Equivalence relation

Step 1: Equivalence classes divide a set into complete non-overlapping parts. Step 2: If two classes share an element, they cannot be different and must be the same. Step 3: Equivalence classes do not partially overlap.

Step 1: (a^2=a^2), so reflexivity holds. Step 2: If (a^2=b^2), then (b^2=a^2), and equality also passes through a chain. Step 3: For equality-based conditions, check the three properties separately.

Step 1: (|-2|=2). Step 2: In the given set, (2) also has absolute value (2), so the class is ({-2,2}). Step 3: In absolute value questions, focus on the magnitude rather than the sign.

Step 1: A person lives in the same city as themselves, so reflexivity is satisfied. Step 2: Living in the same city remains true when order is reversed and within a group. Step 3: A shared property is a good sign of an equivalence relation.

Step 1: No number is greater than itself, so reflexivity fails. Step 2: If (a>b), then generally (b>a) is not true, so symmetry also fails. Step 3: Be careful with comparison relations.

Step 1: The pair ((1,3)) is present. Step 2: For symmetry, the reverse pair ((3,1)) must be present, but it is missing. Step 3: Always check the reverse of every non-identical pair.

Step 1: Symmetry is also necessary for an equivalence relation. Step 2: Without symmetry, all three properties are not satisfied. Step 3: Even if two properties hold, always check the third one.

Step 1: Reflexivity is a compulsory condition for an equivalence relation. Step 2: Since reflexivity is missing, the relation cannot be an equivalence relation. Step 3: It is important to remember all three defining properties.

Step 1: (2) is even and divisible by (2). Step 2: In the given set, (2,4,6) are all divisible by (2). Step 3: While forming a class, include all elements with the same property.

Step 1: (3) is odd and leaves remainder (1) on division by (2). Step 2: (1,3,5) all leave the same remainder. Step 3: Separating even and odd classes is an easy method.

Step 1: An equivalence class is a group of elements related to one another. Step 2: Therefore, two elements in the same class form a pair in the relation. Step 3: For elements inside the same class, write the related pair.

Step 1: Elements from different equivalence classes are not related. Step 2: Therefore, ((a,b)) will not belong to (R). Step 3: If the classes are different, treat the pair as outside the relation.

Step 1: The set (A) has (3) elements. Step 2: (A\times A) has (3\times 3=9) ordered pairs. Step 3: In a universal relation, count all possible ordered pairs.

Step 1: Reflexivity is connected with identity pairs. Step 2: For every element (a), ((a,a)\in R) must be present. Step 3: To check reflexivity, list all identity pairs.

Step 1: In symmetry, the relation remains true after reversing the order. Step 2: So ((a,b)) must come with ((b,a)). Step 3: Identify symmetry by checking the reverse pair.

Step 1: Transitivity forms a link between two pairs. Step 2: From ((a,b)) and ((b,c)), ((a,c)) must follow. Step 3: Identify the common middle element before answering.

Step 1: ((2,1)\in R) and ((1,3)\in R). Step 2: Transitivity requires ((2,3)\in R), but it is missing. Step 3: In transitivity, look for the missing direct pair.

Step 1: On division by (2), there are only two remainders, (0) and (1). Step 2: Remainder (0) gives even numbers, and remainder (1) gives odd numbers. Step 3: Remember modulo (2) through even and odd classes.

Step 1: (0) leaves remainder (0) on division by (3). Step 2: In the given set, (3) also leaves remainder (0). Step 3: While finding the class, choose elements with the same remainder.

Step 1: In a partition, every element belongs to some part. Step 2: Being in the same part is reflexive, symmetric, and transitive. Step 3: A relation formed from a partition is a standard form of an equivalence relation.