include elements of A or omit valid elements.

The complement of A contains every element of the universal set U that is not an element of A. Starting with U, remove 4, 8, and 12, because these are the elements of A. The elements left are 1, 2, 3, 5, 6, 7, 9, 10, and 11. Therefore, Aᶜ = U − A = {1, 2, 3, 5, 6, 7, 9, 10, 11}, so option A is correct. The universal set must always be used as the boundary for a complement.

The complement Aᶜ contains elements that are in U but absent from A. Starting with U = {0, 1, 2, 3, 4, 5, 6}, remove 0, 3, and 6 because they belong to A. The elements left are 1, 2, 4, and 5. Therefore Aᶜ = {1, 2, 4, 5}, so option A is the only correct answer.

Since U contains the natural numbers from 1 through 10, we can write U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. The even elements form A = {2, 4, 6, 8, 10}. The complement contains the elements of U that are not even; these are precisely the odd numbers {1, 3, 5, 7, 9}. Hence Aᶜ = {1, 3, 5, 7, 9}, making option A correct. The answer depends on the stated universal set.

Because N is the set of positive natural numbers and x < 7, the universal set is U = {1, 2, 3, 4, 5, 6}. The condition x ≤ 3 gives A = {1, 2, 3}. The complement Aᶜ consists of elements in U that do not belong to A. Removing 1, 2, and 3 from U leaves {4, 5, 6}. Therefore, option A is correct. Option B is A itself, while option D is the complete universal set.

First form the union by including every element that occurs in A or B: A ∪ B = {1, 2, 4, 5, 6}. The complement of this union contains the elements of U that are absent from it. From U = {1, 2, 3, 4, 5, 6, 7}, only 3 and 7 remain. Thus (A ∪ B)ᶜ = {3, 7}, so option A is correct.

First find the intersection: A ∩ B contains elements common to both A and B, so A ∩ B = {3, 4}. To find its complement, select all elements of U that are not in {3, 4}. From U = {1, 2, 3, 4, 5, 6, 7}, the remaining elements are {1, 2, 5, 6, 7}. Thus (A ∩ B)ᶜ = {1, 2, 5, 6, 7}, and option A is correct. Option B gives the intersection itself, not its complement.

If every element of A is also an element of B, then any element outside B must certainly be outside A. Therefore the complement of B is contained in the complement of A: Bᶜ ⊆ Aᶜ. This reversal of inclusion is called the complement or order-reversing property. The other statements are not generally true because A and B may be unequal and need not exhaust U.

The complement \(A^c\) contains all elements of the universal set \(U\) that are not in \(A\). Taking the complement once more selects exactly the elements that were originally in \(A\), so \((A^c)^c=A\). In contrast, \(A\cap A^c=\varnothing\) and \(A\cup A^c=U\). Thus, only option A is always true.

The complement \(A^c\) is defined relative to the universal set \(U\). It contains every student in the school who is not a member of \(A\), where membership in \(A\) means coming by bus. Therefore, \(A^c\) represents students who use another mode of travel or otherwise do not come by bus. It does not mean that there are no students.

The complement of \(A\) is found by taking all elements of \(U\) and removing the elements that belong to \(A\). The universal set contains red, blue, green, and yellow, while \(A\) contains red and yellow. The remaining elements are blue and green. Hence, \(A^c=\{\text{blue},\text{green}\}\), so option A is correct.