Let's break down each part step by step with explanations and the final answer at the end. --- ### **a. Operation \(*\) on \( \mathbb{R} \) defined by \( a * b = a + b - ab^2 \)** #### **Step 1: Find the inverse element, given the identity element is 0** **Explanation:** - The identity element \( e = 0 \) means \( a * 0 = a \) for all \( a \). - \( a * b = a + b - ab^2 \) - Let \( x \) be the inverse of \( a \) such that \( a * x = 0 \). So, \[ a * x = a + x - a x^2 = 0 \] \[ a + x - a x^2 = 0 \] \[ x - a x^2 = -a \] \[ x(1 - a x) = -a \] Solve for \( x \): If \( a \neq 0 \): \[ x(1 - a x) = -a \] \[ x - a x^2 = -a \] \[ a x^2 - x - a = 0 \] This is a quadratic in \( x \): \[ a x^2 - x - a = 0 \] \[ x = \frac{1 \pm \sqrt{1 + 4a^2}}{2a} \] If \( a = 0 \): \[ 0 * x = 0 + x - 0 = x = 0 \] So, the inverse of 0 is 0. **Step 1 Final Answer:** - The inverse element of \( a \) is \( x = \frac{1 \pm \sqrt{1 + 4a^2}}{2a} \) if \( a \neq 0 \), and 0 if \( a = 0 \). --- #### **Step 2: Determine if it is a semigroup (i.e., is it associative?)** **Explanation:** - Check if \((a * b) * c = a * (b * c)\) for all \(a, b, c \in \mathbb{R}\). Compute both sides: **Left side:** \[ (a * b) * c = [a + b - ab^2] * c = (a + b - ab^2) + c - (a + b - ab^2) c^2 \] \[ = a + b - ab^2 + c - (a + b - ab^2)c^2 \] \[ = a + b - ab^2 + c - a c^2 - b c^2 + ab^2 c^2 \] **Right side:** \[ a * (b * c) = a * (b + c - b c^2) = a + [b + c - b c^2] - a[b + c - b c^2]^2 \] Let’s expand \([b + c - b c^2]^2\): First, \[ b + c - b c^2 = (b + c) - b c^2 \] So, \[ [b + c - b c^2]^2 = (b + c)^2 - 2(b + c) b c^2 + (b c^2)^2 \] \[ = b^2 + 2bc + c^2 - 2b^2 c^2 - 2b c^3 + b^2 c^4 \] So, \[ a * (b * c) = a + b + c - b c^2 - a[b^2 + 2bc + c^2 - 2b^2 c^2 - 2b c^3 + b^2 c^4] \] Comparing both sides, the expressions are not generally equal. Hence, the operation is **not associative**. **Step 2 Final Answer:** - The operation is **not a semigroup** because it is **not associative**. --- ### **b. State Table and DFA** #### **Step 1: Draw the transition diagram** **Explanation:** - States: \( S = \{\sigma_0, \sigma_1, \sigma_2, \sigma_3\} \) - Alphabet: \( I = \{a, b, c\} \) - Start state: \( \sigma_0 \) - Final states: \( F = \{\sigma_1, \sigma_3\} \) **Transition Table:** | S | a | b | c | |--------|--------|--------|--------| | σ₀ | σ₂ | σ₀ | σ₁ | | σ₁ | σ₁ | σ₃ | σ₀ | | σ₂ | σ₃ | σ₂ | σ₀ | | σ₃ | σ₁ | σ₂ | σ₀ | **Transition diagram:** - From σ₀: a→σ₂, b→σ₀, c→σ₁ - From σ₁: a→σ₁, b→σ₃, c→σ₀ - From σ₂: a→σ₃, b→σ₂, c→σ₀ - From σ₃: a→σ₁, b→σ₂, c→σ₀ *Draw as a diagram with arrows as above.* --- #### **Step 2: Check if the strings are accepted** Let’s trace each string starting from σ₀: **String 1: "cccab"** - σ₀ → c → σ₁ - σ₁ → c → σ₀ - σ₀ → c → σ₁ - σ₁ → a → σ₁ - σ₁ → b → σ₃ Ends at σ₃, which is a final state. **Accepted**. --- **String 2: "bcbaa"** - σ₀ → b → σ₀ - σ₀ → c → σ₁ - σ₁ → b → σ₃ - σ₃ → a → σ₁ - σ₁ → a → σ₁ Ends at σ₁, which is a final state. **Accepted**. --- ### **Final Answers:** #### a. i. The inverse of \( a \) is \( x = \frac{1 \pm \sqrt{1 + 4a^2}}{2a} \) if \( a \neq 0 \), and 0 if \( a = 0 \). ii. The operation is **not a semigroup** (not associative). #### b. i. Transition diagram as described above. ii. Both strings **"cccab"** and **"bcbaa"** are **accepted** by the DFA.