This problem involves verifying which axioms of a vector space are violated based on a modified vector addition rule. Let's go through the scenario and analyze it step by step.

Modified Operations

• Vector Addition: Instead of normal vector addition, one unit is added to each coordinate. For example: $(3, 1) + (5, 0) = (9, 2) \quad \text{instead of} \quad (8, 1)$
• Scalar Multiplication: Scalar multiplication operates as expected, so multiplying a vector by a scalar $2 \times (3, 1) = (6, 2)$.

Analyzing Violations

1. Commutativity of Addition: $v + u = u + v$ This is still true even with the modified addition rule because both $v + u$ and $u + v$ will result in the same output after adding 1 to each coordinate.

2. Associativity of Addition: $v + (u + w) = (v + u) + w$ We need to check if the modified addition rule respects associativity. However, because addition consistently adds 1 to each coordinate for each pair, the order in which vectors are added does not affect the result. Thus, associativity holds.

3. Existence of a Neutral Element: The neutral element for addition should satisfy: $v + 0 = v \quad \text{and} \quad 0 + v = v$ However, under this new addition rule, adding the zero vector $(0, 0)$ results in $v + (0, 0) = (v_1 + 1, v_2 + 1)$, which is not equal to $v$. Hence, the existence of a neutral element is violated.

4. Existence of Inverses: For each vector $v$, there should exist an inverse $-v$ such that: $v + (-v) = 0$ Given the modified rule, even if we find a vector $-v$, their sum will still have 1 added to each coordinate, so we can't get back to $(0, 0)$. Thus, the existence of inverses is violated.

5. Scalar Multiplication Properties:

   • $1v = v$: This property still holds since scalar multiplication is not modified.
   • $\alpha(\beta v) = (\alpha \beta) v$: This property holds because scalar multiplication is standard.
   • Distributive properties $\alpha(v + u) = \alpha v + \alpha u$ and $(\alpha + \beta) v = \alpha v + \beta v$: These also hold since scalar multiplication is normal, and the modification only affects addition.

Correct Answers:

• Violation of the existence of a neutral element.
• Violation of the existence of inverses.

Would you like a more detailed explanation or any clarifications?

Questions:

1. Why does the existence of the neutral element fail in this modified vector space?
2. Can we find a vector inverse in this modified space, and if so, why does it fail to satisfy the inverse axiom?
3. Why does scalar multiplication still respect the original axioms despite the addition modification?
4. How would vector subtraction behave in this modified system?
5. What is the significance of the commutative property holding in this context?

Tip:

When checking if a system satisfies vector space axioms, always verify how both addition and scalar multiplication behave relative to the identity elements and inverses.