# Q 13 ) 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 φ is the identity for the operation ∗ and all the elements A of P(X) are invertible with A–1 = A. (Hint : (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ).

G Gautam harsolia

It is given that

$*: P(X) \times P(X) \rightarrow P(X)$  be defined as

$A * B = (A - B) \cup (B - A), A, B \ \epsilon \ P(X).$

Now, let  $A \ \epsilon \ P(X)$.
Then,

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

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

Therefore,

$A * \phi = \phi *A = A , A \ \epsilon \ P(X)$

Therefore, we can say that  $\phi$  is the identity element for the given operation *.

Now, an element A $\epsilon$ P(X) will be invertible if there exists B $\epsilon$ P(X) such that

$A*B = \phi = B*A \ \ \ \ \ \ \ \ \ \ \ \ (\because \phi \ is \ an \ identity \ element)$

Now, We can see that

$%u200B%u200B%u200B%u200B%u200B%u200B%u200BA * A = (A -A) \cup (A - A) = \phi \cup \phi = \phi , A \ \epsilon \ P(X).$$A * A = \left ( A - A \right ) \cup \left ( A-A \right ) = \phi \cup \phi = \phi ,$  such that $A \ \epsilon \ P(X)$

Therefore, by this we can say that all the element A of P(X) are invertible with  $A^{-1}= A$

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