1. **State the problem:** Prove that an irrational number exists. 2. **Explanation:** A number is irrational if it cannot be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. 3. **Common example:** Consider the number $\sqrt{2}$. 4. **Proof by contradiction:** Assume $\sqrt{2}$ is rational, so $\sqrt{2} = \frac{p}{q}$ where $p$ and $q$ are integers with no common factors. 5. Square both sides: $$2 = \frac{p^2}{q^2}$$ 6. Multiply both sides by $q^2$: $$2q^2 = p^2$$ 7. This means $p^2$ is even, so $p$ must be even. Let $p = 2k$ for some integer $k$. 8. Substitute back: $$2q^2 = (2k)^2 = 4k^2$$ 9. Simplify: $$q^2 = 2k^2$$ 10. This means $q^2$ is even, so $q$ is even. 11. But if both $p$ and $q$ are even, they have a common factor of 2, contradicting the assumption. 12. Therefore, $\sqrt{2}$ is irrational. **Final answer:** $\sqrt{2}$ is an irrational number, proving irrational numbers exist.