Given a non-empty set X, let *: P(X) × P(X) → P(X) be defined as A * B = (A − B) ∪ (B − A), &mnForE;


Given a non-empty set $X$, let $^{*}: P(X) \times P(X) \rightarrow P(X)$ be defined as $A$ * $B=(A-B) \cup(B-A), \& m n F o r E ; A, B \in P(X)$. 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 $(A-A) \cup$ $\left.(A-A)=A^{*} A=\Phi\right)$.


It is given that *: P(X) × P(X) → P(X) is defined as

A * B = (A − B) ∪ (B − A) &mnForE; AB ∈ P(X).

Let ∈ P(X). Then, we have:

$A^{*} \Phi=(A-\Phi) \cup(\Phi-A)=A \cup \Phi=A$

$\Phi^{*} A=(\Phi-A) \cup(A-\Phi)=\Phi \cup A=A$

$\therefore A^{*} \Phi=A=\Phi^{\star} A . \& m n F o r E ; A \in \mathrm{P}(X)$

Thus, $\Phi$ is the identity element for the given operation $^{\star}$.

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

$A^{*} B=\Phi=B^{*} A .($ As $\Phi$ is the identity element $)$

Now, we observed that $A * A=(A-A) \cup(A-A)=\phi \cup \phi=\phi \forall A \in \mathrm{P}(X)$.

Hence, all the elements A of P(X) are invertible with A−1 = A.

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