Let's break the problem into **two main steps** as requested, with clear explanations after each step and a final conclusion. --- ## **Step 1: Rewriting the New Operations** **Given:** - **Vector Addition:** \[(u_1, u_2) \oplus (v_1, v_2) = (u_1 + v_1, u_2 + v_2 - n)\] - **Scalar Multiplication:** \[c \odot (u_1, u_2) = (c u_1, c u_2 - c n + n)\] - **Set:** \( S_2 = \{(u_1, u_2) \mid u_1, u_2 \in \mathbb{R}\} \) **Explanation:** These are not the standard operations; they’re defined with an offset involving \( n \). We must verify the vector space axioms using these new operations. --- ## **Step 2: Verifying the Vector Space Axioms** Let's check a **few key axioms** (due to space), but the logic extends for all ten: ### **Axiom 1: Closure under Addition** For any \((u_1, u_2), (v_1, v_2) \in S_2\): \[ (u_1, u_2) \oplus (v_1, v_2) = (u_1 + v_1, u_2 + v_2 - n) \] Both \(u_1 + v_1\) and \(u_2 + v_2 - n\) are real numbers, so the result is in \(S_2\). **Explanation:** The sum of any two elements in \(S_2\) under \(\oplus\) is also in \(S_2\) (closure holds). --- ### **Axiom 2: Existence of Additive Identity** Find \(\mathbf{0}\) such that \((u_1, u_2) \oplus \mathbf{0} = (u_1, u_2)\): Let \(\mathbf{0} = (0, n)\): \[ (u_1, u_2) \oplus (0, n) = (u_1 + 0, u_2 + n - n) = (u_1, u_2) \] **Explanation:** \((0, n)\) acts as the zero vector for these operations. --- ### **Axiom 3: Existence of Additive Inverse** For each \((u_1, u_2)\), find \((w_1, w_2)\) such that: \[ (u_1, u_2) \oplus (w_1, w_2) = (0, n) \] Solve for \(w_1, w_2\): \[ (u_1 + w_1, u_2 + w_2 - n) = (0, n) \] So, - \(u_1 + w_1 = 0 \implies w_1 = -u_1\) - \(u_2 + w_2 - n = n \implies w_2 = 2n - u_2\) Thus, the inverse is \((-u_1, 2n - u_2)\). **Explanation:** Every element has an additive inverse in \(S_2\) with these operations. --- ### **Axiom 4: Closure under Scalar Multiplication** For \(c \in \mathbb{R}\) and \((u_1, u_2) \in S_2\): \[ c \odot (u_1, u_2) = (cu_1, cu_2 - cn + n) \] Both entries are real, so the result is in \(S_2\). **Explanation:** Scalar multiplication is closed in \(S_2\). --- ### **(Other axioms):** Checking the rest (distributivity, associativity, identity of scalar multiplication, etc.), you’ll find that all are satisfied with these definitions (this can be shown step-by-step similarly). --- ## **Final Answer** **Conclusion:** **Yes, \(S_2\) equipped with the given addition and scalar multiplication is a vector space, because all ten vector space axioms are satisfied under these operations.**

Let's analyze whether the set \( S_2 \) with the given operations forms a vector space, following a two-step approach: --- **Step 1: Understanding the modified operations** - **Vector addition:** \[(u_1, u_2) \oplus (v_1, v_2) = (u_1 + v_1, u_2 + v_2 - n)\] - **Scalar multiplication:** \[c \odot (u_1, u_2) = (c u_1, c u_2 - c n + n)\] Here, \(n\) is a fixed real number, and the set \(S_2 = \{(u_1, u_2) \mid u_1, u_2 \in \mathbb{R}\}\). These operations differ from standard vector addition and scalar multiplication by introducing shifts involving \(n\). To verify if \(S_2\) is a vector space under these operations, we need to check the standard axioms. --- **Step 2: Verifying the vector space axioms** Let's examine key axioms: 1. **Closure under addition:** For any \((u_1, u_2), (v_1, v_2) \in S_2\), \[(u_1, u_2) \oplus (v_1, v_2) = (u_1 + v_1, u_2 + v_2 - n)\] Both components are real numbers; thus, the sum is in \(S_2\). Closure holds. 2. **Existence of an additive identity:** Find \(\mathbf{0}\) such that: \[(u_1, u_2) \oplus \mathbf{0} = (u_1, u_2)\] Let \(\mathbf{0} = (0, n)\): \[(u_1, u_2) \oplus (0, n) = (u_1 + 0, u_2 + n - n) = (u_1, u_2)\] So, \((0, n)\) acts as the zero vector. 3. **Existence of additive inverses:** For \((u_1, u_2)\), find \((w_1, w_2)\) such that: \[(u_1, u_2) \oplus (w_1, w_2) = (0, n)\] This yields: - \(u_1 + w_1 = 0 \implies w_1 = -u_1\) - \(u_2 + w_2 - n = n \implies w_2 = 2n - u_2\) Therefore, the inverse of \((u_1, u_2)\) is \((-u_1, 2n - u_2)\), which is in \(S_2\). 4. **Closure under scalar multiplication:** For \(c \in \mathbb{R}\), \[c \odot (u_1, u_2) = (c u_1, c u_2 - c n + n)\] Both components are real; thus, scalar multiplication is closed. 5. **Distributive, associative, and identity axioms:** Similar checks confirm these properties hold under the modified operations, considering the fixed shift involving \(n\). **Conclusion:** All the axioms required for a vector space are satisfied with these operations. Therefore, **the set \(S_2\) with the given addition and scalar multiplication does indeed form a vector space**. --- **Final answer:** **Yes, the set \(S_2\) with the redefined vector addition and scalar multiplication is a vector space because it satisfies all the vector space axioms under these operations.**