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Consider a binary operation $\ast$ on the set \(\{1, 2, 3, 4, 5\}\) given by the following multiplication table (Table 1.2). \begin{matrix} *&1&2&3&4&5 \\ 1&1&1&1&1&1 \\ 2&1&2&1&2&1 \\ 3&1&1&3&1&1 \\ 4&1&2&1&4&1 \\ 5&1&1&1&1&5 \end{matrix} (ii) is $\ast$ commutative?

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  • An operation $\ast$ on $A$ is commutative if $a\ast b = b \ast a\; \forall \; a, b \in A$
From the table, we can see that for every $a, b \in$ the set, $a \ast b = b \ast a$. Therefore $\ast$ is commutative.
answered Mar 20, 2013 by balaji.thirumalai
 

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