1. **Problem statement:** Given two rational numbers $x,y$ with $x < y$, show that there are infinitely many rational numbers $z$ such that $x < z < y$. 2. **Key idea:** Between any two distinct rational numbers, there exists another rational number. This is a fundamental property of rational numbers being dense in the real numbers. 3. **Formula and explanation:** To find a rational number between $x$ and $y$, consider the average: $$z = \frac{x + y}{2}$$ Since $x < y$, it follows that $x < z < y$. 4. **Intermediate step:** We can verify this by comparing: $$x < \frac{x + y}{2} < y$$ which holds true because $x < y$ implies $x < \frac{x + y}{2}$ and $\frac{x + y}{2} < y$. 5. **Constructing infinitely many rationals:** Starting with $x$ and $y$, define a sequence of rationals between them by repeatedly taking averages: $$z_1 = \frac{x + y}{2}, \quad z_2 = \frac{x + z_1}{2}, \quad z_3 = \frac{z_1 + y}{2}, \quad \ldots$$ Each $z_i$ is rational and lies strictly between $x$ and $y$. 6. **Conclusion:** Since this process can be repeated infinitely many times, there are infinitely many rational numbers $z$ such that $x < z < y$. **Final answer:** There are infinitely many rational numbers between any two distinct rational numbers $x$ and $y$ because the rationals are dense in the real numbers.