Simple Verification of a Group

  • Thread starter Bashyboy
  • Start date
  • #1
1,421
5

Homework Statement


Determine whether the set of rational numbers with denominator equal to 1 or 2 is a group under addition.

Homework Equations




The Attempt at a Solution



Please have a look at the closure proof of part 5. I don't quite understand how ##q/2## implies that the denominator of ##p/q## will be 1 or 2, so I tried to conceive of my own.

Either ##2a+d## is even or is odd. If even, then ##\frac{2a+d}{2} = \frac{2k}{2} = \frac{k}{1}## which is therefore in reduced form and puts it in ##G##. Now, if odd, then the numerator and denominator of ##\frac{2a+d}{2}## have no common factors and is therefore already in reduced form. Because the denominator is ##2##, it is by definition in ##G##.

Is there anything wrong with this proof? Although I gave my own, I would like to understand the linked solution.
 

Answers and Replies

  • #2
andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
3,886
1,454
It looks mostly fine. But by using ##2a+d## you have not covered the case where both numerators are odd. You need to cover that case as well to complete the proof.
 

Related Threads on Simple Verification of a Group

  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
9
Views
2K
Replies
1
Views
896
  • Last Post
Replies
6
Views
974
  • Last Post
Replies
1
Views
902
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
2K
Replies
2
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
1K
Top