Let `f: Nuu{0}->Nuu{0}` be defined by `f(n)={n+1,\ if\ n\ i s\ even\,\ \ \n-1,\ if\ n\ i s\ od d` Show that `f` is invertible and `f=f^(-1)` .

Explanation

`f(n)=\{{cc}n+1}` ` n-1 \quad.` if `n` is even


for one - one:-


`{rr}\text { If } n \text { is even. } \text { Jf } n \text { is odd. } ` f(n_{1})=f(n_{2}) f(n_{1})=f(n_{2}) ` n_{1}+1=n_{2}+1 n_{1}-1=n_{2}-1 ` n_{1}=n_{2} n_{1}=n_{2}`



`\Rightarrow f(x)` is one-one function.

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