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- bob bundy
- 2013-02-05 05:22:34
That question has already been answered by Fistfiz in post 3.
Bob
- Still Learning
- 2013-02-05 04:23:57
Ok,but i want to know if the operation has to be always associative or it has to be associative only for the set?
- bob bundy
- 2013-02-04 22:43:16
hi Still Learning
You have to show it obeys the four properties of a group: closure, identity, inverses, asociativity.
I've made a group combination table (see below).
From that it is obvious that closure holds, it has an identity (o) and all members are self inverse.
So what about associativity ? This is often the hardest to prove. You have to show that
a(bc) = (ab)c for all a b and c in the set.
As Stefy has pointed out, commutativity holds (ab = ba) so it is fairly easy to cover all cases by using that property.
I'll use * for a tilda as I cannot see that symbol above, and show one example:
0*(0*n) = 0* n = n
(0*0)*n = 0 * n = n
I'll leave the rest to you.
Bob
- Fistfiz
- 2013-02-04 22:05:13
Still Learning wrote:
Yes,but ~ is associative when you only use 0 and n,does that count?
Ok, I thought that {0,n} was {0,1,...,n}
- anonimnystefy
- 2013-02-04 21:17:00
Yes, that is a group. It is even an Abel's group.
- Still Learning
- 2013-02-04 21:00:14
Yes,but ~ is associative when you only use 0 and n,does that count?
- Fistfiz
- 2013-02-04 20:31:12
~ is not associative: (x~y)~z!=x~(y~z). For example, take x=5, y=3, z=1. You have:
(5~3)~1 = ||5-3|-1| = 1 != 3 = |5-|3-1|| = 5~(3~1)
so ({0,n} ,~) is not a group, if I understood what you meant.
- Still Learning
- 2013-02-04 20:26:21
Sorry there will be "= |" in place of the smiley
- Still Learning
- 2013-02-04 20:20:52
Is ({0,n} ,~) a group where x~y
x-y| and n is any postive real number?