Operations on Sets (Union, Intersection, Difference)

There are floor(60/2) = 30 multiples of 2 and floor(60/3) = 20 multiples of 3 in U. Elements divisible by both 2 and 3 are multiples of 6, and there are floor(60/6) = 10 of them. By inclusion–exclusion, |A ∪ B| = |A| + |B| − |A ∩ B| = 30 + 20 − 10 = 40.

For every pair of sets, A ∩ B is contained in A ∪ B. If the two are equal, then every element that belongs to either set must belong to both sets. Thus every element of A belongs to B, so A ⊆ B; similarly, every element of B belongs to A, so B ⊆ A. Mutual inclusion proves A = B. The other options describe only special cases or contradict the condition.

Use the cardinality formula |A ∪ B| = |A| + |B| − |A ∩ B|. Thus, 50 = 28 + 35 − |A ∩ B|, so |A ∩ B| = 13. The set difference A \ B contains the elements that are in A but not in B, so |A \ B| = |A| − |A ∩ B| = 28 − 13 = 15. Therefore, option A is correct.

The difference A \ B contains the elements of A that are not in B. If removing all elements of B from A leaves A unchanged, then no element of A can have belonged to B. Equivalently, there is no common element between the sets, so A ∩ B = ∅. The condition does not require B to be contained in A, nor does it imply that A is contained in B or that their union equals A.

First find the union of A and B by listing every element that occurs in either set: A ∪ B = {1, 2, 3, 4, 5, 6, 8}. Next, intersect this result with C, which means retain only elements also present in C = {1, 4, 7, 8}. The common elements are 1, 4, and 8. Therefore, (A ∪ B) ∩ C = {1, 4, 8}, so option A is correct.

Because A is a subset of B, every element of B is either already in A or belongs to B but not to A. Thus the two disjoint parts A and B \ A together partition B. Their union contains every element of B and contains nothing outside B. Consequently, A ∪ (B \ A) = B. This identity depends on the given subset condition; without A ⊆ B, it would not generally hold.

The condition A ∩ (B \ C) = ∅ says that no element can simultaneously belong to A and to B without belonging to C. Therefore, every element common to A and B must also be an element of C. Hence A ∩ B ⊆ C. The other options are stronger or unrelated statements and are not forced by the given condition.

First list the integers in A: A = {−3, −2, −1, 0, 1, 2, 3, 4}. For B, the condition x² < 10 means −√10 < x < √10, so the integer elements are B = {−3, −2, −1, 0, 1, 2, 3}. The set difference A \ B keeps elements of A that are not in B. Therefore, only 4 remains, and A \ B = {4}.

Use the inclusion–exclusion formula for three finite sets: |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |B ∩ C| − |C ∩ A| + |A ∩ B ∩ C|. Substitution gives 12 + 18 + 20 − 5 − 7 − 4 + 2 = 36. The pairwise intersections are subtracted because they were counted twice, and the triple intersection is added once because it was then removed too many times.

The equality A ∩ B = A means that intersecting A with B does not remove any element of A. Thus every element of A must already belong to B, which is precisely the statement A ⊆ B. The equality A ∪ B = B expresses the same fact: adding A to B contributes no new elements. Equality of A and B is not required.

To construct B, test each element x of A and check whether x + 1 is also an element of A. For x = 2, x + 1 = 3, which belongs to A, so 2 ∈ B. For 3, 5, 7, and 11, the successors 4, 6, 8, and 12 are not in A. Hence B = {2}. Removing B from A leaves A \ B = {3, 5, 7, 11}.

The union B ∪ C contains every element that belongs to B, to C, or to both. The difference A \ (B ∪ C) therefore removes from A every element found in either B or C. What remains consists of elements that are in A, not in B, and not in C. Equivalently, it can be written as A ∩ Bᶜ ∩ Cᶜ, so option A gives the precise meaning.

First find the intersection A ∩ C, which contains elements common to both sets: A ∩ C = {4, 8, 16}. Next take the difference with B by removing elements that occur in B. Both 4 and 8 belong to B, but 16 does not. Consequently, only 16 remains, so (A ∩ C) \ B = {16}.

The symmetric difference contains elements that belong to exactly one of A and B, so the common elements must be excluded from both sets. Its cardinality is |A △ B| = |A| + |B| − 2|A ∩ B|. Substituting the given values gives 21 + 19 − 2(8) = 40 − 16 = 24. Therefore, option A is correct.

The set (A ∪ B) \ (A ∩ B) consists of elements that belong to exactly one of A or B; it is the symmetric difference. If this set is empty, there are no elements occurring in only one set. Thus every element of A is in B and every element of B is in A, so A = B. The sets need not be empty.

The sets A \ B and B \ A represent the elements belonging exclusively to A and exclusively to B, respectively. They are always disjoint. If two disjoint sets are equal, their common value must be empty; hence A \ B = ∅ and B \ A = ∅. Therefore A has no element outside B and B has no element outside A, so A = B.

The distributive law of sets states that intersection distributes over union: \(A\cap(B\cup C)=(A\cap B)\cup(A\cap C)\). An element belongs to the left side exactly when it is in \(A\) and in at least one of \(B\) or \(C\). This is precisely the condition represented on the right side. Therefore option A is correct; the other expressions use union, difference, or reversed membership conditions and are not generally equivalent.

The primes not exceeding 20 are \(\{2,3,5,7,11,13,17,19\}\). Set \(B\) contains the odd natural numbers, so all the odd primes are removed from \(A\). The only prime that is not odd is 2, because 2 is the unique even prime. Hence \(A\setminus B=\{2\}\), making option A correct.

Because \(A\subseteq C\), every element of \(A\) belongs to \(C\). Similarly, every element of \(B\) belongs to \(C\). Therefore every element of \(A\cup B\) is also in \(C\). Set difference removes elements that are in \(C\), so no element remains: \((A\cup B)\setminus C=\varnothing\). Thus option A is the only correct answer.

An element in \(A\cap B\) must be divisible by both 4 and 10. Such numbers are multiples of \(\operatorname{lcm}(4,10)=20\). The multiples of 20 in \(\{1,\ldots,100\}\) are 20, 40, 60, 80, and 100. There are five numbers, so \(|A\cap B|=5\), making option A correct.