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I want to show that this function is injective -

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

So when $f(x) = f(y)$ I should have $x = y$

$\frac{y}{1 - y^2} = \frac{x}{1 - x^2}$

$x - xy^2 = y - x^2y$

$x - y = xy^2 - x^2y$

$x - y = xy(y - x)$

$\frac{x-y}{y-x} = xy$

$\frac{x-y}{x-y} = -xy$

$-xy = 1$

$x = -\frac{1}{y}$

Where am I going wrong?

sonicboom
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  • Where is it defined? where do you want to prove this to be injective? –  Mar 31 '14 at 10:58

2 Answers2

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In your computations you arrive at $$x - y = xy(y - x);$$ Now, if $y\neq x$, then you can write $$\frac{x-y}{y-x}=xy ~~(*) $$ arriving at $x=-\frac{1}{y}$ as the l.h.s. of $(*)$ is well defined. This is the solution you found. The other case, i.e. $x=y$ is the solution that was apparently missing in the computations you showed.

In any case, this is a partial answer, as we need the domain of definition of $f$ to check injectivity (here we considered $\mathbb R-\{\pm 1\}$).

Avitus
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$x - y = xy(y - x)$

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

$x=y$ or $x=-1/y$.

So the function is not injective.

user85798
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