Let a function $f:N\to N$ be defined by $\mathrm{f}\left(\mathrm{n}\right)=\left[\begin{array}{l}2\mathrm{n},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{n}=2,4,6,8,\dots \\ \mathrm{n}-1,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{n}=3,7,11,15,\dots \\ \frac{\mathrm{n}+1}{2},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{n}=1,5,9,13,\dots \end{array}\right.$

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ccc}4k& ;& \mathrm{n}=2k\\ 4k-2& ;& \mathrm{n}=4k-1\\ 2k-1& ;& \mathrm{n}=4k-3\end{array}\right. \\ \text{ Note that for any element, it will fall into exactly one of these sets. } \\ \begin{array}{l}\{k:y=4k;k\in N\},\{y:y=4k-2;k\in N\},\{y:y=2k-1;k\in N\}\\ \text{ Corresponding to that }\mathrm{y}\text{, we will get exactly one value of n}\\ \therefore \mathrm{f}\text{ is one }-\text{ one }\mathrm{\&}\text{ onto. }\end{array} \end{gathered}$$