General chapter practice

A set is in roster form when its elements are written one by one inside braces. Here 2, 4, 6, and 8 are directly listed, so the representation is roster form. Set-builder form would state a common rule, while the other forms are not shown here.

Odd natural numbers are 1, 3, 5, 7, 9, and so on. Applying the condition x < 10 keeps exactly the odd numbers below 10: 1, 3, 5, 7, and 9. Number 10 is not odd, and 11 is not less than 10, so option A is correct.

The elements a, e, i, o, and u are the standard vowels of the English alphabet. They are not consonants, and they are not the first five alphabet letters, which are a, b, c, d, and e. Thus option A gives the exact common property of the set.

The symbol ∈ means “is an element of.” To test each statement, compare its number with the elements listed in D. Number 3 appears in D, whereas 6, 0, and 7 do not. Therefore, only 3 ∈ D is true, making option A correct.

A membership statement is true only when the stated number occurs in the set. Numbers 2, 4, and 8 are listed in E, so their membership statements are true. Number 5 is absent from E; hence 5 ∈ E is false. Therefore, option C is correct.

Using the usual school convention that natural numbers begin with 1, the natural numbers not exceeding 4 are 1, 2, 3, and 4. The symbol ≤ includes 4 itself. Zero is a whole number, but under this convention it is not a natural number, so option B is correct.

An empty set has no element satisfying its defining condition. Under the usual convention that natural numbers start at 1, there is no natural number less than 1, so option A describes the empty set. The other conditions have elements, such as 1 and 2.

The numbers 2, 3, 5, and 7 are all prime and all are natural numbers less than 10. Option C is too broad because it would also include 1 and 9, while option B and option D describe different collections. Thus the exact description is option A.

The even natural numbers not exceeding 10 are 2, 4, 6, 8, and 10. Counting these distinct elements gives 5. Zero is not included under the usual natural-number convention, and 10 must be included because the condition is ≤ 10. Therefore, option B is correct.

In a set, repetition does not create a new element; each element is counted only once. Removing the repeated 1 and 2 from {1, 1, 2, 2, 3} leaves {1, 2, 3}. Therefore, the set has three distinct elements and option A is correct.

The order in which elements are written does not matter in a set. Both J and K contain exactly the same distinct elements: 1, 2, and 3. Consequently, they are equal and have the same cardinality. Option B incorrectly treats sets like ordered lists, so option A is correct.

Every listed element is an even natural number and each is at most 10. The condition x ≤ 10 includes 10, so it gives exactly 2, 4, 6, 8, and 10. The other options include odd numbers, omit 10, or describe primes. Hence option A is correct.

A week has seven distinct day names: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, and Sunday. Each name is one element of M, so the set contains 7 elements. The answer counts names, not the number of working days or weekdays only; therefore option C is correct.

A finite set has a fixed, limited number of elements. A year has exactly 12 months, so its set of month names is finite. Natural numbers, integers, and rational numbers continue without an end, so those sets are infinite. Therefore, option C is the only finite set.

An infinite set has no final element and cannot be completely counted. Natural numbers continue as 1, 2, 3, 4, and so on without end, so their set is infinite. A class, the months, and the five standard vowels each form finite sets. Thus option C is correct.

The decimal digits are exactly 0 through 9, giving ten digits in all. Zero is a digit even though some definitions of natural numbers begin at 1. Thus the roster form must include 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Option B is correct.

The positive perfect squares below 20 are 1² = 1, 2² = 4, 3² = 9, and 4² = 16. The next square, 5² = 25, is not below 20. Therefore the roster form is {1, 4, 9, 16}, making option A correct.

In roster notation, the entries inside braces are precisely the elements of the set. Q therefore contains 1, 2, and 3, and no additional number is implied. It is not empty, it does contain 3, and 4 is absent. Hence option A is correct.

R contains exactly three listed elements: red, blue, and green. Since its number of elements is limited and countable, R is a finite set. It is not empty because it has members, and it is not infinite. A universal set depends on a stated context, so option A is the appropriate answer.

A prime number has exactly two positive factors, 1 and itself. Among even numbers, only 2 is prime; every larger even number is divisible by 2 and has more than two factors. Therefore the set of even prime numbers is {2}, not an empty set or a collection of all even numbers.