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

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

(ii) a $*$ b $= a^2 + b^2$

(iii) a $*$ b $=$ a + ab

(iv) a $*$ b $=$ (a$- b)^2$

(v) a $*$ b $= \cfrac{{ab}}{4}$

(vi) a $*$ b $= ab^2$

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

(i) For commutativity :

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

Thus, the operation $*$ is not commutative.

For associativity :

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

And, a$*$(b$*$c)$=$a$*$(b$-$c)$=$a$-$(b$-$c)$=$a$-$b + c
$\therefore$ (a$*$b)$*$c$\ne$a$*$(b$*$c)

Thus, the operation $*$ is not associative.

(ii) For commutativity

a $*$ b $- a^2 + b^2 = b^2 + a^2 =$ b $*$ a
Thus, the operation $*$ is commutative.

For associativity

(a $*$ b ) $*$ c $= (a^2 + b^2 ) * c = (a^2 + b^2 ) + c^2$

and a $*$ (b $*$ c) $=$ a $* (b^2 + c^2) = a^2 + (b^2 + c^2 )$

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

Hence, the operation $*$ is not associative.

(iii) For commutativity

a $*$ b $=$ a + ab $=$ a(1 + b) and b $*$ a $=$ b + ba $=$ b(1 + a)

$\therefore$ a $*$ b $\ne$ b $*$ a

Thus, the operation $*$ is not commutative.
For associativity

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

and a $*$ (b $*$ c ) $=$ a $*$ (b + bc) $=$ a + a(b + bc)

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

Hence, the operation $*$ is not associative.

(iv) For commutativity
a $*$ b $=$ (a $- b)^2 =$ (b $- a)^2 =$ b $*$ a.

Thus, the operation $*$ is commutative.

For associativity

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

and (a $*$ b) $*$ c $= (a - b)^2$ * $c$ = $[(a$ - $b)^2 - c]^2$

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

Hence, the operation $*$ is not associative.

(v) For commutativity

a$*$ b$=$ $\cfrac{{ab}}{4} = \cfrac{{ba}}{4}$ $=$ b$*$ a

Thus, the operation $*$ is commutative.

For associativity
a $*$(b $*$ c) $=$ a$*$ $\cfrac{{bc}}{4}$ $=$ $\cfrac{{a\cfrac{{bc}}{4}}}{4} = \cfrac{{abc}}{{16}}$

and (a $*$b) $*$ c $=$ $\cfrac{{ab}}{4}$ $*$c $=$ $\cfrac{{\cfrac{{ab}}{4} \cdot c}}{4} = \cfrac{{abc}}{{16}}$

$\therefore$ a $*$ (b $*$ c ) $=$ (a $*$ b) $*$ c
Thus, the operation $*$ is associative.

(vi) For commutativity

a $*$ b $= ab^2$ and b $*$ a $= ba^2$

$\therefore$ a $*$ b $\ne$ b $*$ a

Thus, the operation $*$ is not commutative.

For associativity

a $*$ ( b $*$ c ) $=$ a $* (bc^2) = a(bc^2)^2 = ab2c4$

and (a$*$ b) $*$ c $= (ab^2) *$ c $= (ab^2)c^2 = ab^2c^2$

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

Thus, the operation $*$ is not associative.