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bob bundy
2013-02-05 05:22:34

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~yx-y| and n is any postive real number?