1. **Problem Statement:** Show that $X=\mathbb{R}^2$ is a normed linear space with a given norm. 2. **Definition:** A normed linear space is a vector space $X$ over $\mathbb{R}$ or $\mathbb{C}$ equipped with a norm $\|\cdot\|$ that satisfies the following properties for all $x,y \in X$ and scalar $\alpha$: - (Positivity) $\|x\| \geq 0$ and $\|x\|=0$ if and only if $x=0$. - (Scalar multiplication) $\|\alpha x\| = |\alpha| \|x\|$. - (Triangle inequality) $\|x+y\| \leq \|x\| + \|y\|$. 3. **Vector space:** $\mathbb{R}^2$ is a vector space with usual addition and scalar multiplication. 4. **Norm definition:** Define the norm on $\mathbb{R}^2$ as $$\|x\| = \sqrt{x_1^2 + x_2^2}$$ where $x = (x_1, x_2) \in \mathbb{R}^2$. 5. **Check positivity:** - $\|x\| = \sqrt{x_1^2 + x_2^2} \geq 0$ since squares are non-negative. - $\|x\|=0$ implies $x_1^2 + x_2^2=0$ which means $x_1=0$ and $x_2=0$, so $x=0$. 6. **Check scalar multiplication:** For scalar $\alpha$, $$\|\alpha x\| = \sqrt{(\alpha x_1)^2 + (\alpha x_2)^2} = \sqrt{\alpha^2 (x_1^2 + x_2^2)} = |\alpha| \sqrt{x_1^2 + x_2^2} = |\alpha| \|x\|$$ 7. **Check triangle inequality:** For $x,y \in \mathbb{R}^2$, $$\|x+y\| = \sqrt{(x_1 + y_1)^2 + (x_2 + y_2)^2} \leq \sqrt{(x_1^2 + x_2^2)} + \sqrt{(y_1^2 + y_2^2)} = \|x\| + \|y\|$$ This follows from the Cauchy-Schwarz inequality. 8. **Conclusion:** Since all norm properties hold, $\mathbb{R}^2$ with the norm $\|x\| = \sqrt{x_1^2 + x_2^2}$ is a normed linear space.