Consider a binary operation & defined by (a&b)=(a-(b+(a*b))) on the integers, with "a" and "b" as integers. One might argue as follows:
By closure (a*b) equals an integer, which we'll call k. Thus, (a-(b+(a*b)))=(a-(b+k). By closure (a+k) equals an integer, which we'll call j. Thus, (a-(b+(a*b)))=(a-(b+k)=(a-j). Since (a-j) means the same thing as (a+i) where i indicates the inverse of j we can write: (a-(b+(a*b)))=(a+i). Thus, (a-(b+(a*b))) or (a&b) consists of an addition of two integers. Addition associates on the integers, meaning that for all x, y, z (x+(y+z))=((x+y)+z). Therefore, & associates on the integers also.
This argument is not valid, since if a=0, b=1, we have
How does the above argument fail?