1. **Problem statement:** Show that $\mathbb{R}^2$ is a normed linear space with the norm defined as $\|x\| = \max(|x_1|, |x_2|)$ for $x = (x_1, x_2) \in \mathbb{R}^2$. 2. **Definition of a norm:** A function $\|\cdot\| : V \to \mathbb{R}$ on a vector space $V$ is a norm if for all $x,y \in V$ and scalar $\alpha$, it satisfies: (i) $\|x\| \geq 0$ and $\|x\| = 0$ if and only if $x = 0$ (positivity). (ii) $\|\alpha x\| = |\alpha| \|x\|$ (homogeneity). (iii) $\|x + y\| \leq \|x\| + \|y\|$ (triangle inequality). 3. **Check positivity:** Since $\|x\| = \max(|x_1|, |x_2|)$, and absolute values are always non-negative, $\|x\| \geq 0$. Also, $\|x\| = 0$ means $\max(|x_1|, |x_2|) = 0$, so $|x_1| = 0$ and $|x_2| = 0$, implying $x = 0$. 4. **Check homogeneity:** For scalar $\alpha$, $$\|\alpha x\| = \max(|\alpha x_1|, |\alpha x_2|) = \max(|\alpha||x_1|, |\alpha||x_2|) = |\alpha| \max(|x_1|, |x_2|) = |\alpha| \|x\|.$$ 5. **Check triangle inequality:** For $x,y \in \mathbb{R}^2$, $$\|x + y\| = \max(|x_1 + y_1|, |x_2 + y_2|).$$ Using the triangle inequality for absolute values, $$|x_i + y_i| \leq |x_i| + |y_i|, \quad i=1,2.$$ Thus, $$\|x + y\| \leq \max(|x_1| + |y_1|, |x_2| + |y_2|) \leq \max(|x_1|, |x_2|) + \max(|y_1|, |y_2|) = \|x\| + \|y\|.$$ 6. **Conclusion:** Since all three norm properties hold, $\mathbb{R}^2$ with the norm $\|x\| = \max(|x_1|, |x_2|)$ is a normed linear space.