Sets and their representations

The positive natural-number multiples of 7 are 7, 14, 21, 28, 35, and so on. The condition x < 30 retains only 7, 14, 21, and 28. The number 35 is excluded because it is greater than 30; 0 is not included under the natural-number convention used here, and 1 is not a multiple of 7. Thus S = {7, 14, 21, 28}.

The positive perfect squares are obtained by squaring natural numbers: 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, and 7² = 49. The next square, 8² = 64, is not less than 50. Under the convention ℕ = {1,2,3,...}, zero is not included. Hence T = {1, 4, 9, 16, 25, 36, 49}.

Because x belongs to the integers, we list only whole-number values, not decimals or fractions. The inequality -2 ≤ x includes -2, while x < 3 excludes 3. Thus the elements are V = {-2, -1, 0, 1, 2}. Counting them gives five elements, so the cardinality of the set is n(V) = 5. The endpoint symbols are essential: the left endpoint is included because of ≤, and the right endpoint is excluded because of <.

The condition |x| < 3 means that the distance of x from zero is less than 3. Therefore, x must lie strictly between −3 and 3. Since x is restricted to integers, the possible values are −2, −1, 0, 1, and 2. The endpoints −3 and 3 are excluded because the inequality is strict, so the roster form is {-2, −1, 0, 1, 2}.

Here x is a natural number satisfying both inequalities 1 ≤ x and x ≤ 5. Because both inequalities are non-strict, the endpoints 1 and 5 are included. Listing every natural number from 1 through 5 gives A₁ = {1, 2, 3, 4, 5}. Zero is not included because the lower bound begins at 1.

The variable x is required to be an integer and must satisfy both 0 < x and x < 1. There is no integer strictly between 0 and 1; numbers such as 0.5 are real numbers, not integers. The endpoints are also excluded by the strict inequalities. Therefore, the set contains no element and is the empty set, ∅.

The positive factors of 15 are the positive integers that divide 15 without a remainder. They are 1, 3, 5, and 15, so B₂ = {1, 3, 5, 15}. Counting these distinct members gives four elements. Negative factors such as -1 and -3 are excluded because the question specifically requires positive factors.

A natural number divisible by both 2 and 3 must be divisible by their least common multiple, lcm(2, 3) = 6. The positive multiples of 6 not exceeding 18 are 6, 12, and 18. Therefore C₂ = {6, 12, 18}. Numbers divisible by only one of 2 or 3 do not satisfy the word both.

The positive natural-number factors of 9 are 1, 3, and 9. Among them, 3 is prime because it has exactly two positive divisors, 1 and 3. The number 1 is not prime, and 9 is composite because it has more than two positive divisors. Therefore, the set C₃ contains only 3, so C₃ = {3}.

The natural-number multiples of 5 are 5, 10, 15, 20, 25, and so on. This sequence continues indefinitely because multiplying 5 by every positive natural number produces another member. There is no greatest natural-number multiple of 5, so D₁ has infinitely many elements and is an infinite set.

The condition −5 < x < 5 restricts x to the integers −4, −3, −2, −1, 0, 1, 2, 3, and 4. From these integers, the even numbers are −4, −2, 0, 2, and 4. Both endpoints −5 and 5 are excluded because the inequalities are strict. Therefore, the roster form is {−4, −2, 0, 2, 4}. Zero is included because zero is an even integer.

The natural numbers strictly between 20 and 30 are 21, 22, 23, 24, 25, 26, 27, 28, and 29. Among them, 23 and 29 are prime: each has exactly two positive divisors, 1 and itself. The other candidates are composite because 21 = 3 × 7, 22 = 2 × 11, 24 is even, 25 = 5 × 5, 26 = 2 × 13, 27 = 3 × 9, and 28 is even. Hence D₃ = {23, 29}.

Assuming the standard school convention ℕ = {1, 2, 3, …}, the odd natural numbers not exceeding 11 are 1, 3, 5, 7, 9, and 11. The symbol ≤ means “less than or equal to,” so the endpoint 11 must be included. The set contains only odd numbers, so the even numbers are excluded. Therefore, the roster form is {1, 3, 5, 7, 9, 11}, which is option B.

The set consists of the names of the twelve months in a year: January through December. Its membership is clearly defined, and no additional month names are possible in the ordinary calendar-year context. Because it has exactly 12 members, it is finite, not infinite, empty, or a singleton. Therefore, option B is correct.

The inclusive interval −4 ≤ x ≤ 4 contains the integers −4, −3, −2, −1, 0, 1, 2, 3, and 4. Selecting only the odd integers removes −4, −2, 0, 2, and 4, leaving −3, −1, 1, and 3. The endpoints −4 and 4 are included in the interval, but they are even and therefore do not belong to the set. Hence the roster form is {−3, −1, 1, 3}, option A.

Solve the equation x² − 1 = 0 by factoring: (x − 1)(x + 1) = 0. Thus the algebraic solutions are x = 1 and x = −1. However, the set-builder condition restricts x to the natural numbers. Under the usual school convention, −1 is not a natural number, whereas 1 is. Therefore only 1 belongs to F₂, so F₂ = {1}. This is a singleton set, making option B correct.

The positive natural-number factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. The additional condition x > 6 removes 1, 2, 3, 4, and 6, because 6 is not greater than 6. The remaining factors are 8, 12, and 24. Therefore G₁ = {8, 12, 24}. Option B incorrectly includes 6, while option D contains numbers that are not all factors of 24.

To find the roster form, solve the condition x² = x. Rearranging gives x² − x = 0, so x(x − 1) = 0. Therefore, x = 0 or x = 1. Both values are integers and satisfy the original equation: 0² = 0 and 1² = 1. Hence the set contains exactly these two distinct elements, so A = {0, 1}.

The positive factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36. We now retain only those factors that are perfect squares. Among them, 1 = 1², 4 = 2², 9 = 3² and 36 = 6². The number 16 is not a factor of 36, and 6 is not a perfect square. Therefore, B = {1, 4, 9, 36}, making option A correct.

The inequality |x − 2| ≤ 1 means that x is at a distance of at most 1 from 2. Equivalently, −1 ≤ x − 2 ≤ 1. Adding 2 throughout gives 1 ≤ x ≤ 3. Since x must be an integer, the possible values are 1, 2 and 3. Thus the correct roster form is C = {1, 2, 3}. The endpoints are included because the inequality is non-strict.

First list the positive divisors of 36: 1, 2, 3, 4, 6, 9, 12, 18 and 36. The prime divisors in this list are 2 and 3, so they must be removed. The number 1 is not prime because a prime number has exactly two positive divisors, whereas 1 has only one. All remaining divisors are composite or 1, giving option A.

Solve the inequality: 2x + 3 ≤ 15 gives 2x ≤ 12, and hence x ≤ 6. Under the convention used here, ℕ denotes the positive natural numbers 1, 2, 3, and so on. Therefore, the natural numbers satisfying x ≤ 6 are 1, 2, 3, 4, 5 and 6. Thus the roster form is option A. Option B incorrectly includes zero, while C omits 6.

Use the standard absolute-value inequality rule: |x − 2| ≤ 3 is equivalent to −3 ≤ x − 2 ≤ 3. Adding 2 to all parts gives −1 ≤ x ≤ 5. The integers in this closed interval are −1, 0, 1, 2, 3, 4 and 5. Counting them gives 7 elements, so option C is correct. The closed endpoints must be included because the original sign is ≤.

The English alphabet has exactly five commonly recognized vowels: a, e, i, o, and u. These elements are clearly specified, so the set is well-defined. Because its elements can be listed completely and their number is limited to five, G is a finite set. It is neither empty nor infinite.

For any number x, the equation x + 5 = x would require subtracting x from both sides, giving 5 = 0. This is impossible, so no natural number satisfies the stated condition. Therefore H contains no elements. A set containing no element is called the empty set, usually denoted by ∅.