Injectivity of a function $\frac{x}{1+|x|}$

Question

Let $f: \mathbb R \to (-1,1)$

$x \in $ $\mathbb R$, $f(x) = \frac{x}{1+|x|} $

Show that f is (1-1)

What I did : $x,y \in $ $\mathbb R$

$f(x)=f(y)$ iff $$\frac{x}{1+|x|} =\frac{y}{1+|y|}\Rightarrow\left| \frac{x}{1+|x|} \right| =\left| \frac{y}{1+|y|} \right| \\\Rightarrow \frac{|x|}{1+|x|} =\frac{|y|}{1+|y|}\Rightarrow |x|=|y|$$ $\Rightarrow$ $x=y$ or $x=-y$

If $x=-y,\frac{y}{1+|y|} =\frac{-y}{1+|y|}\Rightarrow y=-y$

It is only valid for unique real value of $y$ that is $0$ namely $f(x) \neq f(y)$ for other values of $y$. Thus $x=y$

I think it is not sufficient for injectivity, can someone correct me please?

Answer 1

You obtained that $|x|=|y|$ then replacing this in the first equality you get

$$\frac{x}{1+|x|} =\frac{y}{1+|y|} \Longleftrightarrow\frac{x}{1+|x|} =\frac{y}{1+|x|}\Longleftrightarrow x=y$$

Answer 2

Your proof is fine, maybe you just have to write it a bit differently at the end : $$x=y \quad \text{or} \quad x=-y$$ If $x=-y$, then, as you shown, $y=-y$ which implies $y=0$, which implies on its turn $x=-y=0=y$. If $y\neq 0$, $x=y$.

So, in ALL CASES, you have $x=y$. The injectivity is well-proven.