Here’s how to use the **trichotomy law** to prove the cancellation law for inequalities in \(\mathbb{R}\): **Recall:** The trichotomy law says for any two real numbers \(x\) and \(y\), exactly one of \(x = y\), \(x < y\), or \(y < x\) is true. --- ### (i) \(x + y < x + z \implies y < z\) **Proof:** Suppose \(x + y < x + z\). By the trichotomy law applied to \(y\) and \(z\), exactly one of the following is true: 1. \(y = z\) 2. \(y < z\) 3. \(z < y\) - **Case 1:** If \(y = z\), then \(x + y = x + z\), which contradicts \(x + y < x + z\). - **Case 2:** If \(y < z\), this matches the desired conclusion. - **Case 3:** If \(z < y\), then \(x + z < x + y\), contradicting \(x + y < x + z\). So the only possibility is \(y < z\). --- ### (ii) \(xy < xz\) for \(x > 0\) implies \(y < z\) **Proof:** Suppose \(x > 0\) and \(xy < xz\). By the trichotomy law for \(y\) and \(z\), one of: 1. \(y = z\) 2. \(y < z\) 3. \(z < y\) - **Case 1:** If \(y = z\), then \(xy = xz\), contradicting \(xy < xz\). - **Case 2:** If \(y < z\), then, since \(x > 0\), multiplying both sides by \(x\) preserves the inequality: \(xy < xz\), which matches the assumption. - **Case 3:** If \(z < y\), then \(x z < x y\), which contradicts \(xy < xz\). So the only possibility is \(y < z\). --- ### (iii) \(xy < xz\) for \(x < 0\) implies \(y > z\) **Proof:** Suppose \(x < 0\) and \(xy < xz\). By the trichotomy law for \(y\) and \(z\): 1. \(y = z\) 2. \(y < z\) 3. \(z < y\) - **Case 1:** If \(y = z\), then \(xy = xz\), contradicting \(xy < xz\). - **Case 2:** If \(y < z\), then, since \(x < 0\), multiplying both sides by \(x\) reverses the inequality: \(xy > xz\), which contradicts \(xy < xz\). - **Case 3:** If \(z < y\), then multiplying both sides by \(x < 0\) reverses the inequality: \(xz > xy\), which is equivalent to \(xy < xz\). So \(z < y\) is possible, which is \(y > z\). So the only possibility is \(y > z\). --- **Summary Table:** | Statement | Reasoning via Trichotomy Law | |-----------------------------------------|--------------------------------------------------------------------| | \(x+y < x+z \implies y < z\) | Only \(y < z\) avoids contradiction with the inequality | | \(xy < xz\), \(x>0 \implies y < z\) | Only \(y < z\) makes \(xy < xz\) when \(x > 0\) | | \(xy < xz\), \(x<0 \implies y > z\) | Only \(y > z\) makes \(xy < xz\) when \(x < 0\) |