Let `f: W to W : f(n) = {[(n+1) when n is even], [(n-1) when n is odd ]}` Show that `f` is invertible. Find `f^(-1)`

Let `f: W to W : f(n) = {[(n+1) when n is even], [(n-1) when n is odd

1.	Let `f: W to W : f(n) = {[(n+1) when n is even], [(n-1) when n is odd ]}` Show that `f` is invertible. Find `f^(-1)`
Answer» Let `f(n_(1)) =f(n_(2))` Case 1 when `n_(1) ` is odd and `n_(2)` is even In this case `f(n_(1)) =f(n_(2)) rArr n_(1) -1 = n_(2) +1` ` rArr n_(1)- n_(2)=2` If `n_(1) ` is odd and `n_(2)` is even then `(n_(1) - n_(2)) ne 2` Thus we arrive at a contradiction. So , in the case `f(n_(1)) ne f(n_(2))` Similarly when `n_(1)` is even and `n_(2)` is odd then `f(n_(1)) ne f(n_(2))` Case 2 When `n_(1)` and `n_(2)` are both odd In this case `f (n_(1)) =f(n_(2)) rArr n_(1) -1 =n_(2) -1` `rArr n_(1) =n_(2)` Case 3 When `n_(a) " and " n_(2) ` are both even In this case `f(n_(1))= f(n_(2)) rArr n_(1) +1 =n_(2) +1` `rArr n_(1)+ n_(2)` Thus from all the cases we get `f(n_(1)) =f(n_(2)) rArr n_(1) = n_(2)` `:. ` f is one-one Now we show that f is onto . Let `n in W.` Case 1 when n is odd In this case (n-1) is even and f(n-1) =(n-1) +1=n Case 2 When n is even In this case (n+1) is odd and f(n+1) =(n+1) -1=n Thus each `n in W` has its pre-image in W. `:.` f is onto . Thus f is one-one onto and hence invertible clealry we have `f^(-1) (n) = { underset( (n+1) " when n is even ")((n-1) " when n is odd ")`