(i) Let us check the commutativity of *

Let a, b ∈ Q – {-1}

Then a * b = a + b + ab

= b + a + ba

= b * a

So,

a * b = b * a, ∀ a, b ∈ Q – {-1}

Let us prove that associativity 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 + a b + 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 + a c + b c + a b c

So,

a * (b * c) = (a * b) * c, ∀ a, b, c ∈ Q – {-1}

So, * is associative on Q – {-1}.

(ii) Let e be the identity 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 ! = -1]

Thus, 0 is the identity element in Q – {-1} with respect to *.

(iii) Let a 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 ! = -1]

So, -a/1 + a is the inverse of a ∈ Q – {-1}