Thread: 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).

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.

Originally Posted by lovesmath

I can't figure out how to show that it is onto.

Use the definition that for every

Spoiler:

Consider
when we solve for we have . Now we are not sure as to which root to select.

Consider figure 1 shows thw plot of the quadratic where we need the roots in (-1,1) so this requires . but we have f(1) = -1 and f(-1)=1. we choose and . Similarly you can argue from figure 2 that for choose . Hence it is onto.

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).

Can you construct such a bijective map between and ?

Spoiler:

Consider

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.