The Empty Set, Finite and Infinite Sets, Equal Sets

The natural-number multiples of 4 are 4, 8, 12, 16, 20, and so on. For every multiple listed, adding 4 produces another natural-number multiple, so the list never ends. There is no upper bound in the defining condition. Consequently, A contains infinitely many elements and is an infinite set. Option C is therefore correct.

In a set, repetition of an element does not create a new element. The repeated 2s and 3s are therefore written only once when the set is simplified. The distinct elements are 1, 2, and 3, so A = {1, 2, 3} and its cardinality is 3, not 6. It is neither empty nor infinite. Hence option B is correct.

The two-digit natural numbers begin at 10 and end at 99. The even members satisfying the condition are 10, 12, 14, and so on, up to 98. This list has a definite first and last value, so it contains only finitely many elements. It is not empty because 10 is included, and it cannot be infinite. Therefore, option B, finite set, is correct.

The condition requires one natural number to satisfy x > 7 and x < 7 simultaneously. No number can be greater than 7 and less than 7 at the same time. Thus there are no elements satisfying the condition. A set with no elements is called the empty set, written as ∅. Therefore, option A is correct; although the empty set is finite, the most precise listed description is empty set.

The numbers whose square is 36 are 6 and −6, because 6² = 36 and (−6)² = 36. However, the set specifically restricts x to natural numbers, and under the standard school convention natural numbers are positive counting numbers. Therefore −6 is excluded, while 6 is included. Hence A contains exactly one element and A = {6}.

The set can be written as A = {x ∈ ℝ : x > 3}, or in interval notation as (3, ∞). It contains not only whole numbers such as 4, 5, and 6, but also infinitely many decimal and fractional numbers such as 3.1, 3.01, and 7/2. Since there is no greatest real number in this set and its elements never end, A is infinite.

For option A, substitute the allowed natural-number exponents n = 1, 2, 3, and 4. The resulting values are 2¹ = 2, 2² = 4, 2³ = 8, and 2⁴ = 16. Thus option A produces exactly the four elements of the given set. Option B gives 2, 4, 6, 8, option C gives 4, 8, 12, 16, and option D omits 16 and includes 1.

Solving x² = 2 over the real numbers gives x = √2 and x = −√2. Neither value is an integer, because √2 is irrational and cannot be written as a whole-number integer. The defining condition also requires x ∈ ℤ, so no integer satisfies the equation. Therefore A has no elements and is the empty set, written as ∅.

The positive natural-number factors of 15 are 1, 3, 5, and 15, so A = {1, 3, 5, 15}. In set notation, repeated listing of an element does not create a new element; therefore B = {1, 3, 5, 15, 15} is the same as {1, 3, 5, 15}. Since both sets contain exactly the same elements, A = B.

The odd natural numbers not exceeding 11 are 1, 3, 5, 7, 9, and 11. The inequality x ≤ 11 includes 11 itself, so it must be counted. Listing the elements avoids confusing the answer with the count of odd numbers below 11, which would be five. Therefore the set is finite and has 6 elements.

Because x is restricted to integers, only whole numbers and their negatives are considered; fractions and decimals are not included. The double inequality −1 ≤ x ≤ 3 includes both endpoints and every integer between them. Reading in order gives −1, 0, 1, 2, and 3. Thus the correct roster form is {−1, 0, 1, 2, 3}.

The positive multiples of 9 are 9, 18, 27, 36, 45, 54, and so on. The condition x < 50 allows the first five values but excludes 54 and every later multiple. Therefore A = {9, 18, 27, 36, 45}, which contains exactly 5 distinct elements. The strict inequality is important because 50 itself would not be included even if it were a multiple of 9.

The positive natural-number factors of 14 are 1, 2, 7, and 14. The additional condition requires the factor to be odd. Among these factors, 1 and 7 are odd, whereas 2 and 14 are even. Keeping only the values satisfying both conditions gives A = {1, 7}. The word “and” means that each selected element must meet both requirements.

The natural-number factors of 10 are A = {1, 2, 5, 10}. The factors of 20 not exceeding 10 are B = {1, 2, 4, 5, 10}. Although most elements are common, B contains the additional element 4. Since two sets are equal only when they have exactly the same elements, A and B are unequal. Therefore, option B is correct.

A natural number divisible by both 13 and 17 must be a multiple of their product, 13 × 17 = 221, because the numbers are relatively prime. Thus A contains 221, 442, 663, 884, and so on. Multiplying 221 by every positive natural number produces another member, with no upper bound. Hence the set has infinitely many elements, so option C is correct.

Since x² ≤ 9, taking square roots gives -3 ≤ x ≤ 3. The set is restricted to integers, so its elements are {-3, -2, -1, 0, 1, 2, 3}. There are seven elements in total. The endpoints -3 and 3 must be included because the inequality is less than or equal to, not strictly less than.

The positive natural-number factors of 21 are 1, 3, 7, and 21. Every one of these factors is odd; none is divisible by 2. Therefore, there is no natural number that is both a factor of 21 and even. A has no elements, so it is the empty set and option A is correct.

The powers of 2 among the natural numbers include 1, 2, 4, 8, 16, 32, and so on, depending on whether 2⁰ is included. For every nonnegative integer exponent n, 2ⁿ is a natural number, and increasing n gives a new larger value. There is no greatest exponent or last term, so the set is infinite. Option C is correct.

An empty set, written as ∅, contains no elements, so its cardinality is 0. A singleton set contains exactly one element; for example, {0} is a singleton. The braces matter: ∅ is not the same set as {0}, because the latter contains 0. Therefore, option C states the correct distinction between these two types of sets.

The governing concept is selecting members of a finite set using conditions. The positive factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, and 100. Applying x > 25 removes every factor up to and including 25, leaving A = {50, 100}. Therefore A has 2 elements and option A is correct. The strict inequality is why 25 is excluded.

For n = 1, 2, 3, 4, and 5, the expression n² gives 1, 4, 9, 16, and 25 respectively. Therefore, option A produces exactly the same elements as the given set. The cube rule in option B gives different values, the doubling rule in option C gives even values, and option D lists the bases rather than their squares. Hence A is correct.

The number 31 is prime because its only positive divisors are 1 and 31. Consequently, the set of natural-number factors is A = {1, 31}. These are two distinct elements, so the cardinality of A is 2. The number 31 itself is an element, not the number of elements in the set. Therefore, option B is correct.

Because N is defined here as {1, 2, 3, ...}, zero is not considered a natural number for this question. The positive perfect squares less than 6 are 1 = 1² and 4 = 2². The next perfect square, 9 = 3², is greater than 6 and is excluded. Therefore A = {1, 4}, so option A is correct.

The set contains all even integers: ..., −6, −4, −2, 0, 2, 4, 6, .... There is always another even integer greater than any listed member, and the sequence also continues indefinitely in the negative direction. Therefore, the set has infinitely many elements and is an infinite set. Hence, option C is correct.

The governing concept is converting a set-builder description into roster form. We need natural numbers satisfying both strict inequalities 3 < x and x < 8. Checking the integers between the boundaries gives 4, 5, 6, and 7. Neither endpoint is included because both conditions are strict. Therefore A = {4, 5, 6, 7}, so option B is correct; the other options include 3 or 8 incorrectly.