Equivalence relation

Step 1: Dividing (2) by (4) leaves remainder (2). Step 2: In the given set, only (2) and (6) leave remainder (2). Step 3: While writing an equivalence class, do not include numbers outside the given base set.

Step 1: Every triangle is similar to itself, so the relation is reflexive. Step 2: If the first triangle is similar to the second, the second is similar to the first. Step 3: If the first is similar to the second and the second to the third, then the first is similar to the third.

Step 1: In an equivalence relation, related elements lie in the same class. Step 2: If (aRb), anything related to (a) is also related to (b). Step 3: Therefore the equivalence classes are equal, not just partially overlapping.

Step 1: Equivalence classes divide the set into separate blocks. Step 2: If two classes share even one element, the two classes become identical. Step 3: Hence distinct equivalence classes are disjoint; this is the main idea of partition.

Step 1: This is the equality relation, where each element is related only to itself. Step 2: Therefore every element forms its own singleton equivalence class. Step 3: Since (A) has (4) elements, there are (4) equivalence classes.

Step 1: The greatest integer part of (2.4) is (2). Step 2: Real numbers with greatest integer part (2) are greater than or equal to (2) and less than (3). Step 3: In such questions, check the end points carefully, so the class is ([2,3)).

Step 1: The remainder of (-1) modulo (4) is written as (3). Step 2: So all integers related to (-1) leave remainder (3) on division by (4). Step 3: Even for negative numbers, write the remainder in (0,1,2,3) for exam clarity.

Step 1: The set ({1,2}) has (2) elements. Step 2: Its equivalence class contains all subsets of (A) having exactly (2) elements. Step 3: The number of such subsets is (\binom{4}{2}=6), so order should not be counted.

Step 1: For every real number, (a^3=a^3), so the relation is reflexive. Step 2: Equality of cubes works in both directions and is transitive. Step 3: Over real numbers, (a^3=b^3) implies (a=b), so every equivalence class is singleton.

Step 1: A line has the same direction as itself, so the relation is reflexive. Step 2: Same direction is mutual and remains consistent through a third line. Step 3: Hence each equivalence class contains lines with one fixed direction, not necessarily lines through one point.

Step 1: Division by (3) creates three equivalence classes. Step 2: Each class has (3) elements, so each contributes (3^2=9) ordered pairs. Step 3: Total pairs are (9+9+9=27); pairs across different classes are not counted.

Step 1: For every (a), (\sin a=\sin a), so the relation is reflexive. Step 2: If two sine values are equal, equality also holds in the reverse direction. Step 3: A relation based on equal function values is transitive too, so it is an equivalence relation.

Step 1: (|a-b|) being even means (a) and (b) have the same parity. Step 2: Since (4) is even, its related elements must be even. Step 3: In the given set, the even elements are (2) and (4), so the class is ({2,4}).

Step 1: For any polynomial, (p(0)=p(0)), so reflexivity holds. Step 2: If (p(0)=q(0)), then (q(0)=p(0)) also holds. Step 3: Having the same value at zero passes through a third polynomial, so the relation is equivalence.

Step 1: All ((a,a)) pairs are present, so reflexivity holds. Step 2: Symmetry is also visible from the given pairs. Step 3: But from ((2,1)) and ((1,3)), transitivity requires ((2,3)), which is missing; hence it is not an equivalence relation.

Step 1: In an equivalence relation, all ordered pairs inside each block are included. Step 2: The first block gives (2^2=4) pairs and the second gives (3^2=9) pairs. Step 3: Total pairs are (4+9=13); (5^2) pairs would occur only if there were one class.

Step 1: (ab>0) means (a) and (b) are either both positive or both negative. Step 2: So all positive integers form one class and all negative integers form another. Step 3: Zero is not included in the set, so no separate zero class is formed.

Step 1: The radius of any circle is equal to itself. Step 2: If the first circle has the same radius as the second, the reverse is also true. Step 3: Equality of radius is transitive, so this is an equivalence relation.

Step 1: (a+b) is even only when (a) and (b) are both even or both odd. Step 2: In the given set, the odd elements are (1,3,5) and the even elements are (2,4,6). Step 3: Hence the relation forms two classes based on parity.

Step 1: For reflexivity, every ((a,b)) must satisfy ((a,b)S(a,b)). Step 2: This would require (a+a=b+b), meaning (a=b). Step 3: Not every ordered pair has (a=b), for example ((1,2)), so the relation is not equivalence.