The Empty Set, Finite and Infinite Sets, Equal Sets

For n = 1, 2, 3, 4, and 5, the expression n² gives 1, 4, 9, 16, and 25 respectively. Therefore, option A produces exactly the same elements as the given set. The cube rule in option B gives different values, the doubling rule in option C gives even values, and option D lists the bases rather than their squares. Hence A is correct.

The number 31 is prime because its only positive divisors are 1 and 31. Consequently, the set of natural-number factors is A = {1, 31}. These are two distinct elements, so the cardinality of A is 2. The number 31 itself is an element, not the number of elements in the set. Therefore, option B is correct.

Because N is defined here as {1, 2, 3, ...}, zero is not considered a natural number for this question. The positive perfect squares less than 6 are 1 = 1² and 4 = 2². The next perfect square, 9 = 3², is greater than 6 and is excluded. Therefore A = {1, 4}, so option A is correct.

The set contains all even integers: ..., −6, −4, −2, 0, 2, 4, 6, .... There is always another even integer greater than any listed member, and the sequence also continues indefinitely in the negative direction. Therefore, the set has infinitely many elements and is an infinite set. Hence, option C is correct.

The governing concept is equality of sets together with the stated domain restriction. Algebraically, x² = 49 gives x = 7 or x = −7. However, x must belong to the natural numbers, so −7 is excluded and A = {7}. Since B is also the singleton set {7}, A and B have exactly the same element. Thus option A is correct; option C ignores the domain restriction.

The governing concept is describing a set by common conditions: being a factor of 48 and being prime. Since 48 = 2⁴ × 3, its prime divisors are only 2 and 3. Number 1 is not prime, and 4 and 6 are composite, so they cannot enter A. Hence A = {2, 3}, making option A correct. The other choices include at least one non-prime or non-required number.

The governing concept is converting a set-builder description into roster form. We need natural numbers satisfying both strict inequalities 3 < x and x < 8. Checking the integers between the boundaries gives 4, 5, 6, and 7. Neither endpoint is included because both conditions are strict. Therefore A = {4, 5, 6, 7}, so option B is correct; the other options include 3 or 8 incorrectly.

The governing concept is equality of sets. In a set, the order of elements has no significance, and repeating an element does not create an additional distinct member. Therefore B simplifies to {a, b, c, d}, exactly the same collection as A. The sets are equal, so option A is correct. Option B wrongly treats order or repetition as important, while C and D misclassify a set with four distinct elements.

A number that is a multiple of both 12 and 18 must be a multiple of their least common multiple. Since lcm(12, 18) = 36, the positive multiples below 100 are 36 and 72. The next multiple, 108, is not less than 100. Therefore, A = {36, 72} and has 2 elements, so option B is correct.

The governing concept is distinguishing finite and infinite sets. The natural-number factors of a fixed number, 64, are limited: {1, 2, 4, 8, 16, 32, 64}; hence option C is finite. In contrast, natural numbers greater than 1000 continue without end, negative integers continue downward without end, and positive multiples 9, 18, 27, ... are infinite. Therefore option C is the only set that is not infinite.

The equation x + 5 = 5 gives x = 0. In this question, the usual school convention ℕ = {1, 2, 3, ...} is being used, so 0 is not a natural number. Consequently, no natural number satisfies the condition, and A has no elements. Therefore A is the empty set, written as ∅.

Option A has the same four elements in a different order, and order does not matter in a set. Option B repeats 1, but repetition does not add a new element. Option C is the set of positive factors of 8, namely {1, 2, 4, 8}. Option D contains one additional element, 16, so it is not equal to A.

Factor the equation: x² − 4x + 4 = (x − 2)² = 0. Hence x = 2 is the only integer solution. Although the quadratic has a repeated root, a set records each value only once; it does not count multiplicity of roots. Therefore A = {2}, so the set contains exactly one element.

The notation x ∈ N restricts x to natural numbers. Although infinitely many real numbers lie strictly between 2 and 3, there is no natural number in that interval. The endpoints 2 and 3 are also excluded because the inequalities are strict. Therefore no element satisfies every condition, so A = ∅, the empty set.

The condition includes every integer from −3 through 3, including both endpoints. Therefore, B = {−3, −2, −1, 0, 1, 2, 3}. Counting these distinct integers gives 7 elements. Since the interval is bounded and contains only finitely many integers, B is a finite set. Hence option B is correct.

The governing concept is equality of sets after expanding a set-builder description. The natural even numbers less than 10 are 2, 4, 6, and 8, so B = {2, 4, 6, 8}. This is exactly the element collection shown for A; set equality does not depend on the order of writing. Thus option A is correct. B is not empty, and A has only four elements, so it is not infinite.

In a set, repeating an element does not create a new element. Hence C can be written as {1, 2, 3}. Set D contains the same three elements, written in a different order: {3, 2, 1}. Since set equality depends on having exactly the same elements, not on order or repetition, C = D. Therefore, option A is correct.

An empty set is a set that contains no elements. Its cardinality, or number of elements, is zero. It is commonly represented by the symbols ∅ or { }. This is different from a singleton set, which contains exactly one element, and from finite or infinite non-empty sets. Therefore, option A gives the correct definition.

Under the stated convention, the natural numbers begin with 1. Every natural number is therefore at least 1, so none of them can satisfy x < 1. Since the defining condition has no solution in N, the set contains no elements and is the empty set, written as ∅. It is finite with cardinality zero.

The equation x² = 2 has the real solutions √2 and −√2, but neither solution is an integer. Since the domain in the set-builder description is Z, only integer solutions may be included. No integer satisfies the equation, so the set has zero elements and is therefore the empty set, ∅.