Question

Show that $f:N\to N$ = $x-1$, if $x$ is even

is both one-one and onto

Answer 1

The question is

Show that $f:N\to N$, given by $f\left(x\right)=\{\begin{array}{c}x+1;ifxisodd\\ x-1;ifxiseven\end{array}$

For one-one

Case:$I$When $x_1,x_2$ are odd natural number.

$\therefore f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow x_1+1=x_2+1forall{x}_1,x_2\in N$

$\Rightarrow x_1=x_2$

$\therefore f$ is one-one.

Case:$II:$When $x_1,x_2$ are even natural number.

$\therefore f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow x_1-1=x_2-1$

$\Rightarrow x_1=x_2$

$\therefore f$ is one-one

Case$III:$When $x_1$ is odd and $x_2$ is an even number.

$f\left(x_1\right)=f\left(x_2\right)\Rightarrow x_1+1=x_2-1$

$\Rightarrow x_2-x_1=2$

which is never possible as the difference of odd and even number is always an odd number.

Hence in this case $f\left(x_1\right)\ne f\left(x_2\right)$

Thus,$f$ is one-one.

Case$IV:$When $x_1$ is even and $x_2$ is odd natural number.

$f\left(x_1\right)=f\left(x_2\right)\Rightarrow x_1-1=x_2+1$

$\Rightarrow x_1-x_2=2$

which is never possible as the difference of odd and even number is always an odd number.

Hence in this case $f\left(x_1\right)\ne f\left(x_2\right)$

Thus,$f$ is one-one.

For onto:

$\therefore f\left(x\right)=x+1$ if $x$ is odd

$=x-1$ if $x$ is even.

$\Rightarrow$For every even number $y$ of co-domain there exists an odd number $y-1$ in domain and for every odd number $y$ of co-domain there exists evert number $y+1$ in domain.

Thus,$f$ is an onto fucntion.

Hence $f$ is one-one onto function.