6. a) A binary operation * is defined on the set of real numbers as follows: \[\na * b=2^{-a}+b, \quad a, b \in \mathbf{R} \] (i) Is the operation * commutative? If not give a counter example. (ii) Find the value of \( -1 *(0 * 1) \) and \( (-1 * 0) * 1 \), and state whether \( * \) is associative.

To determine if the operation \( * \) is commutative, we need to check if \( a * b = b * a \) for all \( a, b \in \mathbf{R} \). Calculating \( a * b \): \[ a * b = 2^{-a} + b \] Calculating \( b * a \): \[ b * a = 2^{-b} + a \] For the operation to be commutative, both expressions must yield the same result for any \( a \) and \( b \). Let’s take a counter-example with \( a = 0 \) and \( b = 1 \): \[ 0 * 1 = 2^{-0} + 1 = 1 + 1 = 2 \] \[ 1 * 0 = 2^{-1} + 0 = \frac{1}{2} + 0 = \frac{1}{2} \] Since \( 0 * 1 \neq 1 * 0 \), the operation is not commutative. Next, we calculate \( -1 *(0 * 1) \) and \( (-1 * 0) * 1 \). First, find \( 0 * 1 \): \[ 0 * 1 = 2^{-0} + 1 = 1 + 1 = 2 \] Now, find \( -1 * 2 \): \[ -1 * 2 = 2^{-(-1)} + 2 = 2^{1} + 2 = 2 + 2 = 4 \] Next, find \( -1 * 0 \): \[ -1 * 0 = 2^{-(-1)} + 0 = 2^{1} + 0 = 2 + 0 = 2 \] Now calculate \( 2 * 1 \): \[ 2 * 1 = 2^{-2} + 1 = \frac{1}{4} + 1 = 1 + \frac{1}{4} = \frac{5}{4} \] Now we compare \( -1 *(0 * 1) = 4 \) and \( (-1 * 0) * 1 = \frac{5}{4} \). Since \( 4 \neq \frac{5}{4} \), the operation is not associative. So, in summary: (i) The operation \( * \) is not commutative since \( 0 * 1 \neq 1 * 0 \) as shown above. (ii) The operation \( * \) is not associative since \( -1 *(0 * 1) \neq (-1 * 0) * 1 \) because \( 4 \neq \frac{5}{4} \).