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

(i) On Z, define a $*$ b $=$ a$-$ b

(ii) On Q, define a$*$b$=$ ab + 1

(iii) On Q, define a $*$ b $=$ $\cfrac{{ab}}{2}$

(iv) On Z+, define a$*$ b$=$ 2ab

(v) On Z+, define a$*$b$=$ ab

(vi) On R$-\{ - 1\}$, define a $*$ b$=$ $\cfrac{a}{{b + 1}}$

(i) a $*$ b $=$ a$-$b on Z

For commutativity

a $*$ b $=$ a $-$ b and b $*$ a $=$ b $-$ a $=$ $-$ (a $-$ b) $\ne$a $*$ b

$\Rightarrow$ a $*$ b $\ne$ b $*$ a

For associativity

a $*$ (b $*$ c) $=$ a $*$ (b $-$ c) $=$ a $-$ ( b $-$ c ) $=$ (a $-$ b + c )

And (a $*$ b) $*$ c $=$ (a $-$ b) $*$ c$=$ a $-$ b$-$ c.

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

Thus, the operation $*$ is neither commutative nor associative.

(ii) a $*$ b $=$ ab + 1 on Q

For commutativity

a $*$ b $=$ ab + 1 and b $*$ a $=$ ba + 1 $=$ ab + 1.
a $*$ b $=$ b $*$ a

For associativity

a $*$ ( b $*$ c) $=$ a $*$ (bc + 1) $=$ a (bc + 1) + 1 $=$ abc + a + 1

And, (a $*$ b) $*$ c $=$ (ab + 1) $*$ c $=$ (ab + 1) c +1 $=$ abc + c + 1

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

Thus, the operation $*$ is commutative but not associative.

(iii) a$*$ b$=$ $\cfrac{{ab}}{2}$on Q

For commutativity

a$*$ b $=$ $\cfrac{{ab}}{2}$ and b$*$ a $= \cfrac{{ba}}{2} = \cfrac{{ab}}{2}$

$\therefore$ a $*$ b $=$ b $*$ a

For associativity

a$*$ (b$*$ c) $=$ a$*$ $\left( {\cfrac{{bc}}{2}} \right)$ $= \cfrac{{abc/2}}{2} = \cfrac{{abc}}{4}$

and, (a$*$ b)$*$ c $= \left( {\cfrac{{ab}}{2}} \right)$ $*$(c) $= \left( {\cfrac{{abc/2}}{2}} \right) = \cfrac{{abc}}{4}$

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

Thus, the operation $*$ is commutative and also associative,

(iv) a $*$ b $=$ 2ab, on Z+

For commutativity

a $*$ b $=$2ab and b $*$ a $=$ 2ba $=$ 2ab. $\therefore$ a $*$ b $=$ b $*$ a

For associativity

a$*$ (b$*$ c)$=$a$*$(2bc )$=$(2)$^{a \cdot {2^{bc}}}$and (a$*$ b )$*$ c $=$ ( 2ab ) $*$ c$=$ 2$^{{2^{ab}} \times c}$

a $*$ ( b $*$ c ) $\ne$ (a $*$ b ) $*$ c

Hence, the operation $*$ is commutative but not associative.

(v) a $*$ b $=$ ah on Z$^ -$

For commutativity

a $*$ b $=$ ah and b $*$ a $=$ ba. $\therefore$ a $*$ b $\ne$ b $*$ a

For associativity

a$*$ ( b $*$ c)$=$ a$*$ (bc )$=$ (a)$^{{b^c}}$and (a $*$ b) $*$ c$=$ (ab) $*$ c $=$ (ab)c$=$ abc

Thus, a $*$ (b $*$ c ) $\ne$ (a $*$ b) $*$ c

Hence, the operation $*$ is neither commutative nor associative.

(vi) a $*$ b $=$ $\cfrac{a}{{b + 1}}$ on R $- \{$ - $1\}$

For commutativity

a$*$ b $= \cfrac{a}{{a + 1}}$ and b$*$ a $=$ $\cfrac{b}{{a + 1}}$ $\therefore$ $a$ $*$ b $\ne b * a$

For associativity

$a * (b * c) = a * \left( {\cfrac{b}{{c + 1}}} \right) = \cfrac{a}{{\cfrac{b}{{c + 1}} + 1}} = \cfrac{{a(c + 1)}}{{b + c + 1}}$

and $(a * b) * c = \left( {\cfrac{a}{{b + 1}}} \right) * c = \cfrac{{\cfrac{a}{{b + 1}}}}{{c + 1}} = \cfrac{a}{{(b + 1)(c + 1)}}$

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

Hence, the operation $*$ is neither commutative nor associative.