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Consider the following relations:

R=\left \{ (x,y) \left | x,y\; are\; real\; numbers\; and\; x=wy\, for\, some\, rational\, number\, w \right \};

are integers such that

n,q\neq 0\: and \: qm = pn \left \ \right \}

then

 

  • Option 1)

    R is an equivalence relation but S  is not an equivalence relation

  • Option 2)

    neither R nor S is an equivalence relation

  • Option 3)

    Sis an equivalence relation but  R is not an equivalence relation

  • Option 4)

    R andS both are equivalence relations

 

Answers (1)

best_answer

As we learnt in

Equivalence relation -

Any relation which is reflexive, symmetric and transitive is called an equivalence relation

-

 

 S=\left \{\left(\frac{m}{n}, \frac{p}{q} \right ) \right \}

given that qm = pn , \frac{m}{n}=\frac{p}{q}

Is is always true for (a, a) reflexive.

\therefore\ \; f^{-1}(x)=\pm \sqrt{x+1}-1

\therefore\ \; (x+1)^{2} -1=\sqrt{x+1}-1

    (x+1)^{4}=(x+1)

    (x+1)=0\ \Rightarrow\ \; (x+1)^{3}=1, \;x+1=1, \; x=0

    x = - 1

\therefore    {0,  -1}

Correct option is 4.

 

 


Option 1)

R is an equivalence relation but S  is not an equivalence relation

This is an incorrect option.

Option 2)

neither R nor S is an equivalence relation

This is an incorrect option.

Option 3)

Sis an equivalence relation but  R is not an equivalence relation

This is an incorrect option.

Option 4)

R andS both are equivalence relations

This is the correct option.

Posted by

divya.saini

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