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) $$ \sqrt{\frac{32}{4}} $$

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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.

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**b) $$ \sqrt{5} $$**
- **Nature:** Irrational number.
- **Establishing Bounds:**
  - $$ 2^2 = 4 $$ and $$ 3^2 = 9 $$
  - Therefore, $$ 2 < \sqrt{5} < 3 $$

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**c) $$ \sqrt{0} $$**
- **Simplification:** $$ \sqrt{0} = 0 $$
- **Conclusion:** 0 is a rational number.

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**d) $$ \sqrt{25} $$**
- **Simplification:** $$ \sqrt{25} = 5 $$
- **Conclusion:** 5 is a rational number.

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**e) $$ \sqrt{10} $$**
- **Nature:** Irrational number.
- **Establishing Bounds:**
  - $$ 3^2 = 9 $$ and $$ 4^2 = 16 $$
  - Therefore, $$ 3 < \sqrt{10} < 4 $$

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**f) $$ \sqrt{3} $$**
- **Nature:** Irrational number.
- **Establishing Bounds:**
  - $$ 1^2 = 1 $$ and $$ 2^2 = 4 $$
  - Therefore, $$ 1 < \sqrt{3} < 2 $$

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**g) $$ \sqrt{36} $$**
- **Simplification:** $$ \sqrt{36} = 6 $$
- **Conclusion:** 6 is a rational number.

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**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.

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**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.

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**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**.
Here are the simplified 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**.

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