1. ## Bijection

Show that $\displaystyle f:\mathbb R\to(-1,1)$ where $\displaystyle f(x)=\dfrac x{1+|x|}$ is a bijection.

Wonder if there's a way using not so advanced analysis tools to prove this, and not to do it by the old way by proving first that is injective and surjective.

2. Originally Posted by Connected
Show that $\displaystyle f:\mathbb R\to(-1,1)$ where $\displaystyle f(x)=\dfrac x{1+|x|}$ is a bijection.
Wonder if there's a way using not so advanced analysis tools to prove this, and not to do it by the old way by proving first that is injective and surjective.
And what is wrong with that?

3. Haha, sorry, it's not wrong actually.

Well I'm having problems to prove surjectivity, can you help me with that?

4. Originally Posted by Connected
Haha, sorry, it's not wrong actually.

Well I'm having problems to prove surjectivity, can you help me with that?

We can write f in other form.

h:R-->(-1,1)

h(x)={x/(1+x) for x>0; 0 for x=0; x/(1-x) for x<0}

f=h for all x in R.

5. Originally Posted by Connected
Haha, sorry, it's not wrong actually.

Well I'm having problems to prove surjectivity, can you help me with that?
What AsZ said is simpler, but here's another way to think of it.

Its limit at infinity is 1 and its limit at -infinity is -1. It's continuous on a connected set, so its image is connected, i.e. an interval. So its image contains (-1, 1).

6. Originally Posted by Tinyboss
What AsZ said is simpler, but here's another way to think of it.

Its limit at infinity is 1 and its limit at -infinity is -1. It's continuous on a connected set, so its image is connected, i.e. an interval. So its image contains (-1, 1).
And to prove that f is 1-1 you need to use Rolle's theorem.