A) Reflexive only
B) Symmetric only
C) Transitive only
D) An equivalence relation
Correct Answer: D
Solution :
We have \[(a,b)R(a,b)\]for all \[(a,b)\in N\times N\] Since \[a+b=b+a\]. Hence, R is reflexive. R is symmetric for we have\[(a,b)R(c,d)\] Þ \[a+d=b+c\] Þ \[d+a=c+b\] Þ \[c+b=d+a\Rightarrow (c,d)R(e,f).\] Then by definition of R, we have \[a+d=b+c\]and \[c+f=d+e\], whence by addition, we get \[a+d+c+f=b+c+d+e\] or \[a+f=b+e\] Hence,\[(a,b)\,\,R\,(e,f)\] Thus, (a, b)\[R(c,d)\]and \[(c,d)R(e,f)\Rightarrow (a,b)R\,(e,\,\,f)\].
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