# Approve that $f = \frac{x}{1 - x^2}$ is an injective function

Assume that $$f: R \setminus \{-1,1\} \to R$$ and $$f = \frac{x}{1-x^2}$$. Approve that $$f$$ is an injective function.

My solution:

Based on the theory: for each $$x,y \in R \setminus \{-1,1 \}$$ if $$f(x) = f(y)$$ then $$x=y$$

$$x - xy^2 = y -yx^2 \Leftrightarrow x - y = xy^2 - x^2y \Leftrightarrow x-y = xy(y^2 - x^2) \Leftrightarrow x-y = xy(y-x)(y+x) \Leftrightarrow (y-x)[xy(y+x) +1 ) = 0$$

Two cases:

1) $$x = y$$

or

2) $$(xy(y+x) +1 ) = 0$$

Edit:

$$x - xy^2 = y -yx^2 \Leftrightarrow x - y = xy^2 - x^2y \Leftrightarrow x-y = xy(y - x)$$

Two cases:

1) $$x = y$$

or

2) $$xy = -1$$

We reject the second. So, f is injective function

My question:

1) Can we reject the second case? Please explain!

No, we can not reject the second case !

• $xy^2-x^2y = xy(y-x)$ no squares on the right hand side... – Yanko Jan 12 at 12:50
• Thank you for your comment – Dimitris Dimitriadis Jan 12 at 13:10

Going by the very definition, as you did:

$$f(x)=f(y)\iff\frac x{1-x^2}=\frac y{1-y^2}\iff x-xy^2=y-x^2y\iff$$

$$\iff(x-y)=-xy(x-y)\iff\begin{cases}x=y\\or\\xy=-1\end{cases}$$

Thus, any pair of numbers $$\;x,\,y\in\Bbb R\setminus\{-1,1\}\;$$ s.t. $$\;xy=-1\;$$ give you a counterexample to injectivity. For example

$$x=-2,\,y=\frac12\;,\;\text{and certainly:}\;\;f(-2)=\frac{-2}{1-4}=\frac23=\frac{\frac12}{1-\frac14} =f\left(\frac12\right)$$

and etc.

It is not an injective function as we can see from horizontal line test.

• It equals for $x,y$ such that $x\cdot y = -1$. – Yanko Jan 12 at 12:52

There's an error in your computation:

$$x - y = xy^2 - x^2y \iff x-y = xy\color{red}{(y - x)}\iff\begin{cases}x=y \\ xy=-1\iff y=-\frac1x, \; x\ne 0 \end{cases}$$

• I edited. Thank you – Dimitris Dimitriadis Jan 12 at 13:10
• You can not say that $y= \frac{1}{x}$ because $x$ can be $0$ – Dimitris Dimitriadis Jan 12 at 13:12