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Let \(\ast\) be a binary operation on the set \(Q\) of rational numbers as follows: $(ii)\;\; a \ast b = a^2+b^2 $ Find which of the binary operations are commutative and which are associative.

Note: This is part 2 of a 6 part question, split as 6 separate questions here.

1 Answer

  • An operation $\ast$ on $A$ is commutative if $a\ast b = b \ast a\; \forall \; a, b \in A$
  • An operation $\ast$ on $A$ is associative if $a\ast ( b \ast c) = (a \ast b) \ast c\; \forall \; a, b, c \in A$
Given a binary operation $\ast$ in Q defined by $a \ast b=a^2+b^2$:
$\textbf {Step 1: Checking if the operation is Commutative}$:
For an operation $\ast$ to be commutative $a\ast b = b \ast a$.
$\Rightarrow a \ast b = a^2 + b^2 = b^2 + a^2 = b \ast a$.
Since $a\ast b = b \ast a$ the operation $\ast$ is commutative.
$\textbf {Step 2: Checking if the operation is Associative}$:
For an operation $\ast$ on $A$ is associative $a\ast ( b \ast c) = (a \ast b) \ast c\; \forall \; a, b, c \in A$
$\Rightarrow a\ast ( b \ast c) = a \ast (b^2 + c^2) = a^2 + (b^2+c^2)^2$
$\Rightarrow (a \ast b) \ast c = (a^2 + b^2) \ast c = (a^2 + b^2)^2 + c^2$
Clearly, $a\ast ( b \ast c) \neq (a \ast b) \ast c$. Hence the operation $\ast$ is not associative unless $a=b=c=1$.
answered Mar 13, 2013 by sreemathi.v
edited Mar 19, 2013 by balaji.thirumalai

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