1. **Problem Statement:** We have the set $V = \mathbb{R}^3$ with vectors $v = (a_1, a_2, a_3)$ and $w = (b_1, b_2, b_3)$ where $a_i, b_i \in \mathbb{Z}$. The operations are defined as: - Vector addition: $v + w = (a_1 + b_1, a_2 + b_2, -1)$ - Scalar multiplication: $k \cdot v = (k a_1, 2, k a_3)$ for $k \in \mathbb{R}$ We need to determine if $(V, +, \cdot)$ forms a vector space. 2. **Recall Vector Space Axioms:** A vector space must satisfy axioms including: - Existence of additive identity $0$ such that $v + 0 = v$ for all $v$. - Existence of additive inverses. - Compatibility of scalar multiplication with field multiplication. - Distributivity of scalar multiplication over vector addition. - And others. 3. **Check Additive Identity:** Suppose there exists $0 = (x, y, z)$ such that for any $v = (a_1, a_2, a_3)$, $$ v + 0 = (a_1 + x, a_2 + y, -1) = v = (a_1, a_2, a_3) $$ This implies: $$ a_1 + x = a_1 \implies x = 0 $$ $$ a_2 + y = a_2 \implies y = 0 $$ $$ -1 = a_3 $$ The last equality $-1 = a_3$ must hold for all $a_3$, which is impossible. Therefore, **no additive identity exists** in $V$ under this addition. 4. **Check Scalar Multiplication Consistency:** For scalar multiplication, the second component is always 2 regardless of $k$ or $a_2$. This violates the distributive property: $$ k \cdot (v + w) \neq k \cdot v + k \cdot w $$ because the second component of $k \cdot (v + w)$ is always 2, but the sum of second components of $k \cdot v$ and $k \cdot w$ is $2 + 2 = 4$. 5. **Conclusion:** Since the additive identity does not exist and scalar multiplication is not distributive over vector addition, the set $V$ with the given operations **does not form a vector space**. **Final answer:** $$\boxed{\text{The set } V \text{ with the given operations is NOT a vector space.}}$$