The Empty Set, Finite and Infinite Sets, Equal Sets

The governing concept is describing a set by solving its defining equation over the stated domain. From x³ = x, move all terms to one side: x³ − x = 0. Factorisation gives x(x − 1)(x + 1) = 0, so x = 0, 1, or −1. All three values belong to ℤ, and no other roots arise from the factors. Therefore A = {−1, 0, 1}, making option A correct. The other finite options omit one solution, while ℤ includes many values that do not satisfy the equation.

For option A, x² = x becomes x² − x = 0, or x(x − 1) = 0. Thus the integer solutions are x = 0 and x = 1, so the described set is exactly {0, 1}. Option B gives only {−1, 1} if interpreted over integers, but its stated natural domain gives {1}; option C gives {−1, 1}, and option D is infinite.

Using the usual school convention ℕ = {1, 2, 3, ...}, the condition x ≤ 4 gives A = {1, 2, 3, 4}. For B, the integers satisfying 1 ≤ x ≤ 4 are also {1, 2, 3, 4}. Since two sets are equal when they contain exactly the same elements, A = B. The different stated domains do not matter after the conditions are applied.

For option B, x² = 9 gives x = 3 or x = −3. The additional condition x > 0 excludes −3 and leaves only x = 3, so the set is {3}. It is non-empty and has exactly one element, making it a singleton set. Option A is empty, option C has two elements, and option D is infinite.

The set in option A is {0, 1, 2, 3, ...}, which continues without end and is therefore infinite. It is not equal to ℤ because ℤ also contains negative integers such as −1 and −2, while option A does not. Option B is finite, option C equals {−1, 1}, and option D is empty because no integer lies strictly between 0 and 1.

Factor the expression as x² − 4x + 4 = (x − 2)². The equation is zero only when x − 2 = 0, so x = 2. Consequently, A = {2} and has exactly one element. Although 2 is a repeated root of the polynomial, a set lists an element only once; repeated occurrence does not increase the cardinality.

Solving x² − 1 = 0 gives (x − 1)(x + 1) = 0, so x = 1 or x = −1. Both solutions belong to ℤ, hence A = {−1, 1}. Under the usual convention ℕ = {1, 2, 3, ...}, only 1 belongs to the natural numbers, so B = {1}. The domain in a set-builder description determines which solutions are included.

Factor the quadratic expression: x² + 3x + 2 = (x + 1)(x + 2). Therefore, the equation is satisfied when x = -1 or x = -2. Since both values are integers, A = {-1, -2}. Set order is irrelevant, so {-1, -2} = {-2, -1} = B. Hence A = B. The other options either use incorrect roots, claim no solutions, or incorrectly describe a two-element set as infinite.

Complete the square: x² - 2x + 5 = (x - 1)² + 4. For every real number x, (x - 1)² is at least zero, so the expression is at least 4 and can never equal zero. Therefore, the equation has no real solution, and the set of its real solutions is empty: A = ∅.

A natural number divisible by both 2 and 3 is divisible by their least common multiple, 6. Conversely, every multiple of 6 is divisible by both 2 and 3. Thus A and B contain exactly the same numbers: 6, 12, 18, 24, and so on. Since there are endlessly many multiples of 6, both sets are infinite and A = B.

For natural numbers, x² ≤ 30 implies x ≤ √30, and √30 is approximately 5.47. Therefore, the possible natural numbers are 1, 2, 3, 4, and 5. The odd numbers among them are 1, 3, and 5, so A = {1, 3, 5}. This set contains three elements. The correct answer is option B; the count must be made after applying both the square bound and the odd-number condition.

For option A, the natural-number divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. Among them, the prime divisors are only 2 and 3. Therefore A = {2, 3}, which is exactly B. Two sets are equal when they contain the same elements, regardless of how their defining descriptions look. The other options produce different sets.

We solve the absolute-value inequality first: |2x − 1| < 4 implies −4 < 2x − 1 < 4. Adding 1 and dividing by 2 gives −3/2 < x < 5/2. Since x must be an integer, the possible values are −1, 0, 1, and 2. Therefore A = {−1, 0, 1, 2}, which is exactly the set B. Hence the correct conclusion is A = B.

A positive integer x divides 0 because 0 = x × 0, so the quotient is the integer 0. This is true for every positive natural number: 1, 2, 3, and so on. Since the positive natural numbers are infinite, the set of all such divisors of zero is also infinite. Therefore, option C is correct.

The notation ∅ represents the empty set, which has no elements. However, the braces in {∅} place that empty set itself inside another set as one element. Therefore {∅} has exactly one element and is a singleton set. It is not equal to ∅, because ∅ has zero elements while {∅} has one. Option B is correct.

We need the integers whose square is 9. Solving x² = 9 gives x = 3 or x = -3, because both 3² and (-3)² equal 9. Therefore A = {-3, 3}. This is exactly the set B, so the two sets are equal. Hence statement A is correct, while the other statements are false.

The governing concept is identifying the elements satisfying a condition over the real numbers. For every real x, x² is non-negative, so x² ≤ 0 can occur only when x² = 0. Solving x² = 0 gives the single solution x = 0. Therefore the set is the singleton {0}, and option B is correct. It is not empty because 0 satisfies the condition, and it is not all of ℝ because nonzero real numbers have positive squares.

Equality of sets means that both sets must contain exactly the same elements. Sets A and B already share 1 and 4, so the remaining element in A must be 3 in order to match B. Therefore a = 3. Values a = 1 or 4 would merely repeat an existing element and would fail to include 3, while a = 2 would produce a different set.

Solving x² = 2 over the real numbers gives x = √2 and x = −√2. Neither value is an integer, because √2 is irrational. Since the domain in the set-builder notation is ℤ, neither real solution is allowed. Thus no integer satisfies the condition, so the set is empty and option A is correct.

The empty set, written ∅, has no elements, so its cardinality is 0. In contrast, {∅} is a set whose single element is the empty set itself; therefore, its cardinality is 1 and it is a singleton. The braces create a new set and do not leave the set empty. Hence option C is correct; A and B confuse an element with a set, while D is false.

The positive natural-number factors of 16 are 1, 2, 4, 8, and 16. The additional condition says that x must not be even, so all even factors are removed. Among these factors, only 1 is odd. Hence the set contains exactly one element: A = {1}. Therefore option A is the unambiguous correct answer.

Factor the equation as x² − 5x + 6 = (x − 2)(x − 3) = 0. Thus x = 2 or x = 3, so A = {2,3}. This is exactly the same set as B, which means A = B. The value 6 is the constant term, not a solution, and the equation is not without solutions. Therefore, options B, C, and D are false, and option A is correct.

The positive divisors of 18 are 1, 2, 3, 6, 9, and 18. Among them, the even divisors are 2, 6, and 18, because each is divisible by 2. Therefore, \(A=\{2,6,18\}\), which is exactly the set \(B\). The claim that 18 is not even is false, since 18 is divisible by 2. Hence \(A=B\), and option B is correct.

The notation A = {∅} means that A is a set whose only element is the empty set. Therefore, ∅ ∈ A is true. However, A is not itself empty: it has one element. Hence ∅ = A is false, and A ∈ ∅ is impossible because the empty set contains no elements. This example shows the essential difference between ∅, which has zero elements, and {∅}, which has one element.

Equal sets must contain exactly the same elements, although the order of listing does not matter. The left-hand set already contains 4. Since a ≠ 4, the other element a must be 2 in order for the set to match {2, b}. Once a = 2, the right-hand set must contain 4 as its second element, so b = 4. Therefore the ordered pair is (2, 4). Option B violates the given condition a ≠ 4, while C and D do not produce the same two-element set.