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On integers, (aRb) is defined when (a\equiv b \pmod{6}). Which statement is correct?

Answer and explanation

Correct answer: It is transitive

Step 1: (a\equiv b \pmod{6}) means (a-b) is divisible by (6). Step 2: If (b-c) is also divisible by (6), then (a-c=(a-b)+(b-c)) is divisible by (6). Step 3: For congruence relations, prove transitivity by adding differences.

Related tags

Modulo 6CongruenceTransitiveClass 12Transitive RelationRelations And FunctionsMathematicsClass 12 Mcq

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