Complement of a Set and Its Properties

The complement \(A'\) contains the elements of the universal set \(U\) that are not in \(A\). The intersection \(A'\cap A\) would require an element to be in \(A\) and not in \(A\) at the same time, which is impossible. Therefore, the two sets are disjoint and \(A'\cap A=\emptyset\). This is one of the fundamental complement identities; the related union identity is \(A\cup A'=U\). Hence option C is correct.

The complement \(A'\) is formed by selecting elements that belong to the universal set \(U\) but do not belong to \(A\). The elements 1, 3, 5, 7, and 9 are removed from \(U\), leaving 2, 4, and 6. Hence \(A'=\{2,4,6\}\), so option A is correct. Option C is wrong because 8 is not even in the universal set; option B repeats \(A\), and option D contains an element already in \(A\).

The complement \(A'\) contains the elements of the universal set \(U\) that are not in \(A\). Therefore, to recover \(A\), remove the elements of \(A'\) from \(U\): \(A=U\setminus A'=\{1,2,3,4,5,6\}\setminus\{2,6\}=\{1,3,4,5\}\). Hence option B is correct. Option A is the complement itself, while C and D contain incorrect elements or omit required ones.

The complement of \(A\) relative to \(U\) is \(A'=U\setminus A\), meaning all elements that are in the universal set but not in \(A\). Removing apple and mango from \(U=\{apple, banana, grapes, mango\}\) leaves banana and grapes. Therefore \(A'=\{banana, grapes\}\), represented by option B. Option A is the original set, option C incorrectly retains mango, and option D is not empty.

The complement A′ contains precisely those elements of the universal set U that are not elements of A. Removing 2, 3, 4, and 5 from U gives A′ = U − A = {6, 7, 8, 9}. The largest element of this set is 9, so option D is correct. The element 8 is in the complement, but it is not the largest element.

The complement of A with respect to U is defined as A′ = U − A, meaning the elements in U that are not in A. Here A and U are exactly the same set, so every element of U is already in A. No element remains after removing A from U; therefore A′ = ∅ and its cardinality is 0. Thus option A is correct.

The complement A′ contains all elements of the universal set U that are not in A. Starting with U = {blue, green, yellow, pink}, remove green and pink because both belong to A. The elements left are blue and yellow, so A′ = U − A = {blue, yellow}. Therefore option B is correct; options C and D include an element that belongs to A.

The complement T′ consists of elements in U that are not in T. Removing 3, 6, and 9 from U gives T′ = {1, 2, 4, 5, 7, 8}. Among the given options, 1, 2, and 8 belong to T′, whereas 6 belongs to T itself. Therefore 6 is the only element that is not in T′, so option C is correct.

The complement is formed relative to U: A' = U \ A = {2, 3, 5, 7}. Now apply the strict condition “less than 5.” The elements 2 and 3 satisfy it, while 5 does not because it is equal to 5. Therefore, exactly two elements of A' are less than 5. This also shows why counting 5 would produce the incorrect answer 3.

All complements are taken with respect to U. First, A' = U \ A = {4, 6, 8}. Taking the complement again gives (A')' = U \ {4, 6, 8} = {2, 10}. Thus the double-complement law, (A')' = A, is verified. Option B is only the first complement, and option C is U itself, not the double complement.

The complement A' contains every element of the universal set U that is not in A. Starting with U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, remove 2, 4, 6, and 8. The remaining elements are {1, 3, 5, 7, 9}. Zero cannot be included because it is not an element of the stated universal set, so option B is incorrect.

Because B = A', taking the complement of B gives B' = (A')'. The double-complement law states that the complement of a complement is the original set, so (A')' = A. Hence B' = A = {2, 5, 7}. In fact, B itself is {1, 3, 4, 6}; that is why option B represents the single complement and not B'.

The complement A' relative to U consists of elements that are in U but not in A. Remove 10, 20, and 30 from U = {5, 10, 15, 20, 25, 30}. The elements left are 5, 15, and 25, so A' = {5, 15, 25}. Option B is A itself, while options C and D either retain or omit elements incorrectly.

The complement of A with respect to U is A' = U − A = {2, 3, 4, 5}. The set A contains {1, 6}, so A and A' have no common elements. Therefore, their intersection is empty: A' ∩ A = ∅. This is also the standard complement identity A ∩ A' = ∅. Option B is A' itself, not its intersection with A, so option C is correct.

First find the union: A ∪ B = {1, 2, 3, 4, 5}. The complement is taken relative to the universal set U, so remove every element of this union from U. The remaining elements are {6, 7, 8}. Hence (A ∪ B)' = {6, 7, 8}. Option A gives the union itself, while option C incorrectly includes 1 and 2; therefore option B is correct.

The digits in the decimal system form the universal set U = {0,1,2,3,4,5,6,7,8,9}. The complement A′ consists of every element of U that is not an element of A. Removing 2, 4, 6, and 8 from U leaves {0,1,3,5,7,9}. Therefore option A is correct. The zero must be included because 0 is also a digit.

The complement A′ contains the elements of the universal set U that are not in A. From U = {1,2,3,4,5,6}, remove the elements 1, 3, and 5. This gives A′ = {2,4,6}, which contains three elements. Equivalently, because A is a subset of U, n(A′) = n(U) − n(A) = 6 − 3 = 3. Therefore option B is correct.

The complement A′ is defined relative to the universal set U. It contains all elements of U that are not elements of A. Removing 2, 4, 6, and 8 from U gives A′ = {1, 3, 5, 7}. Therefore, 5 belongs to A′. The numbers 2, 4, and 8 are already in A, so none of them can be in its complement.

Assuming N = {1,2,3,...}, the universal set is U = {1,2,3,4,5,6,7,8,9,10}. The primes in U are A = {2,3,5,7}. The complement A′ contains every member of U that is not prime, giving {1,4,6,8,9,10}. Number 1 is neither prime nor composite, so it correctly belongs to the complement. Option B is A itself, not A′.

The complement of A relative to U is written A′ = U \ A. It contains every element of the universal set U that is not an element of A. Removing 0, 2, and 4 from U leaves 1, 3, and 5. Therefore A′ = {1, 3, 5}, which is exactly B. A itself contains the removed elements, U contains too many elements, and the empty set contains none.