For each binary operation * defined below, determine whether * is commutative or associative.

Solution:

(i) On $\mathbf{Z},{ }^{*}$ is defined by $a{ }^{*} b=a-b$.

It can be observed that $1^{*} 2=1-2=1$ and $2^{*} 1=2-1=1$.

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

Hence, the operation * is not commutative.

Also we have:

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

$1^{*}\left(2^{*} 3\right)=1^{*}(2-3)=1^{*}-1=1-(-1)=2$

$\therefore\left(1^{*} 2\right) * 3 \neq 1^{*}(2 * 3)$; where $1,2,3 \in \mathbf{Z}$

Hence, the operation * is not associative.

(ii) On $\boldsymbol{O}^{*}$ is defined by $\boldsymbol{a}^{*} b=a b+1$

It is known that:

$a b=b a \& m n F o r E ; a, b \in \mathbf{Q}$

$\Rightarrow a b+1=b a+1 \& m n$ ForE; $a, b \in \mathbf{Q}$

$\Rightarrow a^{*} b=a^{*} b \& m n F o r E ; a, b \in \mathbf{Q}$

Therefore, the operation * is commutative.

It can be observed that:

$\left(1^{*} 2\right)^{*} 3=(1 \times 2+1)^{*} 3=3 * 3=3 \times 3+1=10$ $1^{*}\left(2^{*} 3\right)=1^{*}(2 \times 3+1)=1^{*} 7=1 \times 7+1=8$

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

Therefore, the operation * is not associative.

(iii) On $\mathbf{Q},{ }^{*}$ is defined by $a^{*} b=\frac{a b}{2}$.

It is known that:

$a b=b a \& m n F o r E ; a, b \in \mathbf{Q}$

⇒ &mnForE; a, b ∈ Q

⇒ a * b = b * a &mnForE; a, b ∈ Q

Therefore, the operation * is commutative.

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

$(a * b) * c=\left(\frac{a b}{2}\right) * c=\frac{\left(\frac{a b}{2}\right) c}{2}=\frac{a b c}{4}$

$a *(b * c)=a *\left(\frac{b c}{2}\right)=\frac{a\left(\frac{b c}{2}\right)}{2}=\frac{a b c}{4}$

$\therefore(a * b) * c=a *(b * c)$

Therefore, the operation * is associative.

(iv) On $\mathbf{Z}^{+},{ }^{*}$ is defined by $a^{*} b=2^{a b}$.

It is known that:

$a b=b a \& m n F o r E ; a, b \in \mathbf{Z}^{+}$

$\Rightarrow 2^{a b}=2^{b a} \& m n F o r E ; a, b \in \mathbf{Z}^{+}$

$\Rightarrow a^{*} b=b^{*} a \& m n F o r E ; a, b \in \mathbf{Z}^{+}$

Therefore, the operation * is commutative.

It can be observed that:

$(1 * 2) * 3=2^{(1 \times 2)} * 3=4 * 3=2^{4 \times 3}=2^{12}$

$1 *(2 * 3)=1 * 2^{2 \times 3}=1 * 2^{6}=1 * 64=2^{64}$

$\therefore\left(1^{*} 2\right)^{*} 3 \neq 1^{*}\left(2^{*} 3\right) ;$ where $1,2,3 \in \mathbf{Z}^{+}$

(v) On $\mathbf{Z}^{+},{ }^{*}$ is defined by $a^{*} b=a^{b}$.

It can be observed that:

$1 * 2=1^{2}=1$ and $2 * 1=2^{1}=2$

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

It can also be observed that:

$(2 * 3) * 4=2^{3} * 4=8 * 4=8^{4}=\left(2^{3}\right)^{4}=2^{12}$

$2 *(3 * 4)=2 * 3^{4}=2 * 81=2^{81}$

$\therefore(2 * 3) * 4 \neq 2 *(3 * 4) ;$ where $2,3,4 \in \mathbf{Z}^{+}$

Therefore, the operation * is not associative.

(vi) On $\mathbf{R},{ }^{*}-\{-1\}$ is defined by $a * b=\frac{a}{b+1}$.

It can be observed that $1 * 2=\frac{1}{2+1}=\frac{1}{3}$ and $2 * 1=\frac{2}{1+1}=\frac{2}{2}=1$.

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

Therefore, the operation * is not commutative.

It can also be observed that:

$(1 * 2) * 3=\frac{1}{3} * 3=\frac{\frac{1}{3}}{3+1}=\frac{1}{12}$

$1 *(2 * 3)=1 * \frac{2}{3+1}=1 * \frac{2}{4}=1 * \frac{1}{2}=\frac{1}{\frac{1}{2}+1}=\frac{1}{\frac{3}{2}}=\frac{2}{3}$

$\therefore\left(1^{*} 2\right)^{*} 3 \neq 1^{*}\left(2^{*} 3\right) ;$ where $1,2,3 \in \mathbf{R}-\{-1\}$

Therefore, the operation * is not associative.