1. **Problem statement:** Show that $\sqrt{12}$ can be written in the form $a + \sqrt{b}$ where $a$ and $b$ are integers. 2. **Recall the simplification rule for square roots:** If $n$ is a positive integer and can be factored as $n = m^2 \times k$ where $m$ and $k$ are integers, then $$\sqrt{n} = \sqrt{m^2 \times k} = m \sqrt{k}.$$ 3. **Apply this to $\sqrt{12}$:** Factor 12 as $12 = 4 \times 3$ where 4 is a perfect square. 4. **Simplify:** $$\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2 \sqrt{3}.$$ 5. **Express in the form $a + \sqrt{b}$:** Here, $2 \sqrt{3}$ can be written as $0 + 2 \sqrt{3}$, so $a = 0$ and $b = 3$. 6. **Final answer:** $$\sqrt{12} = 0 + 2 \sqrt{3}$$ where $a = 0$ and $b = 3$ are integers. This completes the proof that $\sqrt{12}$ can be expressed in the form $a + \sqrt{b}$ with integer $a$ and $b$.