2. Define an operation on the set of real numbers by \( a * b=b^{a} \) i). Is : a binary operation on the set of real numbers? Give reason for you Answer. ii). Is the operation commutative? iii). Evaluate \( (3,2) \cdot-2 \) 3. Consider the binary operation \( a \cdot b=a+b-2 a b \), where \( a \) and \( b \) are real numbers i) Is * a binary operation on the set of real numbers? Give reason for your answ ii) is the operation * commutative? If not give a counter example.

We are given two different binary operations on ℝ. Recall that for an operation “*” on ℝ to be a binary operation it must be defined for every pair of real numbers (i.e. it must be “closed” on ℝ), and then we can ask properties such as commutativity. ────────────────────────────── 1. Operation 1: a * b = bᵃ (a) Is it a binary operation on ℝ? To be a binary operation on ℝ, for every a and b in ℝ the number bᵃ must be a real number. However, the expression bᵃ is usually defined as exp(a·ln(b)), which is valid only when b > 0. (When b is negative, bᵃ is not defined for arbitrary real exponents; for example, if b < 0 and a is irrational, then bᵃ is not a real number.) Thus, since there exist real numbers a and b (namely, b ≤ 0 with a non‐rational exponent) for which bᵃ is not defined in ℝ, the operation is not a binary operation on ℝ. (b) Is the operation commutative? An operation “*” is commutative if a * b = b * a for all a, b in ℝ (when both sides are defined). Here:   a * b = bᵃ   b * a = aᵇ In general, bᵃ is not equal to aᵇ. For a concrete example, take a = 2 and b = 3:   2 * 3 = 3² = 9  but  3 * 2 = 2³ = 8. Since 9 ≠ 8, the operation is not commutative. (c) Evaluate (3, 2) * –2 The notation (3, 2) is meant to represent the ordered pair used in the operation. That is, (3, 2) means first we apply the operation using 3 and 2:   3 * 2 = 2³ = 8. Now we then compute   8 * (–2) = (–2)⁸. Since (–2) raised to the 8th power is a positive number (an even exponent) and (–2)⁸ = 256, we obtain   8 * (–2) = 256. ────────────────────────────── 2. Operation 2: a • b = a + b – 2ab (a) Is it a binary operation on ℝ? We have a • b = a + b – 2ab. For any real numbers a and b, the expressions a, b, and ab are real, and so is a + b – 2ab. Therefore, the operation is closed in ℝ and is a binary operation on ℝ. (b) Is the operation commutative? We check if a • b = b • a. Notice that:   a • b = a + b – 2ab,   b • a = b + a – 2ba. Since addition and multiplication of real numbers are commutative (that is, a + b = b + a and ab = ba), we have a • b = b • a for all real numbers a and b. Thus, this operation is commutative. ────────────────────────────── Summary of Answers: 2. For the operation a * b = bᵃ:  (i) It is not a binary operation on ℝ because for some choices of a and b (for instance, b negative and a irrational) bᵃ is not defined as a real number.  (ii) It is not commutative (e.g. 2 * 3 = 3² = 9 while 3 * 2 = 2³ = 8).  (iii) (3, 2) * –2: First, 3 * 2 = 2³ = 8, and then 8 * (–2) = (–2)⁸ = 256. 3. For the operation a • b = a + b – 2ab:  (i) It is a binary operation on ℝ since a + b – 2ab is real for all real numbers a and b.  (ii) It is commutative because a + b – 2ab = b + a – 2ba for all a, b in ℝ. These are the complete answers for the given problems.