1.1 If $$ k \in\{0: 1: 2: 3\} $$, determine the value of $$ k $$ such that $$ \sqrt{\frac{18}{k-1}} $$ 1.1.1 is an integer $$ K=3 $$ 1.1.2 is irrational $$ k=2 $$ 1.1.3 is undefined $$ K=1 $$

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Let's analyze the expression $$ \sqrt{\frac{18}{k - 1}} $$ for each value of $$ k $$ in the set $$\{0, 1, 2, 3\}$$. We'll determine when the expression is an integer, irrational, or undefined.

### 1.1.1 When is $$ \sqrt{\frac{18}{k - 1}} $$ an Integer?

**Given:** We want $$ \sqrt{\frac{18}{k - 1}} $$ to be an integer.

**Solution:**

1. **For $$ k = 0 $$:**
   $$
   \sqrt{\frac{18}{0 - 1}} = \sqrt{\frac{18}{-1}} = \sqrt{-18}
   $$
   This is **not a real number**.

2. **For $$ k = 1 $$:**
   $$
   \sqrt{\frac{18}{1 - 1}} = \sqrt{\frac{18}{0}}
   $$
   Division by zero is **undefined**.

3. **For $$ k = 2 $$:**
   $$
   \sqrt{\frac{18}{2 - 1}} = \sqrt{\frac{18}{1}} = \sqrt{18} = 3\sqrt{2}
   $$
   $$ 3\sqrt{2} $$ is **irrational**, not an integer.

4. **For $$ k = 3 $$:**
   $$
   \sqrt{\frac{18}{3 - 1}} = \sqrt{\frac{18}{2}} = \sqrt{9} = 3
   $$
   $$ 3 $$ is an **integer**.

**Conclusion:** The expression is an integer when $$ k = 3 $$.

### 1.1.2 When is $$ \sqrt{\frac{18}{k - 1}} $$ Irrational?

**Given:** We want $$ \sqrt{\frac{18}{k - 1}} $$ to be irrational.

**Solution:**

From the previous analysis:

- For $$ k = 2 $$:
  $$
  \sqrt{\frac{18}{2 - 1}} = \sqrt{18} = 3\sqrt{2}
  $$
  $$ 3\sqrt{2} $$ is **irrational**.

**Conclusion:** The expression is irrational when $$ k = 2 $$.

### 1.1.3 When is $$ \sqrt{\frac{18}{k - 1}} $$ Undefined?

**Given:** We want $$ \sqrt{\frac{18}{k - 1}} $$ to be undefined.

**Solution:**

- For $$ k = 1 $$:
  $$
  \sqrt{\frac{18}{1 - 1}} = \sqrt{\frac{18}{0}}
  $$
  Division by zero is **undefined**.

**Conclusion:** The expression is undefined when $$ k = 1 $$.

### Summary

- **Integer:** $$ k = 3 $$
- **Irrational:** $$ k = 2 $$
- **Undefined:** $$ k = 1 $$

**Note:** For $$ k = 0 $$, the expression results in $$ \sqrt{-18} $$, which is a complex number. Since the problem focuses on real numbers, $$ k = 0 $$ is not considered in the integer, irrational, or undefined categories within the real number system.
When $$ k = 3 $$, $$ \sqrt{\frac{18}{k - 1}} $$ is an integer. When $$ k = 2 $$, it is irrational. When $$ k = 1 $$, it is undefined.

1.1 If \( k \in\{0: 1: 2: 3\} \), determine the value of \( k \) such that \( \sqrt{\frac{18}{k-1}} \) 1.1.1 is an integer \( K=3 \) 1.1.2 is irrational \( k=2 \) 1.1.3 is undefined \( K=1 \)

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