Let V=(5,∞)
. For u,v∈V
 and a∈R
 define vector addition by u⊞v:=uv−5(u+v)+30
 and scalar multiplication by a⊡u:=(u−5)a+5
. It can be shown that (V,⊞,⊡)
 is a vector space over the scalar field R
. Find the following: the additive inverse of x
:

⊟x=

We are given a vector space \( (V, \oplus, \otimes) \) with non-standard operations defined as:

1. **Vector addition:** \( u \oplus v := uv - 5(u + v) + 30 \)
2. **Scalar multiplication:** \( a \otimes u := (u - 5)a + 5 \)

We are tasked with finding the **additive inverse** of an element \( x \), denoted \( \ominus x \), which satisfies the equation:

\[
x \oplus (\ominus x) = 0
\]

Here, 0 is the additive identity of the vector space. The additive identity in a vector space is the element such that:

\[
x \oplus 0 = x \quad \text{for all} \quad x \in V
\]

Let’s first determine the additive identity by solving for \( 0 \). We set \( u \oplus 0 = u \) and use the definition of \( \oplus \):

\[
u \oplus 0 = u \cdot 0 - 5(u + 0) + 30 = u
\]

Simplifying:

\[
0 - 5u - 5 \cdot 0 + 30 = u
\]

\[
-5u + 30 = u
\]

\[
30 = 6u
\]

\[
u = 5
\]

Thus, the additive identity is \( 5 \).

### Step 1: Find the Additive Inverse

Now we find \( \ominus x \) such that \( x \oplus (\ominus x) = 5 \). Using the definition of \( \oplus \), we get:

\[
x \oplus (\ominus x) = x \cdot (\ominus x) - 5(x + \ominus x) + 30 = 5
\]

Simplify:

\[
x \cdot (\ominus x) - 5(x + \ominus x) + 30 = 5
\]

\[
x \cdot (\ominus x) - 5(x + \ominus x) = -25
\]

Let \( y = \ominus x \). We now have the equation:

\[
xy - 5(x + y) = -25
\]

Expanding:

\[
xy - 5x - 5y = -25
\]

Rearranging:

\[
xy - 5x - 5y = -25
\]

We need to solve this equation for \( y \) in terms of \( x \). Factoring it can help:

\[
(x - 5)(y - 5) = 25
\]

Thus, we find:

\[
y - 5 = \frac{25}{x - 5}
\]

\[
y = 5 + \frac{25}{x - 5}
\]

Hence, the additive inverse of \( x \) is:

\[
\ominus x = 5 + \frac{25}{x - 5}
\]

---

Would you like a further breakdown of the steps? Here are 5 related questions:

1. How does scalar multiplication behave in this vector space?
2. What are the properties of the non-standard vector addition in this space?
3. How would you find the scalar multiple of a vector using this operation?
4. Can you derive the identity for scalar multiplication in this vector space?
5. How would this operation affect the solution of a system of linear equations?

**Tip**: In unusual vector spaces like this, always start by understanding how the identity elements behave under the given operations.

Let V=(5,∞). For u,v∈V and a∈R define vector addition by u⊞v:=uv−5(u+v)+30 and scalar multiplication by a⊡u:=(u−5)a+5. It can be shown that (V,⊞,⊡) is a vector space over the scalar field R. Find the following: the additive inverse of x: ⊟x=

Learn how to find the additive inverse of an element in a vector space with non-standard operations. This problem covers vector addition, scalar multiplication, and the properties of vector spaces with detailed steps and formulas. Suitable for undergraduate students in linear algebra.

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