InterviewSolution
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InterviewSolution
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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}, definea*b =(a)/(b+1) |
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Answer» SOLUTION :(i) `In Z , a b = a-b` Let `a,b in Z` `therefore a * b = a - b` `""b* a = b-a` `therefore a ** bne b ** a` Therefore, operation is not commutative. LETA, b, c`in`Z. `thereforea** (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 isnot 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,cin Z^(+)` `thereforea**(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 , operationis not associative. (vi) InR-{-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 isnot 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))` `thereforea** (b**c) ne (a**b) **c` Therefore, operation is not associative. |
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