Question -
Answer -
(i) First we have tocheck commutativity of *
Let a, b ∈ Q – {-1}
Then a * b = a + b +ab
= b + a + ba
= b * a
Therefore,
a * b = b * a, ∀ a, b ∈ Q – {-1}
Now we have to proveassociativity of *
Let a, b, c ∈ Q – {-1}, Then,
a * (b * c) = a * (b +c + b c)
= a + (b + c + b c) +a (b + c + b c)
= a + b + c + b c + ab + a c + a b c
(a * b) * c = (a + b +a b) * c
= a + b + a b + c + (a+ b + a b) c
= a + b + a b + c + ac + b c + a b c
Therefore,
a * (b * c) = (a * b)* c, ∀ a, b, c ∈ Q – {-1}
Thus, * is associativeon Q – {-1}.
(ii) Let e be theidentity element in I+ with respect to * such that
a * e = a = e * a, ∀ a ∈ Q – {-1}
a * e = a and e * a =a, ∀ a ∈ Q – {-1}
a + e + ae = a and e +a + ea = a, ∀ a ∈ Q – {-1}
e + ae = 0 and e + ea= 0, ∀ a ∈ Q – {-1}
e (1 + a) = 0 and e (1+ a) = 0, ∀ a ∈ Q – {-1}
e = 0, ∀ a ∈ Q – {-1} [because a not equal to -1]
Thus, 0 is theidentity element in Q – {-1} with respect to *.
(iii) Let a ∈ Q – {-1} and b ∈ Q – {-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
b (1 + a) = – a Q –{-1}
b = -a/1 + a Q – {-1}[because a not equal to -1]
Thus, -a/1 + a is theinverse of a ∈ Q – {-1}