Short Answer Typef : 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
Consider the identity function 1N : N → N defined as lN(x) = x ∀ x ∈ N. Show that although IN is onto but IN + IN : N → N defined as (IN + IN) (x) = IN(x) + IN(x) = x + x = 2 x is not onto.
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i)  f : R → R defined by f (x) = 3 – 4 x
(ii) f : R → R defined by f (x) = 1 + x2.
Let f : N – {1} → N defined by f (n) = the highest prime factor of n. Show that f is neither one-to-one nor onto. Find the range of f.
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