Example 12 - Show that f(x) = { x + 1, if x is odd, x-1 if x is even

Example 12 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. 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 Putting f(x1) = f(x2) x1 + 1 = x2 + 1 x1 = x2 If x1 & x2 are both are even f(x1) = x1 – 1 f(x2) = x2 – 1 If f(x1) = f(x2) x1 – 1 = x2 – 1 x1 = x2 If x1 is odd and x2 is even f(x1) = x1 + 1 f(x2) = x2 – 1 If f(x1) = f(x2) x1 + 1 = x2 – 1 x2 – x1 = 2 which is impossible as difference between even and odd number can never be even Hence, if f(x1) = f(x2) , x1 = x2 ∴ function f is one-one Check onto f(x) = 𝑥+1 , 𝑖𝑓 𝑥 𝑖𝑠 𝑜𝑑𝑑﷮𝑥−1, 𝑖𝑓 𝑥 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ Let f(x) = y , such that y ∈ N x = 𝑦−1 , 𝑖𝑓 𝑦 𝑖𝑠 𝑒𝑣𝑒𝑛﷮𝑦+1, 𝑖𝑓 𝑦 𝑖𝑠 𝑜𝑑𝑑﷯﷯ Hence, if y is a natural number, x will also be a natural number i.e. x ∈ N Thus, f is onto.

Example 12 - Chapter 1 Class 12 Relation and Functions