1. **State the problem:** Prove that $3\sqrt{3}$ (three times the square root of 3) is an irrational number. 2. **Recall definitions and properties:** - A number is **rational** if it can be expressed as $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. - A number is **irrational** if it cannot be expressed as such a fraction. - It is known that $\sqrt{3}$ is irrational. 3. **Use the property of irrational numbers:** - If $\sqrt{3}$ is irrational, then multiplying it by a nonzero rational number (like 3) results in an irrational number. 4. **Proof by contradiction:** - Suppose $3\sqrt{3}$ is rational. - Then $3\sqrt{3} = \frac{p}{q}$ for some integers $p, q$ with $q \neq 0$. - Divide both sides by 3: $\sqrt{3} = \frac{p}{3q}$. - Since $\frac{p}{3q}$ is rational, this implies $\sqrt{3}$ is rational, which contradicts the known fact. 5. **Conclusion:** - Therefore, $3\sqrt{3}$ must be irrational. **Final answer:** $3\sqrt{3}$ is irrational.