On R − 1 , a binary operation * is defined by a * b = a + b − ab. Prove that * is commutative and associative. Find the identity element for * on R − 1 . Also, prove that every element of R − 1 is invertible.

Commutativity:Let, a,b - 1 . Then, a * b = a + b - ab = b + a - ba = b * a Therefore, a * b = b * a, a,b - 1 Thus, * is commutative on R - 1 . Associativity: Let, a,b - 1 . Then, a * (b * c) = a * (b + c - bc) = a + b + c - bc - a(b + c - bc) = a + b + c - bc - ab - ac - abc (a * b) * c = (a + b - ab) * c = a + b - ab + c - (a + b - ab)c = a + b + c - ab - ac - bc + abc Therefore, a * (b * c) = (a * b) * c, a,b,c - 1 Thus, * is associative on R - 1 . Finding identity element: Let e be the element in R - 1 with respect to * such that a * e = a = e * a, a - 1 a * e = a and e * a = a, a - 1 ⇒ a + e - ae = a and e + a - ea = a, a - 1 e(1 - a) = 0, a - 1 e=0 a - 1 , a - 1 [ a 1] Thus, 0 is the identity element in R - 1 with respect to *. Finding inverse: Let a - 1 and b - 1 be the inverse of a. Then, a * b = e = b * a a * b = e and b * a = e ⇒ a + b - ab = 0 and b + a - ba = 0 ⇒ a = ab - b ⇒ a = b(a - 1) = a a-1 Thus, a a-1 is inverse of a - 1 .

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