#### Explain solution RD Sharma class 12 chapter Functions exercise 2.1 question 23 maths

Given:

$f(n)=\left\{\begin{array}{ll} n+1, & \text { if } n \text { is odd } \\ n-1, & \text { if } n \text { is even } \end{array}\right.$

To prove:

Function f is bijection.

Hint:

For any function to be a bijection, the given function should be one-one and onto.

Solution:

Here $f:N \rightarrow N$

First we will check whether the given function is one-one or not.

For one-one,

Case I : If n is odd

Let $x,y\in N$ such that $f(x)=f(y)$

As        $f(x)=f(y)$

$\begin{array}{ll} \Rightarrow & x+1=y+1 \\ \Rightarrow & x=y \end{array}$

Case II : If n  is even

Let $x,y\in N$ such that $f(x)=f(y)$

As        $f(x)=f(y)$

$\begin{array}{ll} \Rightarrow & x-1=y-1 \\ \Rightarrow & x=y \end{array}$

So, f is injective (one-one).

Again we will check whether the given function is onto or not.

For onto,

Case I : If n  is odd

As for every$f:N \rightarrow N$,$n \in N$ , there exists $y=n-1$

$f(y )=f(n-1)$

$=n-1+1$

$=n$

Case II : If n is even

As for every $n \in N$, there exists$y=n+1$

$\Rightarrow f(y)=f(n+1)=n+1-1=n$

So f is surjective (onto).

Hence f is a bijection.