Short Answer TypeLet g1 and g2 be two inverses of f for all y ∈ Y, we have, (f o g1)(y) = y = IY (y) and (f o g 2) (y) = y = IY (y)
⇒ (f o g1) (y) = (f o g2) (y) for all y ∈ Y
⇒ f(g1(y)) = f(g2(y)) for all y ∈ Y
⇒ g1 (y) = g2(y) [∵ f is one-one as f is invertible]
∴ g1 = g2 ∴ inverse of f is unique.
2. Let f : x → Y be an invertible function. Show that the inverse of f–1 is f, i.e.(f–1)–1 = f.
Let S = {a, b, c} and T = {1, 2, 3}. Find F–1 of the following functions F from S to T, if it exists
(i) F = {(a, 3), (b, 2), (c, 1)} (ii) F = {(a, 2), (b, 1), (c, 1)}
Let S = {1, 2, 3}. Determine whether the functions f : S → S defined as below have inverses. Find f–1, if it exists.
(a) f = {1, 1), (2, 2), (3, 3)} (b) f = {(1, 2), (2, 1), (3, 1)}
(c) f = {(1,3),(3,2), (2, 1)}
State with reason whether 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)}
Let P be the set of all subsets of a given set X.
Show that U:PxP→P given by (A, B) ∴ A ∪ B and ∩ : P x P → P given by (A, B) → A ∩ B are binary operations on the set P.
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