# 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 withA–1 = A.(Hint: (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ).

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

A * B = (A – B) (B – A), A, B P(X).

Now, let A ϵ P(X). Then, we get,

A * ф = (A – ф) (ф A) = A ф = A

ф * A = (ф - A) (A - ф) = ф A = A

A * ф = A = ф * A, A ϵ P(X)

Therefore, ф is the identity element for the given operation *.

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

A * B = ф = B * A. (as ф is an identity element.)

Now, we can see that A * A = (A –A) (A A) = ф ф = ф A ϵ P(X).

Therefore, all the element A of P(X) are invertible with A-1 = A.

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