## Abstract Algebra - showing commutativity and associativity

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)
?????

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 Mentor What are you definitions of associativity and commutivity?
 commutative is iff a*b=b*a for all a and b. associative is iff (a*b)*c=a*(b*c)

Mentor

## Abstract Algebra - showing commutativity and associativity

 Quote by kathrynag commutative is iff a*b=b*a for all a and b. associative is iff (a*b)*c=a*(b*c)
Ok, so have you tried to use your operation for * in these expressions? If so, what did you get?

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

Mentor
 Quote by kathrynag That's where I get confused in using the operation. a*b=ab+1 so, b*a=ba+1
Yes, but why? You should state "because multiplication of real numbers in commutative..."

 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.
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.