5 a) Determine whether the binary operation * defined is commutative and whether * is associative i) * defined on \( Z \) by letting \( a * b=a-b \) ii) * defined on \( Q \) by letting \( a * b=a b+1 \) iii) \( \quad \) defined on \( Z^{+} \)by letting \( a * b=2^{a b} \) iv) * defined on \( Z^{+} \)by letting \( a * b=a^{b} \)

To determine whether the binary operations defined are commutative and associative, we will analyze each operation step by step. ### i) Operation defined on \( Z \) by \( a * b = a - b \) **Commutative:** An operation is commutative if \( a * b = b * a \) for all \( a, b \). \[ a * b = a - b \] \[ b * a = b - a \] For \( a * b = b * a \): \[ a - b = b - a \implies a + a = b + b \implies 2a = 2b \implies a = b \] This is not true for all \( a \) and \( b \). Therefore, the operation is **not commutative**. **Associative:** An operation is associative if \( (a * b) * c = a * (b * c) \) for all \( a, b, c \). Calculating \( (a * b) * c \): \[ (a * b) * c = (a - b) * c = (a - b) - c = a - b - c \] Calculating \( a * (b * c) \): \[ b * c = b - c \implies a * (b * c) = a * (b - c) = a - (b - c) = a - b + c \] Since \( a - b - c \neq a - b + c \), the operation is **not associative**. ### ii) Operation defined on \( Q \) by \( a * b = ab + 1 \) **Commutative:** \[ a * b = ab + 1 \] \[ b * a = ba + 1 \] Since multiplication is commutative (\( ab = ba \)), we have: \[ ab + 1 = ba + 1 \] Thus, the operation is **commutative**. **Associative:** Calculating \( (a * b) * c \): \[ (a * b) * c = (ab + 1) * c = (ab + 1)c + 1 = abc + c + 1 \] Calculating \( a * (b * c) \): \[ b * c = bc + 1 \implies a * (b * c) = a * (bc + 1) = a(bc + 1) + 1 = abc + a + 1 \] Since \( abc + c + 1 \neq abc + a + 1 \), the operation is **not associative**. ### iii) Operation defined on \( Z^{+} \) by \( a * b = 2^{ab} \) **Commutative:** \[ a * b = 2^{ab} \] \[ b * a = 2^{ba} \] Since \( ab = ba \), we have: \[ 2^{ab} = 2^{ba} \] Thus, the operation is **commutative**. **Associative:** Calculating \( (a * b) * c \): \[ (a * b) * c = 2^{ab} * c = 2^{(2^{ab})c} = 2^{abc} \] Calculating \( a * (b * c) \): \[ b * c = 2^{bc} \implies a * (b * c) = 2^{a(2^{bc})} = 2^{abc} \] Since both expressions yield \( 2^{abc} \), the operation is **associative**. ### iv) Operation defined on \( Z^{+} \) by \( a * b = a^{b} \) **Commutative:** \[ a * b = a^{b} \] \[ b * a = b^{a} \] In general, \( a^{b} \neq b^{a} \) unless \( a = b \) or both are specific values (like 2 and 4). Thus, the operation is **not commutative**. **Associative:** Calculating \( (a * b) * c \): \[ (a * b) * c = (a^{b}) * c = (a^{b})^{c} = a^{bc} \] Calculating \( a * (b * c) \): \[ b * c = b^{c} \implies a * (b * c) = a^{(b^{c})} \] Since \( a^{bc} \neq a^{(b^{c})} \) in general, the operation is **not associative**. ### Summary: - i) Not commutative, not associative. - ii) Commutative, not associative. - iii) Commutative, associative. - iv) Not commutative, not associative.