QUESTION 1 1.1 For which of the value of $$ x $$ : $$ x \in\{-11 ;-5 ; 0 ; 15\} \text { is } \sqrt{\frac{9}{11-x}} $$ (2) 1.1.1 a rational number 1.1.2 an irrational number 1.1.3 not defined (2) 1.2 Which of the following expressions will always be a natural number if $$ a $$ and $$ t $$ natural number . Give reasons for those that are not natural numbers. 1.2.1 $$ \mathrm{a}+\mathrm{b} $$ (1) $$ 1.2 .2 \mathrm{a}-\mathrm{b} $$ (2) 1.2 .3 ab (1) 1.2.4 $$ \frac{a}{b} $$ 1.3 Without the use a calculator and showing all working , determine between which two Integers $$ \sqrt{39} $$ lies.

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Let's address each part of **Question 1** systematically.

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### **1.1 Evaluate $$\sqrt{\frac{9}{11 - x}}$$ for $$x \in \{-11, -5, 0, 15\}$$**

**Definitions:**
- **Rational Number:** A number that can be expressed as the quotient of two integers.
- **Irrational Number:** A number that cannot be expressed as a simple fraction; its decimal form is non-repeating and non-terminating.
- **Not Defined:** The expression is invalid (e.g., division by zero or square root of a negative number).

Let's evaluate the expression for each value of $$x$$:

1. **For $$x = -11$$:**
   $$
   \sqrt{\frac{9}{11 - (-11)}} = \sqrt{\frac{9}{22}} = \sqrt{\frac{9}{22}} = \frac{3}{\sqrt{22}}
   $$
   - $$\sqrt{22}$$ is irrational, making $$\frac{3}{\sqrt{22}}$$ irrational.

2. **For $$x = -5$$:**
   $$
   \sqrt{\frac{9}{11 - (-5)}} = \sqrt{\frac{9}{16}} = \frac{3}{4}
   $$
   - $$\frac{3}{4}$$ is a **rational number**.

3. **For $$x = 0$$:**
   $$
   \sqrt{\frac{9}{11 - 0}} = \sqrt{\frac{9}{11}} = \frac{3}{\sqrt{11}}
   $$
   - $$\sqrt{11}$$ is irrational, making $$\frac{3}{\sqrt{11}}$$ irrational.

4. **For $$x = 15$$:**
   $$
   \sqrt{\frac{9}{11 - 15}} = \sqrt{\frac{9}{-4}} = \sqrt{-2.25}
   $$
   - The square root of a negative number is **not defined** in the set of real numbers.

**Summary:**
- **1.1.1 Rational Number:** $$x = -5$$
- **1.1.2 Irrational Number:** $$x = -11, 0$$
- **1.1.3 Not Defined:** $$x = 15$$

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### **1.2 Determine Which Expressions Always Yield a Natural Number**

**Definitions:**
- **Natural Numbers:** Positive integers starting from 1 (i.e., 1, 2, 3, ...).

Given $$a$$ and $$b$$ are natural numbers, evaluate each expression:

1. **1.2.1 $$a + b$$:**
   - **Always Natural:** The sum of two natural numbers is always a natural number.
   
2. **1.2.2 $$a - b$$:**
   - **Not Always Natural:** If $$a < b$$, the result is negative, which is not a natural number. Only when $$a \geq b$$ is the result natural.
   
3. **1.2.3 $$a \times b$$:**
   - **Always Natural:** The product of two natural numbers is always a natural number.
   
4. **1.2.4 $$\frac{a}{b}$$:**
   - **Not Always Natural:** The division of two natural numbers does not always result in a natural number unless $$b$$ divides $$a$$ exactly.

**Summary:**
- **Always Natural Numbers:** 
  - **1.2.1 $$a + b$$**
  - **1.2.3 $$ab$$**
  
- **Not Always Natural Numbers:** 
  - **1.2.2 $$a - b$$** (Because $$a$$ might be less than $$b$$)
  - **1.2.4 $$\frac{a}{b}$$** (Because $$a$$ might not be divisible by $$b$$)

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### **1.3 Determine Between Which Two Integers $$\sqrt{39}$$ Lies**

To find between which two integers $$\sqrt{39}$$ lies, identify two consecutive integers $$n$$ and $$n+1$$ such that:
$$
n^2 < 39 < (n+1)^2
$$

1. **Testing $$n = 6$$:**
   $$
   6^2 = 36 < 39
   $$
   
2. **Testing $$n = 7$$:**
   $$
   7^2 = 49 > 39
   $$
   
Since $$36 < 39 < 49$$, it follows that:
$$
6 < \sqrt{39} < 7
$$

**Conclusion:** $$\sqrt{39}$$ lies **between 6 and 7**.

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**1.1 Evaluations:** - **x = -5:** Rational number ($$\frac{3}{4}$$) - **x = -11 and x = 0:** Irrational numbers - **x = 15:** Not defined **1.2 Expressions Always Natural Numbers:** - **a + b** - **ab** **1.3 $$\sqrt{39}$$ Lies Between:** - 6 and 7

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