Question

Let $\ast$ be a binary operation on the set Q of rational number as follows:
$\left(i\right)a\ast b=a-b$

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

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

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

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

$\left(vi\right)a\ast b=a{b}^2$
Find which of the binary operation are commutative and which are associative?

Answer 1

On Q,the operation $\ast$ is defined as $a\ast b=a-b.$
It can be observed that the for $2,3,4\in Q$, we have
$2\ast 3=2-3=-1$ and $3\ast 2=3-2=1$
$2\ast 3\ne 3\ast 2$
Thus, the operation $\ast$ is not commutative.
It can also be observed that
$(2\ast 3)\ast 4=(-1)\ast 4=-1-4=-5$
and $2\ast (3\ast 4)=2\ast (-1)=2-(-1)=3$
$2\ast (3\ast 4)\ne 2\ast (3\ast 4)$
Thus, the operation $\ast$ is not associative.

On Q, the operation $\ast$ is defined as a$\ast b=a^2+b^2.$
For $a,b\in Q$,we have
$a\ast b=a^2+b^2=b^2+a^2=b\ast a$
Therefore, $a\ast b=b\ast a$
Thus, the operation $\ast$ is commutative.
It can be observed that
$(1\ast 2)\ast 3=(1^2+2^2)\ast 3=(1+4)\ast 4=5\ast 4=5^2+4^2=411\ast (2\ast 3)=1\ast (2^2+3^2)=1\ast (4+9)=1\ast 13=1^2+{13}^2=170$
$(1\ast 2)\ast 3\ne 1\ast (2\ast 3)$ where $1,2,3,\in Q$
Thus, the operation $\ast$ is not associative.

On Q, the operation $\ast$ is defined as $a\ast b=a+ab$
It can be observed that
$1\ast 2=1+21\times 2=1+2=3,2\ast 1=2+2\times 1=2+2=4$
$\therefore 1\ast 2\ne 2\ast 1$; where $1,2\in Q$
Thus, the operation \ast is not commutative.
It can also be observed that
$(1\ast 2)\ast 3=(1+1\times 2)\ast 3=3\ast 3=3+3\times 3=3+9=121\ast (2\ast 3)=1\ast (2+2\times 3)=1\ast 8=1+1\times 8=9$
$\therefore (1\ast 2)\ast 3\ne 1\ast (2\ast 3)$ where $1,2,3\in Q$
Thus, the operation $\ast$ is not associative.

On Q, the operation $\ast$ is defined by $a\ast b=(a-b{)}^2$
For $a,b\in Q$,we have
$a\ast b=(a-b{)}^{2}andb\ast a=(b-a{)}^2=[-(a-b){]}^2=(a-b{)}^2$
Therefore, $a\ast b=b\ast a$
Thus, the operation $\ast$ is commutative.

It can be obsreved that
$(1\ast 2)\ast 3=(1-2{)}^2\ast 3=(-1{)}^2\ast 3=1\ast 3=(1-3{)}^2=(-2{)}^2=41\ast (2\ast 3)=1\ast (2-3{)}^2=1\ast (-1{)}^2=1\ast 1=(1-1{)}^2=0$
$\therefore (1\ast 2)\ast 3\ne 1\ast (2\ast 3)$ where $1,2,3\in Q$
Thus, the operation $\ast$ is not associative.

On Q, the operation $\ast$ is defined as $a\ast b=\frac{ab}{4}.$
For $a,b\in Q$, we have a $\ast b=\frac{ab}{4}=\frac{ba}{4}=b\ast a$
Therefore, $a\ast b=b\ast a$
Thus, the operation $\ast$ is commutative.
For $a,b,c\in Q$, we have $(a\ast b)\ast c=\frac{ab}{4}\ast c=\frac{\frac{ab}{4}.c}{4}=\frac{abc}{16}$
$a\ast (b\ast c)=a\ast \frac{bc}{4}=\frac{a.\frac{bc}{4}}{4}=\frac{abc}{16}$
Therefore, $(a\ast b)\ast c=a\ast (b\ast c)$. Thus, the operation $\ast$ is associative.

On Q, the operation $\ast$ is defined as $a\ast b=a{b}^2$
It can be observed that for $2,3\in Q$
$2\ast 3={2.3}^2=18$ and $3\ast 2={3.2}^2=12$
Hence, $2\ast 3\ne 3\ast 2$
Also, $\frac{1}{2}\ast \frac{1}{3}=\frac{1}{2}(\frac{1}{3}{)}^2=\frac{1}{2}.\frac{1}{9}=\frac{1}{18}$
$\frac{1}{3}.\frac{1}{2}=\frac{1}{3}(\frac{1}{2}{)}^2=\frac{1}{3}.\frac{1}{4}=\frac{1}{12}$
$\therefore \frac{1}{2}\ast \frac{1}{3}\ne \frac{1}{3}\ast \frac{1}{2}$ where, $\frac{1}{2},\frac{1}{3}\in Q$
Thus, the operation $\ast$ is not commutative.
It can also be observed that for $1,2,3\in Q$

$(1\ast 2)\ast 3=\left({1.2}^2\right)\ast 3=4\ast 3={4.3}^2=361\ast (2\ast 3)=1\ast \left({2.3}^2\right)=1\ast 18={1.18}^2=324(1\ast 2)\ast 3\ne 1\ast (2\ast 3)$
Also, $(\frac{1}{2}\ast \frac{1}{3})\ast \frac{1}{4}=[\frac{1}{2}.(\frac{1}{3}{)}^2]\ast \frac{1}{4}=\frac{1}{18}\ast \frac{1}{4}=\frac{1}{18}.(\frac{1}{4}{)}^2=\frac{1}{18\times 16}$
$\frac{1}{2}\ast (\frac{1}{3}\ast \frac{1}{4})=\frac{1}{2}\ast \left[\frac{1}{3}\right(\frac{1}{4}{)}^2]=\frac{1}{2}\ast \frac{1}{48}=\frac{1}{18}=\frac{1}{2}(\frac{1}{48}{)}^2=\frac{1}{4608}$
$\therefore (\frac{1}{2}\times \frac{1}{3})\ast \frac{1}{4}\ne \frac{1}{2}\ast (\frac{1}{3}\ast \frac{1}{4})where\frac{1}{2},\frac{1}{3},\frac{1}{4}\ne Q$
Thus, the operation $\ast$ is not associative.
Hence, the operations defined in parts (ii), (iv), (v)are commutative and the operation defined in part (v) is associative.