The Empty Set, Finite and Infinite Sets, Equal Sets

The natural numbers satisfying x > 50 are 51, 52, 53, 54, and so on. After every such number, a larger natural number can be found, so the list never ends. Therefore the set has infinitely many elements. The lower bound 50 only tells us where the elements begin; it does not impose an upper limit or make the set finite.

A three-digit natural number must be at least 100; the three-digit numbers begin with 100 and continue through 999. The additional condition x < 100 contradicts the requirement that x have three digits. Hence there is no natural number satisfying both conditions, so A contains no elements and is the empty set, written as ∅.

The word “Mathematics” contains the letters m, a, t, h, e, m, a, t, i, c, s. In a set, repeated occurrences are written only once, so the distinct-letter set is {m, a, t, h, e, i, c, s}. Set order is irrelevant, and therefore this set is exactly B. Hence option A is correct.

A prime number is a positive integer greater than 1 with exactly two positive divisors: 1 and itself. The positive primes less than 7 are 2, 3, and 5. The number 1 is not prime, and 7 is excluded because the condition says less than 7, not less than or equal to 7. Therefore A = {2, 3, 5} = B.

For the quadratic equation x² + x + 1 = 0, the discriminant is b² − 4ac = 1² − 4(1)(1) = −3. Since the discriminant is negative, the equation has no real roots. Every integer is a real number, so it cannot have any integer solution either. Therefore, no element satisfies the defining condition, and A is the empty set, written as ∅.

Factor the equation as x² − 6x + 9 = (x − 3)² = 0. Thus, its only solution is x = 3, so A = {3}. In set notation, repeated entries are not counted more than once; therefore {3, 3, 3} is also simply {3}. Both sets contain exactly the same element, so A and B are equal. Repetition does not create additional set elements.

Solve the equation x² = 2x by bringing all terms to one side: x² − 2x = 0, so x(x − 2) = 0. Hence x = 0 or x = 2. Both values are integers, and both are even because each is divisible by 2; zero is also an even integer since 0 = 2 × 0. Therefore, the required set is A = {0, 2}.

Under the usual school convention ℕ = {1, 2, 3, ...}, the positive multiples of 20 are 20, 40, 60, and so on. The smallest such multiple is already 20, but the condition requires x < 20. Therefore, no natural number can satisfy both conditions. Hence A contains no elements and is the empty set. Even if a convention includes zero in ℕ, zero is not a positive multiple of 20 in this context.

Set A contains every natural-number multiple of 4: 4, 8, 12, 16, 20, and so on without a final member, so A is infinite. Set B lists only the multiples from 4 through 40, namely ten elements, and therefore is finite. Although the terms begin similarly, B stops at 40 while A continues beyond 40. Since their elements are not the same, A and B are not equal.

The two inequalities must hold simultaneously: x must be at least 0 and at most 0. The only real number satisfying both conditions is x = 0. Therefore, A = {0}. This is a singleton set because it contains exactly one element, and every singleton set is finite. The use of inclusive signs is important; if the inequalities were strict, the result would instead be empty.

The condition requires one real number x to be greater than 0 and less than 0 at the same time. No real number can satisfy both requirements, because a number cannot lie strictly between two equal endpoints. In interval notation, the condition describes the open interval (0, 0), which contains no points. Hence the solution set A is empty, written as A = ∅.

We need integers strictly greater than −1 and strictly less than 1. The only integer in that open interval is 0; the boundary integers −1 and 1 are excluded because the inequalities are strict. Therefore, A = {0}. Since B is also defined as {0}, the two sets contain exactly the same element and are equal. Thus the correct statement is A = B.

The positive natural-number divisors of 15 are 1, 3, 5, and 15, so A = {1, 3, 5, 15}. Set B requires the divisor to be less than 15, so it contains only {1, 3, 5}; the divisor 15 itself is excluded. Thus 15 belongs to A but not to B. Since two sets cannot be equal when one contains an element absent from the other, A ≠ B.

The natural-number elements of A are the positive multiples of 5 not exceeding 100: 5, 10, 15, ..., 100. Each multiple has the form 5k, where k = 1, 2, ..., 20. Equivalently, the number of terms is 100 ÷ 5 = 20. Because 100 itself is divisible by 5 and satisfies x ≤ 100, it is included.

The integer multiples of 10 are ..., −20, −10, 0, 10, 20, ..., 100, .... The strict inequalities 0 < x < 100 exclude 0 and 100, as well as all negative values. The remaining multiples are 10, 20, 30, 40, 50, 60, 70, 80, and 90, so option A gives the complete roster form.

For every real number x, x² is at least 0. Consequently, x² + 1 is at least 1, so it can never be less than or equal to 0. Therefore, no real number satisfies the defining inequality. The set has no elements and is written as A = ∅, so it is the empty set, not a singleton, an infinite set, or the set of all real numbers.

The natural numbers are discrete: consecutive natural numbers do not have another natural number between them. Since 9 and 10 are consecutive, there is no natural number x satisfying 9 < x < 10. The phrase “between 9 and 10” excludes both endpoints, so the set contains no elements and is therefore the empty set ∅.

Every real number is either rational or irrational, but no real number can be both. A rational number can be written as p/q, where p and q are integers and q ≠ 0; an irrational number cannot be represented in that form. These two categories are disjoint, so the condition requiring both properties has no solution. Hence A is the empty set ∅.

A composite number is a natural number greater than 1 that has more than two positive divisors. The natural numbers less than 11 are 1 through 10. Among them, 4, 6, 8, 9, and 10 are composite. The numbers 2, 3, 5, and 7 are prime, while 1 is neither prime nor composite. Therefore A = {4, 6, 8, 9, 10}.

An integer that leaves remainder 1 on division by 2 is odd. The set can be written as A = {2k + 1 : k ∈ Z}, which includes ..., -5, -3, -1, 1, 3, 5, ... . Since k can be any integer and there is no largest or smallest member, the set contains infinitely many elements. Therefore, option A is correct.

In Euclidean division by 4, the remainder must be one of 0, 1, 2, or 3; it must always be less than the divisor. Therefore, no integer can leave remainder 5 when divided by 4. The defining condition has no solution, so A has no elements and is the empty set. Hence, option A is correct.

The conditions x < 100 and x > 100 must hold simultaneously. No number, whether natural, integer, or real, can be both less than 100 and greater than 100 at the same time. Thus, there is no natural number satisfying the definition of A. Consequently, A contains no elements and is the empty set, so option A is correct.

When n = 1, A consists of all one-digit natural numbers. Under the standard school convention N = {1, 2, 3, ...}, these numbers are 1, 2, 3, 4, 5, 6, 7, 8, and 9. The number 0 is not a one-digit natural number in this convention. Therefore, A has 9 elements, making option A correct.

Solving x² = 49 gives x = 7 or x = -7, because both numbers have square 49. The additional condition x > 0 eliminates -7 and retains only 7. Therefore, the set of all integers satisfying both conditions is A = {7}. It is a singleton set, so option A is correct.

The positive factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. Among these factors, the prime numbers are 2, 3, and 5 because each has exactly two positive divisors: 1 and itself. The number 1 is not prime, and 30 is composite because it has more than two positive divisors. Hence the required set in roster form is A = {2, 3, 5}.