Question

1) Let $\ast$ be the binary operation defined on $Q$. Find wherther the given binary operation is commutative.

i) $a\ast b=a-b\forall a,b\in Q$


2) Let $\ast$ be the binary operation defined on $Q$. Find wherther the given binary operation is commutative.

ii) $a\ast b=a^2+b^2\forall a,b\in Q$


3) Let $\ast$ be the binary operation defined on $Q$. Find wherther the given binary operation is commutative.

iii) $a\ast b=a+ab\forall a,b\in Q$


4) Let $\ast$ be the binary operation defined on $Q$. Find wherther the given binary operation is commutative.


iv) $a\ast b=(a-b{)}^2\forall a,b\in Q$

Answer 1

i) Checking commutattivity

Given thet $\ast$ is binary opertation defined on $Q$

Since, $a\ast b=a-b\forall a,b\in Q$

$\Rightarrow b\ast a=b-a\forall a,b\in Q$

so, $a\ast b\ne b\ast a(\because a-b\ne b-a\forall a,b\in Q)$

Hence, $\ast$ is not commutative.

ii) Checking commutattivity

Given thet $\ast$ is binary opertation defined on $Q$

Since, $a,b\in Q,a^2,b^2\in Q$

Now, $a\ast b=a^2+b^2$

$\Rightarrow b\ast a=b^2+a^2$

$\Rightarrow a^2+b^2=b^2+a^2$

$(\because \text{Addition is commutative for rational numbers})$

$\Rightarrow a\ast b=b+a\forall a,b\in Q$

Hence, is commutative.

iii) Checking commutattivity

Given thet $\ast$ is binary opertation defined on $Q$

$a\ast b=a+ab\forall a,b\in Q$

$\Rightarrow b\ast a=b+ba\forall a,b\in Q$

Clearly,

$a+ab\ne b+ba\text{for any}a,b\ni Q$

$\therefore a\ast b\ne b\ast a$

Hence, $\ast$ is not commutative

iv) Checking commutattivity

Given thet $\ast$ is binary opertation defined on $Q$

$a\ast b=(a-b{)}^2\forall a,b\in Q$

$\Rightarrow b\ast a=(b-a{)}^2\forall a,b\in Q$

$\therefore (a-b{)}^2=(b-a{)}^2$

$\therefore a\ast b=b\ast a\forall a,b\in Q$

Hence, $\ast$ is commutative