The Empty Set, Finite and Infinite Sets, Equal Sets

Subtracting 4 from both sides of x² + 4 = 4 gives x² = 0. The only integer, and indeed the only real number, whose square is zero is x = 0. Therefore E₁ contains exactly one element: E₁ = {0}. A set with precisely one element is called a singleton set, so option B is correct; there are not two roots because +0 and −0 are the same number.

For every real number x, x² is at least 0. Consequently, x² + 4 is at least 4, so it can never equal 0. Equivalently, the equation would require x² = −4, which has no real solution. Since F₁ contains real numbers satisfying the condition and there are none, F₁ = ∅, the empty set.

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.

Option A is the set of positive integers {1, 2, 3, …}. It has infinitely many elements because there is no largest positive integer, but it is not the set of all integers because it excludes zero and every negative integer. Option B is finite, option C is finite, and option D equals {−1, 1}, which is also finite. Hence A is the unique answer.

Taking n = 1, 2, 3, 4, and so on produces K₁ = {1, 8, 27, 64, 125, …}. Natural numbers continue without an ending value, and each distinct natural number gives a distinct cube, so the list of elements never terminates. Therefore K₁ is infinite. It is neither empty nor a singleton, and it is not finite because no final cube exists.

Since x belongs to the natural numbers and x² < 10, test the natural numbers in order: 1² = 1, 2² = 4, and 3² = 9, all of which are less than 10. However, 4² = 16, so 4 is excluded. Therefore L₁ = {1, 2, 3}, which is exactly M₁. Two sets are equal when they contain precisely the same elements, regardless of how those elements are written or listed.

For natural numbers, evaluate x² ≤ 10. The values x = 1, 2, and 3 work because their squares are 1, 4, and 9. The next natural number, 4, does not work because 4² = 16, which is greater than 10. Thus N₁ = {1, 2, 3}. This is exactly O₁, so the correct statement is N₁ = O₁. The inclusive sign ≤ does not add 4 because no natural-number square equals 10.

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.

Solving the equation 2x + 1 = 0 gives 2x = −1 and hence x = −1/2. Although this is a solution over the real or rational numbers, it is not an integer. The definition of Q₁ specifically restricts x to ℤ, the set of integers. Therefore no integer satisfies the condition, and Q₁ contains no elements. Hence Q₁ is the empty set, written as ∅.

Solving 2x + 1 = 0 gives x = −1/2. The number −1/2 is rational because it can be expressed as a ratio of two integers with a nonzero denominator. It therefore satisfies the stated domain x ∈ ℚ. Since the equation has exactly one solution, R₁ = {−1/2} and contains one element. A set with exactly one element is called a singleton set.

A natural number leaves remainder 2 on division by 5 precisely when it has the form 5k + 2, where k is a nonnegative integer (with the usual convention that natural numbers include positive values). The resulting members begin 2, 7, 12, 17, 22, and continue indefinitely by adding 5. There is no upper bound on k, so the set has infinitely many elements and is therefore infinite.

The natural numbers that leave remainder 2 upon division by 5 have the form 5k + 2. Up to 40, they are 2, 7, 12, 17, 22, 27, 32, and 37. The next number, 42, exceeds the bound 40 and must be excluded. Thus T₁ contains eight distinct elements, so option B is correct. The upper bound makes this otherwise repeating pattern finite.

By the division algorithm, when a number is divided by a positive divisor 6, the remainder must be one of 0, 1, 2, 3, 4, or 5. A remainder must always be strictly less than the divisor. Therefore remainder 6 is impossible for every natural number. No x satisfies the defining condition, so U₁ has no elements and is the empty set, denoted by ∅.

The integer solutions of x² = 25 are x = 5 and x = −5, so X₁ = {5, −5}. In set notation, repeating an element does not create a new element; duplicates are ignored. Therefore Y₁ = {5, −5, 5} represents the same set as {5, −5}. Both sets contain exactly the two elements 5 and −5, so X₁ = Y₁. Statement A is correct.

A two-digit natural number ranges from 10 through 99, inclusive. The condition x > 99 asks for a two-digit number strictly greater than the largest possible two-digit number, 99. No such natural number exists. Consequently, the defining conditions cannot be satisfied simultaneously, and Z₁ contains no elements. Therefore Z₁ is the empty set, written as ∅; it is not the set containing 99 because the inequality is strict.

The condition 10 ≤ x < 100 selects the natural numbers 10, 11, 12, ..., 99. There is a definite first element, 10, and a definite last element, 99. Therefore, the set contains only 90 elements and is finite. A set may have many elements and still be finite if its total number of elements is limited.

A product is zero when at least one factor is zero. Hence x − 1 = 0, x − 2 = 0, or x − 3 = 0, giving A = {1, 2, 3}. In set notation, repetition does not create a new element, so B = {3, 2, 1, 2} is also {1, 2, 3}. Order and repeated writing do not affect set equality; therefore A = B.

The integers satisfying −5 < x < 5 are −4, −3, −2, −1, 0, 1, 2, 3, and 4. Among these, the odd integers are −3, −1, 1, and 3. Their squares are 9, 1, 1, and 9 respectively. None has x² > 9; the values −3 and 3 only give x² = 9, which is not greater than 9. Therefore, C is the empty set and has 0 elements.