1. **State the problem:** We want to prove by contradiction that if $a$ is rational ($a \in \mathbb{Q}$) and $b$ is irrational ($b \notin \mathbb{Q}$), then $a + 2b$ is also irrational ($a + 2b \notin \mathbb{Q}$). 2. **Assume the opposite (for contradiction):** Suppose $a + 2b$ is rational. That is, assume $a + 2b \in \mathbb{Q}$. 3. **Express in terms of rational and irrational numbers:** Since $a \in \mathbb{Q}$ and by assumption $a + 2b \in \mathbb{Q}$, subtract $a$ from both sides: $$a + 2b - a = 2b \in \mathbb{Q}.$$ 4. **Divide by 2:** Since 2 is rational and $2b$ is rational, it follows that $$b = \frac{2b}{2} \in \mathbb{Q}.$$ 5. **Contradiction:** This contradicts the given condition that $b$ is irrational ($b \notin \mathbb{Q}$). 6. **Conclusion:** Therefore, our assumption that $a + 2b$ is rational must be false. Hence $a + 2b$ is irrational. **Final Answer:** If $a \in \mathbb{Q}$ and $b \notin \mathbb{Q}$, then $a + 2b \notin \mathbb{Q}$.