Equivalence relation

Step 1: Transitivity is the third required condition for an equivalence relation. Step 2: If this property fails, the relation is not an equivalence relation. Step 3: Remember all three properties together.

Step 1: All identity pairs are present, so it is reflexive. Step 2: Both ((1,4)) and ((4,1)) are present, so it is symmetric. Step 3: The classes are ({1,4}), ({2}), and ({3}), so transitivity also holds.

Step 1: We have ((1,2)\in R) and ((2,3)\in R). Step 2: Transitivity requires ((1,3)\in R), but it is absent. Step 3: To test transitivity, form chains of ordered pairs and find missing links.

Step 1: On a non-empty set, reflexivity requires ((1,1),(2,2),(3,3)). Step 2: The empty relation has no identity pairs. Step 3: On a non-empty set, the empty relation is not an equivalence relation.

Step 1: In an empty set, there is no element to check. Step 2: Therefore, reflexive, symmetric, and transitive conditions are not violated. Step 3: In school exams, treat the non-empty set case separately if it is specified.

Step 1: The given (R) contains all possible ordered pairs of (A). Step 2: So it is the universal relation and satisfies all three properties. Step 3: When all possible pairs are present, identify (A\times A) quickly.

Step 1: All elements inside the same class are related to each other. Step 2: The class ({a,b}) gives four pairs, and ({c}) gives one identity pair. Step 3: To form a relation from classes, take each class crossed with itself.

Step 1: (aRb) means (a) and (b) are in the same group. Step 2: In an equivalence relation, elements in the same group have identical classes. Step 3: If two equivalence classes share an element, they are the same.

Step 1: (a-a=0), and (0) is divisible by (5). Step 2: If (a-b) is divisible by (5), then (b-a) is also divisible by (5). Step 3: Transitivity in such divisibility relations follows from adding differences.

Step 1: For reflexivity, (aRa) must be true for every (a). Step 2: (a+a=2a) is always even, not odd. Step 3: Checking reflexivity first can quickly decide many questions.

Step 1: (1\le 2), so ((1,2)\in R). Step 2: But (2\le 1) is false, so ((2,1)\notin R). Step 3: Order relations are often not symmetric, so they are not equivalence relations.

Step 1: For reflexivity, (a<a) must be true for every (a). Step 2: No number is less than itself. Step 3: The less-than relation is not an equivalence relation.

Step 1: (a-b=0) means (a=b). Step 2: Equality is reflexive, symmetric, and transitive. Step 3: First simplify the condition, then check the properties.

Step 1: In equivalence relations, ([a]) denotes the class of (a). Step 2: It contains all elements related to (a). Step 3: Do not confuse ([a]) with the square of (a).

Step 1: Elements in the same class are related to each other. Step 2: (1) and (3) are in different classes, so ((1,3)) will not be in the relation. Step 3: In class-based questions, first check where the two elements belong.

Step 1: The three properties together form the definition of an equivalence relation. Step 2: Therefore, such an (R) is called an equivalence relation. Step 3: In definition-based questions, read the property words carefully.

Step 1: (4) is even, so its remainder is (0). Step 2: In (A), the numbers with the same remainder (0) are (2) and (4). Step 3: While finding a class, do not include elements outside the given set.

Step 1: An equivalence relation is decided by properties, not by its name. Step 2: Checking reflexive, symmetric, and transitive properties is necessary. Step 3: In exams, make a short property checklist instead of guessing.

Step 1: On division by (3), (1) leaves remainder (1). Step 2: In the given set, (4) also leaves remainder (1), so the class of (1) is ({1,4}). Step 3: While finding an equivalence class, use only the elements of the given set.

Step 1: ((1,1),(2,2),(3,3)) are present, so the relation is reflexive. Step 2: ((2,3)) comes with ((3,2)), so symmetry holds, and the small class ({2,3}) also satisfies transitivity. Step 3: In such relations, identifying the formed classes helps you answer quickly.