Power Set and Subsets

A set with n elements has exactly 2^n subsets, because each element can either be included or excluded independently. Since n(A) = 5, the number of elements in the power set is n(P(A)) = 2^5 = 32. Thus, P(A) contains 32 subsets, including the empty set and the original set A itself.

A two-element subset is formed by choosing any two different elements from the five elements of A, with order ignored. Thus the number is C(5, 2) = 5!/(2!3!) = (5 × 4)/2 = 10. For example, {1, 2} and {2, 1} represent the same subset, so they are counted only once. Hence option B is correct.

Using the usual school convention N = {1, 2, 3, ...}, the condition x ≤ 3 gives A = {1, 2, 3}. A proper subset must contain only elements of A and must not be equal to A. The set {1, 2} satisfies both conditions, so it is a proper subset. Option B equals A, option C contains 4, and option D contains 0, which is not in A under this convention.

If a finite set has n elements, its power set has 2^n elements because each element has two independent choices: it may either be included in a subset or not included. Here A has n = 3 elements, so |P(A)| = 2^3 = 8. These eight subsets are the empty set, three one-element subsets, three two-element subsets, and A itself. Therefore option C is correct.

The power set P(A) is the set of all subsets of A. Therefore it contains the empty set ∅, every one-element subset, every two-element subset such as {1, 2}, and A itself, {1, 2, 3}. The ordinary number 4 is not a subset of A and is not one of the listed elements of A. Hence 4 does not belong to P(A), so option D is correct.

Taking N = {1, 2, 3, ...}, the condition x ≤ 4 gives A = {1, 2, 3, 4}. Since 4 must be included, it is fixed. Each of the remaining three elements, 1, 2, and 3, can independently be included or omitted. Therefore the number of valid subsets is 2³ = 8.

Solving x² = 16 over the integers gives x = 4 and x = −4, so A = {−4, 4}. The set C = {4} contains only 4, which belongs to A; therefore C is a subset of A. Since A also contains −4, an element not in C, the sets are not equal. Thus C is a proper subset of A.

Factor the quadratic expression: x² − 5x + 6 = (x − 2)(x − 3). Therefore x = 2 or x = 3. Both values are natural numbers, so they satisfy the restriction x ∈ N. Hence A = {2, 3}, and option A gives the set equal to A. The order of elements does not matter in a set.

The even natural-number divisors of 24 are \(A=\{2,4,6,8,12,24\}\). A set is a subset of A only when every one of its elements belongs to A. Options A and B contain only valid even divisors. Option C contains 10, and 10 does not divide 24 exactly, so C is not a subset of A. The empty set in option D is a subset of every set, including A.

For a finite set with n distinct elements, each element has two independent choices when forming a subset: it may be included or excluded. Therefore, the total number of subsets is 2ⁿ. Here n = 3, so the number is 2³ = 8. These include the empty set, the three one-element subsets, the three two-element subsets, and A itself.

The distinct letters occurring in the word “math” are m, a, t, and h. Therefore the verbal description of set A gives exactly \(A=\{m,a,t,h\}\), which is the same as set B. In set theory, the order in which elements are written does not matter, and repeated elements would also be written only once. Hence A and B have precisely the same elements, so \(A=B\). A proper-subset statement would be false because neither set has an extra element.

The positive odd integers smaller than 8 are 1, 3, 5, and 7. No other positive odd integer satisfies the condition: 9 is not less than 8, while 0 and negative odd integers are not positive. Thus the roster form of B is \(B=\{1,3,5,7\}\), exactly the same as A. Since equal sets contain all and only the same elements, the correct relation is \(A=B\), not a proper-subset relation.

Every element of B, namely 1 and 2, is also an element of A. Therefore \(B\subseteq A\). However, A also contains 3 and 4, which are not in B, so the two sets are not equal. Consequently B is a proper subset of A, written \(B\subset A\) when the symbol denotes a proper subset. Option A reverses the inclusion, option C claims equality, and option D is false because the intersection is \(\{1,2\}\), not empty.

The prime factorization of 12 is \(12=2^2\times3\). Its distinct prime factors are therefore 2 and 3. A set does not record multiplicity, so the repeated factor 2 is listed only once; we do not write \(\{2,2,3\}\). Hence \(A=\{2,3\}=B\). Option C is false because equal sets are not proper subsets of one another, and option D is false because 12 is not an element of either set.

A subset may contain some or all elements of the original set, but every element selected must itself belong to that set. Here A contains 1, 2, and the empty set ∅ as its three elements. Therefore {1, ∅} is a subset of A. Option C, {{1}}, contains the set {1}, which is not an element of A; the element 1 alone is in A.

For a finite set with n elements, every element has two choices in forming a subset: it is either included or excluded. Thus the total number of subsets of A is 2³ = 8. A proper subset is any subset that is not equal to A itself, so we remove the one non-proper subset A. Hence the number of proper subsets is 8 − 1 = 7.

To prove that C is a proper subset of A, two conditions must be checked. First, every element of C must belong to A; here both 2 and 4 are in A. Second, C must not equal A; C lacks 1 and 3, so C ≠ A. Therefore C ⊂ A. Merely having fewer elements does not guarantee inclusion.

The natural numbers satisfying 2 < x ≤ 6 are 3, 4, 5, and 6, so A = {3, 4, 5, 6}. Option B, {3, 5}, contains only elements of A and is not equal to A; therefore it is a proper subset. Option A is A itself, option C contains 2, which is excluded, and option D contains 7, which is not in A.

A set records membership, not the number of times an item appears. The distinct letters in LEVEL are L, E, and V; repeated occurrences of L and E do not create new elements. Thus A = {L, E, V}, which is exactly B. Therefore A = B and option A is correct.

The power set P(A) contains every subset of A as an element. For A = {a, b}, the subsets are the empty set ∅, the singleton sets {a} and {b}, and the set A itself, {a, b}. Thus P(A) = {∅, {a}, {b}, {a, b}}, which is option A. Notice the difference between a and {a}: the first is an element, whereas the second is a singleton subset.

The prime divisors of 24 are 2 and 3, because 24 = 2^3 × 3. The prime divisors of 36 are also 2 and 3, because 36 = 2^2 × 3^2. Therefore, the prime divisors common to both numbers are exactly 2 and 3, so A = {2, 3}. Number 1 is not prime, and 6 or 12 are composite, not prime.

The set A has exactly three elements: {1}, {2}, and {1,2}. Therefore, the set {1,2} itself is an element of A, so option A is correct. The numbers 1 and 2 are elements of the inner sets, but they are not direct elements of A. Option D is another set that is not listed as an element of A. This question tests the important distinction between an element and a set containing elements.

The positive divisors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40. The condition x ≤ 10 allows only 1, 2, 4, 5, 8, and 10. Hence A = {1,2,4,5,8,10}, which is exactly option A. Option B contains divisors greater than 10, while option C omits valid divisors and option D includes numbers that violate the upper-bound condition.

For each element of a set, a subset has two independent choices: the element is either included or excluded. Thus, a set with n elements has 2ⁿ subsets in total. For n = 5, the number is 2⁵ = 2 × 2 × 2 × 2 × 2 = 32. This count includes both the empty set and the original set itself. Therefore, option C, 32, is correct; 25 and 120 come from unrelated calculations such as 5² and 5!.

The power set P(A) contains every subset of A, including the empty set and A itself. If a set has n elements, its power set has 2^n elements because each original element can either be included or excluded from a subset. Here n = 3, so |P(A)| = 2³ = 8. Therefore, option C is correct; 3 counts elements of A, not its subsets.