Q on associative binary operation

[BillyGOTO]

On Thu, Aug 28, 2003 at 11:21:08PM -0700, Sarad AV wrote:
> hi,
> 
> Let ~ represents a relation.
> 
> If a~b and b~a,then
> 
> a~a (by transitivity)
> is an incorrect argument.

> By definition of transitivity, if a~b and b~c implies
> that a~c.

right.

> I was asking on the same lines if (a*d)*d=a*(d*d)=d.

What does that have to do with transitivity?
You didn't mention transitivity when you posed the question.

Ridiculous.  '*' is an operator, not a relation.
Relations can't be parenthesized unless you're going to
make truth or falsehood a symbol to be operated upon.

Tim is right.  Cypherpunks isn't a place to look for help with your
algebra homework.  I like doing interesting math problems, but
you're not even properly asking the questions you want answered.
That makes it a LOT less fun.

> By definition associativity is defined on a,b,c
> element of set S and not two elements of the set.

This is getting stupid.  The '*' operator was defined associative.
The property of associativity applies to ASSOCIATIONS between symbols
(i.e binary operators).

> x*y (ie, left*top) can be followed.

I'm totally done with this.