Question

Let A = ℝ × ℝ and let * be a binary operation on A defined by (a, b) * (c, d) = (ad + bc, bd) for all (a, b), (c, d) ∈ ℝ × ℝ.
(i) Show that * is commutative on A.
(ii) Show that * is associative on A.
(iii) Find the identity element of * in A.

Answer 1

(i) For (a, b), (c, d) ∈ ℝ × ℝ, we have 

\[\left( a, b \right) * \left( c, d \right) = \left( ad + bc, bd \right)\] 

\[= \left( cb + da, db \right)\left[ \because \text { Addition  and  multiplication  are  commutative } \right]\] 

\[= \left( c, d \right) * \left( a, b \right)\]

So, * is commutative on A.

(ii)
For any (a, b), (c, d), (e, f) ∈ A, we have

\[\left\{ \left( a, b \right) * \left( c, d \right) \right\} * \left( e, f \right) = \left( ad + bc, bd \right) * \left( e, f \right)\]

\[ = \left( \left( ad + bc \right)f + \left( bd \right)e, \left( bd \right)f \right)\]

\[ = \left( adf + bcf + bde, bdf \right) . . . . . (i)\]

And,

\[\left( a, b \right) * \left\{ \left( c, d \right) * \left( e, f \right) \right\} = \left( a, b \right) * \left( cf + de, df \right)\]

\[ = \left( a\left( df \right) + b\left( cf + de \right), b\left( df \right) \right)\]

\[ = \left( adf + bcf + bde, bdf \right) . . . . . (ii)\]

From (i) and (ii), we get

\[\left\{ \left( a, b \right) * \left( c, d \right) \right\} * \left( e, f \right) = \left( a, b \right) * \left\{ \left( c, d \right) * \left( e, f \right) \right\} for all \left( a, b \right), \left( c, d \right), \left( e, f \right) \in \mathbb{R} \times \mathbb{R} = A\]

So, * is associative on A.
(iii)
Let (x, y) be the identity element in A, Then,

\[\left( a, b \right) * \left( x, y \right) = \left( a, b \right) \text { for all a, b  }\in \mathbb{R}\]

\[ \Rightarrow \left( ay + bx, by \right) = \left( a, b \right) \text { for all a, b } \in \mathbb{R}\]

\[ \Rightarrow ay + bx =\text {  a and by = b for all a, b } \in \mathbb{R}\]

\[ \Rightarrow x = 0, y = 1\]

But, 0 ∉ ℝ.
Therefore, (0, 1) ∉ ℝ × ℝ = A.
Hence there is no identity element in A with respect to binary operation * on A.