# Invertibility / Group problem

1. Apr 16, 2012

### Alv95

I have to find if the set $[1;+∞[ x [1;+∞[$ with the operation $(x;y)°(v;w) = (x+v-1; yw)$ is a group

I have already proven Closure, associativity and Identity but I have some problems with invertibility :)

The neutral element that I have found is (1;1)

I did $(x;y)°(x1;y1)= (1;1)$ and I have found $x1=-x+2$ and $y1=1/y$
The problem is that $1/y$ is not included in the set if $y>1$...

Any advice? Is it a group? Thank you :)

Last edited: Apr 16, 2012
2. Apr 16, 2012

### LCKurtz

Use the forward slash "/" to close your tex tags.
It looks to me like you have answered your own question. Inverses don't seem to be there.

3. Apr 16, 2012

### Alv95

So it's not a group? :) The text of the homework that the teacher gave us seemed to imply that it was :)

Last edited: Apr 16, 2012
4. Apr 16, 2012

### micromass

Staff Emeritus
Indeed: it is not a group!

5. Apr 16, 2012

### Alv95

Thanks ;)

6. Apr 16, 2012

### micromass

Staff Emeritus
Just so you get it. The problem is that 1/y is not necessarily in our set. But that is not the only problem. There is a problem with -x+2 as well. That also doesn't necessarily lie in our set!! Indeed, if x=2, then -x+2=0 and this is not in our set!

7. Apr 16, 2012

### LCKurtz

C'mon. Show a little confidence in your work. Do you want to build a fence around it and insure it before you are willing to assert the result?

8. Apr 16, 2012

### Alv95

Well, I prefer to be sure about what I write especially if it is on a new topic that I have just learned at school ;) By seeking help and advices on the internet I hope to improve and strenghten my knowledge and thus be more confident. I didn't expect it to be a problem.

Last edited: Apr 16, 2012
9. Apr 16, 2012

### LCKurtz

It isn't a problem; I didn't mean to imply that it was. You nailed it on your first try. I was just trying to encourage you to be brave about what you've done. You do good work, so trust it.

10. Apr 17, 2012

### Alv95

Thanks, I will :)