JEE Main & Advanced AIEEE Solved Paper-2011

  • question_answer
    Let R be the set of real numbers. Statement-1: \[A=\{(x,y)\in R\times R:y-x\] is an integer} is an equivalence relation on R. Statement-2: \[B=\{(x,y)\in R\times R:x=\alpha y\] for some rational number ?} is an equivalence relation on R.   AIEEE  Solved  Paper-2011

    A) Statement-1 is false, Statement-2 is true

    B) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

    C) Statement-1 is true, Statement-2 is true Statement-2 is not a correct explanation for Statement-1

    D) Statement-1 is true, Statement-2 is false

    Correct Answer: D

    Solution :

                 Statement-1 is true We observe that Reflexivity \[xRx\] as \[x-x=0\] is an integer, \[\forall x\in A\] Symmetric Let \[(x,y)\in A\]              \[\Rightarrow y-x\] is an integer \[\Rightarrow x-y\]is also an integer Transitivity Let \[(x,y)\in A\] and \[(y,z)\in A\] \[\Rightarrow y-x\] is an integer and \[z-y\] is an integer \[\Rightarrow y-x+z-y\] is also an integer \[\Rightarrow z-x\] is an integer \[\Rightarrow (x,z)\in A\] Because of the above properties A is an equivalence relation over R Statement 2 is false as 'B' is not symmetric on \[\mathbb{R}\] We observe that \[OBx\] as \[0=0.x\forall x\in \mathbb{R}\] but \[(x,0)\notin B\]


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