Short Answer TypeState whether the following statements are true or false. Justify.
(i) For an arbitrary binary operation * on a set N, a*a = a ∀ a ∈N.
(ii) If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a.
Consider a binary operation * on N defined as a * b = a3 + b3 . Choose the correct answer.
(A) Is * both associative and commutative ?
(B) Is * commutative but not associative?
(C) Is * associative but not commutative?
(D) Is * neither commutative nor associative?
Let A = Q x Q. Let be a binary operation on A defined by (a, b) * (c, d) = (ac, ad + b).
Then
(i) Find the identify element of (A, *)
(ii) Find the invertible elements of (A, *)
Define binary operation ‘*’ on Q as follows : a * b = a + b – ab, a, b∈ Q
(i) Find the identity element of (Q, *).
(ii) Which elements in (Q. *) are invertible?
(i) Let e ∈ Q be identifies element of (Q. *)
∴ a * e = a ∀ a ∈ Q ⇒ a + e – ae = a ⇒ (1 – a) e = 0 ⇒ e = 0 as 1 – a ≠ 0 for all a ∈ Q.
Now a * 0 = a + 0 – a. 0 = a and 0 * a = 0 + a – 0 . a = a ∴ a * 0 = 0 * a = a
∴ 0 is the identity element of (Q, *).
(ii) Let a ∈ Q be invertible. Therefore, there exists b ∈ Q such that a * b = 0 ⇒ a + b – ab ⇒ ab – b = a ⇒ (a – 1) b = a
Also 1 ∈ Q has no inverse, for if b is the inverse of 1, then I * b = 0 ⇒ 1 + b – b = 0 ⇒ 1 = 0, which is absurd each element a ∈ Q except a = 1 is invertible.
is the inverse of a ≠ 0 for the multiplication operation ‘x’ on R.
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