To prove that $-(-v) = v$ for every $v$ in a vector space $V$, we need to use the properties of vectors, particularly the properties of the additive inverse and the identity element in a vector space.

Proof:

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

2. Property of the Additive Identity:
   The additive identity $0$ in $V$ has the property that for any vector $v$ in $V$: $v + 0 = v$ and $0 + v = v$

3. Additive Inverse of $-v$:
   Consider the vector $-(-v)$. By definition, $-(-v)$ is the additive inverse of $-v$, so: $-(-v) + (-v) = 0$

4. Substitute $v$ for $0$:
   From step 1, we know that $v + (-v) = 0$. Substitute this into the equation from step 3: $-(-v) + (-v) = v + (-v)$

5. Cancel $-v$ on both sides:
   Since $v + (-v) = 0$, we have: $-(-v) = v$

Thus, we have shown that $-(-v) = v$ for every vector $v$ in the vector space $V$.

Do you want details or have any questions?

Here are 5 related questions that expand on this proof:

1. How does the proof of $-(-v) = v$ change if $V$ is not a vector space but a module over a ring?
2. Can you provide examples of vectors $v$ in different vector spaces to illustrate $-(-v) = v$?
3. How does this proof work in the context of complex vector spaces?
4. What role does the uniqueness of the additive inverse play in this proof?
5. How can this proof be generalized to other algebraic structures like groups?

Tip: When dealing with vector spaces, always remember the key properties: closure under addition and scalar multiplication, the existence of an additive identity, and the existence of additive inverses. These properties are fundamental to proving various identities.