# Proving a set closed under a binary operation is associative [closed]

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## closed as not a real question by Henning Makholm, William, Michael Greinecker♦, Thomas, AangSep 25 '12 at 1:53

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You'll have to be a bit more specific. What is your set and operation? –  Karolis Juodelė Sep 14 '12 at 15:22
A binary operation $\circ\colon A\times A\to A$ need not be associative. –  Hagen von Eitzen Sep 14 '12 at 15:22
Another such example is the cross product: $(A\times A)\times B=0$, while $A\times(A\times B)$ is not. –  Robert Mastragostino Sep 14 '12 at 15:52
It's easy to contrive a simple example. Consider the set $\{a,b\}$ and the operator $\circ$ with $a\circ a = a\circ b = b\circ b = b, b\circ a = a$. Then $(a\circ b)\circ a \ne a\circ(b\circ a)$. –  Rick Decker Sep 14 '12 at 16:45
An example of a set together with a non-associative binary operator is the integers $\mathbb Z$ with subtraction. For instance, (7 − 2) − 3 = 2, while 7 − (2 − 3) = 8. Because the way you group parentheses gives different results, it follows that subtraction is not associative.