A) reflexive on \[{{R}^{+}}\]
B) symmetric on \[{{R}^{+}}\]
C) antisymmetric on \[{{R}^{+}}\]
D) equivalence relation on \[{{R}^{+}}\]
Correct Answer: D
Solution :
Given \[x\,Sy\,\Leftrightarrow \,{{x}^{2}}-{{y}^{2}}=2(y-x),\] \[\forall \,x,\,\,y\in \,{{R}^{+}}\] \[\because \] \[{{x}^{2}}-{{y}^{2}}=2(x-x),\,\] \[\forall \,x,\,\,y\,\in {{R}^{+}}\] 0 = 0 \[\therefore \] R is reflexive. For symmetric of \[S,\,\,xSy\] \[\Rightarrow \] \[{{x}^{2}}-{{y}^{2}}=2(y-x)\] \[\Rightarrow \] \[-({{y}^{2}}-{{x}^{2}})=-2\,(x-y)\] \[\Rightarrow \] \[{{y}^{2}}-{{x}^{2}}=2\,(x-y)\] \[\Rightarrow \] \[y\,Sx\] \[\therefore \] S is symmetric. Now, by definition of S, \[{{x}^{2}}-{{y}^{2}}=2(y-x)\] and \[{{y}^{2}}-{{z}^{2}}=2(z-y)\] On adding, we get \[({{x}^{2}}-{{z}^{2}})=2(z-x)\] \[\Rightarrow \] \[xSz\] \[\therefore \] S is transitive. Clearly, S is an equivalence relation on\[{{R}^{+}}\].
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