# Show that the function f:R $\rightarrow \{ x \in$ R:-1$<$x$<$1 $\}$ defined by $f(x) = \frac {x} { 1+|\;x\;|}, x \in R$ is one-one and onto function.

## 1 Answer

Toolbox:
• A function $f: X \rightarrow Y$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function.
• A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
Given the function f:R $\rightarrow \{ x \in$ R:-1$<$x$<$1 $\}$ defined by $f(x) = \frac {x} { 1+|\;x\;|}, x \in R$
$\textbf {Step 1: Checking one-one}$
A function $f: X \rightarrow Y$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function.
Let $f(x) = f(y) \Rightarrow \frac {x} { 1+|\;x\;|} = \frac {y} { 1+|\;y\;|}$
$\textit{Case 1: Consider the case where x is positive and y is negative}$
$\Rightarrow \frac {x} { 1+x} = \frac {y} { 1- y} \rightarrow x(1-y) = y(1+x) \rightarrow 2xy = x-y$
Since $x$ is positive and $y$ is negative, $x>y$
We can see that $x-y >0$, but $2xy$ is negative as $y$ is negative.
Therefore, $2xy \neq x-y$. Similarly one can make the argument for when x is negative and y is positive.
$\textit{Case 2: Consider the case where x and y are positive}$
$\Rightarrow \frac {x} { 1+x} = \frac {y} { 1+y} \rightarrow x(1+y) = y(1+x) \rightarrow x+xy = y+xy \rightarrow x=y$
$\textit{Case 3: Consider the case where x and y are negative}$
$\Rightarrow \frac {x} { 1+x} = \frac {y} { 1+y} \rightarrow x(1-y) = y(1-x) \rightarrow x-xy = y-xy \rightarrow x=y$
Therefore, the function $f$ is one-one.
$\textbf {Step 2: Checking onto}$
A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
$x = \frac {y} { 1+|\;y\;|}$
$\textit{Case 1: Consider the case where y is negative}$
$\Rightarrow$ there exisits $x = \frac {y} { 1+|\;y\;|}$ such that $f(x) = f(\frac {y} { 1+ y})$
$\Rightarrow f(x) = \Large ( \frac{\frac{y}{1+y}}{1+|\frac{y}{1+y}|}) = \Large ( \frac{\frac{y}{1+y}}{1+\frac{-y}{1+y}}) = \frac{y}{1+y-y}$$= y \textit{Case 2: Consider the case where y is positive} \Rightarrow there exisits x = \frac {y} { 1+|\;y\;|} such that f(x) = f(\frac {y} { 1- y}) \Rightarrow f(x) = \Large ( \frac{\frac{y}{1-y}}{1+|\frac{y}{1-y}|}) = \Large ( \frac{\frac{y}{1-y}}{1+\frac{y}{1-y}}) = \frac{y}{1-y+y}$$ = y$
Therefore $f$ is onto.
Hence, the function $f$ is one-one and onto.
answered Mar 20, 2013

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