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## #1 2013-02-04 20:20:52

Still Learning
Guest

### Is this a group ?

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

## #2 2013-02-04 20:26:21

Still Learning
Guest

### Re: Is this a group ?

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

## #3 2013-02-04 20:31:12

Fistfiz
Member

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### Re: Is this a group ?

~ 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)

## #4 2013-02-04 21:00:14

Still Learning
Guest

### Re: Is this a group ?

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

## #5 2013-02-04 21:17:00

anonimnystefy
Real Member

Online

### Re: Is this a group ?

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

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #6 2013-02-04 22:05:13

Fistfiz
Member

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### Re: Is this a group ?

#### 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)

## #7 2013-02-04 22:43:16

bob bundy
Moderator

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### Re: Is this a group ?

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

## #8 2013-02-05 04:23:57

Still Learning
Guest

### Re: Is this a group ?

Ok,but i want to know if the operation has to be always associative or it has to be associative only for the set?

## #9 2013-02-05 05:22:34

bob bundy
Moderator

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### Re: Is this a group ?

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