Let A and B be two sets. Show that f :A x B → B x A such that f (a,b) = (b,a) is a bijective function.
f : A x B → B x A is such that f (a,b) = (b,a).
Let (a1, b1) and (a2, b2) any two elements of A x B such that
f (a1, b1) = f (a2, b2)
∴ (b1, a1) = (b2, a2) ⇒ b1= b2 and a1 = a2
⇒ = (a2,b2)
∴ f (a1,b1) = f (a2,b2) ⇒ (a1b1) = (a2,b2) ∀ (a1, b1) (a2, b2) ∈ A x B
∴ f is one-to-one
Again let (b, a) be any element of B X A ∴ b ∈ B and a ∈ A. So (a, b) ∈ A X B ∴ for all (b, a) ∈ B X A, there exists (a, b) ∈ A x B such that f (a, b) = (b, a)
∴ f : A x B → B x A is an onto function ∴ f is one-to-one and onto.