Question

Let $R$ be the set of real numbers and let $ G\subseteq {{R}^{2}} $ be a relation defined by
$ G=\{(a,b),\,(c,d)|\,b-a=d-c\}. $ . Then, $G$ is

• A. reflexive only
• B. symmetric only
• C. transitive only
• D. an equivalence relation

Answer 1

Answer: (D)

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.