Equivalence relation

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.

Step 1: The relation (a>b) is not reflexive because (a>a) is false. Step 2: It is also not symmetric, since (5>3) does not imply (3>5). Step 3: Strict order relations are generally not equivalence relations.

Step 1: (a^2-a^2=0), so the relation is reflexive. Step 2: If (a^2-b^2) is divisible by (3), then (b^2-a^2) is also divisible by (3). Step 3: Equality of square remainders modulo (3) is transitive.

Step 1: (2^2=4) and (5^2=25). Step 2: (25-4=21), which is divisible by (3). Step 3: For options, subtract the squares and test divisibility.

Step 1: Equivalence classes are either equal or completely disjoint. Step 2: Since ([a]\neq[b]), they are not equal. Step 3: Hence their intersection is empty.

Step 1: All elements in the same equivalence class are related to one another. Step 2: (4) and (7) both lie in ({1,4,7}). Step 3: Therefore ((4,7)) is definitely in the relation.

Step 1: A person may be considered acquainted with himself or herself, so reflexivity can hold. Step 2: If one person is acquainted with another, the reverse is also true, so symmetry holds. Step 3: But acquaintance need not pass through a third person, so transitivity can fail.

Step 1: (6) is greater than (4). Step 2: The elements greater than (4) are (5,6,7). Step 3: In a grouping relation, the class of an element is the group containing it.

Step 1: (1) is not prime. Step 2: In the given set, (4) and (6) are also not prime. Step 3: The relation creates two classes: prime and non-prime elements.

Step 1: The condition (a=b) gives four diagonal pairs. Step 2: The condition (a+b=5) gives non-diagonal pairs ((1,4),(4,1),(2,3),(3,2)). Step 3: The total is (4+4=8) pairs.