1. **Stating the problem:** We want to prove that there exist both rational and irrational numbers. 2. **Definition of rational numbers:** A rational number is any number that can be expressed as $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. 3. **Definition of irrational numbers:** An irrational number cannot be expressed as a ratio of two integers. Its decimal expansion is non-terminating and non-repeating. 4. **Example of a rational number:** Consider $\frac{1}{2}$. It is rational because it is a ratio of integers 1 and 2. 5. **Proof that $\sqrt{2}$ is irrational:** - Assume $\sqrt{2}$ is rational, so $\sqrt{2} = \frac{a}{b}$ where $a,b$ are integers with no common factors. - Square both sides: $$2 = \frac{a^2}{b^2}$$ - Multiply both sides by $b^2$: $$2b^2 = a^2$$ - This means $a^2$ is even, so $a$ must be even. Let $a=2k$. - Substitute back: $$2b^2 = (2k)^2 = 4k^2$$ - Simplify: $$b^2 = 2k^2$$ - This means $b^2$ is even, so $b$ is even. - Both $a$ and $b$ are even, contradicting the assumption that $a$ and $b$ have no common factors. 6. **Conclusion:** Our assumption is false, so $\sqrt{2}$ is irrational. Thus, we have shown an example of a rational number ($\frac{1}{2}$) and proved that $\sqrt{2}$ is irrational.