The Empty Set, Finite and Infinite Sets, Equal Sets

Solving the equation x + 4 = 4 gives x = 0. Under the usual school convention, the natural numbers are 1, 2, 3, and so on, so 0 does not belong to N. Consequently, no natural number satisfies the given condition. Therefore S has no elements and S = ∅, the empty set.

The states of India and the days of a week are clearly finite sets. The natural numbers less than 1000 also form a finite set because there is a fixed upper bound. In contrast, prime numbers continue indefinitely; there is no largest prime number. Therefore the set of prime natural numbers is infinite and is not finite, so option C is correct.

A singleton set contains exactly one element. In option A, x² = 4 gives x = −2 or x = 2, but the additional condition x > 0 removes −2 and leaves only x = 2. Thus the set is {2}, which has one member. Option B contains two integers, −2 and 2; option C is empty if N begins at 1; and option D contains 0 and 1. Therefore, only option A is a singleton set.

Evaluate the condition x² < 10 for natural numbers. For x = 1, 2, and 3, the squares are 1, 4, and 9, all less than 10. For x = 4, x² = 16, which is not less than 10; larger natural numbers also fail. Thus A = {1, 2, 3}, exactly the same as B. Therefore A = B, so option A is correct.

Solving x² = 25 over the integers gives two solutions: x = 5 and x = −5. Both numbers belong to Z, so A = {−5, 5}. These are distinct elements, and no other integer has square 25. Therefore the cardinality of A, written |A|, is 2. The negative solution must not be omitted.

Factor the equation as x² + 2x + 1 = (x + 1)² = 0. Its only real and integer solution is x = −1. However, −1 is not a natural number, so it does not satisfy the restriction x ∈ N. Consequently, no permitted value belongs to A, which means A = ∅, the empty set.

For option A, substitute the natural-number values n = 1, 2, 3, and 4 into x = 4n. The resulting elements are 4, 8, 12, and 16, exactly the elements of A. Therefore, option A represents the same set. Equality of sets depends on having precisely the same elements; the order of elements does not matter, but missing or additional elements would make the sets unequal.

The outer braces define the set A. Inside them are three distinct objects: the number 0, the empty set ∅, and the singleton set {0}. Although {0} contains 0 as its own element, it is not the same object as 0. Likewise, ∅ is a set with no elements. Thus A has three elements, so its cardinality is 3.

We need integers whose squares are strictly less than 4. The integers −1, 0, and 1 have squares 1, 0, and 1, respectively, all less than 4. The integers −2 and 2 have square 4, so they are excluded because the inequality is strict. No other integer can satisfy the condition. Therefore, A = {−1, 0, 1}, so option B is correct.

The set contains primes such as 23, 29, 31, 37, and many more. There is no largest prime number; Euclid’s theorem establishes that infinitely many prime numbers exist. Removing the finitely many primes that are at most 20 still leaves infinitely many primes greater than 20. Therefore A is an infinite set, not a finite or singleton set.

The governing concept is equality of sets together with the stated domain restriction. Algebraically, x² = 49 gives x = 7 or x = −7. However, x must belong to the natural numbers, so −7 is excluded and A = {7}. Since B is also the singleton set {7}, A and B have exactly the same element. Thus option A is correct; option C ignores the domain restriction.

The governing concept is describing a set by common conditions: being a factor of 48 and being prime. Since 48 = 2⁴ × 3, its prime divisors are only 2 and 3. Number 1 is not prime, and 4 and 6 are composite, so they cannot enter A. Hence A = {2, 3}, making option A correct. The other choices include at least one non-prime or non-required number.

A number that is a multiple of both 12 and 18 must be a multiple of their least common multiple. Since lcm(12, 18) = 36, the positive multiples below 100 are 36 and 72. The next multiple, 108, is not less than 100. Therefore, A = {36, 72} and has 2 elements, so option B is correct.

The governing concept is distinguishing finite and infinite sets. The natural-number factors of a fixed number, 64, are limited: {1, 2, 4, 8, 16, 32, 64}; hence option C is finite. In contrast, natural numbers greater than 1000 continue without end, negative integers continue downward without end, and positive multiples 9, 18, 27, ... are infinite. Therefore option C is the only set that is not infinite.

Option A has the same four elements in a different order, and order does not matter in a set. Option B repeats 1, but repetition does not add a new element. Option C is the set of positive factors of 8, namely {1, 2, 4, 8}. Option D contains one additional element, 16, so it is not equal to A.

Factor the equation: x² − 4x + 4 = (x − 2)² = 0. Hence x = 2 is the only integer solution. Although the quadratic has a repeated root, a set records each value only once; it does not count multiplicity of roots. Therefore A = {2}, so the set contains exactly one element.

The notation x ∈ N restricts x to natural numbers. Although infinitely many real numbers lie strictly between 2 and 3, there is no natural number in that interval. The endpoints 2 and 3 are also excluded because the inequalities are strict. Therefore no element satisfies every condition, so A = ∅, the empty set.

The equation x² = 2 has the real solutions √2 and −√2, but neither solution is an integer. Since the domain in the set-builder description is Z, only integer solutions may be included. No integer satisfies the equation, so the set has zero elements and is therefore the empty set, ∅.

For every real number x, x² is nonnegative, so x² + 1 is at least 1 and can never equal zero. Alternatively, the equation would give x² = −1, which has no real solution; its solutions are imaginary numbers. Because the domain is R, no solution is admitted, and the set is empty.

The symbol ∅ denotes a set with no elements, so its cardinality is 0. In contrast, {∅} is a set whose sole element is the empty set itself; therefore its cardinality is 1. The braces change the role of ∅ from representing a set to being an element inside another set. Hence A is correct.

In set notation, repetition does not create a new element. Thus A = {1, 2, 3} after removing the repeated 1, and B = {1, 2, 3} after removing the repeated 2 and 3. Since two sets are equal when they contain exactly the same elements, A and B are equal. A has three distinct elements, and B also has three distinct elements.

We need natural-number values of x satisfying x² = 4. The equation has the two integer solutions x = 2 and x = −2, but under the usual school convention N contains positive natural numbers, so −2 is excluded. The only allowed value is x = 2. Therefore, the solution set is A = {2}, making option A correct.

Taking ℕ = {1, 2, 3, ...}, test the natural numbers in order. We have 1² = 1 < 10, 2² = 4 < 10, and 3² = 9 < 10, while 4² = 16 is not less than 10. Thus A = {1, 2, 3}, so it contains exactly three elements. If a convention includes 0 in ℕ, 0² also satisfies the condition; however, the options and standard school convention here use ℕ beginning with 1.

Using the convention N = {1, 2, 3, ...}, there is no natural number less than 1. Therefore B has no elements and B = ∅. Since A is also defined as the empty set, both sets contain exactly the same elements. Hence A = B. The answer depends on the stated convention that natural numbers begin with 1.

Assuming N = {1, 2, 3, ...}, the set A contains all natural numbers from 1 through a. The set B contains exactly 1, 2, 3, and 4, so its largest element is 4. For A and B to be equal, the upper bound must be a = 4. If a were 3, an element would be missing; if a were 5, an extra element would appear.