1. Let's start by understanding the problem: proving that a specific number is not rational means showing it cannot be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. 2. Note that the statement "any number is not rational" is false because many numbers are rational (e.g., $\frac{1}{2}$, $3$, $-5$). 3. However, if you want to prove that a particular number (like $\sqrt{2}$ or $\pi$) is irrational, you can use specific proofs. 4. For example, to prove $\sqrt{2}$ is irrational: - Assume the opposite: $\sqrt{2} = \frac{p}{q}$ with $p,q$ integers having no common factors. - Then $2 = \frac{p^2}{q^2}$, so $p^2 = 2q^2$. - This implies $p^2$ is even, so $p$ is even. - Let $p=2k$, then $4k^2 = 2q^2$ or $2k^2 = q^2$. - So $q^2$ is even, and $q$ is even. - But if both $p$ and $q$ are even, they have a common factor 2, contradicting the assumption. - Therefore, $\sqrt{2}$ is irrational. 5. In general, proving a number is irrational requires a specific argument or contradiction like above. Final answer: You cannot prove "any number" is irrational, but you can prove specific numbers are irrational using contradiction and properties of integers.