# Abstract Algebra - showing commutativity and associativity

1. Feb 11, 2009

### kathrynag

1. The problem statement, all variables and given/known data

I need to determine if a*b=ab+1 is commutative and associative.
* is any arbitrary operation

2. Relevant equations

3. The attempt at a solution
a*b=ab+1 is commutative, but not associative.
I'm getting stuck in showing why.
b*a=ba+1
a*(b*c)=a(b+1)
?????

2. Feb 11, 2009

### cristo

Staff Emeritus
What are you definitions of associativity and commutivity?

3. Feb 11, 2009

### kathrynag

commutative is iff a*b=b*a for all a and b.
associative is iff (a*b)*c=a*(b*c)

4. Feb 11, 2009

### cristo

Staff Emeritus
Ok, so have you tried to use your operation for * in these expressions? If so, what did you get?

5. Feb 11, 2009

### kathrynag

That's where I get confused in using the operation.
a*b=ab+1
so, b*a=ba+1

a*b=ab+1
so (a*b)*c=a*(b*c)
ab+1*c=a*(bc+1)
(ab+1)c+1=a(bc+1)+1
abc+c+1=abc+a+1
So, not associative.

Is this the right idea?

6. Feb 11, 2009

### cristo

Staff Emeritus
Yes, but why? You should state "because multiplication of real numbers in commutative..."

This is a very sloppy way of doing things, since the equals signs don't hold: i.e. the thing on the left is not equal to the thing on the right.

Better would be to set it up something like this:

* is associative if(f) (a*b)*c=a*(b*c). Then,

LHS: (a*b)*c=(ab+1)*c=(ab+1)c+1=abc+c+1 ..(#)
RHS: a*(b*c)=a*(bc+1)=a(bc+1)+1=abc+a+1 ..(§)

Since (#)$\neq$(§), then * is not associative.