1. **Problem:** Suppose $x$ is a rational number and $y$ is an irrational number. Determine which of the following statements is necessarily true: A. $xy$ is rational. B. $\frac{x}{y}$ is irrational. C. $\frac{y}{x}$ is a real number. D. $x - y$ is irrational for any $x$. 2. **Relevant concepts:** - A rational number can be expressed as $\frac{p}{q}$ where $p,q \in \mathbb{Z}$ and $q \neq 0$. - An irrational number cannot be expressed as a ratio of integers. - The product of a nonzero rational and an irrational number is irrational. - The sum or difference of a rational and an irrational number is irrational. - Division by zero is undefined. 3. **Analyze each option:** - A. $xy$ is rational. - If $x=0$ (rational), then $xy=0$ which is rational. - But if $x \neq 0$, then $xy$ is irrational because multiplying a nonzero rational by an irrational yields an irrational. - So $xy$ is not necessarily rational. - B. $\frac{x}{y}$ is irrational. - Since $y$ is irrational and $x$ rational, $\frac{x}{y}$ is irrational unless $x=0$ (then $\frac{0}{y}=0$ rational). - So $\frac{x}{y}$ is not necessarily irrational. - C. $\frac{y}{x}$ is a real number. - Since $y$ is irrational (real) and $x$ is rational (real), and $x \neq 0$ (division by zero undefined), $\frac{y}{x}$ is real. - This is always true if $x \neq 0$. - D. $x - y$ is irrational for any $x$. - The difference of a rational and irrational number is irrational. - This holds for any rational $x$. 4. **Conclusion:** - Option C states $\frac{y}{x}$ is real, which is true for any rational $x \neq 0$. - Option D states $x - y$ is irrational for any $x$, which is true. However, since the problem states $x$ is rational (including zero), option C requires $x \neq 0$ to be true always, but $x$ could be zero. Option D is true for any rational $x$. 5. **Final answer:** **D. $x - y$ is irrational for any $x$.**