Complement of a Set and Its Properties

Complements are taken relative to U. Thus \(A'=\{b,d,f\}\) and \(B'=\{a,d,e\}\). Their only common element is d, so \(A'\cap B'=\{d\}\). Equivalently, De Morgan’s law gives \(A'\cap B'=(A\cup B)'\); since \(A\cup B=\{a,b,c,e,f\}\), the only element of U outside the union is d.

A complement is determined by the universal set. Here the universal set is \(\mathbb{N}\), and \(A\) consists of all odd natural numbers. Every natural number is either odd or even, and no natural number is both. Therefore, the elements of \(\mathbb{N}\) that are not in \(A\) are precisely the even natural numbers: \(A'=\{x\in\mathbb{N}:x\text{ is even}\}\). Option B incorrectly changes the universe to \(\mathbb{Z}\), while C includes only primes.

The set \(A\) contains the positive multiples of 5 from 1 through 50: \(5,10,15,20,25,30,35,40,45,50\). There are \(50/5=10\) such multiples, so \(n(A)=10\). The universal set has 50 elements. Since \(A'\) contains all elements of \(U\) that are not in \(A\), its cardinality is \(n(A')=n(U)-n(A)=50-10=40\). Therefore option A is correct; 10 is the cardinality of A itself.

The two given relations describe a complement: \(A\cup A'=U\) means that together A and \(A'\) contain every element of the universe, while \(A\cap A'=\varnothing\) means that they have no common element. Therefore \(A'\) consists exactly of the elements of \(U\) that do not belong to A, and it can be written as \(U\setminus A\). Option A states this definition. Option B is impossible for a subset A of U, and C and D do not express the complement.

Assuming the standard school convention \(\mathbb{N}=\{1,2,3,\ldots\}\), the universal set contains 25 elements. The square numbers not exceeding 25 are \(1,4,9,16,25\), so \(n(A)=5\). The complement contains all remaining natural numbers from 1 through 25. Therefore \(n(A')=n(U)-n(A)=25-5=20\), making option A correct. Option B counts the square numbers themselves, not the elements of their complement.

The complement of a set is taken with respect to the stated universal set. Here, \(A'=\{2,4,6,8,10\}\), because these are the elements of \(U\) that are not in \(A\). Taking the complement again returns all the original elements of \(A\). Therefore, the double-complement law gives \((A')'=A=\{1,3,5,7,9\}\), so option A is correct. Option B represents only the first complement.

De Morgan’s law states that the complement of a union is the intersection of the complements. Thus, \((A\cup B)'=A'\cap B'\). An element is outside \(A\cup B\) only when it is outside both \(A\) and \(B\), which explains the intersection symbol. Therefore, option A is correct. Option B incorrectly uses a union, and options C and D do not complement both original sets correctly.

The correct De Morgan identity is \((A\cup B)'=A'\cap B'\). The complement of a union contains precisely those elements that belong to neither \(A\) nor \(B\), so they must lie in both complements. Statement A is incorrect because \((A\cap B)'=A'\cup B'\). Also, \((A')'=A\), not the empty set, and \(A\cup A'=U\), not the empty set. Hence option B is the only always-true statement.

A set and its complement are disjoint and together contain every element of the universal set. Hence, \(n(A)+n(A')=n(U)\). Substituting the given values gives \(n(A)+47=120\), so \(n(A)=120-47=73\). Therefore, option A is correct. The value 47 is the cardinality of the complement, while 120 is the cardinality of the entire universal set, not of \(A\).

Within the universal set \(U=\{1,2,\ldots,24\}\), set \(A\) contains exactly the even numbers. Its complement therefore contains every element of \(U\) that is not even, namely the odd numbers \(\{1,3,5,\ldots,23\}\). Set \(B\) is defined as precisely this collection of odd numbers. Consequently, \(A'=B\). Option B is false because the two sets are equal, not a proper subset.

The elements of U divisible by 4 are 4, 8, 12, and 16, so A = {4, 8, 12, 16}. The complement A' consists of every element of U that is not in A. Removing those four multiples of 4 from U leaves {1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15}. Thus option A is correct; option B is A itself.

A contains every number in U that is not prime. Therefore its complement contains exactly the prime numbers in U. The primes from 1 through 25 are 2, 3, 5, 7, 11, 13, 17, 19, and 23. Note that 1 is neither prime nor composite, so it remains in A rather than A'. Hence option A is correct.

For every subset A of a universal set U, the elements of A and the elements outside A together cover all of U. Therefore A union A' = U, which is the complement law of excluded alternatives. In this question, A' contains 6 through 12, and combining it with A gives every element from 1 through 12. Thus option A is correct.

The complement A′ is defined with respect to the universal set U. It contains every element of U that is not present in A. Removing a, e, and h from U leaves b, c, d, and g. The element f must not be included because f is not an element of the given universal set U. Therefore, A′ = {b, c, d, g}.

The complement U′ consists of all elements in the universal framework that are not in U. Because U already contains every element under consideration, its complement is the empty set: U′ = ∅. Therefore, A ∩ U′ = A ∩ ∅ = ∅, since the intersection of any set with the empty set is empty. Thus option A is correct.

The universal set U contains all integers from −10 through 10, including both endpoints, so n(U) = 10 − (−10) + 1 = 21. The condition |x| ≤ 3 means −3 ≤ x ≤ 3, giving A = {−3, −2, −1, 0, 1, 2, 3}, which has 7 elements. Since A' contains the elements of U that are not in A, n(A') = n(U) − n(A) = 21 − 7 = 14. Therefore, option A is correct.

Every multiple of 12 is automatically a multiple of 6, because 12 = 2 × 6. Thus every multiple of 12 belonging to U is already an element of A. The multiples of 12 in U are 12, 24, and 36, and none lies outside A. Consequently, no multiple of 12 belongs to A′, so the required number is 0.

The expression A′ ∩ B represents the elements that belong to B but do not belong to A. Since A is a subset of B, removing all 31 elements of A from the 54 elements of B leaves B − A. Therefore, n(A′ ∩ B) = n(B) − n(A) = 54 − 31 = 23. The value n(U) is not needed for this calculation.

The condition x² ≤ 16 is equivalent to |x| ≤ 4, or −4 ≤ x ≤ 4. Since x must be an integer, A = {−4, −3, −2, −1, 0, 1, 2, 3, 4}, so n(A) = 9. The universal set contains the 21 integers from −10 to 10 inclusive. Therefore, n(A′) = n(U) − n(A) = 21 − 9 = 12.

Because A is a subset of B, the elements in B that are outside A are exactly B − A. The set identity A′ ∩ B = B − A therefore applies. Removing the 64 elements of A from the 97 elements of B gives n(A′ ∩ B) = 97 − 64 = 33. The size of U is extra information and does not affect this difference.

The complement A′ consists of all odd numbers in U. The primes from 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29. Since 2 is even, it is not in A′. Every other prime in this list is odd, so the intersection is {3, 5, 7, 11, 13, 17, 19, 23, 29}.

A′ consists of the elements of U not in A: {b, c, e, f, h}. B′ consists of the elements of U not in B: {a, c, f, g}. Their intersection contains the elements common to both complements, namely c and f. Equivalently, by De Morgan’s law, A′ ∩ B′ = (A ∪ B)′; since A ∪ B = {a, b, d, e, g, h}, its complement in U is {c, f}.

The odd elements of U are B = {1, 3, 5, 7, 9, 11, 13, 15}. The square numbers in U are {1, 4, 9, 16}; among these, the odd square numbers are 1 and 9. A′ contains numbers that are not squares, so intersecting A′ with B removes 1 and 9 from the odd set. The result is {3, 5, 7, 11, 13, 15}, which is option A.

The prime numbers from 1 through 25 are 2, 3, 5, 7, 11, 13, 17, 19, and 23, so there are 9 primes. Set A contains the multiples of 5, including 5 itself. Among the primes, only 5 belongs to A; every other listed prime belongs to A′. Therefore the number of primes in A′ is 9 − 1 = 8. Hence option C, not option A, is correct.

A′ ∩ B′ consists of elements that are neither even nor square. Thus they must be odd, non-square natural numbers not exceeding 40. There are 20 odd numbers from 1 through 40. The odd square numbers in this range are 1, 9, and 25; these must be excluded. Therefore n(A′ ∩ B′) = 20 − 3 = 17, so option A is correct. The square 49 is outside U and is not considered.