Question

A function f from the set of natural numbers to integers defined by

f(n) = {(n-1)/2, when n is odd, (-n/2), when n is even} is

(a) onto but not one-one
(b) one-one and onto both
(c) neither one-one nor onto
(d) one-one but not onto.

Answer 1

(b) : If n is odd, let n = 2k – 1
Let f(2k₁ – 1) = f (2k₂ – 1)

=> (2k₁ - 1 - 1)/2 = (2k₂ - 1 - 1)/2

=> k₁ = k₂
=> f(n) is one-one functions if n is odd
Again, If n = 2k (i.e. n is even)
Let f(2k₁) = f(2k₂)

=> - (2k₁/2) = - (2k₂/2)

=> k₁ = k₂
=> f(n) is oneone if n is even

Again f(n) = (n-1)/2

f '(n) =1/2 > 0∀n ∈ N if n is odd
and f '(n) = -1/2< 0∀n ∈ N if n is even
Now all such function which are either increasing or decreasing in the stated domain are said to be onto function. Finally f (n) is one-one onto function.