The Empty Set, Finite and Infinite Sets, Equal Sets

The natural factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. We must retain only those factors that are also multiples of 4. These are 4, 8, 12, and 24. The word “and” means both conditions must hold simultaneously, so D = {4, 8, 12, 24}. Therefore option A is correct.

Because x is restricted to real numbers, x² is always non-negative. Consequently, x² + 4 is always at least 4 and can never equal zero. Thus the equation has no real solution, so A contains no elements and A = ∅. Since B is also given as the empty set, A and B are equal. Option A is therefore correct.

The natural numbers are discrete: consecutive natural numbers have no natural number between them. Since 5 and 6 are consecutive, there is no natural number x satisfying the strict inequalities 5 < x < 6. The endpoints are excluded as well. Therefore A has no elements, so it is the empty set, written as ∅.

In a set, an element is either present or absent; writing it more than once does not increase the set’s membership. Thus A simplifies to {1, 2, 3}. The order of elements is also irrelevant in set notation, so B = {3, 2, 1} represents the same set {1, 2, 3}. Hence A and B contain exactly the same elements and are equal.

There is no integer whose square is 2. The integers nearest to a possible square root are 1 and 2, but 1² = 1 and 2² = 4; in fact, √2 is irrational and is not an integer. Therefore no integer satisfies the defining condition, so A = ∅. The empty set has cardinality zero and is finite.

A natural number is a factor of 24 if it divides 24 exactly. The positive factors are 1, 2, 3, 4, 6, 8, 12, and 24. This list is precisely the roster description of B. Since two sets are equal when they contain exactly the same elements, A = B. The set is finite, not infinite, and 24 is correctly included.

The interval (0, 1) contains infinitely many real numbers, so A is infinite. It also contains infinitely many rational numbers, such as 1/2, 1/3, 1/4, and so on, so B is infinite. However, A contains irrational numbers such as √2/2, while B contains only rational numbers. Therefore B is a proper subset of A, and the two sets are not equal.

The symbol ∅ denotes a set with no elements, whereas {∅} denotes a set whose only element is the empty set itself. Therefore, {∅} contains exactly one element and is a singleton set. It is not equal to ∅, because ∅ has zero elements while {∅} has one element. This distinction is fundamental in set notation.

Factor the quadratic equation: x² − 5x + 6 = (x − 2)(x − 3) = 0. Hence x = 2 or x = 3. Both values are integers, so the set defined by the condition is A = {2, 3}. Since B is also {2, 3}, the two sets contain exactly the same elements and therefore A = B. The order of elements does not matter in a set.

The set contains all prime natural numbers not exceeding 100. There are only finitely many natural numbers from 1 through 100, so the set can have only finitely many elements. It is also non-empty because numbers such as 2, 3, 5, 7, and 11 are prime and satisfy the condition. Therefore, A is finite and non-empty.

Because x must be an integer strictly greater than −3 and strictly less than 3, the boundary values −3 and 3 are excluded. The integers satisfying the condition are −2, −1, 0, 1, and 2. Thus, A has five distinct elements, so its cardinality is n(A) = 5. The notation n(A) means the number of elements in set A.

Rewrite the equation as x² − x = 0 and factor it: x(x − 1) = 0. By the zero-product property, x = 0 or x = 1. Both are real numbers, so A = {0, 1}. Since B is defined as {0, 1}, A and B have exactly the same elements. Therefore, A = B, and the set is finite with two elements.

A set records membership, not the number of times an element appears in the original word. Although LEVEL has five positions, its distinct letters are only L, E, and V. Therefore, A = {L, E, V}. This is exactly the set B. Repeated letters are written only once in a set, so A and B are equal and A has three elements, not five.

The natural-number multiples of 3 include 3, 6, 9, 12, 15, and so on. For every multiple 3k, where k is a natural number, another larger multiple 3(k+1) can be found. Therefore, the sequence never ends and there is no greatest element. Hence A is an infinite set, not merely the singleton {3}.

Using the usual school convention ℕ = {1, 2, 3, …}, no natural number is less than 1, so A is empty. Also, there is no integer strictly between 0 and 1; consecutive integers have no integer between them. Hence B is empty as well. Therefore, A = B = ∅. The strict inequalities exclude both endpoints.

For any real number x, |x| < 4 is equivalent to −4 < x < 4. Since x must be an integer, the possible values are −3, −2, −1, 0, 1, 2, and 3. The endpoints −4 and 4 are excluded because the inequality is strict. Thus A contains exactly the seven integers listed in option A, so that set is equal to A.

To determine A, solve x² = 9 over the real numbers. Taking square roots gives x = 3 or x = -3, because both 3² and (-3)² equal 9. Therefore A = {-3, 3}. This is exactly the same collection of elements as B, so A = B. The negative root must not be omitted when solving a square equation.

Every positive natural number divides 0 because 0 = x × 0 for every natural number x. Thus 1, 2, 3, 4, and infinitely many other natural numbers satisfy the condition. Therefore A contains infinitely many elements and is an infinite set. This is different from division by zero, which is undefined; here we are asking whether a number divides zero.

For every integer x, the square x² is non-negative: it is either positive when x ≠ 0 or equal to 0 when x = 0. Therefore no integer can satisfy x² < 0. The set A has no elements, so A is the empty set. The cardinality of the empty set is n(A) = 0, making option A correct.

Subtracting x from both sides of the condition x = x + 1 gives 0 = 1. This is a contradiction and cannot be true for any real number x. Consequently, there is no real number belonging to A. A set containing no elements is called the empty set, denoted by ∅, so A = ∅.