The Empty Set, Finite and Infinite Sets, Equal Sets

The condition 1 < x < 2 describes numbers strictly between 1 and 2. Although real numbers exist in that interval, no natural number lies strictly between two consecutive natural numbers 1 and 2. Therefore no natural number satisfies the defining condition, so A has no elements and is the empty set, A = ∅.

The only natural number less than 2 is 1, assuming the usual positive-natural-number convention. However, 1 is not prime: a prime number has exactly two distinct positive divisors, whereas 1 has only one positive divisor. Thus no natural number satisfies both conditions, so A is the empty set, ∅.

The natural numbers less than 10 that are even are 2, 4, 6, and 8. Among these numbers, only 2 is prime, because a prime number has exactly two distinct positive divisors, 1 and itself. Therefore, the conditions “even,” “prime,” and “less than 10” are satisfied simultaneously only by 2. Hence A is the singleton set {2}, making option A correct.

Two sets are equal precisely when every element of the first set belongs to the second set and every element of the second belongs to the first. In other words, their elements must be exactly the same. Equal cardinality alone does not guarantee equality; for example, {1, 2} and {a, b} have the same size but different elements. The order and names used to write sets are irrelevant.

The three-digit natural numbers begin at 100 and end at 999, with both endpoints included. The number of integers in this inclusive interval is 999 − 100 + 1 = 900. Therefore, A is a finite set with 900 elements. The plus-one is necessary because both 100 and 999 are counted.

The natural numbers satisfying 100 < x < 999 are 101, 102, 103, ..., 998. This is a bounded list of consecutive natural numbers, so it contains only finitely many elements. In fact, its cardinality is 998 − 101 + 1 = 898. The strict inequalities exclude 100 and 999, but they do not make the set empty or infinite.

A natural number divisible by 11 must be a positive multiple of 11, such as 11, 22, 33, and so on. The smallest positive multiple is 11 itself, but the condition x < 11 excludes 11 and every larger multiple. Therefore, no natural number satisfies both conditions, so A is the empty set, written as ∅.

In a set, the order in which elements are written is irrelevant. The sets {p, q, r} and {r, p, q} both contain exactly the same elements: p, q, and r. Therefore, they are equal. Option B is false because the second set has one additional element, option C is false because {0} contains 0 whereas the empty set contains nothing, and option D is false because the elements are different.

The natural numbers not exceeding 8 are 1, 2, 3, 4, 5, 6, 7, and 8. Among these, the prime numbers are 2, 3, 5, and 7. Removing the primes leaves 1, 4, 6, and 8, so A = {1, 4, 6, 8}. Notice that 1 is neither prime nor composite, but it is certainly not prime, so it must be included.

A product is zero when at least one factor is zero. Thus (x − 2)(x + 3) = 0 gives x − 2 = 0 or x + 3 = 0, so x = 2 or x = -3. Both values belong to the integers, and hence A = {2, -3}. Since the order of elements does not affect a set, {2, -3} = {-3, 2} = B. Therefore, A = B. The set has only two elements, so it is finite, not infinite.

The powers of 2 not exceeding 64 are obtained from 2⁰ through 2⁶: 2⁰ = 1, 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32, and 2⁶ = 64. Thus A = {1, 2, 4, 8, 16, 32, 64}. The value 1 must be included because 1 = 2⁰, so option B is correct. The set is finite because the condition x ≤ 64 stops the powers.

Since 0.25 = 1/4, the equation x² = 0.25 becomes x² = (1/2)². A square equation has both the positive and negative square roots, so x = 1/2 or x = -1/2. Both values are rational and must be included. Hence A = {-0.5, 0.5}, a finite set containing two elements.

Among the positive natural numbers less than 12, the number 1 is neither prime nor composite. A prime number has exactly two distinct positive divisors, while a composite number has more than two; 1 has only one positive divisor, itself. Every other positive natural number below 12 is either prime or composite. Therefore A = {1}.

Using the standard school convention N = {1, 2, 3, ...}, every natural number is at least 1. Consequently, there is no natural number satisfying x < 1. A set that contains no element is called the empty set and is written as ∅. Therefore the given set is empty, not a singleton or an infinite set.

For every real number x, x² is non-negative, so x² + 1 is at least 1. It can therefore never equal zero. Equivalently, the equation x² + 1 = 0 would require x² = -1, which has no real solution. Since no real number satisfies the condition, the set contains no elements and is empty.

In set notation, repetition of an element does not create a new element, and the order in which elements are written is irrelevant. Thus A = {2, 3, 4}, because the repeated 3 is counted only once. Set B also contains exactly 2, 3, and 4. Therefore A and B are equal, while A does not have four distinct elements.

For x² < 10, we need −√10 < x < √10. Since √10 is approximately 3.16 and x must be an integer, the permitted values are −3, −2, −1, 0, 1, 2, and 3. There are seven distinct integers in this set. The set is finite because its elements are bounded between −√10 and √10. Therefore, its cardinality is 7, so option C is correct.

For every natural number n, the set contains the square n². Its beginning is 1, 4, 9, 16, 25, and so on. As n can be chosen arbitrarily large, the squares continue without an ending largest element. Also, different natural numbers give different squares, so infinitely many distinct elements occur. Therefore the set is infinite.

The notation {0} represents a set whose only element is the number 0, so its cardinality is 1 and it is a singleton set. The symbol ∅ represents a set with no elements at all, so its cardinality is 0 and it is the empty set. Zero is an element in {0}; it is not the same as having no elements. Hence option C is correct.

Two sets are equal only when they contain exactly the same elements; the order of elements does not matter. Set A already contains 2, so its other element, a + 1, must correspond to 3 in set B. Therefore, a + 1 = 3, and subtracting 1 from both sides gives a = 2. Hence option B is correct.

Interpreting “between 10 and 50” as strictly between the two endpoints, the multiples of 5 are 15, 20, 25, 30, 35, 40, and 45. The numbers 10 and 50 are excluded because they are the endpoints. This gives seven distinct elements, so the set is finite and option A is correct.

The ellipsis on the left indicates that the sequence continues with further negative integers, such as -6, -7, -8, and so on. There is no final first element at the negative end, so the list does not terminate and its elements cannot all be counted. Therefore, the set has infinitely many elements and is an infinite set. Option C is correct.

The set contains every real number strictly between 0 and 1. There are infinitely many such numbers; for example, 1/2, 1/3, 1/4, and infinitely many others. More generally, between any two distinct real numbers there are infinitely many real numbers. Hence the set is infinite, and option C is correct.

In a set, the order in which elements are listed does not matter, and repeating an element does not produce an additional distinct element. Thus B has only the distinct elements 1, 2, and 3, exactly as A does. Since both sets contain the same elements, A = B. Therefore, option A is correct.

The only natural number strictly between 14 and 16 is 15. However, 15 is not prime because it has factors other than 1 and itself; specifically, 15 = 3 × 5. Therefore no natural number satisfies both conditions, so the set has no elements. It is the empty set, and option A is correct.