Bob

]]>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

]]>Yes,but ~ is associative when you only use 0 and n,does that count?

Ok, I thought that {0,n} was {0,1,...,n}

]]>(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.

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