# Example 12 - Chapter 1 Class 12 Relation and Functions (Term 1)

Last updated at Jan. 28, 2020 by Teachoo

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Chapter 1 Class 12 Relation and Functions (Term 1)

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Last updated at Jan. 28, 2020 by Teachoo

Example 12 Show that f : N β N, given by f(x) = {β(π₯+1 , ππ π₯ ππ πππ@π₯β1, ππ π₯ ππ ππ£ππ)β€ is both one-one and onto. Check one-one There can be 3 cases x1 & x2 both are odd x1 & x2 both are even x1 is odd & x2 is even If x1 & x2 are both odd f(x1) = x1 + 1 f(x2) = x2 + 1 Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 Putting f(x1) = f(x2) x1 + 1 = x2 + 1 x1 = x2 If x1 & x2 are both are even f(x1) = x1 β 1 f(x2) = x2 β 1 If f(x1) = f(x2) x1 β 1 = x2 β 1 x1 = x2 Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 If x1 is odd and x2 is even f(x1) = x1 + 1 f(x2) = x2 β 1 If f(x1) = f(x2) x1 + 1 = x2 β 1 x2 β x1 = 2 which is impossible as difference between even and odd number can never be even Hence, if f(x1) = f(x2) , Then x1 = x2 β΄ function f is one-one Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 Check onto f(x) = {β(π₯+1 , ππ π₯ ππ πππ@π₯β1, ππ π₯ ππ ππ£ππ)β€ Let f(x) = y , such that y β N x = {β(π¦β1 , ππ π¦ ππ ππ£ππ@π¦+1, ππ π¦ ππ πππ)β€ If x is odd f(x) = x + 1 y = x + 1 y β 1 = x x = y β 1 If x is odd, y is even If x is even f(x) = x β 1 y = x β 1 y + 1 = x x = y + 1 If x is even, y is odd Hence, if y is a natural number, x will also be a natural number i.e. x β N Thus, f is onto.