Prove or disprove that there is a rational bijective function f : R to (0; 1)

[hangainlover]

Homework Statement

Prove or disprove that there is a rational bijective function f : R to (0; 1)

Homework Equations

i found a bijective map from (0,1) to R (y=(2x-1)/(2x^2-2x)

The Attempt at a Solution

Im just stuck and i was thinking since it has to be a rational function, denominaotor should be defined on all R ... so nothing funky in the denominator..
Somehow i have to play with the numerator to limit the map to (0,1) which i think is impossible...

 

[Chaos2009]

I don't really know the answer to this, but if you consider a function such as:

f \left( x \right) = \frac{x^2}{x^2+1}

The range is \left[ 0, 1 \right), but it is not bijective.

 

[morphism]

A rational function is the ratio of two polynomials. You observed if there's anything "funky" in the denominator, then certainly R won't get mapped to (0,1). Now just think about what "funky" really means.. Remember, the denominator is a polynomial.

[STRIKE]Edit: Sorry--that's not really helpful.[/STRIKE] Actually, I think it can be made to work. One thing the denominator can't be is a polynomial of odd degree (why?). Keeping this in mind, consider the derivative of the rational function, and try to think about turning points. You will also need to think about the degree of the numerator.

Edit2: There's a much easier approach: consider limits as $x\to\pm\infty$.

 

[hangainlover]

What i mean by funky is that the denominator should be defined on the enitre R.(for example, x-2 won't be defined at x=0, we don't want the denominator to have any roots in R)
The function that I am trying to find has to be continuous so that as x gets sent to infinity, it has to go to either 0 or 1

I don't get why you said the denominator cannot be of odd degree. odd degree polynomial means one end is up and the other end is down.

 

[morphism]

A polynomial of odd degree always has a real root, and the denominator can't have a root if we want the rational function to be defined on all of R.

 

[hangainlover]

i have plugged in some random rational functions with the denominator beign a polynomial of degree 2. I can get it to be restricted between 0 and 1 but it is not injective.
I don't think you can get it to be injective.
In order for one end to go up (approaching 1) and for the other to go near 0, the numerator has to be odd. But then f will be 0 as x goes to -,+ infinity

 

[morphism]

Sounds like you've pretty much got it. Care to summarize your argument in one post, so we can make sure there are no gaps?

 

[hangainlover]

The reason why i think it is impossible to find a bijective function going from R to (0,1) is
1, the denominator has to be a polynomial of even
2, one end of the function has to approach 1 asymptotically and the other end has to approach 0 and you can only achieve something like this and this means the numerator has to be a polynomial of odd
3, in order to restrict it in the interval (0,1) , the degree of my numerator cannot exceed the degree of the numerator
4, if it is the same degree (even), then it will both approach positive 1 and will not be injective
5, if the degree of my numerator is less than that of the denominator, it will approach 0 both ways and will not be injective.