Question

Show that $f:N\to N$, given by

$f\left(x\right)=\{\begin{array}{c}x+1,\text{if}x\text{is}\text{odd}\\ x-1,\text{if}x\text{is}\text{even}\end{array}$

is both one-one and onto.

Answer 1

For one-one:

If $f\left(a\right)=f\left(b\right)$

$\Rightarrow a+1=b+1$ or $a-1=b-1$, if both are even or odd.

Note : $f\left(a\right),f\left(b\right)$ are not equal if one is even and other is odd, since if $a$ is even and $b$ is odd, $a-1$ is odd and $b+1$ is even.

$\Rightarrow a=b$.

So, one-one function

Now, for onto:

For all $x\in$ odd nos. $f\left(x\right)$ gives all the even nos.

And all $x\in$ even nos. $f\left(x\right)$ gives all the odd nos.
$\Rightarrow \text{range}=\text{codomain}$

So, onto function.