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  • Given a non-empty set X, let star defined from P(X) cross P(X) to P(X) be defined as A star B equals A minus B union B minus A, for all A, B in P(X). Show that the empty set phi is the identity for the operation ∗ and all the elements A of

 

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).

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best_answer

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

???????A * 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

Posted by

Gautam harsolia

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