General chapter practice

A set is written in roster form when its individual elements are listed explicitly inside braces and separated by commas. In A = {2, 4, 6, 8}, every element is directly displayed, so this is roster form. Set-builder form would describe the elements through a condition, such as the even numbers less than 10. Therefore option A is correct.

The set-builder condition requires x to be both a natural number and odd, with x less than 10. The odd natural numbers below 10 are 1, 3, 5, 7, and 9, so the roster form is B = {1, 3, 5, 7, 9}. Option B contains even numbers, option C includes non-odd values, and option D includes 11, which fails x < 10. Thus A is correct.

A set is described by the common property shared by all its distinct elements. The symbols a, e, i, o, and u are the standard vowels of the English alphabet, so option A gives the exact description. They are not consonants. Also, the first five alphabet letters are a, b, c, d, and e, so option C does not match the complete set.

The symbol ∈ means “is an element of” or “belongs to.” To check a membership statement, compare the number on the left with every element listed in the set D. The number 3 appears in D, so 3 ∈ D is true. The numbers 6, 0, and 7 are absent, making options B, C, and D false. Therefore, option A is correct.

The governing concept is set membership. The symbol ∈ means that the element on its left is an exact member of the set on its right. The set E is written explicitly as {2, 4, 6, 8}; therefore 2, 4, 6, and 8 belong to E. This makes 4 ∈ E, 2 ∈ E, and 8 ∈ E true statements. The number 5 is not listed, so 5 ∈ E is false. Since the question asks which statement is false, option C is the unambiguous answer. Membership is not decided by whether a number is close to a listed element, whether it is even or odd, or whether it lies in the numerical range from 2 to 8. Only the exact elements displayed in the set count.

Roster form lists the individual elements of a set. Using the usual school convention that natural numbers begin with 1, the natural numbers satisfying x ≤ 4 are 1, 2, 3, and 4. The symbol ≤ includes 4 itself. Zero is generally treated as a whole number rather than a natural number in this convention, so option B is correct.

An empty set contains no element that satisfies its defining condition. Under the usual school convention that natural numbers start with 1, no natural number is less than 1. Thus option A represents the empty set. Option B contains 1, 2, 3, and 4; option C contains vowels; and option D contains even numbers such as 2, 4, 6, and 8, so those are not empty.

List the even natural numbers that satisfy x ≤ 10: H = {2, 4, 6, 8, 10}. Counting the distinct entries gives 5 elements. The endpoint 10 is included because the inequality is ≤, not <. Under the usual school convention, 0 is not counted as a natural number, so it does not add another element. Therefore, option B is correct.

The defining rule of a set is that repetition does not create a new element. In I, the first 1 and second 1 represent one element, and the two occurrences of 2 also represent one element. Removing repetitions leaves I = {1, 2, 3}. Therefore, it has three distinct elements and option A is correct; options B and D count repetitions incorrectly.

Two sets are equal when they contain exactly the same distinct elements, regardless of the order in which those elements are written. Both J and K contain 1, 2, and 3, so J = K and each set has cardinality 3. Option B incorrectly treats a set like an ordered list. Options C and D are also false because neither set has more elements.

Set-builder form states the rule that generates all members of a set. Every element of L is an even natural number and is less than or equal to 10, so option A produces exactly 2, 4, 6, 8, and 10. Option B gives odd numbers, option C includes numbers such as 1 and 3 and excludes 10, and option D gives primes rather than all even numbers.

A set contains each distinct object only once. The days of a week are Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, and Sunday, giving seven different names. Therefore, M has 7 elements. The number 5 counts weekdays only, while 6 and 8 do not represent the complete set of weekly day names, so option C is correct.

A finite set has a definite number of elements that can be counted completely. The months of a year are January through December, so there are exactly 12 of them. Natural numbers, integers, and rational numbers extend without an ending and are infinite sets. Hence the set of months is finite, making option C correct.

An infinite set has endlessly many elements and no last element. The natural numbers continue as 1, 2, 3, 4, and so on, so they cannot be completely listed or counted. A particular class has a limited number of students, a year has 12 months, and English has five standard vowels. Therefore, option C is correct.

The governing concept is conversion from set-builder notation to roster notation. The expression N = {x : x is a digit} describes the set of all decimal digits. In the decimal system, the complete list is 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Roster form requires writing every member explicitly, with no omission and no extra value. Therefore N = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, so option B is correct. Option A omits zero, even though zero is a digit. Option C omits four valid digits, and option D includes 10, which is a two-digit number rather than a single digit. The distinction between a digit and a number is essential here.

The governing concept is roster notation for a set. Curly braces list the elements that belong to the set, and the absence of an ellipsis or a range means that no additional elements are implied. Hence Q has exactly the three distinct elements 1, 2, and 3. Option A states this correctly, including the important word “only.” Option B incorrectly adds 4, option C contradicts the presence of three listed elements, and option D falsely says that 3 is absent even though it appears explicitly. Therefore option A is the only correct reading.

The set R explicitly lists three distinct elements: red, blue, and green. Since its members are limited and can be completely counted, R is a finite set with cardinality 3. It is not empty because it has members, and it is not infinite. A universal set can be identified only after a surrounding context is stated, so option A is correct.

A prime number has exactly two positive factors: 1 and the number itself. The only even prime is 2. Every even number greater than 2 is divisible by both 1 and 2 and has additional factors, so it is composite. Consequently, the set of even prime numbers contains only 2, and its roster form is {2}; option B is correct.

The governing concept is translating an inclusive bounded condition into roster form. The inequality 3 ≤ x ≤ 7 says that x must be at least 3 and at most 7. Because both inequalities include equality, the endpoints 3 and 7 must be included. The natural numbers satisfying the condition are therefore 3, 4, 5, 6, and 7. Hence T = {3, 4, 5, 6, 7}, making option A correct. Option B leaves out both endpoints, option C omits the upper endpoint 7, and option D omits the lower endpoint 3. No other natural number can satisfy the stated interval. The inclusion signs are the key detail: replacing either ≤ by < would change the corresponding endpoint.

Every member of U is a decimal digit, every member is even, and every member is less than 10. Thus U is exactly the set of even digits below 10. Option B describes odd digits, and C excludes most listed values because they are not prime. Option D is too broad because it would include odd natural numbers as well. Hence A is correct.

The governing concept is the number of elements in a set. The English alphabet contains exactly 26 letters, so V has a fixed and limited number of elements. A set with a definite limited cardinality is called a finite set. It is not empty because letters exist, not infinite because the number is bounded, and the description clearly defines its members. Therefore, option A is correct.

The governing concept is that a set must be well-defined: membership should be decidable without personal opinion. A list of 40 students identifies a definite collection, so every member can be recognized objectively. The terms beautiful, good, and tall may be judged differently by different people unless a precise standard is supplied. Thus only the collection in option A is clearly a set.

A set must have a clear membership rule. ‘Tasty’ depends on personal taste, so people may disagree about which fruits belong to that collection; it is therefore not well-defined. The states of India, days of a week and natural numbers below 10 can be identified objectively. Hence option C is correct.

First solve the condition x² = 9. The two integer solutions are x = 3 and x = −3. The second condition restricts x to natural numbers; under the standard school convention, −3 is not natural, while 3 is natural. Therefore the set contains only one element, and its roster form is W = {3}. Option B is correct; option A ignores the natural-number restriction.

The governing idea is roster representation: list every element satisfying the stated rule. Integers include negative numbers, zero, and positive numbers. Because both inequalities are inclusive, the endpoints −2 and 2 must be included. Listing all integers from −2 through 2 gives −2, −1, 0, 1, and 2. Hence option A is the complete and correct roster form.