Question

For each operation * defined below, determine whether * is binary, 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 `=(ab)/(2)`
(iv) On `Z^(+)`, define a*b `= (a)/(b+1)`
(v) On `Z^(+)`, define a*b =`a^(b)`
(vi) On R- {-1}, define a*b `=(a)/(b+1)`

Answer 1

(i) `In Z , a b = a-b`
Let `a,b in Z`
`therefore a * b = a - b`
`" "b* a = b-a`
`therefore a ** b ne b ** a`
Therefore, operation is not commutative.
Let a, b, c`in`Z.
`therefore a** (b**c) =a **(b-c) =a -(b-c) = a- b +c`
` = a- b+c`
` and (a**b) ** c = (a-b)**c = a-b-c`
`therefore a** b(b**c) ne (a**b) **c`
Therefore, operation is not associative .
(ii) In Q, a*b = ab +1
Let a,b `in` Q
`therefore a* b= ab + 1=ab +1 = b*a`
Therefore,operation is commutative .
Let a,b, `in` Q
`therefore" " a**(b**C) = a ** (bc +1)`
`= a(bc + 1) **c`
`= (ab+1) c+1 = abc + c+ 1`
`ne a**(b**c)`
Therefore, operation is not associative.
(iii) In Q `" "a**b=(ab)/(2)`
Let `a,b, in Q`
`therefore" "a**b = (ab)/(2) = (ba)/(2) = b**a`
Therefore , operation is commutative.
Let a,b, `in` Q
` therefore a **(b**c) = a**((bc)/(2)) = (a((bc)/(2)))/(2) = (((ab)/(c))c)/(2)`
`= ((a**b)c)/(2) = (a**b)**c`
Therefore, operation is commutative.
Let `a,bc,in, Z^(+)`
`because a**(b**c) = a**(2^(bc)) = 2^(ba) = b**a`
and `(a **b) ** c = (2^(ab)) ** c = 2^(ab_(c)) ne a ** (b**c)`
Therefore, operation is not associative.
(v) In `Z^(+), " "a**b = a^(b)`
Let `a,b in Z^(+)`
`therefore" " a**b = a^(b) ne b^(a) ne b**a`
Therefore, operation is not commutative .
Let `a, b,c in Z^(+)`
`therefore a**(b**c)= a **(b^(c)) ** (b^(c)) = a^((b^(c))`
and `(a**b) **c = (a^(b)) **c = (a^(b))^(c) = a^(bc)`
`therefore " "a**(b**c) ne (a**b) **c`
Therefore , operation is not associative.
(vi) In R-{-1} a, `b = (a)/(b+1)`
Let a, b `in R - {-1}`
`therefore " "a**b = (a)/(b+1) ne (a)/(a+1) ne b *a`
Therefore, operation is not commutative.
Let a,b,c `in R - {-1}`
`therefore a**(b**c) = a**((b)/(c+1)) = (a)/(((b)/(c+1))) = (a(c+1))/(b+c+1)`
and `(a**b)**c ((a)/(b+1)) **c ((a)/(b+1))/(c+1) = (a)/((b+1)(c+1))`
`therefore a** (b**c) ne (a**b) **c`
Therefore, operation is not associative.