Equivalence relation

Step 1: Every element is related to itself, so the relation is reflexive. Step 2: Since only identical ordered pairs are present, symmetry and transitivity also hold. Step 3: In exams, always check all three properties for an equivalence relation.

Step 1: An equivalence relation is identified by three properties. Step 2: These are reflexive, symmetric, and transitive. Step 3: If even one property fails, the relation is not an equivalence relation.

Step 1: The relation has all identity pairs, so it is reflexive. Step 2: The pair ((1,2)) has ((2,1)), and the small block ({1,2}) is closed under transitivity. Step 3: Check the pair block carefully before rejecting the relation.

Step 1: Since (a-a=0) is even, the relation is reflexive. Step 2: If (a-b) is even, then (b-a) is also even, and the sum of two even differences is even. Step 3: Relations based on parity often form equivalence relations.

Step 1: Every number belongs to the same parity class as itself, so the relation is reflexive. Step 2: If (a) and (b) are in the same class, then (b) and (a) are also in the same class. Step 3: Class-based relations usually make transitivity easy to verify.

Step 1: (A\times A) contains every possible ordered pair. Step 2: So all identity pairs, reverse pairs, and required transitive pairs are present. Step 3: Remember the universal relation as a common example of an equivalence relation.

Step 1: In the equality relation, every element is related to itself. Step 2: If (a=b), then (b=a), and if (a=b) and (b=c), then (a=c). Step 3: The equality relation is the simplest example of an equivalence relation.

Step 1: The symmetric property says that if ((a,b)\in R), then ((b,a)\in R). Step 2: Every equivalence relation must be symmetric. Step 3: Looking for the reverse pair is the easiest way to test symmetry.

Step 1: In transitivity, the second element of the first pair matches the first element of the second pair. Step 2: Therefore, ((a,b)) and ((b,c)) imply ((a,c)). Step 3: In transitivity questions, focus on the middle element.

Step 1: For reflexivity, every element (a) of (A) must have ((a,a)\in R). Step 2: Here, ((3,3)) is missing for the element (3). Step 3: When checking equivalence, first check all identity pairs.

Step 1: Every element is in the same part as itself, so the relation is reflexive. Step 2: If two elements are in the same part, reversing the order keeps them in the same part. Step 3: A relation formed by partitions is usually an equivalence relation.

Step 1: An equivalence relation groups elements with the same property. Step 2: Thus, the set is divided into distinct equivalence classes. Step 3: Thinking in terms of classes helps solve these questions faster.

Step 1: The equivalence class of (a) is the set of all elements related to (a). Step 2: In an equivalence relation, this class forms a clear separate block. Step 3: When asked for a class, first understand the condition of the relation.

Step 1: Every integer is congruent to itself because (a-a=0). Step 2: Having the same remainder is preserved when order is reversed and through a chain. Step 3: Same-remainder modulo relations are standard examples of equivalence relations.

Step 1: (a+a=2a) is always even, so the relation is reflexive. Step 2: If (a+b) is even, then (b+a) is also even, so it is symmetric. Step 3: If two numbers have the same parity as a middle number, they have the same parity with each other.

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

Step 1: Identity pairs like ((a,a)) are connected with the reflexive property. Step 2: Such a pair must exist for every element. Step 3: Reflexivity does not need comparison with another element.

Step 1: In the symmetric property, the relation remains true when the order is reversed. Step 2: So ((a,b)) must come with ((b,a)). Step 3: Remember it as the reverse-pair rule.

Step 1: Transitivity forms a chain of two related pairs. Step 2: From ((a,b)) and ((b,c)), the direct pair ((a,c)) is required. Step 3: Identify it by matching the middle element (b).

Step 1: On division by (2), (1) and (3) have the same remainder. Step 2: So the class of (1) contains (1) and (3). Step 3: To find an equivalence class, look for elements with the same remainder as the given element.

Step 1: On division by (3), (2) leaves remainder (2). Step 2: (5) also leaves remainder (2), so the class of (2) is ({2,5}). Step 3: In modulo questions, making a remainder table is an easy method.

Step 1: A person is born in the same year as themselves. Step 2: If one person has the same birth year as another, the reverse is also true, and the chain condition also holds. Step 3: Conditions based on sameness usually indicate equivalence relations.

Step 1: (a>a) is never true, so greater than is not reflexive. Step 2: If (a>b), then (b>a) is not true, so it is not symmetric. Step 3: For comparison relations, check all equivalence properties carefully.

Step 1: Every natural number is equal to itself, so the relation is reflexive. Step 2: Equality remains true when the order is reversed. Step 3: The equality relation is always a safe example of an equivalence relation.

Step 1: (|a|=|a|) is always true, so it is reflexive. Step 2: If (|a|=|b|), then (|b|=|a|) is also true. Step 3: Equality of values is also transitive, so this is an equivalence relation.