[tex]f:\mathbb{N}\to \mathbb{N},\, f(n) = n+(-1)^n[/tex]
[tex]f(n) =\begin{cases} n+1,\quad n\text{ par} \\ n-1,\quad n\text{ impar}\end{cases} \\ \\ \Rightarrow f(n) =\begin{cases} \text{impar},\quad n\text{ par} \\ \text{par},\quad n\text{ impar}\end{cases}[/tex]
[tex]\Rightarrow f(n) =\begin{cases} \{\overset{n=0}{1},\overset{n=2}{3},\overset{n=4}{5},\overset{n=6}{7},\overset{n=8}{9},\overset{n=10}{11},...\},\quad n\text{ par} \\ \\ \{\overset{n=1}{0},\overset{n=3}{2},\overset{n=5}{4},\overset{n=7}{6},\overset{n=9}{8},\overset{n=11}{10},...\},\quad n\text{ impar}\end{cases}\\ \\ \text{}\quad\quad\quad\quad\quad\quad\quad\quad\Downarrow\\ \\f(n) = \{0,1,2,3,4,5,...\} = \mathbb{N}\\ \\ x_1\neq x_2 \Rightarrow f(x_1)\neq f(x_2),\quad \forall x_1,x_2\in \mathbb{N}[/tex]
=> funcția este bijectivă.
[tex]f(n) = n+(-1)^n \\ \\ (f \circ f)(n) = f\Big(f(n)\Big) = f\Big(n+(-1)^n\Big) = \\ \\ = \Big(n+(-1)^n\Big)+(-1)^{n+(-1)^n} = n+(-1)^n+(-1)^n\cdot (-1)^{(-1)^n} = \\ \\ = n+(-1)^n+(-1)^n\cdot(-1)^n =\\ \\= n+(-1)^n+(-1)^{n+1} = n\pm 1 \mp 1 = n[/tex]
Sau demonstrația din imagine.
