Question

For each binary operation * defined below, determine whether * is 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 = \(\frac { a }{ 2 }\)

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

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

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

Answer 1

(i) On Z, operation * is defined as
(a) a * b = a – b
⇒ b * a = b – a
But a – b + b – a ⇒ a * b # b * a

(b) a – (b – c) * (a-b)-c
Binary operation defined as # is not associative.

(ii) On Q, operation * is defined as a * b = ab + 1.
(a) ab + 1 = ba + 1, a * b = b * a
∴ Defined binary operation is commutative.

(b) 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
⇒ a * (b * c) # (a * b) * c
∴ Binary operation defined as * is not associative.

(iii) (a) On Q, operation * is defined as a * b = \(\frac { ab }{ 2 }\)
∴ a * b = b * a
∴ Operaion binary defined as * is commutative.

(b) a * (b * c) = a* \(\frac { bc }{ 2 }\) = \(\frac { abc }{ 2 }\).
and (a * b) * c = \(\frac { ab }{ 2 }\) * c = \(\frac { abc }{ 4 }\).
⇒ Defined binary operation * is associative.

(iv) On Z+, operation * is defined as a * b = 2ab
(a) a * b = 2ab, b * a = 2ab = 2ab
⇒ a * b = b * a
∴ Binary operation defined as * is commutative.

(b) a * (b * c) = a * 2ab = 2a.2b
(a * b) * c = 2ab * c = 22ab
Thus, (a * b) * c ≠ a * (b * c).
∴ Binary operation defined as * is not associative.

(v) On Z+, a * b = ab
(a) b * a = ba
∴ ab ≠ ba ⇒ a * b # b * a
* is not commutative.

(b) (a * b) * c = ab * c = (ab)c = abc
a * (b * c) = a * bc = abc
Thus, (a * b) * c # (a * b * c)
∴ Operation * is not associative.

(vi) On Z+, operation * is defined as
a + b = \(\frac { a }{ b+1 }\), b # 1
∴ b * a = \(\frac { b }{ a+1 }\)

(a) a * b * b * a
Binary operation defined above is not commutative.

⇒ Binary operation defined above is not associative.