Question -
Answer -
(i) Given 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
⇒ f is notone-one.
⇒ f is not abijection.
So, f does not have an inverse.
(ii) Given g: {5, 6,7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
from the question it is clear that g (5) = g (7) = 4
⇒ f is notone-one.
⇒ f is not abijection.
So, f does not have an inverse.
(iii) Given 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.
⇒ h is one-one.
Also, each element in the co-domain has a pre-image in the domain.
⇒ h is onto.
⇒ h is abijection.
Therefore h inverseexists.
⇒ h has an inverseand it is given by
h-1 = {(7, 2), (9, 3), (11, 4), (13, 5)}