January 14th, 2013, 05:43 PM  #1 
Newbie Joined: Jan 2013 From: west bengal,india Posts: 22 Thanks: 0  Binary composion
The composition which satisfied the closure prop is called the binary copposition. But if there exit any composition which does not satisfied the closure prop but it satisfied all other prop? If there exit any prop of such type then by which keyword we can call them? 
January 14th, 2013, 09:20 PM  #2 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88  Re: Binary composion
the usual term is "binary operation". by definition, a binary operation on S is a function SxS>S for a function SxS>T to land "outside of S" means we no longer have "closure". there are LOTS of functions like this: for example, suppose f(a,b) is the "average of a and b", that is: f(a,b) = (a+b)/2. if a, and b are integers, their average need not be: f(3,4) = 7/2 is not an integer. 
January 15th, 2013, 12:22 AM  #3 
Newbie Joined: Jan 2013 From: west bengal,india Posts: 22 Thanks: 0  Re: Binary composion
But f(a,b)=(a+b)/2 does not satisfied the associative prop. Then how it will be example of such oparation which satisfied all prop except closure of a commutative group(Like the property associative,inverse,identity commutative etc)?

January 15th, 2013, 04:20 AM  #4  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Binary composion Quote:
Take the average of a,b and c : f(a,b,c) = (a+(b+c))/3 = ((a+b)+c)/3  
January 15th, 2013, 12:14 PM  #5 
Senior Member Joined: Nov 2011 Posts: 595 Thanks: 16  Re: Binary composion
Yes you are right this law is not associative and can not define a group.

January 17th, 2013, 02:26 AM  #6 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88  Re: Binary composion
there is going to be a problem defining associativity for an operation that does NOT satisfy closure: suppose *:SxS>T for associativity to even make sense we need two MORE functions: *':SxT>U *":TxS>U and these functions are going to have to agree with * on S?T x S?T. when we say: (a*b)*c = a*(b*c) we are already assuming a*b is the same TYPE of object as c, and that b*c is the same TYPE of object as a. since a,b,c are allowed to be arbitrary elements of some set S. associativity, then, is a property of binary operations on S (the two inputs, and the single output are of the same type: that is, belong to the same set). to talk of "associativity" outside of this context is to talk about very strange objects, indeed. but....it CAN be done. let S = {1,0,1}, and define * on S by a*b = a+b (addition of integers). then: 1) * is associative (since addition on the integers is associative) 2) * has an identity, 0 3) every element has a *inverse in S. however, * is not CLOSED on S, since 1+1 is not in S. do you see the problem? * is not a mapping from SxS>S, but a mapping from SxS>Z, and to prove associativity, we have to consider sums which lie in Z, and not in S. for example: (1+1) + 1 = 2 + 1 <evaluation in Z, not S = 3 <evaluation in Z again. 1 + (1+1) = 1 + 2 <evaluation in Z = 3 <again, evaluation in Z. the equality that holds, is an equality in Z, as (a+b)+c = a+(b+c) cannot even be evaluated entirely within S. 

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