Q on associative binary operation
Sarad AV
jtrjtrjtr2001 at yahoo.com
Thu Aug 28 23:21:08 PDT 2003
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 at 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?
>
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