Let * be a binary operation on the set Q of rational numbers as follows:
Question:

Let * be a binary operation on the set of rational numbers as follows:

(i) $a^{*} b=a-b$ (ii) $a^{*} b=a^{2}+b^{2}$

(iii) $a^{*} b=a+a b$ (iv) $a^{*} b=(a-b)^{2}$

(v) $a^{*} b=\frac{a b}{\text { (vi) } a^{*}} b=a b^{2}$

Find which of the binary operations are commutative and which are associative.

Solution:

(i) On Q, the operation * is defined as * b = a − b.

It can be observed that:

$\frac{1}{2} * \frac{1}{3}=\frac{1}{2}-\frac{1}{3}=\frac{3-2}{6}=\frac{1}{6}$ and $\frac{1}{3} * \frac{1}{2}=\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=\frac{-1}{6}$

$\therefore \frac{1}{2} * \frac{1}{3} \neq \frac{1}{3} * \frac{1}{2} ;$ where $\frac{1}{2}, \frac{1}{3} \in \mathbf{Q}$

Thus, the operation * is not commutative.

It can also be observed that:

$\left(\frac{1}{2} * \frac{1}{3}\right) * \frac{1}{4}=\left(\frac{1}{2}-\frac{1}{3}\right) * \frac{1}{4}=\frac{1}{6} * \frac{1}{4}=\frac{1}{6}-\frac{1}{4}=\frac{2-3}{12}=\frac{-1}{12}$

$\frac{1}{2} *\left(\frac{1}{3} * \frac{1}{4}\right)=\frac{1}{2} *\left(\frac{1}{3}-\frac{1}{4}\right)=\frac{1}{2} * \frac{1}{12}=\frac{1}{2}-\frac{1}{12}=\frac{6-1}{12}=\frac{5}{12}$

$\therefore\left(\frac{1}{2} * \frac{1}{3}\right) * \frac{1}{4} \neq \frac{1}{2} *\left(\frac{1}{3} * \frac{1}{4}\right) ;$ where $\frac{1}{2}, \frac{1}{3}, \frac{1}{4} \in \mathbf{Q}$

Thus, the operation * is not associative.

(ii) On $\mathbf{Q}$, the operation * is defined as $a^{*} b=a^{2}+b^{2}$.

For $a, b \in \mathbf{Q}$, we have:

$a * b=a^{2}+b^{2}=b^{2}+a^{2}=b * a$

$\therefore a^{*} b=b^{*} a$

Thus, the operation * is commutative.

It can be observed that:

$\left(1^{*} 2\right)^{*} 3=\left(1^{2}+2^{2}\right)^{*} 3=(1+4)^{*} 3=5^{*} 3=5^{2}+3^{2}=25+9=34$

$1^{*}\left(2^{*} 3\right)=1^{*}\left(2^{2}+3^{2}\right)=1 *(4+9)=1 * 13=1^{2}+13^{2}=1+169=170$

$\therefore\left(1^{*} 2\right)^{\star} 3 \neq 1^{*}\left(2^{*} 3\right)$, where $1,2,3 \in \mathrm{Q}$

Thus, the operation * is not associative.

(iii) On $Q$, the operation * is defined as $a^{\star} b=a+a b$.

It can be observed that:

$\therefore 1^{*} 2 \neq 2 * 1 ;$ where $1,2 \in \mathbf{Q}$

Thus, the operation * is not associative.

(iv) On $\mathbf{Q}$, the operation * is defined by $a^{*} b=(a-b)^{2}$.

For $a, b \in \mathbf{Q}$, we have:

$a^{*} b=(a-b)^{2}$

$b^{*} a=(b-a)^{2}=[-(a-b)]^{2}=(a-b)^{2}$

∴ * b = b * a

Thus, the operation * is commutative.

It can be observed that:

$(1 * 2) * 3=(1-2)^{2} * 3=(-1)^{2} * 3=1 * 3=(1-3)^{2}=(-2)^{2}=4$

$1 *(2 * 3)=1 *(2-3)^{2}=1 *(-1)^{2}=1 * 1=(1-1)^{2}=0$

$\therefore(1 * 2) * 3 \neq 1 *(2 * 3)$; where $1,2,3 \in \mathbf{Q}$

Thus, the operation * is not associative.

(v) On $\mathbf{Q}$, the operation * is defined as $a * b=\frac{a b}{4}$.

For $a, b \in \mathbf{Q}$, we have:

$a * b=\frac{a b}{4}=\frac{b a}{4}=b * a$

$\therefore a^{*} b=b^{*} a$

Thus, the operation * is commutative.

For a, b, c ∈ Q, we have:

$(a * b) * c=\frac{a b}{4} * c=\frac{\frac{a b}{4} \cdot c}{4}=\frac{a b c}{16}$

$a *(b * c)=a * \frac{b c}{4}=\frac{a \cdot \frac{b c}{4}}{4}=\frac{a b c}{16}$

∴(* b) * c = a * (* c)

Thus, the operation * is associative.

(vi) On $\mathbf{Q}$, the operation * is defined as $a^{*} b=a b^{2}$

It can be observed that:

$\frac{1}{2} * \frac{1}{3}=\frac{1}{2} \cdot\left(\frac{1}{3}\right)^{2}=\frac{1}{2} \cdot \frac{1}{9}=\frac{1}{18}$

$\frac{1}{3} * \frac{1}{2}=\frac{1}{3} \cdot\left(\frac{1}{2}\right)^{2}=\frac{1}{3} \cdot \frac{1}{4}=\frac{1}{12}$

$\therefore \frac{1}{2} * \frac{1}{3} \neq \frac{1}{3} * \frac{1}{2} ;$ where $\frac{1}{2}, \frac{1}{3} \in \mathbf{Q}$

Thus, the operation * is not commutative.

It can also be observed that:

$\left(\frac{1}{2} * \frac{1}{3}\right) * \frac{1}{4}=\left[\frac{1}{2} \cdot\left(\frac{1}{3}\right)^{2}\right] * \frac{1}{4}=\frac{1}{18} * \frac{1}{4}=\frac{1}{18} \cdot\left(\frac{1}{4}\right)^{2}=\frac{1}{18 \times 16}$

$\frac{1}{2} *\left(\frac{1}{3} * \frac{1}{4}\right)=\frac{1}{2} *\left[\frac{1}{3} \cdot\left(\frac{1}{4}\right)^{2}\right]=\frac{1}{2} * \frac{1}{48}=\frac{1}{2} \cdot\left(\frac{1}{48}\right)^{2}=\frac{1}{2 \times(48)^{2}}$

$\therefore\left(\frac{1}{2} * \frac{1}{3}\right) * \frac{1}{4} \neq \frac{1}{2} *\left(\frac{1}{3} * \frac{1}{4}\right) ;$ where $\frac{1}{2}, \frac{1}{3}, \frac{1}{4} \in \mathbf{Q}$

Thus, the operation * is not associative.

Hence, the operations defined in (ii), (iv), (v) are commutative and the operation defined in (v) is associative.

 

 

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