Bijection

• Mar 26th 2011, 05:17 PM
Connected
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.
• Mar 26th 2011, 05:29 PM
Plato
Quote:

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?
• Mar 26th 2011, 05:33 PM
Connected
Haha, sorry, it's not wrong actually.

Well I'm having problems to prove surjectivity, can you help me with that?
• Mar 26th 2011, 07:58 PM
Also sprach Zarathustra
Quote:

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.
• Mar 26th 2011, 09:16 PM
Tinyboss
Quote:

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).
• Mar 27th 2011, 11:01 AM
Also sprach Zarathustra
Quote:

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.