A) S is an equivalence relation on R but T is not
B) T is an equivalence relation on R but S is not
C) Neither S nor T is an equivalence relation on R
D) Both S and T are equivalence relations on R
Correct Answer: B
Solution :
For S, \[y=x+1\] for reflexive \[x=x+1\,\,\,\Rightarrow \,\,\,0=1\] \[\Rightarrow \] S is not reflexive. So S can not be equivalence. For T, \[x-y\in I\], then \[x-x=0\in I\Rightarrow \]T is reflexive. \[x-y\in I\], then \[y-x\in I\Rightarrow \]T is symmetric also. Now \[x-y\in I\] and \[\Rightarrow x-z\in I\Rightarrow \] T is transitive also. Hence T is an equivalence relation.You need to login to perform this action.
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