Equivalence relation

Step 1: Every element has the same parity as itself, so the relation is reflexive. Step 2: Same parity remains true when order is reversed and through a chain. Step 3: In such questions, first form the classes.

Step 1: (4) leaves remainder (1) on division by (3). Step 2: In the given set, (1) also leaves the same remainder, so the class is ({1,4}). Step 3: Put only elements with the same remainder in the equivalence class.

Step 1: All three identity pairs are present, so the relation is reflexive. Step 2: ((a,b)) comes with ((b,a)), and the group ({a,b}) satisfies transitivity. Step 3: Identifying small groups helps check equivalence quickly.

Step 1: ((1,2)\in R) and ((2,3)\in R). Step 2: Transitivity requires ((1,3)\in R), but it is missing. Step 3: In transitivity, join the middle element to find the required pair.

Step 1: The definition of an equivalence relation is based on three properties. Step 2: Reflexive, symmetric, and transitive properties are all required together. Step 3: Do not decide from only one property in exams.

Step 1: (a-a=0), and (0) is divisible by (4), so the relation is reflexive. Step 2: If (a-b) is divisible by (4), then (b-a) is also divisible by (4). Step 3: Transitivity also follows from adding differences.

Step 1: Reflexivity requires the identity pair of every element. Step 2: Here, ((4,4)) is not present for the element (4). Step 3: While checking equivalence, first match the list of identity pairs.

Step 1: An equivalence relation must be symmetric. Step 2: Therefore, if ((x,y)\in R), then the reverse pair ((y,x)\in R). Step 3: The reverse pair is the key sign of symmetry.

Step 1: In ((p,q)) and ((q,r)), the middle element (q) matches. Step 2: By transitivity, ((p,r)\in R) must hold. Step 3: For transitivity, read the chain of pairs carefully.

Step 1: (5) leaves remainder (0) on division by (5). Step 2: In the given set, only (5) leaves remainder (0). Step 3: Do not include (0) outside the given set in the equivalence class.

Step 1: (a+a=2a) is always even, so the relation is reflexive. Step 2: If (a+b) is even, then (b+a) is even, and same parity gives transitivity. Step 3: An even sum means both numbers have the same parity.

Step 1: An equivalence relation groups similar elements together. Step 2: These groups divide the set into separate parts. Step 3: It is easy to understand equivalence classes as parts of a partition.

Step 1: ([a]) means the equivalence class of (a). Step 2: It contains all elements related to (a) according to (R). Step 3: To find a class, directly apply the condition of the relation.

Step 1: (|a|=|a|) is true for every (a). Step 2: Equality of absolute values remains true when order is reversed and through a chain. Step 3: Relations based on equality of values often show all three properties clearly.

Step 1: On division by (4), the possible remainders are (0,1,2,3). Step 2: All four remainders occur in the given set, so four classes are formed. Step 3: In modulo questions, counting remainders is a quick method.

Step 1: A universal relation contains all possible ordered pairs. Step 2: Therefore, identity pairs, reverse pairs, and transitive pairs are all present. Step 3: On seeing (A\times A), quickly identify all three properties.

Step 1: On a non-empty set, reflexivity requires ((1,1)) and ((2,2)). Step 2: The empty relation has no identity pair. Step 3: Do not treat the empty relation on a non-empty set as an equivalence relation.

Step 1: In (R), every element is related only to itself. Step 2: Such a relation is reflexive, symmetric, and transitive. Step 3: The equality relation is a key example of an equivalence relation.

Step 1: All elements in the same class are related to one another. Step 2: (3) and (4) both belong to ({2,3,4}), so ((3,4)\in R). Step 3: Before checking a pair, identify the class of both elements.

Step 1: Elements from different classes are not related. Step 2: (1) is in ({1,5}), while (3) is in ({2,3,4}), so ((1,3)\notin R). Step 3: If the classes are different, the pair is not in the 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.