Question

Let $\ast$ be a binary operation on the set $Q$ of rational numbers as follows:
(i) $a\ast b=a-b$ (ii) $a\ast b=a^2+b^2$
(iii) $a\ast b=a+ab$ (iv) $a\ast b=(a-b{)}^2$
(v) $a\ast b=\frac{ab}{4}$ (vi) $a\ast b=a{b}^2$
Find which of the binary operations are commutative and which are associative

• A. $ii,iv,v$ are commutative and $v$ associative
• B. $ii,iv,v$ are not commutative and $v$ associative
• C. $iii,iv,v$are commutative and $v$ associative
• D. $vi,iv,v$are commutative and $v$ associative

Answer 1

The correct option is A $ii,iv,v$ are commutative and $v$ associative

$\left(i\right)$ $a\ast b=a-b$

Check commutative is

$a\ast b=b\ast a$

$a\ast b=a-b$

$b\ast a=b-a$

Since, $a\ast b\ne b\ast a$

$\ast$ is not commutative.

Check associative

$\ast$ is associative if

$(a\ast b)\ast c=a\ast (b\ast c)(a\ast b)\ast c={(a-b)}^{\ast }c=(a-b)-c=a-b-ca\ast (b\ast c)=a\ast (b-c)=a-(b-c)=a-b+c$

Since $(a\ast b)\ast c\ne a\ast (b\ast c)$

$\ast$ is not an associative binary operation.

$\left(ii\right)$ $a\ast b=a^2+b^2$

Check commutative

$\ast$ is commutative if $a\ast b=b\ast a$

$a\ast b=a^2+b^{2}b\ast a=b^2+a^2=a^2+b^2$

Since $a\ast b=b\ast a\forall a,bϵQ$

$\ast$ is commutative.

Check associative

$\ast$ is associative if

$(a\ast b)\ast c=a\ast (b\ast c)(a\ast b)\ast c=(a^2+b^2)\ast c={(a^2+b^2)}^2+c^{2}a\ast (b\ast c)=a\ast (b^2+c^2)=a^2+{(b^2+c^2)}^2$

Since $(a\ast b)\ast c\ne a\ast (b\ast c)$

$\ast$ is not an associative binary operation.

$\left(iii\right)$ $a\ast b=a+b$

Check commutative

$\ast$ is commutative is $a\ast b=b\ast a$

$a\ast b=a+ab;b\ast a=b+ba$

Since $a\ast b\ne b\ast a$

$\ast$ is not commutative.

$\left(iv\right)$ $a\ast b={(a-b)}^2$

Check commutative

$\ast$ is commutative if $a\ast b=b\ast a$

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

Since $a\ast b=b\ast a\forall a,bϵQ$

$\ast$ is commutative.

Check associative

$\ast$ if

$(a\ast b)\ast c=a\ast (b\ast c)(a\ast b)\ast c={(a-b)}^2\ast c={[{(a-b)}^2-c]}^{2}a\ast (b\ast c)=a\ast {(b-c)}^2={[a-{(b-c)}^2]}^2$

Since $(a\ast b)\ast c\ne a\ast (b\ast c)$

$\ast$ is not an associative binary operation.

$\left(v\right)$ $a\ast b=\frac{ab}{4}$

Check commutative.

$\ast$ is commutative if $a\ast b=b\ast a$

$a\ast b=\frac{ab}{4};b\ast a=\frac{ba}{4}=\frac{ab}{4}$

Since $a\ast b=b\ast a\forall a,bϵQ$

$\ast$ is commutative.

Check associative.

$\ast$ is association if $(a\ast b)\ast c=a\ast (b\ast c)$

$(a\ast b)\ast c=\left(\frac{\frac{ab}{4}\ast c}{4}\right)=\frac{abc}{16}a\ast (b\ast c)=a\ast \left(\frac{bc}{4}\right)=\frac{a\times \frac{bc}{4}}{4}=\frac{abc}{16}$

Since $(a\ast b)\ast c=a\ast (b\ast c)\forall a,b,cϵQ$

$\ast$ is an associative binary operation.

$\left(vi\right)$ $a\ast b={ab}^2$

check commutative.

$\ast$ is commutative if $a\ast b=b\ast a$

$a\ast b={ab}^2;b\ast a={ba}^2$

Since $a\ast b\ne b\ast a$

$\ast$ is not commutative.

Check associative

$\ast$ is associative if $(a\ast b)\ast c=a\ast (b\ast c)$

$(a\ast b)\ast c={ab}^2\ast c=\left({ab}^2\right)c^2=a{b}^2{c}^2.a\ast (b\ast c)=a\ast {bc}^2=a{\left({bc}^2\right)}^2=a{b}^2{c}^4$

Since $(a\ast b)\ast c\ne a\ast (b\ast c)$

$\ast$ is not an associate binary operation.