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Show that f : N → N, given by f(x) = {(x + 1, if x is odd), (x - 1, if x is even) is both one-one and onto. |
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Answer» For one-one Case I : When x1, x2 are odd natural number. ∴ f(x1) = f(x2) ⇒ x1 + 1 = x2 + 1 ∀ x1 , x2 ∈ N ⇒ x1 = x2 i.e., f is one-one. Case II : When x1, x2 are even natural number ∴ f(x1) = f(x2) ⇒ x1 – 1 = x2 – 1 ⇒ x1 = x2 i.e., f is one-one. Case III : When x1 is odd and x2 is even natural number f(x1) = f(x2) ⇒ x1 + 1 = x2 – 1 ⇒ x2 – x1 = 2 which is never possible as the difference of odd and even number is always odd number. Hence in this case f (x1) ≠ f(x2) i.e., f is one-one. Case IV: When x1 is even and x2 is odd natural number Similar as case III, We can prove f is one-one For onto: ∴ f(x) = x +1 if x is odd = x – 1 if x is even ⇒ For every even number ‘y’ of codomain ∃ odd number y - 1 in domain and for every odd number y of codomain ∃ even number y +1 in Domain. i.e. f is onto function. Hence f is one-one onto function. |
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