EXERCISE 2 1. Consider each of the roots below. If the root is a rational number, simplify it fully. If it is an irrational number, establish the two integers it lies between. a) \( \sqrt{1} \) b) \( \sqrt{5} \) c) \( \sqrt{0} \) d) \( \sqrt{25} \) e) \( \sqrt{10} \) f) \( \sqrt{3} \) g) \( \sqrt{36} \) h) \( \sqrt{\frac{9}{4}} \) i) \( \sqrt{\frac{18}{50}} \) j) \( \sqrt{\frac{(12+4)}{4}} \) k) \( \sqrt{\frac{32}{4}} \)

Certainly! Let's go through each part of Exercise 2, Part 1, step by step. ### 1. Simplifying Roots: **a) \( \sqrt{1} \)** - **Simplification:** \( \sqrt{1} = 1 \) - **Conclusion:** 1 is a rational number. --- **b) \( \sqrt{5} \)** - **Nature:** Irrational number. - **Establishing Bounds:** - \( 2^2 = 4 \) and \( 3^2 = 9 \) - Therefore, \( 2 < \sqrt{5} < 3 \) --- **c) \( \sqrt{0} \)** - **Simplification:** \( \sqrt{0} = 0 \) - **Conclusion:** 0 is a rational number. --- **d) \( \sqrt{25} \)** - **Simplification:** \( \sqrt{25} = 5 \) - **Conclusion:** 5 is a rational number. --- **e) \( \sqrt{10} \)** - **Nature:** Irrational number. - **Establishing Bounds:** - \( 3^2 = 9 \) and \( 4^2 = 16 \) - Therefore, \( 3 < \sqrt{10} < 4 \) --- **f) \( \sqrt{3} \)** - **Nature:** Irrational number. - **Establishing Bounds:** - \( 1^2 = 1 \) and \( 2^2 = 4 \) - Therefore, \( 1 < \sqrt{3} < 2 \) --- **g) \( \sqrt{36} \)** - **Simplification:** \( \sqrt{36} = 6 \) - **Conclusion:** 6 is a rational number. --- **h) \( \sqrt{\frac{9}{4}} \)** - **Simplification:** \[ \sqrt{\frac{9}{4}} = \frac{\sqrt{9}}{\sqrt{4}} = \frac{3}{2} \] - **Conclusion:** \( \frac{3}{2} \) is a rational number. --- **i) \( \sqrt{\frac{18}{50}} \)** - **Simplification:** \[ \frac{18}{50} = \frac{9}{25} \quad (\text{Simplified by dividing numerator and denominator by 2}) \] \[ \sqrt{\frac{9}{25}} = \frac{\sqrt{9}}{\sqrt{25}} = \frac{3}{5} \] - **Conclusion:** \( \frac{3}{5} \) is a rational number. --- **j) \( \sqrt{\frac{12 + 4}{4}} \)** - **Simplification:** \[ \frac{12 + 4}{4} = \frac{16}{4} = 4 \] \[ \sqrt{4} = 2 \] - **Conclusion:** 2 is a rational number. --- **k) \( \sqrt{\frac{32}{4}} \)** - **Simplification:** \[ \frac{32}{4} = 8 \] \[ \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \] - **Nature:** Irrational number. - **Establishing Bounds:** - \( 2^2 = 4 \) and \( 3^2 = 9 \) - Therefore, \( 2 < \sqrt{8} < 3 \) --- ### Summary of Answers: **a)** \( \sqrt{1} = 1 \) (rational) **b)** \( \sqrt{5} \) is irrational and lies between **2** and **3**. **c)** \( \sqrt{0} = 0 \) (rational) **d)** \( \sqrt{25} = 5 \) (rational) **e)** \( \sqrt{10} \) is irrational and lies between **3** and **4**. **f)** \( \sqrt{3} \) is irrational and lies between **1** and **2**. **g)** \( \sqrt{36} = 6 \) (rational) **h)** \( \sqrt{\frac{9}{4}} = \frac{3}{2} \) (rational) **i)** \( \sqrt{\frac{18}{50}} = \frac{3}{5} \) (rational) **j)** \( \sqrt{\frac{12 + 4}{4}} = 2 \) (rational) **k)** \( \sqrt{\frac{32}{4}} = \sqrt{8} \) is irrational and lies between **2** and **3**.