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Question
Given a non-empty set X, let *: P(X) × P(X) → P(X) be defined as A * B = (A − B) ∪ (B −A), &mnForE; 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) = Aand (A − A) ∪ (A − A) = A * A = Φ).
Solution
It is given that *: P(X) × P(X) → P(X) is defined as
A * B = (A − B) ∪ (B − A) &mnForE; A, B ∈ P(X).
Let A ∈ P(X). Then, we have:
A * Φ = (A − Φ) ∪ (Φ − A) = A ∪ Φ = A
Φ * A = (Φ − A) ∪ (A − Φ) = Φ ∪ A = A
∴A * Φ = A = Φ * A. &mnForE; A ∈ P(X)
Thus, Φ is the identity element for the given operation*.
Now, an element A ∈ P(X) will be invertible if there exists B ∈ P(X) such that
A * B = Φ = B * A. (As Φ is the identity element)
Now, we observed that A * A = (A - A) ∪ (A -A) = Φ∪Φ = Φ ∀ A ∈ P(X).
Hence, all the elements A of P(X) are invertible with A−1 = A.
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