Equal sets and Subsets

The statement A ⊆ B says that every element of A belongs to B. The condition B − A = ∅ says that B has no element outside A; therefore every element of B also belongs to A, so B ⊆ A. Since both A ⊆ B and B ⊆ A hold, the two sets are equal. Thus A = B, and option A is correct. No information requires A to be empty.

The positive multiples of 4 are 4, 8, 12, 16, 20, and so on. The condition “less than 20” excludes 20 and every larger multiple, while “positive” excludes 0 and negative numbers. Thus the members of A are exactly 4, 8, 12, and 16. Therefore option A represents a set equal to A. Option B includes 0, option C includes 20, and option D omits 4.

The prime factorization of 30 is 2 × 3 × 5, so the set of its prime divisors is A = {2, 3, 5}. A proper subset must contain only elements of A but must not be equal to A. The set {2, 5} satisfies both conditions: its elements belong to A and it omits 3. Therefore option A is the proper subset. Option B equals A, while C and D contain elements that are not in A.

Solving x² = 16 over the integers gives x = 4 or x = −4. Therefore A = {−4, 4}. The set B is also defined as {−4, 4}, so A and B contain exactly the same elements and are equal. Thus option A is true. Option B omits −4, option C incorrectly claims a proper inclusion even though the sets are equal, and option D is false because A has two elements.

The set A has exactly three elements: 1, the set {2}, and 3. Therefore {2} is an element of A, so option A is true. However, 2 itself is not an element of A; it is inside the nested set {2}. Options C and D are also false because both require 2 to be an element of A, which it is not. This example shows why membership (∈) and subset (⊆) must be distinguished carefully, especially when a set contains another set as one of its elements.

Because A = {1,2} appears explicitly as one complete element inside B, A ∈ B is true. Also, the two elements of A, namely 1 and 2, both occur in B. Therefore every element of A belongs to B, so A ⊆ B is also true. Notice the distinction: A is present as a single set-element in B, while 1 and 2 are present individually, which establishes the subset relation. Thus both statements in option A are correct.

A proper subset of B means that every element of A belongs to B, but A and B are not equal. For finite sets, this necessarily gives n(A) < n(B). Since n(A) = 6, B must contain at least one additional element and therefore must have more than six elements. Hence n(B) = 6 is impossible. Values 7, 8, and 10 are all possible because B can be formed by adding one, two, or four new elements to A, respectively.

For finite sets, if A ⊆ B, then every element of A is already in B. If B had even one additional element, its cardinality would be greater than that of A. Here both sets have cardinality 4, so B cannot contain any extra element. Consequently, A and B contain exactly the same elements and are equal. Option B is false because the inclusion is not proper, option C is false because B − A is empty, and option D is false because their intersection is A, not the empty set.

A set is a subset of A only when every one of its elements is also an element of A. The elements of A are 1, 2, 5, and 10. Options A, B, and C contain only elements from this list, so each is a subset of A. Option D contains 4, and 4 is not a divisor of 10 and is not an element of A. Therefore {2,4} is not a subset, making option D correct.

By definition, X ⊆ A means that every element of X is also an element of A. The empty set has no elements, so it is impossible to find an element of ∅ that lies outside A. Thus the universal condition is automatically true for every set A; this is commonly called vacuous truth. The reason states exactly why the empty set is a subset of every set, so both the assertion and reason are true and the reason correctly explains the assertion.

Equality of sets means that A and B contain exactly the same elements. Therefore, every element of A is necessarily an element of B, which is precisely the definition of A ⊆ B. The assertion is consequently true. The reason is also true because identical membership is what set equality means, and it directly explains why the subset relation follows. In fact, if A = B, then both A ⊆ B and B ⊆ A hold.

The assertion is false because equal cardinality only tells us that two sets have the same number of elements; it does not tell us that the elements themselves are identical. For example, A = {1,2} and B = {3,4} both have cardinality 2, but A ≠ B. The reason is also false for the same reason: equal size does not imply equal membership. To conclude A = B, we must show that every element of A is in B and every element of B is in A.

The elements 1 and 3 are odd, while 2 and 4 are even. A subset has an even total when it contains either zero or two odd elements. Among the 16 subsets of A, exactly half have an even sum and half have an odd sum. Therefore, the number of subsets with even sum is 16 ÷ 2 = 8. The empty subset is included because its sum is 0, which is even.

For each element of A, the condition X ⊆ Y allows exactly three possibilities: the element is in neither X nor Y, it is in Y but not X, or it is in both X and Y. It cannot be in X alone because every element of X must also belong to Y. Since A has three independent elements, the total number of ordered pairs is 3 × 3 × 3 = 3³ = 27.

The positive natural-number divisors of 18 are 1, 2, 3, 6, 9, and 18. Therefore, the set-builder description of A gives A = {1, 2, 3, 6, 9, 18}, which is exactly the roster description of B. Since two sets are equal when they contain precisely the same elements, A = B. Neither set is a proper subset of the other, and their intersection is certainly not empty.

The element 2 is required, so it is fixed as included in every acceptable subset. The element 4 is forbidden, so it is fixed as excluded. The remaining elements 1, 3, and 5 are unrestricted; each may be selected or not selected independently. Thus there are 2 choices for each of three free elements, giving 2³ = 8 valid subsets. Therefore, option B is correct.

The complement A′ is defined relative to the universal set U. It contains every element of U that does not belong to A. Since U includes all students in the school and A includes the Class 11 students, A′ represents all school students who are not in Class 11. This could include students from several other classes, not necessarily only Class 12 students.

A has 7 elements, and a subset containing exactly 6 elements can be formed by choosing 6 of them. Therefore, the number is C(7,6) = 7. An even simpler view is that each such subset is obtained by omitting exactly one element from A, and there are 7 possible elements to omit. Hence option B is correct. The value 21 is C(7,2), not C(7,6).

The elements 1 and 2 are required, so they occupy two of the four positions. We must choose the remaining two elements from {3,4,5,6}, which has four elements. Using combinations, the count is C(4,2) = 4!/(2!2!) = 6. Thus option B is correct. Choosing C(4,1), C(5,2), or C(6,2) would not enforce both required elements correctly.

The equality A ∩ B = A says that intersecting B with A leaves all of A unchanged. This can happen only when every element of A is also contained in B. Therefore, A is a subset of B, written A ⊆ B. The other options do not follow from the given condition and may hold only in special cases.