Operation * is on the set Q.
(i) Defined as a * b = a – b
(a) Now b * a = b – a. But a – b ≠ b – a
∴ a * b ≠ b * a
∴ Operation * is not commutative.
(b) a * (b * c) = a * (b-c)
= a – (b-c)
= a – b + c
(a * b) * c = (a – b) * c
= a – b – c
Thus a * (b * c) ≠ (a * b) * c
The operation *, as defined above is not associative.
(ii) (a) a * b = a² + b²
b * a = b² + a² = a² + b²
∴ a * b = b * a
∴ This binary operation * is commutative.
(b) a * (b * c) = a * (b² + c²) = a² + (b² + c²)²
(a * b) * c = (a² + b²) * c = (a² + b²) + c²
⇒ a * (b * c) ≠ (a * b) * c
∴ The given operation * is not associaive.
(iii) Operation * is defined as a * b = a + ab
(a) b * a = b + ba
∴ a * b = b * a
∴ This operation * is not commutative.
(b) a * (b * c) = a * (b + bc)
= a + a(b + be)
= a +ab + abc
(a * b) * c = (a + ab) * c
= a + ab + (a + ab).c
= a + ab + ac + abc
⇒ a * (b * c) # (a * b) * c
⇒ The given binary operation * is not associative.
(iv) The binary operation * is defined as
a * b = (a – b)²
(a) b * a = (b – a)² = (a – b)²
⇒ a * b = b * a
This binary operation * is commutative.
(b) a * (b * c) = a * (b – c)²
= [a – (b – c)²]²
(a * b) * c = (a – b)² * c
= [(a-b)²-c]²
⇒ a * (b * c) # (a * b) * c
∴ The operation * is not associative.
(v) Binary operaion * is defined as
a * b = \(\frac { ab }{ 4 }\)
∴ The operation * is commutative.
(b) a * (b * c) = a * \(\frac { ab }{ 4 }\) = \(\frac { a }{ 4 }\)( \(\frac { bc }{ 4 }\)) = \(\frac { abc }{ 16 }\)
(a * b) * c = \(\frac { ab }{ 4 }\) * c = \(\frac { ab }{ 4 }\) * c = \(\frac { ab }{ 4 }\).\(\frac { c }{ 4 }\) = \(\frac { abc }{ 16 }\)
⇒ (a * b) * c = a * (b * c)
Thus, the given operation * is associative.
(vi) Binary operation * is defined as
a * b = ab²
(a) b * a = ba² ≠ ab²
∴ a * b * b * a.
∴ The operation * is not commuative.
(b) a * (b * c) = a * bc² = a(bc²)² = abc²
(a * b) * c = ab² *c = (ab²)c² = ab²c²
∴ a * (b * c) * (a * b) * c.
∴ Given binary operation * is not associative.
