Question

Show that function f: R ${\displaystyle \to }$ {x ∈ R : −1 < x < 1} defined by f(x) = ${\displaystyle \frac{x}{1+\left|x\right|}}$, x ∈ R is one-one and onto function.

Answer 1

It is given that f: R ${\displaystyle \to }$ {x ∈ R: −1 < x < 1} is defined as f(x) =${\displaystyle \frac{x}{1+\left|x\right|}}$, x ∈ R.

Suppose f(x) = f(y), where x, y ∈ R.

${\displaystyle \Rightarrow \frac{x}{1+\left|x\right|}=\frac{y}{1-\left|y\right|}}$

=> 2xy =  x - y

${\displaystyle \Rightarrow \frac{x}{1+x}=\frac{y}{1-y}}$

⇒ x + xy = y + xy

⇒ x = y

Since x is positive and y is negative:

x > y 

⇒ x − y > 0

But, 2xy is negative.

Then, ${\displaystyle 2xy\ne x-y}$.

Thus, the case of x being positive and y being negative can be ruled out.

Under a similar argument, x being negative and y being positive can also be ruled out

∴ x and y have to be either positive or negative.

When x and y are both positive, we have:

${\displaystyle \Rightarrow f\left(x\right)=f\left(y\right)=\frac{x}{1+\left|x\right|}=\frac{y}{1-\left|y\right|}}$

${\displaystyle f\left(x\right)=f\left(y\right)\Rightarrow \frac{x}{1+x}=\frac{y}{1+y}}$

=> x + xy = y + xy

=> x = y

When x and y are both negative, we have:

${\displaystyle f\left(x\right)=f\left(y\right)\Rightarrow \frac{x}{1-x}=\frac{y}{1-y}}$

=> x - xy = y - yx 

=> x = y

∴ f is one-one.

Now, let y ∈ R such that −1 < y < 1.

If x is negative, then there exists ${\displaystyle x=\frac{y}{1+y}\in R}$ such that

${\displaystyle f\left(x\right)=f\left(\frac{y}{1+y}\right)}$

${\displaystyle =\frac{\left(\frac{y}{1+y}\right)}{1+\left|\frac{y}{1+y}\right|}}$

${\displaystyle =\frac{\frac{y}{1+y}}{1+\frac{-y}{1+y}}}$

${\displaystyle =\frac{y}{1+y-y}}$

= y

If x is positive, then there exists ${\displaystyle x=\frac{y}{1-y}\in R}$ such that

${\displaystyle f\left(x\right)=f\left(\frac{y}{1-y}\right)=\frac{\frac{y}{1-y}}{1+\left|\left(\frac{y}{1-y}\right)\right|}}$

${\displaystyle =\frac{\frac{y}{1-y}}{1+\frac{y}{1-y}}}$

${\displaystyle =\frac{y}{1-y+y}}$

= y

∴ f is onto.

Hence, f is one-one and onto.