To prove that $-(-v) = v$ for any vector $v$ in a vector space $V$, we can use the properties of vector spaces, specifically the existence of additive inverses and the definition of the additive identity. Here is the proof:

Proof:

1. Additive Inverse:
   For any vector $v$ in $V$, there exists a vector $-v$ in $V$ such that: $v + (-v) = 0$ where $0$ is the additive identity in $V$.

2. Double Negation:
   Consider the vector $-(-v)$ in $V$. We want to show that $-(-v)$ is equal to $v$.

3. Using Additive Inverse:
   By the definition of the additive inverse, $-(-v)$ is the vector that, when added to $-v$, gives the additive identity $0$. Therefore: $-(-v) + (-v) = 0$

4. Substitute $v$ for $-v$:
   From step 1, we know that $v + (-v) = 0$. By substituting $-v$ with $v$ in the equation: $-(-v) + (-v) = 0$

5. Add $v$ to Both Sides:
   Add $v$ to both sides of the equation to get: $(-(-v) + (-v)) + v = 0 + v$ Simplifying, we use the associative property of addition: $-(-v) + (-v + v) = v$

6. Simplify the Expression:
   Since $(-v + v) = 0$ (by the definition of the additive inverse): $-(-v) + 0 = v$ And since adding the additive identity $0$ to any vector does not change the vector: $-(-v) = v$

This completes the proof. Therefore, $-(-v) = v$ for any vector $v$ in $V$.

Would you like to see more details on any specific part of this proof or have any questions?

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Here are five related questions you might find interesting:

1. How does the proof of $-(-v) = v$ relate to other properties of vector spaces?
2. Can this proof be extended to other algebraic structures like groups?
3. What is the significance of the additive identity in vector spaces?
4. How does this concept apply to matrices or functions as elements of vector spaces?
5. How does the concept of the additive inverse differ in different fields, such as real numbers vs. complex numbers?

Tip: When working with vector spaces, always keep the basic properties in mind (e.g., the existence of an additive identity and inverses). These fundamental concepts often simplify more complex proofs.