I know that there are many algebraic associative operations which are commutative and which are not commutative.
for example multiplications of matrices as associative operation is not commutative.
I need to know about inverse of this!
I mean is there any algebraic commutative operation which is not associative?
can you show me sample?
On
Nice question. First of all, it's important to note that commutativity does not imply associativity, i.e. one can construct a counterexample. Wikipedia has one, but - even simpler - consider $\mathbb R$ with the operation
$$ a \circ b := ab - (a+b). $$
Clearly $\circ$ is commutative, however e.g. $$1\circ (2\circ 3) = -1 \neq -5 = (1\circ 2)\circ 3.$$
Another question is whether such structures do arise naturally somewhere in mathematics. For most parts, associativity is a very fundamental property and we usually require commutativity on top of that (groups -> abelian groups, commutative rings ...)
On
Two counterexamples have already been given, but here's another that seems a bit simpler and that comes up "naturally". Consider the set of all points on a line (or a plane, or 3-dimensional space --- it won't matter for the example) and let the operation send each pair $(P,Q)$ of points to the midpoint of the segment $PQ$.
On
Extending the example in Rahul Narain's comment, if "$+$" is the addition operation in an Abelian group, then the operation $a\circ b=a+a+b+b$ is commutative but not associative if the group contains two elements $a$ and $b$ such that $a+a\not=b+b$ (so that $(a\circ 0)\circ b\not=a\circ(0\circ b)$).
Consider the set $M_n(\mathbb{R})$ and the binary operation $A * B = \frac{1}{2}(AB+BA)$
This isn't associative as
$A * (B * C) = A * \frac{1}{2}(BC + CB) = \frac{1}{4}(ABC + ACB + BCA + CBA)$
Yet
$(A * B) * C = \frac{1}{4}(ABC + BAC + CAB + CBA)$
and $BAC + CAB \neq ACB + BCA$ for all $A,B,C \in M_n(\mathbb{R})$
However, $A * B = \frac{1}{2}(AB+BA) = \frac{1}{2}(BA+AB) = B * A$
So it is commutative.