1. **Problem statement:** Determine if $(\mathbb{R}^2, +, \cdot)$ with addition defined by $(a,b)+(c,d)=(a+c,b+d)$ and scalar multiplication defined by $\lambda \cdot (a,b) = (\lambda a, 0)$ is a vector space over $\mathbb{R}$. 2. **Recall vector space axioms:** A vector space over $\mathbb{R}$ must satisfy axioms including closure under addition and scalar multiplication, existence of additive identity and inverses, distributivity, associativity, and scalar identity. 3. **Check closure under scalar multiplication:** For any $\lambda \in \mathbb{R}$ and $(a,b) \in \mathbb{R}^2$, $\lambda \cdot (a,b) = (\lambda a, 0)$ is in $\mathbb{R}^2$, so closure holds. 4. **Check additive identity:** The zero vector must satisfy $(a,b)+(0,0)=(a,b)$. Here, $(0,0)$ is the additive identity. 5. **Check scalar identity:** For all $(a,b)$, $1 \cdot (a,b) = (1 \cdot a, 0) = (a,0)$. This must equal $(a,b)$ for scalar identity to hold. 6. Since $1 \cdot (a,b) = (a,0)$ differs from $(a,b)$ unless $b=0$, scalar identity axiom fails. 7. **Check distributivity over vector addition:** $\lambda \cdot ((a,b)+(c,d)) = \lambda \cdot (a+c,b+d) = (\lambda (a+c), 0)$. 8. On the other hand, $\lambda \cdot (a,b) + \lambda \cdot (c,d) = (\lambda a, 0) + (\lambda c, 0) = (\lambda a + \lambda c, 0)$. 9. Both sides equal $(\lambda (a+c), 0)$, so distributivity over vector addition holds. 10. **Check distributivity over scalar addition:** $(\lambda + \mu) \cdot (a,b) = ((\lambda + \mu) a, 0)$. 11. $\lambda \cdot (a,b) + \mu \cdot (a,b) = (\lambda a, 0) + (\mu a, 0) = (\lambda a + \mu a, 0)$. 12. Both sides equal $((\lambda + \mu) a, 0)$, so distributivity over scalar addition holds. 13. **Check associativity of scalar multiplication:** $\lambda (\mu (a,b)) = \lambda \cdot (\mu a, 0) = (\lambda \mu a, 0)$. 14. $(\lambda \mu) \cdot (a,b) = ((\lambda \mu) a, 0)$, so associativity holds. 15. **Check additive inverses:** For $(a,b)$, additive inverse is $(-a,-b)$ since $(a,b)+(-a,-b)=(0,0)$. 16. **Conclusion:** The scalar identity axiom fails because $1 \cdot (a,b) \neq (a,b)$ unless $b=0$. Therefore, $(\mathbb{R}^2, +, \cdot)$ is **not** a vector space over $\mathbb{R}$. **Final answer:** No, $(\mathbb{R}^2, +, \cdot)$ with the given operations is not a vector space over $\mathbb{R}$ because the scalar identity axiom fails.