
verify all vector space axioms and determine, if the set of all triples of real numbers (x,y,z) is a vector space under the given operations (x,y,z) + (x’,y’,z’) = (x+x’+1,y+y’+2,z+z’+3) and k(x,y,z) = (kx,ky,kz)

To determine whether the set of all triples of real numbers \((x, y, z)\) is a vector space under the given operations, we need to verify if it satisfies the vector space axioms. The operations defined are:

1. **Vector Addition**: \((x, y, z) + (x', y', z') = (x + x' + 1, y + y' + 2, z + z' + 3)\)
2. **Scalar Multiplication**: \(k(x, y, z) = (kx, ky, kz)\)

Let's go through the axioms:

### 1. **Closure under Addition**
For any \((x, y, z)\) and \((x', y', z')\) in the set, \((x, y, z) + (x', y', z')\) should also be in the set.

\[
(x + x' + 1, y + y' + 2, z + z' + 3)
\]

This is a triple of real numbers, so the operation is closed under addition.

### 2. **Closure under Scalar Multiplication**
For any scalar \(k \in \mathbb{R}\) and any vector \((x, y, z)\) in the set, \(k(x, y, z)\) should also be in the set.

\[
k(x, y, z) = (kx, ky, kz)
\]

This is also a triple of real numbers, so the operation is closed under scalar multiplication.

### 3. **Associativity of Addition**
For all \((x_1, y_1, z_1)\), \((x_2, y_2, z_2)\), and \((x_3, y_3, z_3)\) in the set:

\[
[(x_1, y_1, z_1) + (x_2, y_2, z_2)] + (x_3, y_3, z_3) = (x_1, y_1, z_1) + [(x_2, y_2, z_2) + (x_3, y_3, z_3)]
\]

Let's compute both sides:

- Left-hand side:
\[
[(x_1, y_1, z_1) + (x_2, y_2, z_2)] + (x_3, y_3, z_3) = (x_1 + x_2 + 1, y_1 + y_2 + 2, z_1 + z_2 + 3) + (x_3, y_3, z_3)
\]
\[
= (x_1 + x_2 + x_3 + 2, y_1 + y_2 + y_3 + 4, z_1 + z_2 + z_3 + 6)
\]

- Right-hand side:
\[
(x_1, y_1, z_1) + [(x_2, y_2, z_2) + (x_3, y_3, z_3)] = (x_1, y_1, z_1) + (x_2 + x_3 + 1, y_2 + y_3 + 2, z_2 + z_3 + 3)
\]
\[
= (x_1 + x_2 + x_3 + 2, y_1 + y_2 + y_3 + 4, z_1 + z_2 + z_3 + 6)
\]

Both sides match, so the addition is associative.

### 4. **Existence of an Additive Identity**
There must be an element \(\mathbf{0} = (0, 0, 0)\) such that for every \((x, y, z)\):

\[
(x, y, z) + (0, 0, 0) = (x, y, z)
\]

Let's compute:

\[
(x, y, z) + (0, 0, 0) = (x + 0 + 1, y + 0 + 2, z + 0 + 3) = (x + 1, y + 2, z + 3)
\]

This does not yield \((x, y, z)\), so there is no additive identity. The required identity would be \((-1, -2, -3)\), but this contradicts the standard vector space axiom since it's not \((0, 0, 0)\).

### 5. **Existence of Additive Inverses**
For each \((x, y, z)\), there must exist \((-x, -y, -z)\) such that:

\[
(x, y, z) + (-x, -y, -z) = (0, 0, 0)
\]

Let's compute:

\[
(x, y, z) + (-x, -y, -z) = (x - x + 1, y - y + 2, z - z + 3) = (1, 2, 3)
\]

The result is \((1, 2, 3)\), not \((0, 0, 0)\), so there are no additive inverses.

### 6. **Distributive Property of Scalar Multiplication over Vector Addition**
For a scalar \(k\) and vectors \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\):

\[
k[(x_1, y_1, z_1) + (x_2, y_2, z_2)] = k(x_1 + x_2 + 1, y_1 + y_2 + 2, z_1 + z_2 + 3)
\]
\[
= (k(x_1 + x_2 + 1), k(y_1 + y_2 + 2), k(z_1 + z_2 + 3))
\]

On the other hand:

\[
k(x_1\[(k(x_1, y_1, z_1)) + k(x_2, y_2, z_2)\]
\[
= (kx_1, ky_1, kz_1) + (kx_2, ky_2, kz_2)
\]
\[
= (kx_1 + kx_2 + 1, ky_1 + ky_2 + 2, kz_1 + kz_2 + 3)
\]

To verify the distributive property:

- The left-hand side is:
\[
k[(x_1 + x_2 + 1), (y_1 + y_2 + 2), (z_1 + z_2 + 3)] = (k(x_1 + x_2 + 1), k(y_1 + y_2 + 2), k(z_1 + z_2 + 3))
\]

- The right-hand side is:
\[
(kx_1 + kx_2 + 1, ky_1 + ky_2 + 2, kz_1 + kz_2 + 3)
\]

These two expressions are not equal unless \(k = 1\), meaning the distributive property fails in general.

### 7. **Distributive Property of Scalar Multiplication over Scalar Addition**
For scalars \(k\) and \(m\) and a vector \((x, y, z)\):

\[
(k + m)(x, y, z) = ((k + m)x, (k + m)y, (k + m)z)
\]
\[
k(x, y, z) + m(x, y, z) = (kx, ky, kz) + (mx, my, mz)
\]
\[
= (kx + mx + 1, ky + my + 2, kz + mz + 3)
\]

The left-hand side \(((k + m)x, (k + m)y, (k + m)z)\) does not equal the right-hand side unless \(k + m = 1\), which is not generally true. Thus, the distributive property also fails.

### 8. **Associativity of Scalar Multiplication**
For any scalars \(k\) and \(m\) and a vector \((x, y, z)\):

\[
k(m(x, y, z)) = (k(mx), k(my), k(mz)) = ((km)x, (km)y, (km)z)
\]

This holds true as it satisfies the associativity of scalar multiplication.

### 9. **Identity Element of Scalar Multiplication**
For the scalar \(1\) and any vector \((x, y, z)\):

\[
1(x, y, z) = (1x, 1y, 1z) = (x, y, z)
\]

This holds true, so the identity element of scalar multiplication exists.

### Conclusion
The set of all triples of real numbers \((x, y, z)\) with the given operations **is not** a vector space. The failures occur due to the lack of a proper additive identity, lack of additive inverses, and violations of the distributive properties. These essential properties are necessary for a set to be considered a vector space, and without them, the set cannot be classified as such.

---

Would you like further details on any specific part of this verification process or have any questions?

Here are five related questions you might explore next:
1. What are examples of vector spaces under non-standard operations?
2. How does changing the operations affect the properties of a vector space?
3. Can a vector space exist with non-standard definitions of addition or scalar multiplication?
4. How does the concept of subspaces relate to the verification of vector space axioms?
5. What are the implications of failing specific vector space axioms in practical applications?

**Tip:** When verifying vector space properties, always start with the basic axioms like closure under addition and scalar multiplication, as these are foundational to determining if a set is a vector space.

Explore whether the set of all triples of real numbers (x, y, z), under the operations (x, y, z) + (x', y', z') = (x + x' + 1, y + y' + 2, z + z' + 3) and k(x, y, z) = (kx, ky, kz), forms a vector space. Analyze closure under addition, scalar multiplication, associativity, existence of additive identity and inverses, and distributive properties.

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