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

In summary, proving or disproving the existence of a rational bijective function from R to (0,1) involves finding a function with a polynomial denominator of even degree, a polynomial numerator of odd degree, and one end approaching 1 while the other approaches 0. This is impossible as it would result in a non-injective function.
  • #1
hangainlover
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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...
 
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  • #2
I don't really know the answer to this, but if you consider a function such as:

[itex]f \left( x \right) = \frac{x^2}{x^2+1}[/itex]

The range is [itex]\left[ 0, 1 \right)[/itex], but it is not bijective.
 
  • #3
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##.
 
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  • #4
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.
 
  • #5
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.
 
  • #6
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
 
  • #7
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?
 
  • #8
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.
 

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

1. What is a rational bijective function?

A rational bijective function is a mathematical function that maps one set of numbers to another, where both sets are rational numbers (numbers that can be expressed as a ratio of two integers). In simpler terms, it is a function that takes in a rational number and outputs another rational number, with no two input numbers having the same output number.

2. How do you prove the existence of a rational bijective function from R to (0; 1)?

To prove the existence of a rational bijective function from R to (0; 1), we need to show that there is a one-to-one correspondence between the two sets. This means that for every rational number in R, there is a unique rational number in (0; 1) that it maps to, and vice versa. We can do this by defining a function f(x) = 1/(1+x), which maps any real number to a number between 0 and 1. This function is both rational and bijective, proving the existence of a rational bijective function from R to (0; 1).

3. Can you provide an example of a rational bijective function from R to (0; 1)?

As mentioned in the previous answer, f(x) = 1/(1+x) is a rational bijective function from R to (0; 1). For example, if we input the rational number 3, the function will output the rational number 1/4, which is also in the set (0; 1).

4. How does a rational bijective function differ from a regular bijective function?

A rational bijective function differs from a regular bijective function in the type of numbers it works with. A regular bijective function can work with any type of number, such as real, irrational, or complex numbers. However, a rational bijective function specifically works with rational numbers, which are numbers that can be expressed as a ratio of two integers.

5. Is there a way to disprove the existence of a rational bijective function from R to (0; 1)?

Yes, it is possible to disprove the existence of a rational bijective function from R to (0; 1). One way to do this is by showing that there is no one-to-one correspondence between the two sets. This can be done by finding two different rational numbers in R that map to the same number in (0; 1) or by showing that there is a rational number in (0; 1) that does not have a corresponding rational number in R.

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