I'm struggling with the following homework assignment.
Let $X = \{x \in \mathbb{R}: -1 < x < 1\}$. We define $g: X \rightarrow \mathbb{R}$ by $g(x) = \frac{x}{1-|x|}$.
We must prove that $g$ is a bijection.
First, we prove $g$ is injective.
By definition of injection, $g$ is injective iff whenever $x_1,x_2 \in X$ and $g(x_1) = g(x_2)$, then $x_1 = x_2$.
Let $x_1,x_2 \in X$ and $g(x_1) = g(x_2)$. Want to show $x_1 = x_2$.
We have $\frac{x_1}{1-|x_1|} = \frac{x_2}{1-|x_2|} \Leftrightarrow x_1(1-|x_2|) = x_2(1-|x_1|)$
$\Leftrightarrow x_1-x_1|x_2|=x_2-|x_1|x_2$.
For cases $x_1,x_2 > 0$ and $x_1,x_2 < 0$ we get $x_1 = x_2$
But for the remaining two cases, how should I proceed?
Thanks in advance!
Best Answer
$\frac x{1-|x|}$ always has the same sign as $x$ when $x\in X$ (since the denominator is positive), so if $x_1$ and $x_2$ have different signs, then $g(x_1)$ and $g(x_2)$ also have different signs -- in particular they can't be equal.