The Empty Set, Finite and Infinite Sets, Equal Sets

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 set contains every real number strictly between 0 and 1. There are infinitely many such numbers; for example, 1/2, 1/3, 1/4, and infinitely many others. More generally, between any two distinct real numbers there are infinitely many real numbers. Hence the set is infinite, and 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.

In a set, the order in which elements are listed does not matter, and repeating an element does not produce an additional distinct element. Thus B has only the distinct elements 1, 2, and 3, exactly as A does. Since both sets contain the same elements, A = B. Therefore, option A is correct.

The only natural number strictly between 14 and 16 is 15. However, 15 is not prime because it has factors other than 1 and itself; specifically, 15 = 3 × 5. Therefore no natural number satisfies both conditions, so the set has no elements. It is the empty set, and option A 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.

Solve the inequality: 2x + 1 < 20 gives 2x < 19, so x < 9.5. Since x belongs to the natural numbers and, in the usual school convention, ℕ = {1, 2, 3, ...}, the possible values are 1 through 9. Therefore, the set is {1, 2, 3, 4, 5, 6, 7, 8, 9} and has 9 elements. Hence, option B is correct.

The set contains all rational numbers strictly between 0 and 1. It includes 1/2, 1/3, 1/4, 2/5, and infinitely many other rational numbers. In fact, for every positive integer n, the number 1/(n+1) is rational and lies between 0 and 1, giving infinitely many distinct members. Therefore, the set is infinite, so option D is correct.

The equation x² = 4 has solutions x = 2 and x = −2 over the integers. However, the condition specifies x ∈ ℕ, so −2 is excluded. The only natural-number solution is x = 2; therefore A = {2}. Since B is also {2}, both sets contain exactly the same element and are equal. Option A is correct.

Set equality requires the two sets to contain exactly the same elements. A contains the numbers 1, 2, and 3, while B contains the symbols a, b, and c. Thus the sets are not equal. However, each set has three elements, so they have the same cardinality and are equivalent in size. Therefore, option C is correct.

For every integer x, x² is non-negative, so x² + 1 is at least 1 and can never equal 0. Thus the equation x² + 1 = 0 has no integer solution, which means B contains no elements. Therefore B is the empty set, just like A. Since two sets are equal when they have exactly the same elements, A = B. Hence, option A is correct.

The empty set, written as ∅ or {}, is the set containing no elements. Its cardinality is therefore zero: n(∅) = 0. This does not mean that 0 is an element of the empty set; the number 0 describes how many elements it has. The empty set is also finite because its number of elements is limited. Hence option B is correct.

Factor the equation: x² − 2x + 1 = (x − 1)². Setting this equal to zero gives (x − 1)² = 0, so x = 1 is the only real solution. Therefore, the set is {1}, which contains exactly one element. A set with exactly one element is called a singleton set. Hence, option B is the correct answer.

Taking n = 1, 2, 3, … produces the elements 2, 4, 6, 8, … . For every natural number n, another even number 2n can be produced, so the list never terminates. Repetition does not occur for different natural values of n. Hence the set contains infinitely many elements and is an infinite set. Option C is correct.

The condition requires x to be a natural number strictly greater than 5 and strictly less than 6. There is no integer, and therefore no natural number, between two consecutive integers 5 and 6. No value satisfies the condition, so the set has no elements. Hence it is the empty set, and option A is correct.

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.

Because x is an integer and both inequalities are strict, −3 and 3 are excluded. The integers strictly between them are −2, −1, 0, 1, and 2. Counting this list gives five distinct elements. Therefore, the cardinality of the set is 5, so option B is the correct answer.

The empty set contains no elements and is denoted by ∅ or by a pair of empty braces, {}. In contrast, {0} contains the element 0, {1} contains the element 1, and {∅} contains the empty set as one element. Therefore only option C represents a set with no elements.

The natural-number factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. This list contains eight elements and is complete because a positive integer has only finitely many positive divisors. Therefore, the set is non-empty and finite, not infinite or a singleton. Option B is correct.

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.