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**Still Learning****Guest**

Is ({0,n} ,~) a group where x~y=|x-y| and n is any postive real number?

**Still Learning****Guest**

Sorry there will be "= |" in place of the smiley

**Fistfiz****Member**- Registered: 2012-07-20
- Posts: 33

~ 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.

30+2=28 (Mom's identity)

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**Still Learning****Guest**

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

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,919

Yes, that is a group. It is even an Abel's group.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**Fistfiz****Member**- Registered: 2012-07-20
- Posts: 33

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}

30+2=28 (Mom's identity)

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,914

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Still Learning****Guest**

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****Moderator**- Registered: 2010-06-20
- Posts: 6,914

That question has already been answered by Fistfiz in post 3.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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