The Empty Set, Finite and Infinite Sets, Equal Sets

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.

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 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.

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.

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.

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.

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.

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.

Since x belongs to ℕ, take the positive natural numbers under the convention used in the options: 1, 2, 3, and so on. The inequality x² ≤ 49 gives x ≤ 7 for natural numbers. Thus the possible values are 1 through 7, so A = {1, 2, 3, 4, 5, 6, 7}. Negative integers are excluded because the domain is ℕ.

The equation x² = 2 has the real solutions x = √2 and x = −√2. However, √2 is irrational, and its negative is also irrational; neither number belongs to ℚ, the set of rational numbers. Since the defining domain is ℚ, not ℝ, there are no admissible solutions. Therefore A contains no elements and A = ∅.

The inequality x² ≤ 4 is equivalent to |x| ≤ 2, or −2 ≤ x ≤ 2. The integers in this closed interval are −2, −1, 0, 1, and 2. Hence A = {−2, −1, 0, 1, 2}, which is exactly the set B. The endpoints are included because the inequality is ≤, and all intermediate integers must also be counted.

No natural number can be greater than every natural number. If x is any natural number, then x + 1 is also a natural number and is greater than x. Thus x fails the required condition. This argument works for every possible natural number, so the set has no elements and is the empty set: A = ∅.

An element of A would have to be less than every natural number, including itself. However, no number is less than itself because the relation x < x is always false. Therefore, regardless of whether a textbook begins N with 0 or 1, no natural number satisfies the condition, so A is the empty set.

The positive divisors of 36 can be found in pairs: 1 and 36, 2 and 18, 3 and 12, 4 and 9, and 6 and 6. Removing the repeated middle entry gives the nine distinct divisors 1, 2, 3, 4, 6, 9, 12, 18, and 36. These are exactly the elements listed in B. Therefore A contains nine elements, so n(A) = 9, and because both sets have the same elements, A = B.

We solve x² − 1 < 0 by adding 1 to both sides, obtaining x² < 1. For real numbers, this is equivalent to −1 < x < 1. Therefore, A is the open interval (−1, 1). Every non-empty real interval contains infinitely many real numbers, so A is an infinite set. The endpoints −1 and 1 are excluded because the inequality is strict.

Complete the square: x² − 4x + 5 = (x − 2)² + 1. For every integer, and indeed every real number, (x − 2)² is at least zero, so the expression is at least 1. It can never equal zero. Therefore, the equation has no integer solution and the set A contains no elements; hence A = ∅.

Rewrite the inequality as x² − 2x < 0, or x(x − 2) < 0. Over the real numbers, this holds when 0 < x < 2. The natural numbers in this open interval depend on the usual school convention ℕ = {1, 2, 3, …}; only x = 1 belongs to it. Therefore, A = {1}. Neither 0 nor 2 satisfies the strict inequality.

Move all terms to one side: x² − x ≤ 0, which factors as x(x − 1) ≤ 0. The real solution interval is 0 ≤ x ≤ 1. Among integers, the only values in this closed interval are 0 and 1, so A = {0, 1}. Therefore, the number of elements is n(A) = 2. Both endpoints are included because the inequality is non-strict.

The phrase “strictly between 10 and 20” means 10 < x² < 20. The only perfect square in this range is 16, because 3² = 9 is too small and 5² = 25 is too large. Solving x² = 16 gives x = 4 or x = −4. Hence A = {−4, 4}, not the four-element set shown in option A. Therefore, the original options contain no correct answer; option A must be corrected to A = {−4, 4}.

The inequality x² < 30 implies −√30 < x < √30, so the possible integers range from −5 to 5. Among these, the integers divisible by 3 are −3, 0, and 3. Therefore A = {−3, 0, 3}, which has three distinct elements. Zero is included because 0 = 3 × 0, so it is divisible by 3. Hence option A is correct; the other counts omit or add valid elements.

The outer set A has two elements: the empty set ∅ and the singleton set {∅}. These are different objects. The symbol ∅ denotes a set with no elements, whereas {∅} denotes a set whose one element is the empty set. Since the two outer elements are distinct, n(A) = 2.

The governing concept is finding common elements of two divisor sets and then comparing sets. The positive divisors of 12 are 1, 2, 3, 4, 6, and 12, while those of 18 are 1, 2, 3, 6, 9, and 18. Their common divisors are therefore 1, 2, 3, and 6. Thus A = {1, 2, 3, 6}, exactly the same elements as B. Option B lists only one common divisor, option C includes non-common divisors, and option D is false.

If an integer is divisible by 6, then it can be written as 6k = 2(3k), so it is divisible by both 2 and 3. Conversely, if an integer is divisible by both 2 and 3, then because 2 and 3 are coprime, it is divisible by their product 6. Thus the two conditions describe exactly the same integers, so A = B. The multiples ..., −12, −6, 0, 6, 12, ... continue without end, making both sets infinite.

The governing concept is determining the cardinality of a set defined by an integer inequality. Starting with x² − 9 < 0 gives x² < 9, which is equivalent to −3 < x < 3. Since x must be an integer, the allowed values are −2, −1, 0, 1, and 2. There are 5 elements. The endpoints −3 and 3 are excluded because the inequality is strict, so options B, C, and D overcount or undercount the set.