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The law of demand always holds if the walrasian demand function $x(\mathbf{p},w)$ satisfies the weak axiom of revealed preference (WARP), Walras' law, homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth.

Suppose that the Walrasian demand function $x(\mathbf{p},w)$ satisfies the weak axiom of revealed preference (WARP), Walras' law, homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth $w$. We want to prove that, for any price change from $\mathbf{p}$ to $\mathbf{p}'$ (without change in wealth), we have \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',w) - x(\mathbf{p},w)] \leq 0.\tag1 \end{align*} By homogeneity of degree 1 in $w$, we shall prove \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] \leq 0\tag2 \end{align*} for all $\mathbf{p}$ and $\mathbf{p}'$. We consider the following cases:

Case 1: Suppose that $x(\mathbf{p},1) = x(\mathbf{p}',1)$. Then the inequality $(2)$ always holds with $(\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] = 0$.

From now on, suppose that $x(\mathbf{p},1) \neq x(\mathbf{p}',1)$.

Case 2: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) > 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) > 1$. Notice that, by Walras' law, $1 = \mathbf{p} \cdot x(\mathbf{p},1) = \mathbf{p}' \cdot x(\mathbf{p}',1)$. Then, \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] &= \mathbf{p}' \cdot x(\mathbf{p}',1) - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + \mathbf{p} \cdot x(\mathbf{p},1)\\ &= 1 - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + 1\\ &< 0. \end{align*}

Case 3: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) \leq 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) > 1$.

Case 4: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) > 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) \leq 1$.

The law of demand always holds if the walrasian demand function $x(\mathbf{p},w)$ satisfies the weak axiom of revealed preference (WARP), Walras' law, homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth.

Suppose that the Walrasian demand function $x(\mathbf{p},w)$ satisfies the weak axiom of revealed preference (WARP), Walras' law, homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth $w$. We want to prove that, for any price change from $\mathbf{p}$ to $\mathbf{p}'$ (without change in wealth), we have \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',w) - x(\mathbf{p},w)] \leq 0.\tag1 \end{align*} By homogeneity of degree 1 in $w$, we shall prove \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] \leq 0\tag2 \end{align*} for all $\mathbf{p}$ and $\mathbf{p}'$. We consider the following cases:

Case 1: Suppose that $x(\mathbf{p},1) = x(\mathbf{p}',1)$. Then the inequality $(2)$ always holds with $(\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] = 0$.

From now on, suppose that $x(\mathbf{p},1) \neq x(\mathbf{p}',1)$.

Case 2: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) > 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) > 1$. Notice that, by Walras' law, $1 = \mathbf{p} \cdot x(\mathbf{p},1) = \mathbf{p}' \cdot x(\mathbf{p}',1)$. Then, \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] &= \mathbf{p}' \cdot x(\mathbf{p}',1) - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + \mathbf{p} \cdot x(\mathbf{p},1)\\ &= 1 - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + 1\\ &< 0. \end{align*}

Case 3: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) \leq 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) > 1$.

Case 4: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) > 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) \leq 1$.

The law of demand always holds if the walrasian demand function $x(\mathbf{p},w)$ satisfies the weak axiom of revealed preference (WARP), Walras' law, homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth.

Suppose that the Walrasian demand function $x(\mathbf{p},w)$ satisfies the weak axiom of revealed preference (WARP), homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth $w$. We want to prove that, for any price change from $\mathbf{p}$ to $\mathbf{p}'$ (without change in wealth), we have \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',w) - x(\mathbf{p},w)] \leq 0.\tag1 \end{align*} By homogeneity of degree 1 in $w$, we shall prove \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] \leq 0\tag2 \end{align*} for all $\mathbf{p}$ and $\mathbf{p}'$. We consider the following cases:

Case 1: Suppose that $x(\mathbf{p},1) = x(\mathbf{p}',1)$. Then the inequality $(2)$ always holds with $(\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] = 0$.

From now on, suppose that $x(\mathbf{p},1) \neq x(\mathbf{p}',1)$.

Case 2: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) > 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) > 1$. Notice that, by Walras' law, $1 = \mathbf{p} \cdot x(\mathbf{p},1) = \mathbf{p}' \cdot x(\mathbf{p}',1)$. Then, \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] &= \mathbf{p}' \cdot x(\mathbf{p}',1) - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + \mathbf{p} \cdot x(\mathbf{p},1)\\ &= 1 - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + 1\\ &< 0. \end{align*}

Case 3: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) \leq 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) > 1$.

Case 4: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) > 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) \leq 1$.

The law of demand always holds if the walrasian demand function $x(\mathbf{p},w)$ satisfies the weak axiom of revealed preference (WARP), Walras' law, homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth.

Suppose that the Walrasian demand function $x(\mathbf{p},w)$ satisfies the weak axiom of revealed preference (WARP), Walras' law, homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth $w$. We want to prove that, for any price change from $\mathbf{p}$ to $\mathbf{p}'$ (without change in wealth), we have \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',w) - x(\mathbf{p},w)] \leq 0.\tag1 \end{align*} By homogeneity of degree 1 in $w$, we shall prove \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] \leq 0\tag2 \end{align*} for all $\mathbf{p}$ and $\mathbf{p}'$. We consider the following cases:

Case 1: Suppose that $x(\mathbf{p},1) = x(\mathbf{p}',1)$. Then the inequality $(2)$ always holds with $(\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] = 0$.

From now on, suppose that $x(\mathbf{p},1) \neq x(\mathbf{p}',1)$.

Case 2: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) > 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) > 1$. Notice that, by Walras' law, $1 = \mathbf{p} \cdot x(\mathbf{p},1) = \mathbf{p}' \cdot x(\mathbf{p}',1)$. Then, \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] &= \mathbf{p}' \cdot x(\mathbf{p}',1) - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + \mathbf{p} \cdot x(\mathbf{p},1)\\ &= 1 - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + 1\\ &< 0. \end{align*}

Case 3: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) \leq 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) > 1$.

Case 4: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) > 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) \leq 1$.

The law of demand always holds if the walrasian demand function $x(\mathbf{p},w)$ satisfies the weak axiom of revealed preference (WARP), Walras' law, homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth.

Suppose that the Walrasian demand function $x(\mathbf{p},w)$ satisfies the weak axiom of revealed preference (WARP), homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth $w$. We want to prove that, for any price change from $\mathbf{p}$ to $\mathbf{p}'$ (without change in wealth), we have \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',w) - x(\mathbf{p},w)] \leq 0.\tag1 \end{align*} By homogeneity of degree 1 in $w$, we shall prove \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] \leq 0\tag2 \end{align*} for all $\mathbf{p}$ and $\mathbf{p}'$. We consider the following cases:

Case 1: Suppose that $x(\mathbf{p},1) = x(\mathbf{p}',1)$. Then the inequality $(2)$ always holds with $(\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] = 0$.

From now on, suppose that $x(\mathbf{p},1) \neq x(\mathbf{p}',1)$.

Case 2: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) > 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) > 1$. Notice that, by Walras' law, $1 = \mathbf{p} \cdot x(\mathbf{p},1) = \mathbf{p}' \cdot x(\mathbf{p}',1)$. Then, \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] &= \mathbf{p}' \cdot x(\mathbf{p}',1) - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + \mathbf{p} \cdot x(\mathbf{p},1)\\ &= 1 - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + 1\\ &< 0. \end{align*}

Case 3: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) \leq 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) > 1$.

Case 4: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) > 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) \leq 1$.

The law of demand always holds if the walrasian demand function $x(\mathbf{p},w)$ satisfies the weak axiom of revealed preference (WARP), Walras' law, homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth.

Suppose that the Walrasian demand function $x(\mathbf{p},w)$ satisfies the weak axiom of revealed preference (WARP), homogeneity of degree 0, and homogeneity of degree 1 with respect to wealth $w$. We want to prove that, for any price change from $\mathbf{p}$ to $\mathbf{p}'$ (without change in wealth), we have \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',w) - x(\mathbf{p},w)] \leq 0.\tag1 \end{align*} By homogeneity of degree 1 in $w$, we shall prove \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] \leq 0\tag2 \end{align*} for all $\mathbf{p}$ and $\mathbf{p}'$. We consider the following cases:

Case 1: Suppose that $x(\mathbf{p},1) = x(\mathbf{p}',1)$. Then the inequality $(2)$ always holds with $(\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] = 0$.

From now on, suppose that $x(\mathbf{p},1) \neq x(\mathbf{p}',1)$.

Case 2: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) > 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) > 1$. Notice that, by Walras' law, $1 = \mathbf{p} \cdot x(\mathbf{p},1) = \mathbf{p}' \cdot x(\mathbf{p}',1)$. Then, \begin{align*} (\mathbf{p}' - \mathbf{p}) \cdot [x(\mathbf{p}',1) - x(\mathbf{p},1)] &= \mathbf{p}' \cdot x(\mathbf{p}',1) - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + \mathbf{p} \cdot x(\mathbf{p},1)\\ &= 1 - \mathbf{p}' \cdot x(\mathbf{p},1) - \mathbf{p} \cdot x(\mathbf{p}',1) + 1\\ &< 0. \end{align*}

Case 3: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) \leq 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) > 1$.

Case 4: Suppose that $\mathbf{p}' \cdot x(\mathbf{p},1) > 1$ and $\mathbf{p} \cdot x(\mathbf{p}',1) \leq 1$.