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 values for both x*y (ie, left*top) as well as y*x (top*left)? For most objects x*y will not equal 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
On Thursday, August 28, 2003, at 12:14 AM, 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?
This looks like a homework assignment for a class.
If a, b, c, and d are all arbitrary symbols, most substitutions (such as a = 2, b = 5, etc.) certainly will _not_ give (a*d)*d=a*(d*d)=d as a true statement. Only special values of a and d will make that true. (For example, a = 1 and d = 1 makes (1*1)*1=1*(1*1)=1. Other values may as well. Finding them is up to you.
Lastly, your English is again unclear. "So can (a*d)*d=a*(d*d)=d considered as associative over * for this case as per definition?" isn't a proper English sentence.
Why do you keep posing these problems to the list? Are they homework problems? Do you think the list is just too quiet and needs ill=phrased puzzlers to keep it occupied?
Did you ever do the coin flip experiment we suggested you do?
Are you an AI which has failed the Turing Test and escaped?
--Tim May
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