Given a non-empty set X, let *: P(X) x P(X) → P(X) be defined as A * B = (A - B) υ (B - A), ∀A, B ∈ P (X). Show that the empty set F is the identity for the operation * and all the elements A of P (X) are invertible with A^{- 1} = A.
(Hint: (A - Φ) υ (Φ - A) = A and (A - A) υ (A - A) = A * A = Φ)

Solution:

In general, a function is invertible as long as each input features a unique output. That is, every output is paired with exactly one input.

That way, when the mapping is reversed, it'll still be a function.

A monotonic function i.e. bijection function is usually invertible.

It is given that * :

P (X) x P(X) → P (X) be defined as

A * B = (A - B) υ (B - A), ∀A, B ∈ P (X).

A ∈ P (X) then,

A * Φ = (A - Φ) υ (Φ - A) = A υ Φ = A

Φ * A = (Φ - A) υ ( A - Φ) = Φ υ A = A

⇒ A * Φ = A = Φ * A for all A ∈ P (X)

F is the identity for the operation *.

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]

A * A = (A - A) υ (A - A)

= Φ υ Φ = Φ for all A ∈ P (X).

All the elements A of P (X) are invertible with A^{- 1} = A

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NCERT Solutions for Class 12 Maths - Chapter 1 Exercise ME Question 13

Given a non-empty set X, let *: P(X) x P(X) → P(X) be defined as A * B = (A - B) υ (B - A), ∀A, B ∈ P (X). Show that the empty set F is the identity for the operation * and all the elements A of P (X) are invertible with A^{- 1} = A .(Hint: (A - Φ) υ (Φ - A) = A and (A - A) υ (A - A) = A * A = Φ)

Summary:

Given that P(X) x P(X) → P(X) be defined as A * B = (A - B) υ (B - A), ∀A, B ∈ P (X). F is the identity for the operation *. All the elements A of P (X) are invertible with A^{- 1} = A