A binary operation \['*'\] is defined on the set |
\[X=R-\{-\,1\}\] by \[x*y=x+y+xy,\] \[\forall x,\] \[y\in X.\] Check whether \['*'\] is commutative and associative. Find its identity element and also find the inverse of each element of X. |
OR |
If N denotes the set of all natural numbers and R be the relation on \[N\times N\] defined by (a, b)R |
(c, d), if \[ad(b+c)=bc(a+d).\]Show that R is an equivalence relation. |
Answer:
We have \[x*y=x+y+xy,\]\[X=R-\{-\,1\}\] Commutative law Let\[x,\,\,y\in R-\{1\}.\]Then, \[x*y=x+y+xy=y+x+yx\] [\[\because \]\[(+)\] and \[(\cdot )\]are commutative on \[R-\,\{-\,1\}\]] Hence, ?*? is commutative. Associative law Let \[x,\,\,y,\,\,z\in R-\{-\,1\},\] Then, \[(x*y)*z=(x+y+xy)*z\] \[=(x+y+xy)+z+(x+y+xy)\,z\] \[=(x+y+z)+(xy+yz+zx)+xyz\] And \[x*(y*z)=x*(y+z+yz)\] \[=x+(y+z+yz)+x\,(y+z+yz)\] \[=(x+y+z)+x\,(xy+yz+xyz)\] \[\therefore \] \[(x*y)*z=x*(y*z)\] Hence, ?*? is associative. Existence of identity element Let e be the identity element Then, for all \[x\in R-\{-\,1\},\] we have \[x*e=x\Rightarrow x+e+xe=x\] \[\Rightarrow \] \[e\,(1+x)=0\Rightarrow e=0\in R-\{1\}\] Now, \[x*0=x+0+x\times 0=x\] And \[0*x=0+x+0\times x=x\] Thus, 0 is the identity element in\[R-\{-\,1\}\]. Existence of inverse Let\[x\in R-\{1\}\]and\[{{x}^{-\,1}}=y\]. Then, \[x*y=0\Rightarrow x+y+xy=0\] \[\Rightarrow \] \[x=-xy-y=-y\,(x+1)\] \[\Rightarrow \] \[y=\frac{-\,x}{x+1}\in R-\{-\,1\}\Rightarrow {{x}^{-\,1}}=\frac{-\,x}{x+1}\in R-\{-\,1\}\] \[\therefore \]Each \[x\in R-\{-\,1\}\]has its inverse in \[R-\{-\,1\}\]. Or We have, \[ad\,(b+c)=bc\,(a+d)\] Reflexive Let\[(a,\,\,b)\in N\times N\]such that \[ab\,(a+b)=ab\,\,(a+b)\,\,\forall \,\,a,\,\,n\in N\] \[\Rightarrow \] \[ab\,(b+a)=ba\,(a+b)\Rightarrow R\] Symmetric For \[(a,\,\,b),\,\,(c,\,\,d)\in N\times N\]such that \[(a,\,\,b)\,\,R\,\,(c,\,\,d)\] \[\Rightarrow \]\[ad\,(b+c)=bc\,(a+d)\Rightarrow bc\,(a+d)=ad\,(b+c)\] \[\Rightarrow \] \[cb\,(d+a)=da\,(c+b)\Rightarrow (c,\,\,d)\,\,R\,\,(a,\,\,b)\] \[\Rightarrow \]R is symmetric. Transitive For \[(a,\,\,b),\,\,(c,\,\,d),\,\,(e,\,\,f)\in N\times N\]such that \[ad\,(b+c)=bc\,(a+d)\] And \[cf\,(d+e)=de\,(c+f)\] \[\Rightarrow \] \[adb+adc=bca+bcd\] ?(i) And \[cfd+cfe=dec+def\] ?(ii) On multiplying Eqs. (i) and (ii) by ef and ab respectively, and then adding, we get \[adbef+adcef+cfdab+cfeab\] \[=bcaef+bcdef+decab+defab\] \[\Rightarrow \] \[adcef+adcfb=bcdea+bcdef\] \[\Rightarrow \] \[adcf\,(e+b)=bcde\,(a+f)\] \[\Rightarrow \] \[af\,(b+e)=be\,(a+f)\Rightarrow (a,\,\,b)\,R\,(e,\,\,f)\] So, R is transitive. Hence, R is an equivalence relation. Hence proved.
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