Equivalence relation

Step 1: In the universal relation, every element is related to every element. Step 2: Hence it equals (A\times A). Step 3: The universal relation easily satisfies all three equivalence properties.

Step 1: An equivalence relation must be reflexive. Step 2: Reflexivity gives ((a,a)\in R) for every (a). Step 3: Use the word reflexivity whenever self-pairs are required.

Step 1: The reverse of ((4,7)) is ((7,4)). Step 2: An equivalence relation is symmetric, so the reverse pair must be present. Step 3: Reverse-pair questions usually use symmetry.

Step 1: In ((1,2)) and ((2,5)), the middle element (2) matches. Step 2: By transitivity, ((1,5)) must be present. Step 3: Take the first and last elements to form the transitive pair.

Step 1: Modulo (2), (0) has remainder (0). Step 2: Integers with remainder (0) are even integers. Step 3: So ([0]) is the class of even integers.

Step 1: (1) has remainder (1) when divided by (2). Step 2: Integers with this remainder are odd. Step 3: Modulo (2) gives two classes: even and odd.

Step 1: (1) leaves remainder (1) on division by (3). Step 2: (4) also leaves remainder (1). Step 3: Elements with the same remainder belong to the same equivalence class.

Step 1: (2) leaves remainder (2) when divided by (3). Step 2: (5) also leaves remainder (2). Step 3: While forming the class, use only elements from the given set.

Step 1: (3) leaves remainder (0) when divided by (3). Step 2: (6) also leaves remainder (0). Step 3: Put all elements with remainder (0) in one class.

Step 1: All self-pairs are present, so the relation is reflexive. Step 2: The non-self pairs occur with their reverse pairs. Step 3: It forms the classes {(1,3)} and {(2,4)}, so it is an equivalence relation.

Step 1: Equivalence classes together cover the original set. Step 2: Combining {(1,2)} and {(3)} gives {(1,2,3)}. Step 3: The union of all classes is the original set.

Step 1: An equivalence relation divides a set into separate groups. Step 2: Two distinct classes have no common element. Step 3: If two classes share even one element, they are the same class.

Step 1: In the identity relation, an element is related only to itself. Step 2: Therefore, the class of (a) contains only (a). Step 3: Identity relation gives singleton equivalence classes.

Step 1: In the universal relation, every element is related to every element. Step 2: So the class of any (a) is the whole set. Step 3: Universal relation gives one large equivalence class.

Step 1: Equivalence classes are either equal or disjoint. Step 2: If ([a]\neq[b]), they have no common element. Step 3: Distinct equivalence classes have empty intersection.

Step 1: (a\le a) makes the relation reflexive. Step 2: But (2\le 3) is true while (3\le 2) is false, so symmetry fails. Step 3: Order relations often fail symmetry.

Step 1: Reflexivity would require (a<a). Step 2: No number is less than itself, so reflexivity fails. Step 3: Strict inequality relations are not equivalence relations.

Step 1: Real numbers include negative numbers and zero. Step 2: If (a) is negative, (aRa) is false because (a) is not positive. Step 3: Reflexivity must hold for every element of the set.

Step 1: Every number has the same last digit as itself. Step 2: Same last digit is symmetric and transitive. Step 3: Think of last digit as remainder classes modulo (10).

Step 1: The definition of an equivalence relation is based on three properties. Step 2: These are reflexive, symmetric, and transitive. Step 3: For definition questions, remember all three together.

Step 1: Same parity gives one class of odd numbers and one class of even numbers. Step 2: In (A), the classes are {(1,3)} and {(2,4)}. Step 3: To count classes, first form the groups clearly.

Step 1: Every integer has the same absolute value as itself, so the relation is reflexive. Step 2: Equality of absolute values is symmetric and transitive. Step 3: Relations based on equality usually become equivalence relations after all three checks.

Step 1: The relation holds only when both elements are in the same class. Step 2: (3) and (5) both lie in ({3,5}), so ((3,5)\in R). Step 3: In class-based questions, first locate the two elements in the given groups.

Step 1: All self-pairs are present and reverse pairs are also visible. Step 2: But ((1,2)) and ((2,3)) should imply ((1,3)), which is missing. Step 3: For transitivity, look for pairs whose middle element matches.

Step 1: (4) leaves remainder (0) when divided by (4). Step 2: In the given set, (8) also leaves remainder (0), so (4) and (8) are in the same class. Step 3: In remainder-class questions, use only elements from the given set.