Question

Define a bijective function. Show that $f:N\to N$ given by \[f\left( x \right)=\left\{ \begin{align}
  & x+1,\text{ if x is odd} \\
 & x-1,\text{ if x is even} \\
\end{align} \right.\] is a bijective function.

Answer 1

Hint: A function is called bijective if it is both one-one and onto, it is one-one if $f:A\to B$ has $f\left( a \right)=f\left( b \right)\Rightarrow a=b\text{ }\forall \text{a}\in \text{A and }\forall b\in B$. The function $f:A\to B$ is onto if $\exists b\in B\exists a$ unique $a\in A$ such that $f\left( a \right)=b$. We will consider cases for a and b being even or odd and then check for all that, they are one-one and onto or not.

Complete step-by-step answer:
Let us define a bijective function first.
A function $f:A\to B$ is called bijective if it is both one-one and onto.
One-one function: A function $f:A\to B$ is one to one if $f\left( {{a}_{1}} \right)=f\left( {{a}_{2}} \right)\Rightarrow {{a}_{1}}={{a}_{2}}$ where ${{a}_{1}}\in A,{{a}_{2}}\in A$.
Here, A and B are both sets.
Onto function: A function $f:A\to B$ is onto function if $\exists b\in B\exists a\in A$ unique such that $f\left( a \right)=b$. Basically, for every image of f there exists a preimage of f.
Hence, we have defined a bijective function. We have f as $f:N\to N$ given by \[f\left( x \right)=\left\{ \begin{align}
  & x+1,\text{ if x is odd} \\
 & x-1,\text{ if x is even} \\
\end{align} \right.\] is a bijective function.
Let us first show f is one-to-one.
Let us assume for $a,b\in N,f\left( a \right)=f\left( b \right)$ we have to show that a = b.
Consider cases when a, b are odd, even etc.
Case I:
\[\begin{align}
  & a\to \text{odd} \\
 & b\to \text{even} \\
 & \text{then }f\left( a \right)=f\left( b \right) \\
 & \text{As a is odd}\Rightarrow f\left( a \right)=a+1 \\
 & \text{As b is even}\Rightarrow f\left( b \right)=b-1 \\
 & \Rightarrow f\left( a \right)=f\left( b \right) \\
 & \Rightarrow a+1=b-1 \\
 & \Rightarrow b-a=2 \\
\end{align}\]
Now, because a and b are odd and even and hence the difference of them can never be equal to 2. So, Case I is not possible as $b-a\ne 2$ for any b-even and a-odd.
Case II:
\[\begin{align}
  & a\to \text{odd} \\
 & b\to \text{odd} \\
 & \Rightarrow f\left( a \right)=f\left( b \right) \\
 & \Rightarrow a+1=b+1 \\
 & \Rightarrow a=b \\
\end{align}\]
So, \[\Rightarrow f\left( a \right)=f\left( b \right)\Rightarrow a=b\]
Consider Case III:
\[\begin{align}
  & a\to \text{even} \\
 & b\to \text{even} \\
 & \Rightarrow f\left( a \right)=f\left( b \right) \\
 & \Rightarrow a-1=b-1 \\
 & \Rightarrow a=b \\
\end{align}\]
So, \[\Rightarrow f\left( a \right)=f\left( b \right)\Rightarrow a=b\]
So, from above cases we have \[\Rightarrow f\left( a \right)=f\left( b \right)\Rightarrow a=b\]
Therefore, function f is one-one . . . . . . . . . . . . . . . . . (i)
Now, we will check for it. We have,
\[f\left( x \right)=\left\{ \begin{align}
  & x+1,\text{ if x is odd} \\
 & x-1,\text{ if x is even} \\
\end{align} \right.\]
Let y=f(x)
Case I:
\[\begin{align}
  & x\to \text{odd} \\
 & \text{when }x\to \text{odd} \\
 & \Rightarrow f\left( x \right)=x+1 \\
 & \Rightarrow y=x+1 \\
 & \Rightarrow x=y-1 \\
\end{align}\]
So, for x even we will get y as even as y=x+1.
Case II:
\[\begin{align}
  & x\to \text{even} \\
 & \text{when }x\to \text{even} \\
 & \Rightarrow f\left( x \right)=x-1 \\
 & \Rightarrow y=x-1 \\
 & \Rightarrow x=y+1 \\
 & \text{when }x\to \text{even}=y\to \text{odd} \\
\end{align}\]
So, we have
\[x=\left\{ \begin{align}
  & y-1;y\to \text{even} \\
 & \text{y+1;y}\to \text{odd} \\
\end{align} \right.\]
Now, as we had $f:1N\to 1N$ so for any number $b\in 1N$ we have a unique $a\in N$ such that $f\left( a \right)=b$ as
\[x={{f}^{-1}}\left( y \right)=\left\{ \begin{align}
  & y-1;y\to \text{even} \\
 & y+1;y\to \text{odd} \\
\end{align} \right.\]
So, all even and odd are covered.
Hence, the function f is onto. . . . . . . . . . . . . . . . . . . . . . . . (ii)
Therefore, from (i) and (ii) we have \[f\left( x \right)=\left\{ \begin{align}
  & x+1,\text{x}\to \text{odd} \\
 & x-1,\text{x}\to \text{even} \\
\end{align} \right.\] is a bijective function.

Note: To have a better understanding, students can always assume values and check whether f given as $f:N\to N$ as $f\left( x \right)=\left\{ \begin{align}
  & x+1,\text{x}\to \text{odd} \\
 & x-1,\text{x}\to \text{even} \\
\end{align} \right.$ is one-one and onto.
For confusion in Case I of one-one as a = odd and b = even, any odd number is of the form $a=2n+1\Rightarrow a=2n-1$ and any even b is of the form $b=2n$ then their difference
\[b-a=\left( 2n \right)-\left( 2n-1 \right)=+1\Rightarrow \left( 2n \right)-\left( 2n+1 \right)=-1\]
So, this is never 2. This is for consecutive numbers. Hence, b-a is never 2, even if consecutive numbers are not taken. So, case I am rejected.