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.
Range of f = {(f (a1) , f (a2), ...., f (an)}
Since f : A → A is an onto function
∴ range of f = A
⇒ {f (a1),f(a2),...f(an)} = (a1,a2,....,an}
But A is a finite set consisting of n elements
∴ f(an), f(a2), ... ,f(an) are district elements of A.
∴ f is one-to-one.