# Find a bijection f : R --> ( R - {0})

• Jul 4th 2011, 06:51 PM
ikurwae89
Find a bijection f : R --> ( R - {0})
I have no idea what the question is about. I know what bijection means but have no idea how to answer it.
• Jul 4th 2011, 07:22 PM
FernandoRevilla
Re: Find a bijection f : R --> ( R - {0})
For example $f(x)=\begin{Bmatrix} x+1 & \mbox{ if }& x\in \mathbb{N}=\{0,1,2,\ldots\}\\x & \mbox{if}& x\not\in\mathbb{N}\end{matrix}$
• Jul 4th 2011, 07:31 PM
Re: Find a bijection f : R --> ( R - {0})
Quote:

Originally Posted by ikurwae89
I have no idea what the question is about. I know what bijection means but have no idea how to answer it.

Well, you know the R is infinite, so you don't have to worry about upper and lower bounds for the space you are working with. All you need is a function which is 1-1 and onto. (by definition)

In this case, you need to biject the real numbers to the real number set without 0. so each number in R (your domain) must have an unique image (your range) which is not 0.

Finding a function does take some creativity, so verify this suggestion to see if it matches the criteria.

Let f(x) be a function such that

f(x) = x + 1 if x is 0, 1, 2, ...
f(x) = x otherwise.

EDIT:// haha too late. we got the same function too. XD
• Jul 4th 2011, 08:33 PM
ikurwae89
Re: Find a bijection f : R --> ( R - {0})
I understand.. so f(x) = x^2 and x^2+1/2 is also doable right ?

What about f--> R to I (irrational real numbers)

???

this one requires only 1 equation right ??
• Jul 4th 2011, 08:37 PM