let
$g(x) = \frac{x}{1-|x|}$ for all $x \in (-1,1)$
a) show that $g$ is a bijection from $(-1,1)$ to $\mathbb{R}$
Here is what I have done.
Say $g(x)=y$ so $$y = \frac{x}{1-|x|}$$
iff $$y(1-|x|)=x$$
Therefore, for each y, there exists an x such that $f(x)=y). So $f$ is a surjection.
Suppose $g(x)= g(z)$ for some $x,z \in (-1,1)$, then
$$\frac{x}{1-|x|}=\frac{z}{1-|z|}$$
iff
$$x-z=x|z|-z|x|$$
Then there are four cases, one $x,z>0$ and $x,z<0$ where obviously $x=z$, but for the other two cases, I am only concerned with figuring out one, because obviously the other one follows, but I have gotten to the point where
$$x-z=x(2z)$$ for $x>0$, $y<0$. Am I correct thus far? Where do I go next? Thank you for any help.
Answer
For the injection part, is $\dfrac{x}{1-|x|}=\dfrac{z}{1-|z|}$ really possible if $x$ and $z$ have different signs ? One of the two terms is $\geq0$ and the other $\leq 0$.
For the surjection part, I would advise you to separate the cases where $y$ is positive, and $y$ is negative.
If $y$ is positive then $x$ must be positive, so $|x| = x$. You have $y = \dfrac{x}{1-x}$, so $y(1-x)-x = 0$ thus $-x(1+y)+y = 0$ and $x = \dfrac{y}{1+y}$. You have know found a $x$ that works. Note that the final step is legit because as $y$ is positive, $(1+y)$ can't be $0$.
Don't forget the case $y < 0$ now !
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