The Empty Set, Finite and Infinite Sets, Equal Sets

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.

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.

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.

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 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.

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 ∅.

There are infinitely many rational numbers strictly between 1 and 2. For example, 3/2, 4/3, and 5/4 are distinct members of A. More generally, rational numbers such as (n + 1)/(n) lie between 1 and 2 for every integer n greater than 1, giving infinitely many different elements. Therefore A is infinite, not a singleton or a finite 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 real number is rational exactly when it can be expressed as p/q with integers p and q, q ≠ 0. The decimal expansion of every rational number terminates or repeats periodically. Conversely, every terminating decimal can be converted to a fraction with a power of 10 as denominator, and every repeating decimal is also rational. Thus A and B contain exactly the same numbers, so A = B.