1. **Problem statement:** Given that $\sqrt{5}$ is irrational, prove that $3 - 2\sqrt{5}$ is irrational. 2. **Recall the definition:** A number is irrational if it cannot be expressed as a ratio of two integers. 3. **Assume the contrary:** Suppose $3 - 2\sqrt{5}$ is rational. Then there exist integers $a$ and $b \neq 0$ such that $$3 - 2\sqrt{5} = \frac{a}{b}.$$ 4. **Isolate the irrational part:** Rearranging, $$2\sqrt{5} = 3 - \frac{a}{b} = \frac{3b - a}{b}.$$ 5. **Divide both sides by 2:** $$\sqrt{5} = \frac{3b - a}{2b}.$$ 6. **Conclusion:** The right side is a ratio of integers, hence rational. This contradicts the given fact that $\sqrt{5}$ is irrational. 7. **Therefore,** our assumption is false, and $3 - 2\sqrt{5}$ must be irrational. **Final answer:** $3 - 2\sqrt{5}$ is irrational.