Let f : W be defined as f(n) = n -1, & if n is odd n+1, &if n is … - Vidyadip

Question

Let $\text{f : W}\rightarrow\text{W}$ be defined as
$\text{f(n)} = \begin{cases} \text{n -1}, & \text{if n is odd} \\ \text{n+1}, &\text{if n is even} \end{cases}$
Show that f is invertible and find the inverse of f. Here, W is the set of all whole numbers.

✓

Answer

$\text{One - One : - Case I : when x and y are even}$
$\text{f(x) = f(y)}\Rightarrow \text{x + 1 = y + 1} \Rightarrow\text{x = y}$
$\text{Case II : when x and y are odd}$
$\text{f(x) = f(y)} \Rightarrow \text{x - 1 =y - 1} \Rightarrow \text{x = y}$
$\text{Case III : one of them is even and one of hem is odd}$
$\text{f(x)}\neq \text{f(y)} \Rightarrow \text{x + 1}\neq \text{y - 1} \Rightarrow \text{x}\neq\text{y}$
$\text{Onto : Let y}\in \text{W}$
$\text{f(y - 1) = y if y is odd}$
$\text{f(y + 1) = y if y is even}$
$\text{So}\forall \text{y} \in \text{W, there exist some element vin domain of f}$
$\Rightarrow \text{f is invertible}$
$\text{f}^{-1}{(\text{x})} = \begin{cases} \text{x -1}, & \text{x is odd} \\ \text{x+1}, &\text{x is even}\end{cases}$

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