Given a non -empty set X, let *: P(X) × P(X) → P(X) be defined as A

1.	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).
Answer» It is given that : P(X) × P(X) → P(X) is defined as A B = (A − B) ∪ (B − A) ∀ 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 for all 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 for all A ∈ P(X). Hence, all the elements A of P(X) are invertible with A⁻¹ = A.

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).

Answer»

It is given that *: P(X) × P(X) → P(X) is defined as A * B = (A − B) ∪ (B − A) ∀ 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 for all 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 for all A ∈ P(X).
Hence, all the elements A of P(X) are invertible with A⁻¹ = A.

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