, 2003 2:36 PM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: Re: Q on associative binary operation
Yeah, kinda bizarre.
There's also an ambiguity that prevents one from saying Q is
associative. Is
the table defined for both directions of *? In other words, is the table
meant to imply value
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 tw
l y*x (indeed, one may be undefined as in
the case of matrix multiplication). Anyone remember the group theoretic term
for these kinds of groups?
-TD
From: Tim May <[EMAIL PROTECTED]>
To: [EMAIL PROTECTED]
Subject: Re: Q on associative binary operation Date: Thu, 28 Aug 2003
10:41:51 -0700
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 conside
hi,
how do we complete this table
Table shown may be completed to define 'associative'
binary operation * on S={a,b,c,d}. Assume this is
possible and compute the missing entries
*|a|b|c|d
-
a|a|b|c|d
-
b|b|a|c|d
-
c|c|d|c|d
-
d| | | |
Its clear for commutativit
