Give an example of a map(i) which is one-one but not onto(ii) which is

1.	Give an example of a map(i) which is one-one but not onto(ii) which is not one-one but onto(iii) which is neither one-one nor onto.
Answer» (i) Let f: N → N, be a mapping defined by f (x) = x² For f (x₁) = f (x₂) Then, x₁² = x₂² x₁ = x₂ (Since x₁ + x₂= 0 is not possible) Further ‘f’ is not onto, as for 1 ∈ N, there does not exist any x in N such that f (x) = 2x + 1. (ii) Let f: R → [0, ∞), be a mapping defined by f(x) = \|x\| Then, it’s clearly seen that f (x) is not one-one as f (2) = f (-2). But \|x\| ≥ 0, so range is [0, ∞]. Therefore, f (x) is onto. (iii) Let f: R → R, be a mapping defined by f (x) = x² Then clearly f (x) is not one-one as f (1) = f (-1). Also range of f (x) is [0, ∞). Therefore, f (x) is neither one-one nor onto.

Give an example of a map(i) which is one-one but not onto(ii) which is not one-one but onto(iii) which is neither one-one nor onto.

Answer»

(i) Let f: N → N, be a mapping defined by f (x) = x²

For f (x₁) = f (x₂)

Then, x₁² = x₂²

x₁ = x₂ (Since x₁ + x₂= 0 is not possible)

Further ‘f’ is not onto, as for 1 ∈ N, there does not exist any x in N such that f (x) = 2x + 1.

(ii) Let f: R → [0, ∞), be a mapping defined by f(x) = |x|

Then, it’s clearly seen that f (x) is not one-one as f (2) = f (-2).

But |x| ≥ 0, so range is [0, ∞].

Therefore, f (x) is onto.

(iii) Let f: R → R, be a mapping defined by f (x) = x²

Then clearly f (x) is not one-one as f (1) = f (-1). Also range of f (x) is [0, ∞).

Therefore, f (x) is neither one-one nor onto.