Question

State with reasons whether the following functions have inverse:
(i) f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}

Answer 1

(i) f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
We have:
f (1) = f (2) = f (3) = f (4) = 10
$\Rightarrow$f is not one-one.
$\Rightarrow$f is not a bijection.
So, f does not have an inverse.

(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
g (5) = g (7) = 4
$\Rightarrow$f is not one-one.
$\Rightarrow$f is not a bijection.
So, f does not have an inverse.

(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}
Here, different elements of the domain have different images in the co-domain.
$\Rightarrow$h is one-one.
Also, each element in the co-domain has a pre-image in the domain.
$\Rightarrow$h is onto.
$\Rightarrow$h is a bijection.
$\Rightarrow$h has an inverse and it is given by
h^-1={(7, 2), (9, 3), (11, 4), (13, 5)}