We have,
a * b = a + b – ab ∀ a, b ∈ Q – {1}, then
(i) Commutative : Let a, b ∈ Q – {1}
Now, a * b = a + b – ab
b * a = b + a – ba = a + b – ab
[∵ Commutative law holds for + & ×]
Hence, a * b = b * a
i.e., ‘*’ is commutative.
(ii) Associative : Let a, b, c ∈ Q – {1}
Now,
(a * b) * c = (a + b – ab) * c = a + b – ab + c – ac – bc + abc
a * (b * c) = a * (b + c – bc)
i.e., (a * b) * c = a * (b * c)
Hence, ‘*’ is associative.
(iii) Identity : Let e be the identity element.
Then, ∀ a ∈ Q – {1}, we have
a * e = a
⇒ a + e – ae = a
⇒ (1 – a) e = 0
⇒ e = 0 ∈ Q – {1} [∵ a ≠ 1 ⇒ 1 – a ≠ 0]
Now, a * 0 = a + 0 – a × 0 = a
0 * a = 0 + a – 0 × a = a
Thus, 0 is the identity element in Q – {1}.
(iv) Inverse : Let b be the inverse element of a,
for each a ∈ Q – {1}.
Then a * b = e = 0
⇒ a * b = 0
⇒ b(a – 1) = a
⇒ \(b=\frac{a}{a-1}∈Q-\){1}
Therefore, for each a the corresponding inverse element is
\(\frac{a}{a-1}∈Q-\){1}.
