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.
Let A = {a1,a2, ...an} where n is finite.
Since f is one-to-one.
∴ f(a1), f(a2)....,f (an) district element of A.
∴ A = {f (a1), f(a2).....,f (an)}
Let b be any element of A. Then
b = f(ai) for some i, 1 ≤ i ≤ n. ∴ f is onto.