1. **Stating the problem:** We want to disprove the statement: "If $a,b \notin \mathbb{Q}$ (irrational numbers), then $a^b \notin \mathbb{Q}$ (irrational)." 2. **Understanding the statement:** The claim says that raising an irrational number $a$ to the power of another irrational number $b$ always results in an irrational number. To disprove it, we need to find a counterexample where $a$ and $b$ are irrational but $a^b$ is rational. 3. **Counterexample:** Consider $a = \sqrt{2}$ (which is irrational) and $b = \sqrt{2}$ (also irrational). Now look at $a^b = (\sqrt{2})^{\sqrt{2}}$. 4. **Two cases:** - If $(\sqrt{2})^{\sqrt{2}}$ is rational, then we have found our counterexample directly. - If $(\sqrt{2})^{\sqrt{2}}$ is irrational, then consider $$\left((\sqrt{2})^{\sqrt{2}}\right)^{\sqrt{2}} = (\sqrt{2})^{\sqrt{2} \times \sqrt{2}} = (\sqrt{2})^2 = 2,$$ which is rational. 5. **Conclusion:** In the second case, $a = (\sqrt{2})^{\sqrt{2}}$ is irrational (by assumption), $b = \sqrt{2}$ is irrational, but $a^b = 2$ is rational. This disproves the original statement. **Final answer:** The statement is false because there exist irrational numbers $a,b$ such that $a^b$ is rational.