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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!

Answer of the Community

No, we can not reject the second case !

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

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

enter image description here

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  • $\begingroup$ It equals for $x,y$ such that $x\cdot y = -1$. $\endgroup$ – Yanko Jan 12 at 12:52
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There's an error in your computation:

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

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

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