Given a non-empty set X, let * : P(X) × P(X) P(X) be defined as A * B = (A - B) (B - A), A, B P(X). Show that the empty set is the identity for the operation * and all the elements A of P(X)are invertible with A - 1 = A. (Hint : (A - ) ( - A) = A and (A - A ) (A - A) = A * A = ).

To show : is the identity For every A P(X) , we have * A = ( - A) (A - ) = A = A and A * = (A - ) ( - A) = A = A is the identity element for the operation * on P(X). Also, A * A = (A - A) ( A - A ) = = Every element A of P(X) is invertible with A - 1 = A

Relations and Functions — Class 12 Maths Solution

ncert misc SA NCERT Misc.,Q.13, Page 30

Question

Given a non-empty set X, let$*$ : P(X) × 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 $=$ $\phi$).

Step-by-step Solution

To show : $\phi$ is the identity

For every A $\in P(X)$, we have

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

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

$\Rightarrow$ $\phi$ is the identity element for the operation $*$ on P(X).

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

$\Rightarrow$ Every element A of P(X) is invertible with ${A^{ - 1}} = A$

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