A) reflexive only
B) symmetric only
C) transitive only
D) an equivalence relation
Correct Answer: D
Solution :
Given, R is the set of real numbers. A relation defined by \[G=\{(a,b),(c,d):b-a=d-c\}\] and \[G\subseteq \,{{R}^{2}}\] For reflexive \[(a,b)\,\,\,G\,\,(a,b)\] \[\Rightarrow \] \[b-a=b-a\] \[\Rightarrow \]G is reflexive For symmetric \[(a,b)\,\,G\,\,(c,d)\] \[\Rightarrow \] \[b-a=d-c\] \[\Rightarrow \] \[d-c=b-a\] \[\Rightarrow \] \[(c,d)\,G\,(a,b)\] \[\Rightarrow \]G is symmetric. For transitive \[\{(a,b),(c,d)\}\in G\]and \[\{(c,d),(e,f)\}\in G\] \[\Rightarrow \] \[b-a=d-c\] and \[d-c=f-e\] \[\Rightarrow \] \[b-a=f-e\] \[\Rightarrow \] \[\{(a,b),(e,f)\}\in G\] \[\Rightarrow \] G is transitive. So, G is reflexive, symmetric and transitive. Hence, G is an equivalence relation.You need to login to perform this action.
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