# Find a Bijection $$\left[ 0,1 \right] \rightarrow \Re$$

1. Jan 4, 2009

### rosh300

1. The problem statement, all variables and given/known data

Find a Bijection $$\left( 0,1 \right) \rightarrow$$ R

2. Relevant equations
Detention of bijection

3. The attempt at a solution

let r be a number in the interval $$\left[ 0,1 \right]$$

r = a b1c1b2c2 ... bncn where a, b,c are digits between [0, 9]

f(r) =
{ - b1b2b3 ...... bn . c1 c2 ..... cn if a<5
{+ b1b2b3 ...... bn . c1 c2 ..... cn if a>=5

i know its incorrect as it is not injective. any help in trying to get the plus minus part.

2. Jan 4, 2009

### HallsofIvy

Staff Emeritus
Another problem with your function is that not every number in [0, 1] can be written that way!

You need a one-to-one function that maps 0 into negative infinity and 1 into positive infinity.

You might consider something like 1/(x-1)- 1/x.

3. Jan 4, 2009

### rosh300

thank u for ur reply and understand how it will work.

would : f(x) = 1/(2x-1) work

as if x < 1/2 it will be -ve and if x > 1/2 it will be +ve
as x --> 1/2 form 0 it will go to -ve infinity
as x --> 1/2 form 1 it will go to +ve infinity

or f(x) = arctanh(2x - 1)

which will transform the arctanh(x) from the (-1, 1) interval to (0, 1) interval as required, and is a 1 to 1 and as you said a function that maps 0 into negative infinity and 1 into positive infinity is needed

Last edited: Jan 4, 2009
4. Jan 4, 2009

### Dick

1/(2x-1) isn't onto. For example, it's never 0. The second example works fine.

5. Jan 4, 2009

### rosh300

just for completeness the second part of the question was to: Find a Bijection [a,b] --> R where a, b are real numbers and b > a

this can easily be done by modifying the second example as shown:
let g(x) = x /(b-a) (to transform it into the inteval (0,1) )

and f(x) = arctanh(2x - 1) (the first example)

and have the final function as: f o g(x) ( f(g(x)) )

Last edited: Jan 4, 2009