# Bijections

• Sep 11th 2012, 08:19 AM
lovesmath
Bijections
Show that the function f(x)=x/(x^2-1) is a bijection between (-1,1) and the real numbers.

I was able to show that the function is one-to-one and y=x, but I can't figure out how to show that it is onto.

How can I also show that given any a and b, which are elements of the real numbers with a<b, there is a bijection between (a,b) and (-1,1).
• Sep 11th 2012, 08:56 AM
kalyanram
Re: Bijections
Quote:

Originally Posted by lovesmath
I was able to show that the function is one-to-one and y=x

This statement is not clear. what do you mean by y=x.

Quote:

Originally Posted by lovesmath
I can't figure out how to show that it is onto.

Use the definition that for every $\alpha \in \mathbb{R} \exists \beta \in (-1,1) \ni f(\beta) = \alpha$
Spoiler:
Consider
$\frac{x}{x^2-1} = \alpha \implies f(x) = \alpha x^2 - x -\alpha = 0$ when $\alpha = 0, x= 0$ $\alpha \neq 0$ we solve for $x$ we have $x = \frac{1 \pm \sqrt{1 + 4 \alpha^2}}{2\alpha}$. Now we are not sure as to which root to select.
Attachment 24769
Consider $\alpha > 0$ figure 1 shows thw plot of the quadratic where we need the roots in (-1,1) so this requires $f(-1)> 0 , f(1)> 0$. but we have f(1) = -1 and f(-1)=1. $\therefore$ we choose $\beta = \frac{1 - \sqrt{1 + 4 \alpha^2}}{2\alpha}$ $\ni \beta \in (-1,1) , f(\beta) > 0$ and $f(\beta) = \alpha$. Similarly you can argue from figure 2 that for $\alpha < 0$ choose $\beta = \frac{1 + \sqrt{1 + 4 \alpha^2}}{2\alpha}$. Hence it is onto.
• Sep 11th 2012, 09:11 AM
kalyanram
Re: Bijections
Quote:

Originally Posted by lovesmath
How can I also show that given any a and b, which are elements of the real numbers with a<b, there is a bijection between (a,b) and (-1,1).

Attachment 24770
Can you construct such a bijective map between $(a,b)$ and $(-1,1)$?
Spoiler:

Consider $f(c) = 2.\frac{c-a}{b-a}-1, \forall a
• Sep 11th 2012, 09:26 AM
HallsofIvy
Re: Bijections
Quote:

Originally Posted by lovesmath
Show that the function f(x)=x/(x^2-1) is a bijection between (-1,1) and the real numbers.

I was able to show that the function is one-to-one and y=x, but I can't figure out how to show that it is onto.

How can I also show that given any a and b, which are elements of the real numbers with a<b, there is a bijection between (a,b) and (-1,1).

To show "1 to 1" let y be any real number and show there exist at least one x such that x/(x^2- 1)=y. You can do that by solving for x. You will, of course, need to show that the value of x you get is between -1 and 1.