Can someone double-check this simple binary relation proof?

  • Thread starter Syrus
  • Start date
  • #1
214
0

Homework Statement




*attached



Homework Equations





The Attempt at a Solution




Let a,b ∈ H. Then (∀x ∈ S)(a*x = x*a) and (∀x ∈ S)(b*x = x*b). It is easy to see, then, that a*b = b*a. Now let c ∈ S. Then (a*b)*c = c*(a*b) by the associativity of *.

Q.E.D.
 

Attachments

Answers and Replies

  • #2
22,089
3,293

Homework Statement




*attached



Homework Equations





The Attempt at a Solution




Let a,b ∈ H. Then (∀x ∈ S)(a*x = x*a) and (∀x ∈ S)(b*x = x*b). It is easy to see, then, that a*b = b*a. Now let c ∈ S. Then (a*b)*c = c*(a*b) by the associativity of *.

Q.E.D.
The last statement does not follow from associativity. Associativity merely says that

[tex](a*b)*c=a*(b*c)[/tex]

I also see no reason to take c in S. You need to prove that a*b in S. That is, you need to show for all c that (a*b)*c=c*(a*b). You don't only need to prove it for the c in S.
 
  • #3
214
0
Hey micromass. In order to show that a*b is an element of H, isn't the statement to prove: (for all c in S)((a*b)*c = c*(a*b))? Am i mistaken here?

*By the way i am taking c in S to be an arbitrary element
 
  • #4
22,089
3,293
Hey micromass. In order to show that a*b is an element of H, isn't the statement to prove: (for all c in S)((a*b)*c = c*(a*b))? Am i mistaken here?

*By the way i am taking c in S to be an arbitrary element
Oops, I had my notation mixed up :frown: Yes, that is correct!
 
  • #5
214
0
Heh. So wait... is the proof correct?
 
  • #6
22,089
3,293
Heh. So wait... is the proof correct?
You still need to explain why (a*b)*c = c*(a*b). It doesn't follow from associativity.
 
  • #7
214
0
Can i say that since a,b are elements of S (since H is a subset of S), then a*b is in S. But also, since c is in S, then (a*b)*c is in S. and since * is associative on S, a*(b*c) is in S?
 
  • #8
22,089
3,293
How does thart prove (a*b)*c=c*(a*b)??

You need to use that a and b are in H.
 
  • #9
214
0
ahhh, finally Figured it out micromass! I wasn't using the fact that a and b were in H to commute the result. That is, since (a*b)*c = a*(b*c), for some unknown reason I thought this was the goal. But now we may use that a is in H to write (b*c)*a and continue to use the fact that a,b are in H along with associativity to obtain our result.

I appreciate your help
 

Related Threads on Can someone double-check this simple binary relation proof?

  • Last Post
Replies
3
Views
1K
Replies
6
Views
3K
Replies
2
Views
409
Replies
0
Views
1K
  • Last Post
Replies
17
Views
2K
Replies
18
Views
2K
Replies
1
Views
845
Replies
9
Views
933
Top