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. I was asking on the same lines if (a*d)*d=a*(d*d)=d. By definition associativity is defined on a,b,c element of set S and not two elements of the set. x*y (ie, left*top) can be followed. Regards Sarath. --- BillyGOTO <billy@dadadada.net> wrote:
On Thu, Aug 28, 2003 at 12:14:20AM -0700, Sarad AV wrote:
hi,
Table shown is completed to define 'associative' binary operation * on S={a,b,c,d}.
*|a|b|c|d --------- a|a|b|c|d --------- b|b|a|c|d --------- c|c|d|c|d --------- d|d|c|c|d
The operation * is associative iff (a*b)*c=a*(b*c) for all a,b,c element of set S.
So can (a*d)*d=a*(d*d)=d considered as associative over * for this case as per definition?
a d d d \ / \ / d d a d \ / \ / d = d
What's the problem?
__________________________________ Do you Yahoo!? Yahoo! SiteBuilder - Free, easy-to-use web site design software http://sitebuilder.yahoo.com