Is the Set [1;+∞[ x [1;+∞[ with the Operation (x;y)°(v;w) a Group?

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Homework Help Overview

The discussion revolves around determining whether the set [1;+∞[ x [1;+∞[ with the operation (x;y)°(v;w) = (x+v-1; yw) forms a group. The original poster has already established closure, associativity, and identity but is struggling with the concept of invertibility.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find the inverse elements and questions whether the operation satisfies group properties, particularly noting that 1/y is not included in the set if y > 1. Other participants highlight additional concerns regarding the values of -x + 2 not being within the set.

Discussion Status

Participants are exploring the implications of the findings regarding invertibility. Some express uncertainty about the group's status, while others confirm that the set does not meet the criteria for a group based on the identified issues.

Contextual Notes

There is an indication that the homework prompt may have suggested the set could be a group, leading to some confusion among participants regarding the validity of their conclusions.

Alv95
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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:
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Alv95 said:

Homework Statement


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


Homework Equations


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


The Attempt at a Solution


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 :)

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.
 
So it's not a group? :) The text of the homework that the teacher gave us seemed to imply that it was :)
 
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Alv95 said:
So it's not a group? :)

Indeed: it is not a group!
 
Thanks ;)
 
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!
 
Alv95 said:
So it's not a group? :) The text of the homework that the teacher gave us seemed to imply that it was :)

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?
 
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.
 
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Alv95 said:
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.

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
Thanks, I will :)
 

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