Question

Check the commutativity and associativity of the following binary operation:
${}^{\prime }{\odot }^{\prime }$ on Q defined by $a\odot b=a^2+b^2$ for all $a,b\in Q$.

Answer 1

Hint – Commutativity property means, $a\odot b=b\odot a$, where ${}^{\prime }{\odot }^{\prime }$ is a binary operation and associative property is $(a\odot b)\odot r=a\odot (b\odot r)$.
Complete step-by-step answer:
Given that ${}^{\prime }{\odot }^{\prime }$is a binary operation on Q defined by $a\odot b=a^2+b^2$ for all $a,b\in Q$.
We know-
Commutative property means, $a\odot b=b\odot a$.
Let’s check the commutative property of given binary operation:
Given, $a\odot b=a^2+b^2$
$$\begin{gathered} \Rightarrow a\odot b=a^2+b^2 \\ \Rightarrow b\odot a=b^2+a^2 \\ \Rightarrow a\odot b=b\odot a \end{gathered}$$
Therefore, the commutative property holds for a given binary operation on ‘Q’.
Now, we know –
Associative property is $(a\odot b)\odot r=a\odot (b\odot r)$.
Let’s check the associative property of given binary operation:
Given, $a\odot b=a^2+b^2$
$$\begin{gathered} \Rightarrow (a\odot b)\odot r=(a^2+b^2)\odot r \\ \Rightarrow (a\odot b)\odot r=(a^2+b^2{)}^2+r^2-(1) \end{gathered}$$
Now, $$\begin{gathered} a\odot (b\odot r)=a\odot (b^2+r^2) \\ \Rightarrow a\odot (b\odot r)=a^2+(b^2+r^2{)}^2-(2) \end{gathered}$$
From equation (1) and (2), we can clearly say that associativity property doesn’t hold for the binary operation on ‘Q’.
Therefore, the given binary operation is commutative but not associative on Q.

Note – Whenever such types of questions appear, then first check for the commutative property of the binary operation given in the question and then check for the associativity property on Q. Proceed step by step to avoid any mistake.