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Given a non-empty set X. Let * : P(X) × P(X) → P(X) 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.

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Here operation ‘*‘ is defined as

* : P(X) × P(X) → P(X) such that 

A * B = (A – B) ∪ (B – A) ∀ A, B ∈ P(X)

Existence of identity : 

Let E ∈ P(X) be identity for ‘*‘ in set P(X)

⇒ A * E = A = E * A

⇒ (A – E) ∪ (E – A) = A = (E – A) ∪ (A – E)

It is possible only when E = Φ, Because

(A – Φ) ∪ (Φ – A) = A ∪ Φ = A and

(Φ – A) ∪ (A – Φ) = Φ ∪ A = A

Hence, Φ is identity element.

Existence of inverse :

Let A–1 be the inverse of A for ‘*‘ on set P(X).

∴ A * A–1 = Φ = A–1 * A

⇒ (A – A–1 ) ∪ (A–1 – A) = Φ

⇒ A – A–1 = Φ = A–1 – A = Φ

⇒ A ⊂ A–1 and A–1 ⊂ A

⇒ A = A–1

Hence, each element of P(X) is inverse of itself.

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