Question 683768
we check that (a*b)*c = a*(b*c)


(a*b)*c = (a+b+ab/2)*c = 

a+b+ab/2 +c + ((a+b+ab/2)c)/2=




a*(b*c) = a*(b+c+bc/2) =

a+(b+c+bc/2)+(a((b+c+bc/2))/2



comparing...

(a+b+ab/2 +c + ((a+b+ab/2)c)/2) - 
(a+(b+c+bc/2)+(a((b+c+bc/2))/2)

which is 0


the operation is associative


:)