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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

(0&(0&1))=(0&-1)=1 and


How does the above argument fail?

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The operation "$-$" consists of adding two integers ($a-b = a + (-b)$), but it isn't associative. The integers you're adding when computing $a\&b$ are not $a$ and $b$ themselves, so there is no reason to think that associativity would carry over. – Ben Mar 23 '13 at 14:31
The sentence "Therefore, & associates on the integers also" does not follow from anything you said earlier. – Michael Joyce Mar 23 '13 at 14:33
The flaw is thinking that the $i$ you obtained by examining $a\& b$ depends on $a$ and $b$ in an associative way. – J. Loreaux Mar 23 '13 at 14:36
Remark that the operation becomes associative if you replace the subtraction by addition, yielding $\rm\: a*b = a+b+ab.\:$ Then it amounts to labeling each element by its successor, i.e. transporting the ring structure along the bijection $\rm\: x\mapsto x+1.\ \ $ – Math Gems Mar 23 '13 at 14:45

The operation $x \star y = x + \sin y$ consists of the addition of real numbers, namely $x$ and $\sin y$, but isn't associative. It has to be the addition of the arguments of the binary operation in order for associativity to be inherited.

You should check associativity, or lack thereof, by direct calculation.

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It is not true that it has to be addition to be associative. For example $\rm\: x \star y = x + f(y)\:$ is associative for any idempotent linear $\rm\:f,\:$ e.g. $\rm\:f(y) = ey,\:$ for $\rm\:e\:$ idempotent, e.g. $\rm\ x + 5y\ (mod\ 10).\ \ $ – Math Gems Mar 23 '13 at 15:02

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