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" "+" "a b` (iv) `a" "*" "b" "=(a-b)^2` (v) `a*b=(a b)/4` (vi) `a" "*" "b" "=a b^2` . Find wh

Explanation

(i) IN Q a * b a-b
Let a, b `in ` Q
`therefore` a*b =a - b `in` b - a `in ` b*a
`rArr` Operation * is not commutatve.
`therefore` 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
Therefore, operation * is not associative.
(ii) In Q a*b `= a^(2) + b^(2)`
Let a,b `in` Q
`therefore` a* b `= a^(2) + b^(2)`
`= b^(2) + a^(2) = ` b*a
`rArr` Operation * is commutative.
Let a,b,c `in` Q
`therefore ` a*(b*c) = `a"*"(b^(2) + c^(2))`
`=a^(2) +(b^(2) + c^(2))^(2)`
and `(a"*"b) "*" c = (a^(2) + b^(2))"*"c`
`= (a^(2) +b^(2))+c^(2)`
`therefore` a* (b*c) `in` (a*b) *c
`rArr` Operation * is not associative.
(iii) In Q , a* b = a+ ab
Let a, b `in` Q
`therefore` a* b = a+ ab and b* a = b+ba
Therefore, a*b `ne` b*a
`rArr` Operation * is not commutative
Let a,b,c `in` Q
`therefore` a*(b*c) = a* (b+ bc)
= a+a(b +bc)= a+ ab + abc
and (a*b) * c = (a+ab) *c
=a+ ab + ac + abc
`therefore` a* (b*c) `ne` (a * b) * c
`rArr` OPeration * is not associative .
(iv) In Q, a* b `= (a-b)^(2)`
`therefore` a* b `= (a-b)^(2)`
`= (b-a)^(2) = ` b * a
`rArr` Operation * is commutative .
Let a,b,c `in` Q
`therefore ` 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`
`rArr` Operatin * is not associative .
(v) In Q, `"a * b" =(ab)/(4)`
Let a,b`in `Q
`therefore` a * ( b*c) = a* `((bc)/(4)) = (a((bc)/(4)))/(4)`
`(((ab)/(4))c)/(4)= ((ab)/(c))"*"c = "(a*b)*c"`
`rArr` Operation * is associtative.(iv) In Q, `"a * b " = ab^(2)`
Let a,b,c `in` Q
`therefore` a* (b*c) `= a"*" (bc)^(2)`
`=a(bc)^(2) = a ab^(2)c^(2)`
and (a*b) *c `= (ab)^(2) "*"c = ab^(2)c^(2)`
`therefore` a* b (b*c) `ne` (a*b) *c
`rArr` operation * is not associative .