Homework Help: 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

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