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Class 12MathematicsRelations and FunctionsTransitive relationExpert
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On real numbers, (aRb) is defined when (a+b=0). Is this relation transitive?

Answer and explanation

Correct answer: No, because (1R(-1)) and ((-1)R1) hold but (1R1) does not

Step 1: (1+(-1)=0), so (1R(-1)), and ((-1)+1=0), so ((-1)R1). Step 2: Transitivity requires (1R1), but (1+1=2), so it is not in the relation. Step 3: When reverse or opposite pairs appear, check the required self-pair.

Related tags

Additive InverseNon TransitiveReal NumbersClass 12Transitive RelationRelations And FunctionsMathematicsClass 12 Mcq
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10 more MCQs
  1. 1Which transitivity requirement fails for R={(1,1),(2,2),(3,3),(4,4),(1,2),(2,3),(3,4),(1,3),(2,4)} on A={1,2,3,4}?
  2. 2On A={1,2,3}, what is the smallest addition to R={(1,2),(2,3)} that makes R transitive?
  3. 3If R and S are transitive relations on the same set, which statement about their union R∪S is always valid?
  4. 4For R={(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(2,4),(4,2)} on A={1,2,3,4}, which missing ordered pair directly witnesses that R is not transitive?
  5. 5Which relation on A={1,2,3} is transitive but not symmetric?
  6. 6If R and S are transitive relations on A, which implication proves that R∩S is transitive?
  7. 7Which pair of transitive relations gives a union that is not transitive?
  8. 8Which is the correct definition of a transitive relation?
  9. 9On (A={1,2,3}), (R={(1,2),(2,3),(1,3)}) is given. Is (R) transitive?
  10. 10On (A={1,2,3}), (R={(1,2),(2,3)}). Why is (R) not transitive?
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