1. Let \( R \) be the set of real numbers. Define an operation * on real numbers fy \( a \cdot b=a-2 b \) for all \( a, b \in \mathbb{R} \). (ii is the operation binary on \( R \) ? (ii) Calculate \( 2 \cdot 1 \) (ii) is the operation associative?

We are given an operation ⋅ on the set of real numbers ℝ defined by   a ⋅ b = a − 2b  for all a, b ∈ ℝ. Now we answer each part: 1. Is the operation binary on ℝ?   A binary operation on ℝ requires that for every pair of real numbers a and b, the result a − 2b is also a real number. Since subtraction and scalar multiplication (multiplication by 2) are operations that yield a real number when applied to real numbers, a − 2b is a real number.   Answer: Yes, the operation is binary on ℝ. 2. Calculate 2 ⋅ 1   Using the definition:   2 ⋅ 1 = 2 − 2(1) = 2 − 2 = 0.   Answer: 2 ⋅ 1 = 0. 3. Is the operation associative?   An operation is associative if for all a, b, c ∈ ℝ, we have:     (a ⋅ b) ⋅ c = a ⋅ (b ⋅ c).   Let's compute both sides:   Left-hand side:     (a ⋅ b) ⋅ c = (a − 2b) ⋅ c       = (a − 2b) − 2c       = a − 2b − 2c.   Right-hand side:     a ⋅ (b ⋅ c) = a ⋅ (b − 2c)       = a − 2(b − 2c)       = a − 2b + 4c.   Since a − 2b − 2c is not equal to a − 2b + 4c for all c (unless 6c = 0, which only happens when c = 0), the operation is not associative.   Answer: The operation is not associative. Summary:   (i) The operation is binary on ℝ.   (ii) 2 ⋅ 1 = 0.   (iii) The operation is not associative.