question_answer

         

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.