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Question

If R is the real line.

Consider the following subsets of the plane R×R;S={(x,y):y=x+1;0<x<2};T={(x,y):xyisaninteger} Which one of the following is correct?


A

T is an equivalence relation on R but S is not

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B

Neither S nor T is an equivalence relation on R

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C

Both S and T are equivalence relations on R

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D

S is an equivalence relation on R but T is not

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Solution

The correct option is A

T is an equivalence relation on R but S is not


Explanation for the correct option:

Step 1: Check whether the given relation T={(x,y):xyisaninteger} is an equivalence relation or not.

If a relation is reflexive, symmetric, and transitive, it is said to be an equivalence relation.

Reflexive: A relation is reflexive, if a,aR, for every aR.

Symmetric: A relation is symmetric, if a,bR, then b,aR.

Transitive: A relation is transitive if a,bR and b,cR, then a,cR.

  • Let xR, then xx=0 is an integer,

Which implies if function (x,x)T, which shows T is reflexive;

  • Let (x,y)R , if xy is an integer, then y-x is also an integer, which implies (y,x)R , then

The function T is symmetric relation.

  • Assume (x,y,z)R
  • Let (x,y)T and (y,z)T then xyI1 is an integer and y-zI2 is also an integer

Then add both the above relation, then

I1+I2=xy+y-z=x-zI,aninteger

This implies (x,z)T, which shows that T is transitive.

Therefore, the relation T is an equivalence relation, because T is reflexive, symmetric, and transitive.

Step 2: Check whether the given relation S={(x,y):y=x+1;0<x<2} is an equivalence relation or not.

If xR, assume (x,x)S, then

x=x+1 is not possible if 0<x<2,

Therefore relation S is not reflexive which implies it is not an equivalence relation.

Hence, the correct option is (A).


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