Equivalence relation

Step 1: In the given set, (2,3,5) are prime. Step 2: (4) is not prime, so its class is ({1,4,6}). Step 3: For classification relations, first separate the groups clearly.

Step 1: In an equivalence relation, only elements within the same class are related. Step 2: (2) is in ({1,2}), while (5) is in ({3,4,5}). Step 3: Elements from different classes do not form related ordered pairs.

Step 1: The degree of (x^2+1) is (2). Step 2: All non-zero polynomials of degree (2) are related to it. Step 3: To find an equivalence class, identify the key property of the chosen element.

Step 1: The first component of ((1,2)) is (1). Step 2: In the given set, the ordered pairs with first component (1) are ((1,1)) and ((1,2)). Step 3: In component-based relations, keep the specified component the same to form the class.

Step 1: Every integer related to (25) must differ from (25) by a multiple of (9). Step 2: (25) gives remainder (7) on division by (9), so integers with the same remainder (7) form its class. Step 3: In modulo questions, first find the remainder.

Step 1: (7) gives remainder (3) on division by (4). Step 2: In the given set, only (3) and (7) give remainder (3). Step 3: For a finite set, include only elements from that set.

Step 1: Dividing (4) by (3) gives quotient (1). Step 2: The numbers (3,4,5) all have quotient (1). Step 3: Read carefully whether the relation is based on quotient or remainder.

Step 1: (1,3,5) are mutually related, so they form one class. Step 2: (2) and (4) are related only to themselves, so they form singleton classes. Step 3: Identify connected groups from the ordered pairs.

Step 1: The pairs ((1,2)) and ((2,3)) are in the relation. Step 2: Transitivity requires ((1,3)). Step 3: Identify the middle element and check the pair from the first to the third element.

Step 1: Reflexivity requires every element of the set to be related to itself. Step 2: (3) is in the set, but ((3,3)) is missing. Step 3: If even one diagonal pair is absent, the relation cannot be an equivalence relation.

Step 1: The equivalence class of an element is the block containing that element. Step 2: (3) lies in ({1,3}). Step 3: When a partition is given, directly choose the block containing the element.

Step 1: A class of size (m) contributes (m^2) ordered pairs. Step 2: The total is (4^2+2^2=16+4). Step 3: No pairs are added between different classes, so the answer is (20).

Step 1: A partition must have non-empty, non-overlapping blocks whose union is the whole set. Step 2: The first option has disjoint blocks covering all of (A). Step 3: Equivalence classes always form such a partition.

Step 1: In a partition, different blocks must not share an element. Step 2: In the first option, (2) appears in both ({1,2}) and ({2,3}). Step 3: Equivalence classes do not overlap.

Step 1: (-\frac{3}{5}) is negative. Step 2: (\frac{a}{b}>0) holds when both numbers have the same sign. Step 3: Hence its class is the set of all negative rational numbers.

Step 1: (5.3-2.3=3). Step 2: Since (3) is an integer, (5.3) belongs to the class of (2.3). Step 3: Subtract from the given element and check whether the difference is an integer.

Step 1: (\lfloor \frac{5}{3}\rfloor=1). Step 2: For (3,4,5), the value of (\lfloor \frac{a}{3}\rfloor) is also (1). Step 3: In floor-based relations, elements with the same floor value form one class.

Step 1: The last digit of (42) is (2). Step 2: Numbers such as (2,12,22,32,42) have the same last digit. Step 3: This class is decided by the last digit, not just by parity.

Step 1: (\frac{1}{a}) has the same sign as itself, so reflexivity holds. Step 2: Having the same sign is symmetric. Step 3: If two reciprocal signs are equal through a middle element, the first and third signs are also equal.

Step 1: The second component of ((1,2)) is (2). Step 2: The pairs with second component (2) are ((1,2),(2,2),(3,2)). Step 3: In component-based relations, all elements with the specified same component form the class.

Step 1: The (y)-coordinate of ((3,5)) is (5). Step 2: All points with the same (y)-coordinate (5) belong to its class. Step 3: In coordinate relations, note carefully which coordinate must be equal.

Step 1: (|a-2|=|a-2|), so the relation is reflexive. Step 2: Equality remains true when the order is reversed, so it is symmetric. Step 3: If two distances equal the same middle distance, they are equal to each other.

Step 1: The distance of (5) from (2) is (3). Step 2: The real numbers at distance (3) from (2) are (5) and (-1). Step 3: In distance-based classes, include points at the same distance from the center.

Step 1: Every number has the same digit sum as itself. Step 2: Equality of digit sum is symmetric. Step 3: If two numbers have the same digit sum as a middle number, they have the same digit sum as each other.

Step 1: The digit sum of (23) is (2+3=5). Step 2: Its class contains numbers whose digit sum is (5). Step 3: An equivalence class is formed by the defining condition of the relation.