Let $$R$$ be a relation over the $$N \times N$$  and it is defined by $$\left( {a,\,b} \right)R\left( {c,\,d} \right) \Rightarrow a + d = b + c.$$      Then, $$R$$ is :

Solution :
$$\eqalign{ & {\text{We have }}\left( {a,\,b} \right)R\left( {a,\,b} \right)\,{\text{for all }}\left( {a,\,b} \right)\, \in \,N \times N \cr & {\text{As}}\,a + b = b + a{\text{. Hence }}R\,{\text{is reflexive}} \cr & R\,{\text{is symmetric for we have }}\left( {a,\,b} \right)R\left( {c,\,d} \right) \cr & \Rightarrow a + d = b + c\,\,\,\,\,\, \Rightarrow d + a = c + b \cr & \Rightarrow c + b = d + a\,\,\,\,\,\, \Rightarrow \left( {c,\,d} \right)R\left( {e,\,f} \right) \cr & {\text{Then, by defination of }}R,\,{\text{we have}} \cr & a + d = b + c{\text{ and }}c + f = d + e \cr & {\text{So, by addition, we get}} \cr & a + d + c + f = b + c + d + c{\text{ or }}a + f = b + e \cr & {\text{Hence, }}\left( {a,\,b} \right)R\left( {e,\,f} \right) \cr & {\text{Thuse, }}\left( {a,\,b} \right)R\left( {c,\,d} \right){\text{ and }}\left( {c,\,d} \right)R\left( {e,\,f} \right) \cr & \Rightarrow \,\left( {a,\,b} \right)R\left( {e,\,f} \right) \cr} $$