Let S be the set of all real numbers except – 1 and let ‘*’ be an operation defined by a*b = a + b + ab for all ab∈S. Determine whether ‘*’ is a binary operation on ‘S’. if yes, Check its commutativity and associativity. Also, solve the equation (2*x)*3 = 7.
Given that ‘*’ is an operation that is valid on the set S which consists of all real numbers except – 1 i.e., R – { – 1} defined as a*b = a + b + ab
Let us assume a + b + ab = – 1
⇒ a + ab + b + 1 = 0
⇒ a(1 + b) + (1 + b) = 0
⇒ (a + 1)(b + 1) = 0
⇒ a = – 1 or b = – 1
But according to the problem, it is given that a≠ – 1 and b≠ – 1 so,
a + b + ab≠ – 1, so we can say that the operation ‘*’ defines a binary operation on set ‘S’.
We know that commutative property is p*q = q*p, where * is a binary operation.
Let’s check the commutativity of given binary operation:
⇒ a*b = a + b + ab
⇒ b*a = b + a + ba = a + b + ab
⇒ b*a = a*b
∴ Commutative property holds for given binary operation ‘*’ on ‘S’.
We know that associative property is (p*q)*r = p*(q*r)
Let’s check the associativity of given binary operation:
⇒ (a*b)*c = (a + b + ab)*c
⇒ (a*b)*c = a + b + ab + c + ((a + b + ab)×c)
⇒ (a*b)*c = a + b + c + ab + ac + bc + abc ...... (1)
⇒ a*(b*c) = a*(b + c + bc)
⇒ a*(b*c) = a + b + c + bc + (a×(b + c + bc))
⇒ a*(b*c) = a + b + c + ab + bc + ac + abc ...... (2)
From (1) and (2) we can clearly say that associativity holds for the binary operation ‘*’ on ‘N’.
We need to also solve for x in the given expression:
⇒ (2*x)*3 = 7
⇒ (2 + x + 2x)*3 = 7
⇒ (2 + 3x)*3 = 7
⇒ 2 + 3x + 3 + ((2 + 3x)×3) = 7
⇒ 5 + 3x + 6 + 9x = 7
⇒ 11 + 12x = 7
⇒ 12x = – 4
⇒
⇒
∴ the value of x is .