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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?

Community
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usereb
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2 Answers2

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

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

Guy Fsone
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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.

Netchaiev
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  • The question only asks for injectivity! – jonsno Jan 15 '18 at 11:16
  • Yes, for the question of esarabasar, but within the question : "Show that f is (1-1)" (one-to-one? so that we have bijectivity too?) – Netchaiev Jan 15 '18 at 11:19
  • @Netchaiev A one-to-one function $f$ is by definition injective. See https://en.wikipedia.org/wiki/One-to-one – user236182 Jan 15 '18 at 11:20
  • @user236182 : ok! So one-to-one function = injective ; one-to-one correspondance = bijective function. I edited my answer .... (but here, the function of interest still bijective :p ) – Netchaiev Jan 15 '18 at 11:40