question_answer

If \[{{R}_{1}}\] and \[{{R}_{2}}\] be two equivalence relations on a set A, prove that \[{{R}_{1}}\cap {{R}_{2}}\] is also an equivalence relation on A.

OR

Let X be a non-empty set and P(X) be its power set. Let '*' be an operation defined on elements    of P(X) by \[A*B=A\cap B,\] \[\forall A,\] \[B\in P(X).\] then,

(i) Prove that ?*? is a biliary operation in P(X),

(ii) Is * commutative?

(iii) Is * associative?

(iv) Find the identity element in P(X) w.r.t. ?*?.

(v) If o is another binary operation defined on P(X) as \[AoB=A\cup B,\] then verify that O distributes itself over ?*?.

Answer:

Let \[{{R}_{1}}\] and \[{{R}_{2}}\] be two equivalence relations on a set A.

Then, \[{{R}_{1}}\subseteq A\times A,\] \[{{R}_{2}}\subseteq A\times A\] \[\Rightarrow \]

\[({{R}_{1}}\cap {{R}_{2}})\subseteq A\times A.\]

So, \[({{R}_{1}}\cap {{R}_{2}})\] is a relation on A.

This relation on A satisfies the following properties.

(i) Reflexivity

\[{{R}_{1}}\] is reflexive and \[{{R}_{2}}\] is reflexive

\[\Rightarrow \] \[(a,\,\,a)\in {{R}_{1}}\] and \[(a,\,\,a)\in {{R}_{2}}\] for all \[a\in A\]

\[\Rightarrow \] \[(a,\,\,a)\in {{R}_{1}}\cap {{R}_{2}}\] for all \[a\in A\]

\[\Rightarrow \] \[{{R}_{1}}\cap {{R}_{2}}\] is reflexive.

(ii) Symmetry

Let (a, b) be an arbitrary element of \[{{R}_{1}}\cap {{R}_{2}}.\]then,

\[(a,\,\,b)\in {{R}_{1}}\cap {{R}_{2}}\]

\[\Rightarrow \]   \[(a,\,\,b)\in R,\] and \[(a,\,\,b)\in {{R}_{2}}\]

\[\Rightarrow \]   \[(b,\,\,a)\in {{R}_{1}}\] and \[(b,\,\,a)\in {{R}_{2}}\]

\[[\because \,\,{{R}_{1}}\]is symmetric and \[{{R}_{2}}\] is symmetric]

\[\Rightarrow \]   \[(b,\,\,a)\in {{R}_{1}}\cap {{R}_{2}}\]

This shows that \[{{R}_{1}}\cap {{R}_{2}}\] is symmetric

(iii) Transitivity

\[(a,\,\,b)\in {{R}_{1}}\cap {{R}_{2}}\,\,and\,\,(b,\,\,c)\in {{R}_{1}}\cap {{R}_{2}}\]

\[\Rightarrow \]   \[(a,\,\,b)\in {{R}_{1}},\] \[(a,b)\in {{R}_{2}}\] and \[\,(b,\,\,c)\in {{R}_{1}},\]

\[\,(b,\,\,c)\in {{R}_{2}}\]

\[\Rightarrow \]   \[\{(a,\,\,b)\in {{R}_{1}},(b,\,\,c)\in {{R}_{1}}\},\]

and \[\{(a,\,\,b)\in {{R}_{2}},(b,\,\,c)\in {{R}_{2}}\]

\[\Rightarrow \]   \[\{a,\,\,c\}\in {{R}_{1}}\,\,and\,\,(a,\,\,c)\in {{R}_{2}}\]

\[[\because \,\,{{R}_{1}}\]is transitive and \[{{R}_{2}}\] is transitive]

\[\Rightarrow \]   \[(a,\,\,c)\in {{R}_{1}}\cap {{R}_{2}}\]

This shows that \[({{R}_{1}}\cap {{R}_{2}})\] is transitive.

Thus, \[{{R}_{1}}\cap {{R}_{2}}\] is reflexive, symmetric and transitive.

Hence, \[{{R}_{1}}\cap {{R}_{2}}\] is an equivalence relation.

OR

We have, \[A*B=A\cap B,\] \[\forall A,\] \[B\in P(X)\]

(i) Let \[A,\,\,B\in P(X)\] be any arbitrary elements.

Then, \[A\subset X\,\,\text{and}\,\,B\subset X\] \[\Rightarrow \] \[A\cap B\subset X\]

\[\Rightarrow \] \[A\cap B\in P(X)\] \[\Rightarrow \] \[A*B\in P(X)\]

\[\therefore \] \[*\] is a binary operation in P(X)

(ii) Let \[A,\,\,B\in P(X)\] be any arbitrary elements. Then,

\[A*B=A\cap B=B\cap A=B*A\]

Thus,    \[A*B=B*A\,\,\forall \,\,A,\] \[B\in P(X)\]

Hence, \[*\] is commutative.

(iii) Let \[A,\,\,B,\,\,C\in P(X)\] be any arbitrary elements. Then, \[(A*B)*C=(A\cap B)\cap C\]

\[=A\cap B\cap C=A\cap (B\cap C)=A*(B*C)\]

Thus, \[=(A*B)*C=A*(B*C),\]

\[\forall A,\,\,B,\,\,C\in P(X)\]

Hence, \[*\] is commutative.

(iv) Let, if possible, E be an identity element in P(X).

Then, \[A*E=A=E*A,\] \[\forall \,\,A\in P(X)\]

\[\Rightarrow \] \[4x+3y<110,\] \[\forall \,\,A\in P(X)\]

\[\Rightarrow \] \[A\subset E\,\,\forall \,\,A\in P(X)\]

This is possible only when E = X.

Also, \[X\in P(X)\]

Thus, X is the identity element.

(v) Let A be an invertible element of P(X). Then, there exist an element B in P(X) such that

\[A\cap B=X=B\cap A\]

This is possible only when A = B = X

Thus, X is the only invertible element in P(X) and \[{{X}^{-1}}=X\]

(vi) Let \[A,\,\,B,\,\,C\in P(X)\] be any arbitrary elements.

To verify  \[Ao(B*C)\text{ }=\text{ (}AoB\text{)}*(AoC)\]

Consider \[LHS=Ao(B*C)\text{ }=\text{ A}\cup \text{(}B\cap C\text{)}\]

\[=(A\cup B)\cap (A\cup C)\]

\[=(AoB)*(AoC)=RHS\]

Hence, 'o' distributes itself over *.