To prove that \(-(-v) = v\) for every \(v\) in a vector space \(V\), we will use the properties of vector spaces, particularly the property of additive inverses.

### Step 1: Understand the definition of the additive inverse
In a vector space \(V\), for any vector \(v\), there exists a vector \(-v\) such that:
\[
v + (-v) = 0
\]
where \(0\) is the zero vector in \(V\).

### Step 2: Apply the definition of the additive inverse to \(-v\)
Now consider the vector \(-v\). By the definition of the additive inverse, there exists a vector \(-(-v)\) such that:
\[
(-v) + (-(-v)) = 0
\]

### Step 3: Add \(v\) to both sides of the equation
To isolate \(-(-v)\), we add \(v\) to both sides of the equation:
\[
v + \left[(-v) + (-(-v))\right] = v + 0
\]

### Step 4: Use the associative property of vector addition
The associative property of vector addition states that:
\[
(v + (-v)) + (-(-v)) = v
\]
Since \(v + (-v) = 0\), the equation simplifies to:
\[
0 + (-(-v)) = v
\]

### Step 5: Simplify the equation
Adding the zero vector to any vector \(w\) in \(V\) results in the vector \(w\):
\[
-(-v) = v
\]

### Conclusion
We have shown that \(-(-v) = v\) for every \(v\) in the vector space \(V\), as required.

Would you like further details or have any questions? Here are some related problems to consider:

1. Prove that the additive inverse of the zero vector is the zero vector itself.
2. Show that the additive inverse in a vector space is unique.
3. Prove that for any vector \(v\) in a vector space \(V\), the equation \(v + v = 2v\) holds.
4. Demonstrate that the scalar multiplication \(0 \cdot v = 0\) for any vector \(v\) in a vector

Explore the proof that -(-v) = v for every vector v in a vector space V. Understand the properties of vector spaces and additive inverses with a detailed step-by-step solution. Suitable for advanced high school students and beyond.

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