1. **Problem Statement:** Determine whether $\sqrt{2}$ is a rational or irrational number. 2. **Definition:** A rational number can be expressed as $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. An irrational number cannot be expressed as such a fraction. 3. **Assumption:** Suppose $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{p}{q}$ where $p$ and $q$ are integers with no common factors (in simplest form). 4. **Square both sides:** $$\sqrt{2} = \frac{p}{q} \implies 2 = \frac{p^2}{q^2} \implies 2q^2 = p^2$$ 5. **Implication:** $p^2$ is even (since it equals $2q^2$), so $p$ must be even. 6. **Let $p=2k$ for some integer $k$:** $$2q^2 = (2k)^2 = 4k^2 \implies q^2 = 2k^2$$ 7. **Conclusion:** $q^2$ is even, so $q$ is even. 8. **Contradiction:** Both $p$ and $q$ are even, contradicting the assumption that $\frac{p}{q}$ is in simplest form. 9. **Therefore,** $\sqrt{2}$ is irrational. **Final answer:** $\sqrt{2}$ is an irrational number.