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

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- Jul 4th 2011, 06:51 PMikurwae89Find 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 PMFernandoRevillaRe: Find a bijection f : R --> ( R - {0})
For example $\displaystyle f(x)=\begin{Bmatrix} x+1 & \mbox{ if }& x\in \mathbb{N}=\{0,1,2,\ldots\}\\x & \mbox{if}& x\not\in\mathbb{N}\end{matrix}$

- Jul 4th 2011, 07:31 PMMacstersUndeadRe: Find a bijection f : R --> ( R - {0})
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 PMikurwae89Re: 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 PMMacstersUndeadRe: Find a bijection f : R --> ( R - {0})
EDIT://

No. the f(x) you've given is not onto, and not 1-1. (can you see why? hint: try plugging x = -1 and x= 1, for example. then notice the range has no negative elements.)

Every question asking for a bijection only requires one such function to be given, unless stated otherwise.

Try the question you've posed on your own, keeping in mind that you need to consider all the elements in your domain and range and that images have to be unique.