1. **Problem statement:** Show that $X=\mathbb{R}^2$ is a normed linear space with a given norm. 2. **Definition of normed linear space:** A normed linear space is a vector space $V$ over $\mathbb{R}$ or $\mathbb{C}$ equipped with a norm $\|\cdot\|$ that satisfies for all $x,y \in V$ and scalar $\alpha$: - $\|x\| \geq 0$ and $\|x\|=0$ iff $x=0$ - $\|\alpha x\| = |\alpha| \|x\|$ - $\|x+y\| \leq \|x\| + \|y\|$ (triangle inequality) 3. **Vector space:** $\mathbb{R}^2$ is a vector space with usual addition and scalar multiplication. 4. **Norm definition:** Consider the norm $\|x\| = \sqrt{x_1^2 + x_2^2}$ for $x=(x_1,x_2) \in \mathbb{R}^2$. 5. **Check norm properties:** - Non-negativity: $\|x\| = \sqrt{x_1^2 + x_2^2} \geq 0$ and $\|x\|=0$ iff $x_1=0$ and $x_2=0$ so $x=0$. - Scalar multiplication: $\|\alpha x\| = \sqrt{(\alpha x_1)^2 + (\alpha x_2)^2} = \sqrt{\alpha^2 (x_1^2 + x_2^2)} = |\alpha| \sqrt{x_1^2 + x_2^2} = |\alpha| \|x\|$. - Triangle inequality: For $x,y \in \mathbb{R}^2$, $\|x+y\| = \sqrt{(x_1 + y_1)^2 + (x_2 + y_2)^2} \leq \sqrt{(x_1^2 + x_2^2)} + \sqrt{(y_1^2 + y_2^2)} = \|x\| + \|y\|$ by Minkowski inequality. 6. **Conclusion:** Since $\mathbb{R}^2$ with this norm satisfies all norm properties, it is a normed linear space. Final answer: $\mathbb{R}^2$ with norm $\|x\| = \sqrt{x_1^2 + x_2^2}$ is a normed linear space.