Short Answer TypeLet f : x → Y be an invertible function. Show that the inverse of f–1 is f,
i.e.(f–1)–1 = f.
Since inverse of a bijection is also a bijection.
∴ f–1 : Y → X is a bijection and hence invertible.
Since f–1 : Y → X is a bijection.
∴ (f–1)–1 : X → Y is also a bijection.
Let x be an arbitrary element of X such that f(x) = y. Then.
f–1 (y) = x [∵ f–1 is the inverse of f ]
⇒ (f–1)–1 (x) = y [∵(f–1)–1 is the inverse of f–1]
(f–1)–1 (x) = f(x) [∴ f(x) = y]
Since x is an arbitrary element of X.
∴ (f–1)–1 (x) = f(x) for all x ∈ X
∴ (f–1)–1 = f .
Let f : N → Y be function defined as f (x) = 4 x + 3, where, Y = {y ∈N : y = 4 x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse.
Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose the correct answer. (A) (2, 4) ∈ R (B) (3, 8) ∈ R (C) (6, 8) ∈ R (D) (8, 7) ∈ R
Show that the function f : R. → R. defined by 
is one-one and onto, where R. is the set of all non-zero real numbers. Is the result true, if the domain R. is replaced by N with co-domain being same as R.?
Check the injectivity and surjectivity of the following functions:
f : N → N given by f(x) = x2
Check the injectivity and surjectivity of the following functions:
f : Z → Z given by f(x) = x2
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