A) Symmetric only
B) Reflexive only
C) Transitive only
D) An equivalence relation
Correct Answer: D
Solution :
| For \[\left( a,b \right),\,\left( c,d \right)\in N\times N\] |
| \[\left( a,b \right)R\left( c,d \right)\Rightarrow ad\,\left( b+c \right)=bc\left( a+d \right)\] |
| Reflexive: Since \[ab\left( b+a \right)=ba\left( a+b \right)\forall ab\,\in N,\] |
| \[\therefore \,\,\left( a,b \right)R\left( a,b \right)\,\,\,\,\,\Rightarrow R\] is reflexive. |
| Symmetric: For \[\left( a,b \right),\left( c,d \right)\in N\times N,\] let \[\left( a,b \right)R\left( c,d \right)\] |
| \[\therefore \,\,ad\left( b+c \right)=bc\left( a+d \right)\Rightarrow bc\left( a+d \right)=ad\left( b+c \right)\]\[\Rightarrow \,\,cb\,\left( d+a \right)=da\left( c+b \right)\Rightarrow \left( c,d \right)R\left( a,b \right)\] |
| \[\Rightarrow \] R is symmetric |
| Transitive: For \[\left( a,b \right),\left( c,d \right),\left( e,f \right)\in N\times N,\] Let \[\left( a,b \right)\,R\left( c,d \right),\left( c,d \right)R\left( e,f \right)\] |
| \[\therefore \,\,ad\left( b+c \right)=bc\left( a+d \right),cf\left( d+e \right)=de\left( c+f \right)\] |
| \[\Rightarrow \,\,\,adb+adc=bca+bcd\] ...(i) |
| and \[cfd+cfe=dec+def\] ...(ii) |
| \[\left( i \right)\times ef+\left( ii \right)\times ab\] gives, \[adbef+adcef+cfdab+cfeab\]\[=bcaef+bcdef+decab+defab\] |
| \[\Rightarrow \,\,adcf\left( b+e \right)=bcde\left( a+f \right)\] |
| \[\Rightarrow \,\,af\left( b+e \right)=be\left( a+f \right)\] |
| \[\Rightarrow \left( a,b \right)R\left( e,f \right)\Rightarrow R\] is transitive. |
| Hence R is an equivalence relation. |
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