To show that the set $E = \{(x, y) \in \mathbb{R}^2 \mid x \geq 0 \text{ and } y \geq 0\}$ is not a vector space, we must verify whether $E$ satisfies the axioms of a vector space over the field $\mathbb{R}$. If any axiom is violated, then $E$ is not a vector space.

Axiom: Closed under addition

For $E$ to be closed under addition, the sum of any two vectors in $E$ must also belong to $E$.

• Let $u = (x_1, y_1) \in E$ and $v = (x_2, y_2) \in E$, where $x_1, x_2 \geq 0$ and $y_1, y_2 \geq 0$.
• The sum of $u$ and $v$ is: $u + v = (x_1 + x_2, y_1 + y_2)$ Since $x_1 + x_2 \geq 0$ and $y_1 + y_2 \geq 0$, the sum $u + v \in E$. Thus, $E$ is closed under addition.

Axiom: Closed under scalar multiplication

For $E$ to be closed under scalar multiplication, multiplying any vector in $E$ by a scalar $\alpha \in \mathbb{R}$ must result in a vector that also belongs to $E$.

• Let $u = (x, y) \in E$, where $x \geq 0$ and $y \geq 0$.

• Consider $\alpha \in \mathbb{R}$, and the scalar multiplication is: $\alpha u = (\alpha x, \alpha y)$ If $\alpha > 0$, then $\alpha x \geq 0$ and $\alpha y \geq 0$, so $\alpha u \in E$.

  However, if $\alpha < 0$, then $\alpha x < 0$ and/or $\alpha y < 0$, which violates the condition $x \geq 0$ and $y \geq 0$. Thus, $\alpha u \notin E$ for $\alpha < 0$.

Conclusion:

Since $E$ is not closed under scalar multiplication (specifically for negative scalars), $E$ does not satisfy all the axioms of a vector space. Therefore, $E$ is not a vector space.

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Would you like more details about vector space axioms or examples to expand on this? Here are some questions to explore further:

1. What are the other axioms of a vector space, and how do they apply to this problem?
2. How does scalar multiplication specifically fail for $\alpha < 0$?
3. Could $E$ form a vector space under different constraints or operations?
4. What would the set $E$ look like geometrically, and why does it fail to meet vector space requirements visually?
5. Can you provide a formal proof that $\mathbb{R}^2$ is a vector space?

Tip: Always check scalar multiplication and additive identity when testing whether a set is a vector space! These are common sources of failure.