Complement of a Set and Its Properties

Students who do not learn music are represented by the complement Aᶜ of the set A of music learners. Assuming A is a subset of the universal group U, the number of students in the complement is n(Aᶜ) = n(U) − n(A). Substituting the given values gives 60 − 18 = 42. Therefore, 42 students do not learn music.

The complement Aᶜ contains all elements of the universal set U that do not belong to A. Therefore, n(Aᶜ) means the number of students in the class who are not included in A. Numerically, n(Aᶜ) = n(U) − n(A) = 45 − 28 = 17. Thus, option A gives the correct meaning as well as the resulting count.

A proper subset A of U is contained in U but is not equal to U. Consequently, at least one element of U is missing from A. That missing element, or those missing elements, belong to the complement Aᶜ. Hence Aᶜ must contain at least one element and cannot be empty. Also, Aᶜ is always a subset of U, so option D is false.

A complement is not determined by A alone; it is defined with respect to a specified universal set. Relative to U₁, Aᶜ = U₁ − A = {1, 3}. Relative to U₂, Aᶜ = U₂ − A = {1, 3, 5}. Thus the extra element 5 appears when the universal set changes. Set A itself remains unchanged, so option A is correct.

For B to be the complement of A in the universal set U, A and B must satisfy two conditions. First, they must be disjoint, so A ∩ B = ∅. Second, together they must contain every element of U, so A ∪ B = U. These conditions ensure that B contains exactly the elements of U that are outside A. Therefore, option A is the correct test.

The complement of a set A, written as Aᶜ, consists of all elements in the universal set U that are not elements of A. Here, U contains 10, 11, 12, 13, 14, and 15, while A contains 11, 13, and 15. Removing the elements of A from U leaves 10, 12, and 14. Therefore, Aᶜ = {10, 12, 14}.

The multiples of 5 in U are 5, 10, and 15, so A = {5, 10, 15}. The complement is found by taking U − A. Removing these three multiples from the numbers 1 through 15 leaves {1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14}. Therefore option A is correct; option B lists A itself, not its complement.

The complement Aᶜ contains the elements of U that are not in A. Set A contains all five listed vowels, while U also contains the letters b, c, and d. Removing the vowels from U leaves exactly {b, c, d}. Therefore option A is correct. This example also shows why the universal set must always be specified before finding a complement.

First form the union by collecting every element appearing in A or B: A ∪ B = {1, 2, 3, 7, 8}. The complement of this union contains the elements of U that are absent from the union. Removing 1, 2, 3, 7, and 8 from U leaves {4, 5, 6}. Hence (A ∪ B)ᶜ = {4, 5, 6}, so option A is correct.

The intersection A ∩ B contains elements common to both sets. The only common element is 6, so A ∩ B = {6}. To find its complement relative to U, remove 6 from U. The result is {1, 2, 3, 4, 5, 7, 8, 9}. Therefore option A is correct, while option B gives the intersection before taking its complement.

A and its complement Aᶜ are disjoint, and together they make the universal set U. Therefore their cardinalities satisfy n(A) + n(Aᶜ) = n(U). Substituting the given values gives n(A) + 31 = 72, so n(A) = 72 − 31 = 41. Thus option A is correct. The other values represent either a sum or one of the given cardinalities.

The complement Aᶜ contains all elements of the universal set U that are not in A. Since A = (3, 8], the number 3 is excluded from A and therefore belongs to Aᶜ. The number 8 is included in A, so it must be excluded from the complement. Thus, within U = [0, 12], the complement is [0, 3] ∪ (8, 12].

First list the integers in the universal set: U = {-3,-2,-1,0,1,2,3,4}. The complement Aᶜ consists of elements of U that are absent from A. Removing -2, 0, 2, and 4 leaves {-3,-1,1,3}. Therefore option A is correct. The universal set is essential because a complement is always defined relative to it.

Students who do not like mathematics form the complement Aᶜ of A. When A is a subset of the universal set U, the cardinality rule is n(Aᶜ) = n(U) - n(A). Substituting the given values gives n(Aᶜ) = 55 - 37 = 18. Thus 18 students do not like mathematics; 37 is the size of A and 55 is the total class size.

By definition, the complement B = Aᶜ contains every element of the universal set U that is not in A. Therefore A and B have no common element, so A ∩ B = ∅. Together, A and its complement contain every element of U, so A ∪ B = U. These two identities characterize a complement and make option A the only valid statement.

The complement consists of the points in U that do not belong to A. In A = (1, 4], the endpoint 1 is excluded, so 1 belongs to the complement. The endpoint 4 is included, so it does not belong to the complement; the interval resumes immediately after 4. Therefore, relative to U = [-2, 6], Aᶜ = [-2, 1] ∪ (4, 6].

The complement \(A^c\) consists of exactly those elements of the universal set \(U\) that are not elements of \(A\). Therefore, no element can belong to both \(A\) and \(A^c\), so their intersection is empty: \(A\cap A^c=\varnothing\). The related identity is \(A\cup A^c=U\), not \(A\). Thus, option A is the only universally valid statement.

The universal set contains the 50 natural numbers from 1 to 50. Its elements divisible by 5 are \(5,10,15,20,25,30,35,40,45,50\), so \(n(A)=10\). A and its complement partition U, meaning \(n(U)=n(A)+n(A^c)\). Therefore, \(n(A^c)=50-10=40\), making option A correct.

The integers from -5 through 5 give 11 elements in U. Solving \(x^2=9\) gives \(x=3\) or \(x=-3\), both of which belong to U; therefore \(A=\{-3,3\}\) and \(n(A)=2\). The complement has the remaining elements, so \(n(A^c)=n(U)-n(A)=11-2=9\). Thus option A is correct.

The complement Aᶜ contains all students in the universal set who do not belong to A. Since A and Aᶜ are disjoint and together make U, their cardinalities satisfy n(A) + n(Aᶜ) = n(U). Substituting the given values gives n(A) + 34 = 90, so n(A) = 90 − 34 = 56. Therefore, 56 students like science, making option A correct.

The perfect squares in U from 1 through 25 are 1, 4, 9, 16, and 25. Thus A has 5 elements. The complement Aᶜ consists of every element of U that is not a perfect square. Because U has 25 elements, n(Aᶜ) = n(U) − n(A) = 25 − 5 = 20. Hence, option A is the only correct answer.

Aᶜ is defined relative to the stated universal set U. It contains every element of U that is not an element of A. Starting with U = {1, 2, 3, 4, 5, 6} and removing A = {1, 2, 3} leaves {4, 5, 6}. Therefore, Aᶜ = {4, 5, 6}, so the set B satisfying B = Aᶜ is option A. The universal set must always be used when determining a complement.

The complement Aᶜ contains all elements of U that are absent from A. Since A = {a, c, e}, removing these elements from U = {a, b, c, d, e, f} leaves {b, d, f}. This remaining set is exactly B. Hence Aᶜ = {b, d, f} = B, so statement A is correct. In fact, A and B are disjoint and their union is U, which confirms that they are complements.

The digits are U = {0,1,2,3,4,5,6,7,8,9}. The even digits are A = {0,2,4,6,8}. The complement Aᶜ therefore contains the digits that are not even, namely the odd digits {1,3,5,7,9}. This set has 5 elements, so option A is correct. The digit 0 is even and is correctly included in A, not in its complement.

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.