The Empty Set, Finite and Infinite Sets, Equal Sets

Factor the quadratic expression: x² − 5x + 6 = (x − 2)(x − 3). Therefore, the equation is satisfied when x = 2 or x = 3. Since x is restricted to real numbers and both values are real, the solution set contains exactly these two elements. Hence the given set is {2, 3}. The order of elements does not affect a set.

We need natural numbers n such that n² = 25. Over the integers, the equation has two solutions, n = 5 and n = −5. However, the question restricts n to natural numbers, and −5 is not natural. Since 5² = 25, the only allowed element is 5. Therefore, A = {5}, a singleton set.

For every integer x, the square x² is non-negative: it is either zero or positive. Therefore, no integer can have a square equal to −4. Although complex numbers can solve the equation, the condition specifically restricts x to ℤ, the integers. Hence the defining condition has no allowed solution, so the set is empty and is written as ∅.

Solving x² = 4 gives x = 2 or x = −2. However, the set specifically restricts x to natural numbers. Under the usual convention that natural numbers are positive counting numbers, 2 is natural but −2 is not. Consequently B = {2}, which has exactly one element and is a singleton set.

To find the elements, solve the defining equation x² = 16. Taking square roots gives x = 4 or x = −4, since both 4² = 16 and (−4)² = 16. Both values are integers and therefore satisfy the domain condition x ∈ Z. Thus R = {−4, 4}. Option A omits one valid solution, option C confuses the square with the value of x, and option D wrongly claims that no solution exists.

Numbers divisible by both 2 and 3 are multiples of 6. The positive multiples of 6 that are less than 20 are 6, 12, and 18. There are only three members because the condition x < 20 gives an upper bound. Hence Z is a finite set, even though the unrestricted set of multiples of 6 is infinite.

Solving x² = 2 over the real numbers gives x = √2 and x = −√2. Neither value is an integer, because √2 is irrational. The condition x ∈ ℤ restricts us to integers only, so no value satisfies both requirements. Therefore A₁ contains no elements and is the empty set, A₁ = ∅.

Sets are equal when they contain exactly the same elements; the order of listing does not matter, and repeating an element does not create a new element. Options A, B, and D therefore describe equal sets. In option C, the first set contains 0 while the second contains 6, so the sets are not equal.

The condition -4 < x ≤ 1 requires x to be an integer greater than -4 and less than or equal to 1. Therefore, the permitted integers are -3, -2, -1, 0, and 1. The endpoint -4 is excluded because the inequality is strict, while 1 is included because the inequality is non-strict. Hence B has exactly five elements.

Factor the quadratic equation: x² - 5x + 6 = (x - 2)(x - 3) = 0. Thus x = 2 or x = 3. Both values are natural numbers, so D = {2, 3}. Since E is also {2, 3}, the two sets contain exactly the same elements and are equal. The order or form in which elements are written does not affect set equality.

Solve the defining equation: x² - 1 = 0 can be factored as (x - 1)(x + 1) = 0. Therefore, x = 1 or x = -1. Both values belong to the integers, so A = {-1, 1}. This is exactly the set B. Set elements may be written in either order, so the different possible ordering does not make the sets unequal.

Factor the equation x² - 7x + 12 = 0 as (x - 3)(x - 4) = 0. Its solutions are x = 3 and x = 4. Both are natural numbers, so both satisfy the defining condition of C. Therefore C = {3, 4}, which contains exactly two distinct elements. Repeated or equivalent descriptions would not increase the cardinality, but here the two roots are distinct.

The equation x² + 2x + 1 = 0 is (x + 1)² = 0, so its only algebraic solution is x = -1. However, the set definition requires x to be a natural number. Since -1 is not a natural number, it is rejected and no permitted element remains. Thus D contains no elements and is the empty set, written as ∅.

The condition |x + 1| ≤ 2 means that x + 1 lies between −2 and 2, inclusive. Therefore, −2 ≤ x + 1 ≤ 2. Subtracting 1 from all three parts gives −3 ≤ x ≤ 1. Since x must be an integer, the possible values are −3, −2, −1, 0, and 1. There are 5 values, so the set F has 5 elements. The endpoints are included because the inequality is ‘less than or equal to’.

The positive natural-number factors of 18 are 1, 2, 3, 6, 9, and 18. Among these, only 2 and 3 are prime numbers because each has exactly two positive divisors. The number 1 is not prime, and 6, 9, and 18 are composite. Therefore the set satisfying both conditions is G = {2, 3}.

The condition -2 ≤ x < 3 includes the integer -2 because the left inequality is inclusive, and excludes 3 because the right inequality is strict. The integers in this interval are -2, -1, 0, 1, and 2. This is exactly the roster given for O. Since two sets are equal when they have precisely the same elements, N = O.

The positive natural-number factors of 36 are obtained by checking which natural numbers divide 36 exactly: 1, 2, 3, 4, 6, 9, 12, 18, and 36. This list contains every element written in Q and no other elements. Therefore P and Q have exactly the same members, so P = Q. The number 8 is not a factor because 36 is not divisible by 8.

The positive natural-number multiples of 36 are 36, 72, 108, 144, and so on, so R continues indefinitely and is infinite. In contrast, Q is explicitly listed and contains only nine elements, so it is finite. The two sets are not equal: Q lists factors of 36, whereas R contains multiples of 36. Hence option B is correct.

The condition defining U is x² = x. Rearranging gives x² − x = 0, which factors as x(x − 1) = 0. Therefore, x = 0 or x = 1. Both values are integers, so both satisfy the stated domain condition x ∈ ℤ. Hence U contains exactly the two elements 0 and 1, and U = {0, 1}.

Factor the defining equation: x² = x gives x² − x = 0, so x(x − 1) = 0. The algebraic solutions are x = 0 and x = 1. Here ℕ is taken as {1, 2, 3, ...}, so 0 is not in the stated domain. The only permitted element is 1, and therefore V = {1}. If zero were included in the convention for ℕ, option A would apply, but not under the convention used here.

The common natural-number factors of 12 and 18 are 1, 2, 3, and 6. The additional condition requires x to be greater than 6, but none of these common factors satisfies that inequality. Hence no element belongs to D₁, and the correct description is D₁ = ∅, the empty set. It is not {6}, because 6 is not greater than 6.

Over the real numbers, x² = 2 has the solutions x = √2 and x = −√2. However, √2 is irrational, and its negative is also irrational. Because the definition restricts x to ℚ, neither real solution is allowed. Thus G₁ has no rational element and is the empty set, G₁ = ∅. The two real solutions do not make it a two-element rational set.

Solving x² = 2 over the real numbers gives x = √2 or x = −√2. Therefore, the solution set is H₁ = {−√2, √2}. This is exactly the set named I₁, so H₁ = I₁. Set equality depends on having the same elements, not on how the set is written or whether its elements contain radical signs. Both sets are finite and have two elements.

Option A contains all natural numbers greater than 100, so it is infinite. Option B is empty because no integer lies strictly between 2 and 3. Option C contains the natural divisors of 24, namely 1, 2, 3, 4, 6, 8, 12, and 24; therefore it is finite and non-empty. Option D is the set of all integers and is infinite.

The condition x² < 10 is satisfied by the integers whose absolute value is less than √10, approximately 3.16. Therefore the possible integers are −3, −2, −1, 0, 1, 2, and 3. The numbers −4 and 4 are excluded because their squares are 16. Counting the listed values gives 7 elements, so P₁ is a finite set with cardinality 7.