# Q. 13   Given a non-empty set X, let$* : P(X) \times P(X) \rightarrow P(X)$ be defined as $A * B = (A - B) \cup (B -A), \;\forall A, B \in P(X).$ Show that the empty set $\phi$ is the identity for the operation ∗ and all the elements A of P(X) are invertible with $A^{-1} =A$. (Hint : $(A - \phi) \cup (\phi - A) = A$and $(A -A) \cup (A - A) = A * A = \phi$).

Let  $* : P(X) \times P(X) \rightarrow P(X)$ be defined as  $A * B = (A - B) \cup (B -A), \;\forall A, B \in P(X).$

Let  $A \in P(X)$. Then

$A * \phi = (A - \phi ) \cup (\phi -A) = A\cup \phi =A$

$\phi * A = (\phi - A ) \cup (A -\phi ) = \phi \cup A =A$

$\therefore \, \, \, A*\phi =A= \phi * A$      for all  $A \in P(X)$

Thus, $\phi$ is identity element for operation *.

An element   $A \in P(X)$  will be invertible if there exists   $B\in P(X)$,

such that $A*B=A=B*A$.     (here $\phi$ is identity element)

$A * A = (A - A ) \cup (A -A) = \phi \cup \phi =\phi$  $\forall A \in P(X)$

Hence, all elements A  of P(X) are invertible with $A^{-1}=A.$

Exams
Articles
Questions